Org Chem JEE Main Cheatsheet

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### General Organic Chemistry (GOC) - Electronic Effects #### Inductive Effect (I-Effect) - **Definition:** Permanent displacement of $\sigma$-electrons along a carbon chain due to difference in electronegativity. - **Characteristics:** - Operates through single bonds. - Decreases rapidly with distance (negligible after 3-4 carbon atoms). - Permanent effect. - **Types:** - **+I Effect (Electron Releasing Groups):** Alkyl groups (e.g., $-\text{CH}_3$, $-\text{CH}_2\text{CH}_3$), $-\text{COO}^-$, $-\text{O}^-$. Increases electron density. - **-I Effect (Electron Withdrawing Groups):** $-\text{NO}_2$, $-\text{CN}$, $-\text{F}$, $-\text{Cl}$, $-\text{Br}$, $-\text{I}$, $-\text{COOH}$, $-\text{CHO}$, $-\text{OR}$, $-\text{OH}$. Decreases electron density. - **Order of -I effect:** $-\text{NR}_3^+ > -\text{NH}_3^+ > -\text{NO}_2 > -\text{CN} > -\text{SO}_3\text{H} > -\text{CHO} > -\text{COOR} > -\text{COOH} > -\text{F} > -\text{Cl} > -\text{Br} > -\text{I} > -\text{OR} > -\text{OH} > -\text{C}\equiv\text{CH} > -\text{C}_6\text{H}_5 > -\text{CH}=\text{CH}_2 > -\text{H}$ - **Order of +I effect:** $-\text{O}^- > -\text{COO}^- > -\text{C}(\text{CH}_3)_3 > -\text{CH}(\text{CH}_3)_2 > -\text{CH}_2\text{CH}_3 > -\text{CH}_3$ - **Application:** Explains stability of carbocations/carbanions, acidity/basicity. #### Resonance Effect (R-Effect or Mesomeric Effect, M-Effect) - **Definition:** Delocalization of $\pi$-electrons or lone pair electrons within a conjugated system. - **Characteristics:** - Operates through $\pi$-bonds (conjugation). - Permanent effect. - Leads to resonance structures (canonical forms), none of which describes the molecule completely. The actual molecule is a resonance hybrid. - Resonance hybrid is more stable than any single resonance structure. - Greater number of equivalent resonance structures, greater the stability. - **Conditions for Resonance:** - Conjugated system: $\pi-\sigma-\pi$, $\pi-\sigma-\text{lone pair}$, $\pi-\sigma-\text{positive charge}$, $\pi-\sigma-\text{negative charge}$, $\pi-\sigma-\text{free radical}$. - **Types:** - **+R or +M Effect (Electron Releasing Groups):** Groups that donate electrons to the conjugated system. They have lone pairs or negative charge. (e.g., $-\text{OH}$, $-\text{OR}$, $-\text{NH}_2$, $-\text{NR}_2$, $-\text{SH}$, $-\text{SR}$, $-\text{Cl}$, $-\text{Br}$, $-\text{I}$, $-\text{F}$, $-\text{O}^-$, $-\text{NHCOR}$). - **-R or -M Effect (Electron Withdrawing Groups):** Groups that withdraw electrons from the conjugated system. They have a multiple bond to a more electronegative atom. (e.g., $-\text{NO}_2$, $-\text{CN}$, $-\text{CHO}$, $-\text{COOH}$, $-\text{COOR}$, $-\text{CONH}_2$, $-\text{SO}_3\text{H}$). - **Resonance Rules:** 1. All resonance structures must be valid Lewis structures. 2. All atoms in resonance structures must maintain their positions (only electrons move). 3. The number of unpaired electrons must remain the same. 4. Major contributors are those with: - More covalent bonds. - Less charge separation. - Negative charge on more electronegative atom and positive charge on less electronegative atom. - Complete octets for all atoms (except H). - **Application:** Explains stability, reactivity, bond lengths, dipole moments, acidity/basicity. #### Hyperconjugation (No-Bond Resonance or Baker-Nathan Effect) - **Definition:** Delocalization of $\sigma$-electrons of a C-H bond of an alkyl group directly attached to an unsaturated system (double bond, triple bond, aromatic ring) or to an atom with an unshared p-orbital (carbocation, free radical). - **Mechanism:** Overlap of $\sigma$-orbital of C-H bond with adjacent empty p-orbital (carbocation), half-filled p-orbital (free radical), or $\pi$-orbital (alkene/alkyne). - **Conditions:** Presence of $\alpha$-hydrogens (hydrogens on carbon adjacent to the unsaturated system/charged atom). - **Characteristics:** - Permanent effect. - Weaker than resonance but stronger than inductive effect. - Greater the number of $\alpha$-hydrogens, greater the hyperconjugation, greater the stability. - **Application:** Explains: - Stability of alkenes (more substituted alkene is more stable). - Stability of carbocations (tertiary > secondary > primary). - Stability of free radicals (tertiary > secondary > primary). - Reactivity of alkylbenzenes in electrophilic substitution. - Orienting influence of alkyl groups. - **Order of hyperconjugation:** $\text{C}(\text{CH}_3)_3 > \text{CH}(\text{CH}_3)_2 > \text{CH}_2\text{CH}_3 > \text{CH}_3$ (due to number of $\alpha$-hydrogens). #### Summary Table of Electronic Effects | Effect | Mechanism | Bond Type | Permanent/Temporary | Distance Dependent | Application | |----------------|------------------------|----------------|---------------------|--------------------|---------------------------------------------------| | Inductive (I) | $\sigma$-electron disp. | $\sigma$-bond | Permanent | Yes (dec. rapidly) | Acidity/Basicity, Stability of intermediates | | Resonance (R/M)| $\pi$-electron/LP deloc.| $\pi$-bond | Permanent | No | Acidity/Basicity, Stability, Reactivity, Bond lengths | | Hyperconjug. | $\sigma$-electron deloc.| $\sigma$-bond | Permanent | No (but $\alpha$-H) | Stability of alkenes/carbocations/free radicals | ### General Organic Chemistry (GOC) - Acidity and Basicity #### Acidity - **Definition:** Tendency to donate a proton ($\text{H}^+$). - **Factors Affecting Acidity:** 1. **Stability of Conjugate Base:** Stronger acid forms a more stable conjugate base. - **Electronegativity:** Acidity increases across a period (e.g., $\text{CH}_4 \text{sp}^2 > \text{sp}^3$. More s-character means electrons are held closer to nucleus, making C-H bond more polar and $\text{H}^+$ easier to remove (e.g., terminal alkynes are acidic). 4. **Aromaticity:** Aromatic conjugate bases are highly stable, leading to high acidity (e.g., cyclopentadiene). - **Comparison:** - **Carboxylic acids > Phenols > Alcohols > Alkynes > Alkanes** - **Phenols:** EWG (like $-\text{NO}_2$) at *ortho* and *para* positions increase acidity due to resonance stabilization of phenoxide ion. EDG (like $-\text{CH}_3$) decrease acidity. *Ortho*-effect can sometimes be anomalous. - **Carboxylic Acids:** EWG increase acidity, EDG decrease acidity. Fluorine substituted acetic acids: $\text{F}_3\text{CCOOH} > \text{F}_2\text{CHCOOH} > \text{FCH}_2\text{COOH} > \text{CH}_3\text{COOH}$. #### Basicity - **Definition:** Tendency to accept a proton ($\text{H}^+$) or donate a lone pair of electrons (Lewis base). - **Factors Affecting Basicity:** 1. **Availability of Lone Pair:** Greater the availability of lone pair, stronger the base. 2. **Stability of Conjugate Acid:** Stronger base forms a more stable conjugate acid. 3. **Electronic Effects:** - **+I Effect:** EDG increase electron density on the basic atom, making lone pair more available, thus increasing basicity. - **-I Effect:** EWG decrease electron density, decreasing basicity. - **+R Effect:** If the lone pair is involved in resonance, its availability decreases, thus decreasing basicity (e.g., aniline is less basic than cyclohexylamine). - **-R Effect:** If the group withdraws electrons via resonance, it decreases basicity. 4. **Hybridization:** Basicity order: $\text{sp}^3 > \text{sp}^2 > \text{sp}$. More s-character means lone pair is held tighter, less available. 5. **Steric Hindrance:** In aqueous solution, solvation of conjugate acid plays a role. Steric hindrance can reduce solvation, affecting basicity order for amines. - **Aliphatic Amines (aqueous):** - Secondary (2°) > Primary (1°) > Tertiary (3°) > $\text{NH}_3$ (for ethyl groups) - Secondary (2°) > Tertiary (3°) > Primary (1°) > $\text{NH}_3$ (for methyl groups) - **Aliphatic Amines (gas phase):** Tertiary (3°) > Secondary (2°) > Primary (1°) > $\text{NH}_3$ (only electronic effects considered). 6. **Aromaticity:** Lone pair involved in aromaticity (e.g., pyrrole) is not available for donation, making it a very weak base. - **Comparison:** - **Aliphatic amines > $\text{NH}_3$ > Aromatic amines** - **Guanidine** is exceptionally strong base due to resonance stabilization of its conjugate acid. #### Memory Tricks & Common Mistakes - **Acidity:** Think "EWG increases acidity (stabilizes conjugate base), EDG decreases acidity." - **Basicity:** Think "EDG increases basicity (makes LP more available), EWG decreases basicity." - **Common Mistake:** Confusing gas phase vs. aqueous phase basicity orders for amines. Remember solvation effects in aqueous phase. - **Ortho Effect:** For substituted benzoic acids and anilines, *ortho*-substituted compounds often show anomalous acidity/basicity regardless of nature of substituent. #### Solved JEE Main Example **Q1:** Arrange the following in increasing order of acidity: Phenol, *p*-Nitrophenol, *m*-Nitrophenol, *o*-Nitrophenol. **A1:** The correct order is Phenol (\text{CH}_3)_2\text{NH} > \text{CH}_3\text{NH}_2 > \text{NH}_3$ (due to increasing +I effect of alkyl groups). ### General Organic Chemistry (GOC) - Stability of Intermediates #### Carbocations - **Definition:** Carbon bearing a positive charge and having only six valence electrons (electron deficient, electrophilic). Hybridization is $\text{sp}^2$, trigonal planar geometry. - **Factors Affecting Stability:** 1. **Inductive Effect (+I):** Alkyl groups are electron-donating (+I effect), they disperse the positive charge, stabilizing the carbocation. More alkyl groups = more stable. 2. **Hyperconjugation:** $\alpha$-hydrogens (hydrogens on carbons adjacent to the positively charged carbon) can delocalize electron density into the empty p-orbital via hyperconjugation. More $\alpha$-hydrogens = more stable. 3. **Resonance Effect (+R):** If the positive charge is in conjugation with a lone pair or $\pi$-bond, resonance stabilizes the carbocation significantly. - Example: Allyl carbocation ($\text{CH}_2=\text{CH}-\text{CH}_2^+$), Benzyl carbocation ($\text{C}_6\text{H}_5-\text{CH}_2^+$). - Carbocations adjacent to N, O, S atoms with lone pairs are highly stable ($\text{R}-\text{O}^+=\text{CH}_2$). - **Order of Stability:** - Aromatic carbocations (e.g., Tropylium ion) > Benzyl ($\text{C}_6\text{H}_5-\text{CH}_2^+$) $\approx$ Allyl ($\text{CH}_2=\text{CH}-\text{CH}_2^+$) > Tertiary (3°) > Secondary (2°) > Primary (1°) > Methyl ($\text{CH}_3^+$) > Vinyl ($\text{CH}_2=\text{CH}^+$) > Phenyl ($\text{C}_6\text{H}_5^+$). - **Note:** Vinyl and phenyl carbocations are highly unstable due to positive charge on $\text{sp}^2$ carbon. - **Rearrangements:** Carbocations can rearrange to more stable forms via: - **Hydride shift (H- shift):** A hydrogen atom with its bonding electrons moves to the adjacent positively charged carbon. - **Alkyl shift (R- shift):** An alkyl group with its bonding electrons moves to the adjacent positively charged carbon. - **Phenyl shift ($\text{C}_6\text{H}_5$- shift):** A phenyl group with its bonding electrons moves. #### Carbanions - **Definition:** Carbon bearing a negative charge and having eight valence electrons (electron rich, nucleophilic). Hybridization is $\text{sp}^3$ (pyramidal) or $\text{sp}^2$ (if stabilized by resonance). - **Factors Affecting Stability:** 1. **Inductive Effect (-I):** Electron-withdrawing groups (EWG) disperse the negative charge, stabilizing the carbanion. 2. **Resonance Effect (-R):** If the negative charge is in conjugation with an electron-withdrawing group (like $-\text{NO}_2$, $-\text{CN}$, $-\text{C}=\text{O}$), resonance stabilizes the carbanion. 3. **Hybridization:** Greater s-character means electrons are held closer to the nucleus, stabilizing the negative charge. Order of stability: $\text{sp} > \text{sp}^2 > \text{sp}^3$. - **Order of Stability:** - Benzyl ($\text{C}_6\text{H}_5-\text{CH}_2^-$) $\approx$ Allyl ($\text{CH}_2=\text{CH}-\text{CH}_2^-$) > Primary (1°) > Secondary (2°) > Tertiary (3°) > Methyl ($\text{CH}_3^-$). - **Note:** Opposite to carbocation stability order for alkyl carbanions. - **Common Mistake:** Thinking +I effect stabilizes carbanions. No, EDG destabilize carbanions. #### Free Radicals - **Definition:** Carbon bearing an unpaired electron (paramagnetic, highly reactive). Hybridization is usually $\text{sp}^2$ (planar) or nearly planar. - **Factors Affecting Stability:** 1. **Inductive Effect (+I):** Alkyl groups donate electron density, stabilizing the radical (similar to carbocations). 2. **Hyperconjugation:** $\alpha$-hydrogens stabilize free radicals through hyperconjugation. More $\alpha$-hydrogens = more stable. 3. **Resonance Effect:** Delocalization of the unpaired electron into a conjugated system stabilizes the free radical. - **Order of Stability:** - Benzyl ($\text{C}_6\text{H}_5-\text{CH}_2^\cdot$) $\approx$ Allyl ($\text{CH}_2=\text{CH}-\text{CH}_2^\cdot$) > Tertiary (3°) > Secondary (2°) > Primary (1°) > Methyl ($\text{CH}_3^\cdot$) > Vinyl ($\text{CH}_2=\text{CH}^\cdot$) > Phenyl ($\text{C}_6\text{H}_5^\cdot$). - **Note:** Similar to carbocation stability order. #### Solved JEE Main Example **Q1:** Compare the stability of the following carbocations: (I) $\text{CH}_3-\text{CH}_2^+$ (II) $(\text{CH}_3)_2\text{CH}^+$ (III) $(\text{CH}_3)_3\text{C}^+$ (IV) $\text{CH}_2=\text{CH}-\text{CH}_2^+$ **A1:** The correct order of stability is (IV) > (III) > (II) > (I). **Explanation:** - (I) Primary carbocation, 3 $\alpha$-hydrogens. - (II) Secondary carbocation, 6 $\alpha$-hydrogens. - (III) Tertiary carbocation, 9 $\alpha$-hydrogens. - (IV) Allyl carbocation, stabilized by resonance. Resonance stabilization is generally stronger than hyperconjugation. Therefore, Allyl > Tertiary > Secondary > Primary. #### Practice Questions 1. Arrange the following free radicals in increasing order of stability: $\text{CH}_3^\cdot$, $(\text{CH}_3)_2\text{CH}^\cdot$, $(\text{CH}_3)_3\text{C}^\cdot$, $\text{CH}_2=\text{CH}^\cdot$. 2. Which of the following carbanions is most stable? (A) $\text{CH}_3-\text{CH}_2^-$ (B) $\text{CH}_2=\text{CH}^-$ (C) $\text{HC}\equiv\text{C}^-$ (D) $(\text{CH}_3)_3\text{C}^-$ 3. Explain why benzyl carbocation is more stable than tertiary butyl carbocation. 4. Draw the resonance structures for the allyl carbocation and explain its stability. 5. Predict the major product when 3-methyl-2-butanol reacts with $\text{HBr}$ (consider carbocation rearrangement). **Answers:** 1. $\text{CH}_2=\text{CH}^\cdot \text{sp}^2 > \text{sp}^3$. The carbon in $\text{HC}\equiv\text{C}^-$ is sp hybridized. 3. Benzyl carbocation is stabilized by resonance (delocalization of positive charge over the benzene ring). Tertiary butyl carbocation is stabilized by hyperconjugation (9 $\alpha$-hydrogens) and +I effect of three methyl groups. Resonance stabilization is generally more effective than hyperconjugation. 4. $\text{CH}_2=\text{CH}-\text{CH}_2^+ \leftrightarrow ^+\text{CH}_2-\text{CH}=\text{CH}_2$. The positive charge is delocalized over two carbon atoms, making it highly stable. 5. The initial carbocation formed is a secondary carbocation at C2. It can rearrange via a hydride shift from C3 to C2 to form a more stable tertiary carbocation at C3. Bromide then attacks the tertiary carbocation. $\text{CH}_3-\text{CH}(\text{OH})-\text{CH}(\text{CH}_3)_2 \xrightarrow{\text{H}^+} \text{CH}_3-\text{CH}^+-\text{CH}(\text{CH}_3)_2 \xrightarrow{\text{1,2-hydride shift}} \text{CH}_3-\text{CH}_2-\text{C}^+(\text{CH}_3)_2 \xrightarrow{\text{Br}^-} \text{CH}_3-\text{CH}_2-\text{CBr}(\text{CH}_3)_2$ (2-bromo-2-methylbutane). ### General Organic Chemistry (GOC) - Electrophiles and Nucleophiles #### Electrophiles (Electron-Loving Species) - **Definition:** Electron-deficient species that accept an electron pair. They are Lewis acids. - **Characteristics:** - Positively charged ions (e.g., $\text{H}^+$, $\text{NO}_2^+$, $\text{R}_3\text{C}^+$). - Neutral molecules with empty orbitals (e.g., $\text{BF}_3$, $\text{AlCl}_3$, $\text{SO}_3$). - Neutral molecules with polar multiple bonds where the atom connected to the more electronegative atom is electron-deficient (e.g., carbon in $\text{C}=\text{O}$ of aldehydes/ketones, carbon in $\text{CO}_2$). - Free radicals can act as electrophiles (due to unpaired electron and incomplete octet). - **Examples:** $\text{H}^+$, $\text{H}_3\text{O}^+$, $\text{NO}_2^+$, $\text{NO}^+$, $\text{Br}^+$, $\text{Cl}^+$, $\text{R}_3\text{C}^+$, $\text{BF}_3$, $\text{AlCl}_3$, $\text{FeCl}_3$, $\text{SO}_3$, $\text{CH}_3\text{CO}^+$, $\text{R}-\text{X}$ (alkyl halides), $\text{R}-\text{CHO}$ (carbonyl carbon). #### Nucleophiles (Nucleus-Loving Species) - **Definition:** Electron-rich species that donate an electron pair. They are Lewis bases. - **Characteristics:** - Negatively charged ions (e.g., $\text{OH}^-$, $\text{CN}^-$, $\text{R}-\text{O}^-$, $\text{R}-\text{S}^-$, $\text{X}^-$). - Neutral molecules with lone pairs (e.g., $\text{H}_2\text{O}$, $\text{NH}_3$, $\text{R}-\text{OH}$, $\text{R}-\text{NH}_2$, $\text{R}-\text{S}-\text{R}$). - Molecules with $\pi$-bonds (e.g., alkenes, alkynes, benzene) can act as nucleophiles. - **Examples:** $\text{OH}^-$, $\text{CN}^-$, $\text{RO}^-$, $\text{RS}^-$, $\text{NH}_2^-$, $\text{R}_3\text{N}$, $\text{H}_2\text{O}$, $\text{R}-\text{OH}$, $\text{R}-\text{SH}$, $\text{R}-\text{NH}_2$, $\text{C}_6\text{H}_6$. - **Ambident Nucleophiles:** Nucleophiles with two or more potential donor sites (e.g., $\text{CN}^-$, $\text{NO}_2^-$, $\text{SCN}^-$). #### Nucleophilicity vs. Basicity - **Basicity:** Thermodynamic property, measured by equilibrium constant for proton abstraction. Related to stability of conjugate acid. - **Nucleophilicity:** Kinetic property, measured by rate of reaction with an electrophile (usually carbon). - **Trends:** - **Across a period:** Basicity and nucleophilicity generally decrease with increasing electronegativity ($\text{NH}_2^- > \text{OH}^- > \text{F}^-$). - **Down a group (protic solvent):** Nucleophilicity increases with increasing size (polarizability) due to better solvation of smaller ions. Basicity decreases. - Example: $\text{I}^- > \text{Br}^- > \text{Cl}^- > \text{F}^-$ (Nucleophilicity in protic solvent). - Example: $\text{F}^- > \text{Cl}^- > \text{Br}^- > \text{I}^-$ (Basicity). - **Down a group (aprotic solvent):** Nucleophilicity generally follows basicity, as solvation effects are minimal. $\text{F}^- > \text{Cl}^- > \text{Br}^- > \text{I}^-$ (Nucleophilicity). - **Steric Hindrance:** Bulky nucleophiles are less nucleophilic due to steric hindrance (e.g., *tert*-butoxide is a strong base but poor nucleophile). - **Charge:** Anionic nucleophiles are generally stronger than neutral ones (e.g., $\text{OH}^- > \text{H}_2\text{O}$). #### Solved JEE Main Example **Q1:** Identify the electrophile and nucleophile in the following reaction: $\text{CH}_3-\text{Br} + \text{OH}^- \rightarrow \text{CH}_3-\text{OH} + \text{Br}^-$ **A1:** - **Electrophile:** $\text{CH}_3-\text{Br}$ (specifically, the carbon atom bonded to bromine, which is partially positive due to the polarity of the C-Br bond). - **Nucleophile:** $\text{OH}^-$ (hydroxide ion, which has a negative charge and lone pairs of electrons). #### Practice Questions 1. Which of the following is an ambident nucleophile? (A) $\text{Cl}^-$ (B) $\text{CN}^-$ (C) $\text{H}_2\text{O}$ (D) $\text{NH}_3$ 2. Arrange the following in increasing order of nucleophilicity in a protic solvent: $\text{F}^-$, $\text{Cl}^-$, $\text{Br}^-$, $\text{I}^-$. 3. Identify the electrophilic center in a carbonyl compound ($\text{R}_2\text{C}=\text{O}$). 4. Explain why $\text{BF}_3$ acts as an electrophile. 5. Is an alkene an electrophile or a nucleophile? Justify your answer. **Answers:** 1. (B) $\text{CN}^-$. It can attack through carbon (forming nitriles) or nitrogen (forming isonitriles). 2. $\text{F}^- ### General Organic Chemistry (GOC) - Types of Bond Fission #### Homolytic Fission (Homolysis) - **Definition:** Breaking of a covalent bond where each atom retains one of the shared electrons, leading to the formation of free radicals. - **Characteristics:** - Occurs symmetrically. - Requires high energy (heat, light, peroxides). - Results in species with unpaired electrons (free radicals). - Favored when the bond is nonpolar or when the resulting radicals are stable. - **Mechanism (using fishhook arrows):** $\text{A} \stackrel{\wedge}{:} \stackrel{\vee}{\text{B}} \rightarrow \text{A}^\cdot + \text{B}^\cdot$ (Each atom takes one electron from the shared pair) - **Examples:** - $\text{Cl}-\text{Cl} \xrightarrow{h\nu} \text{Cl}^\cdot + \text{Cl}^\cdot$ - $\text{CH}_3-\text{CH}_3 \xrightarrow{\Delta} \text{CH}_3^\cdot + \text{CH}_3^\cdot$ - **Reactions involving homolysis:** Free radical substitution (e.g., chlorination of alkanes), free radical addition (e.g., anti-Markovnikov addition). #### Heterolytic Fission (Heterolysis) - **Definition:** Breaking of a covalent bond where one atom retains both shared electrons, leading to the formation of ions (a carbocation and a carbanion, or other ions). - **Characteristics:** - Occurs asymmetrically. - Favored when there is a significant electronegativity difference between the bonded atoms. - Often assisted by polar solvents or catalysts. - Results in charged species (ions). - **Mechanism (using curved arrows):** $\text{A} \stackrel{\curvearrowright}{:} \text{B} \rightarrow \text{A}^+ + \text{B}^-$ (If B is more electronegative) $\text{A} : \stackrel{\curvearrowleft}{\text{B}} \rightarrow \text{A}^- + \text{B}^+$ (If A is more electronegative) (The more electronegative atom takes both electrons) - **Examples:** - $\text{CH}_3-\text{Cl} \rightarrow \text{CH}_3^+ + \text{Cl}^-$ (in polar solvent) - $\text{CH}_3-\text{MgBr} \rightarrow \text{CH}_3^- + \text{MgBr}^+$ - **Reactions involving heterolysis:** Ionic reactions like nucleophilic substitution, electrophilic addition, electrophilic aromatic substitution, elimination reactions. #### Comparison Table | Feature | Homolytic Fission | Heterolytic Fission | |------------------|----------------------------|-------------------------------| | Electron Movement| One electron per atom | Both electrons to one atom | | Products | Free radicals | Ions (carbocations/carbanions)| | Conditions | High temp, light, peroxides| Polar solvents, acids/bases | | Bond Polarity | Nonpolar or weakly polar | Polar | | Arrow Notation | Fishhook arrows ($\wedge$) | Curved arrows ($\curvearrowright$)| #### Solved JEE Main Example **Q1:** Which type of bond fission occurs during the chlorination of methane in the presence of UV light? What are the initial products? **A1:** Homolytic fission occurs. **Explanation:** In the presence of UV light, the $\text{Cl}-\text{Cl}$ bond undergoes homolytic fission to produce two chlorine free radicals ($\text{Cl}^\cdot$). These radicals then initiate the free radical substitution reaction of methane. $\text{Cl}-\text{Cl} \xrightarrow{h\nu} 2\text{Cl}^\cdot$ #### Practice Questions 1. What type of bond fission leads to the formation of carbocations? 2. Explain why $\text{CH}_3\text{CH}_2\text{Cl}$ undergoes heterolytic fission more readily in a polar solvent like water than in a nonpolar solvent like hexane. 3. Draw the products of homolytic fission of $\text{CH}_3-\text{O}-\text{OCH}_3$. 4. What is the role of peroxides in initiating free radical reactions? 5. When a C-C bond undergoes heterolytic fission, what are the two possible types of fragments formed? **Answers:** 1. Heterolytic fission. 2. In a polar solvent like water, the generated ions ($\text{CH}_3\text{CH}_2^+$ and $\text{Cl}^-$) can be solvated, which stabilizes them and lowers the activation energy for the fission. Nonpolar solvents cannot effectively stabilize ions. 3. $\text{CH}_3-\text{O}-\text{OCH}_3 \xrightarrow{\text{homolysis}} \text{CH}_3-\text{O}^\cdot + ^\cdot\text{OCH}_3$ (two methoxy free radicals). 4. Peroxides (like $\text{R}-\text{O}-\text{O}-\text{R}$) have weak $\text{O}-\text{O}$ bonds that readily undergo homolytic fission upon heating or light exposure, generating highly reactive free radicals that can initiate chain reactions. 5. Carbocation and carbanion. ### General Organic Chemistry (GOC) - Reaction Mechanisms Basics #### Definition of Reaction Mechanism - **Definition:** A detailed step-by-step description of how a chemical reaction occurs, showing the movement of electrons, bond breaking and formation, and the formation of intermediates. #### Types of Organic Reactions 1. **Substitution Reactions:** An atom or group in a molecule is replaced by another atom or group. - **Nucleophilic Substitution ($\text{S}_{\text{N}}$):** A nucleophile replaces a leaving group. - $\text{S}_{\text{N}}1$ (unimolecular): Two steps, carbocation intermediate, rate depends on only one reactant. - $\text{S}_{\text{N}}2$ (bimolecular): One step, concerted, transition state, rate depends on two reactants. - **Electrophilic Substitution ($\text{S}_{\text{E}}$):** An electrophile replaces an atom (often H). - Common in aromatic compounds (Electrophilic Aromatic Substitution). - **Free Radical Substitution ($\text{S}_{\text{R}}$):** A free radical replaces an atom (often H). - Example: Halogenation of alkanes. 2. **Addition Reactions:** Two or more molecules combine to form a larger single molecule. - **Electrophilic Addition ($\text{A}_{\text{E}}$):** Electrophile adds first. Common for alkenes and alkynes. - **Nucleophilic Addition ($\text{A}_{\text{N}}$):** Nucleophile adds first. Common for aldehydes and ketones. - **Free Radical Addition ($\text{A}_{\text{R}}$):** Free radical adds first. Example: HBr addition to alkenes in presence of peroxides. 3. **Elimination Reactions:** A small molecule (e.g., $\text{H}_2\text{O}$, $\text{HX}$) is removed from a larger molecule, leading to the formation of unsaturation (double or triple bond). - $\text{E}1$ (unimolecular): Two steps, carbocation intermediate. - $\text{E}2$ (bimolecular): One step, concerted. 4. **Rearrangement Reactions:** Atoms or groups within a molecule migrate to form a new isomer. - Often involve carbocation rearrangements. #### Key Concepts in Mechanism - **Curved Arrows:** Used to show the movement of electron pairs. - Tail starts at electron source (lone pair, bond). - Head points to electron sink (atom forming new bond, atom taking electrons from a breaking bond). - **Fishhook Arrows:** Used to show the movement of single electrons in free radical reactions. - **Leaving Group:** A group that departs with the bonding electrons. Good leaving groups are weak bases (e.g., $\text{Cl}^-$, $\text{Br}^-$, $\text{I}^-$, $\text{TsO}^-$, $\text{H}_2\text{O}$). - **Transition State:** A high-energy, unstable arrangement of atoms where bonds are partially broken and partially formed. It's a fleeting species. - **Reaction Intermediate:** A species formed during the reaction that is relatively stable but rapidly consumed in subsequent steps. (e.g., carbocations, carbanions, free radicals). - **Rate Determining Step (RDS):** The slowest step in a multi-step reaction mechanism, which determines the overall rate of the reaction. #### Energy Diagram of a Reaction - **Reactants:** Starting materials. - **Products:** Final molecules. - **Activation Energy ($\text{E}_{\text{a}}$):** Energy difference between reactants and transition state. Determines reaction rate. - **$\Delta\text{H}$ (Enthalpy Change):** Energy difference between reactants and products. Determines if reaction is exothermic ($\Delta\text{H} 0$). - **Multi-step Reaction:** Each step has its own transition state and activation energy. Intermediates are valleys between transition states. #### Memory Trick - **SN1/E1:** Think "One" means one molecule in rate-determining step, carbocation intermediate, two steps. - **SN2/E2:** Think "Two" means two molecules in rate-determining step, concerted, one step. #### Common Mistakes - Incorrect use of curved arrows (e.g., tail starting from a positive charge). - Forgetting to consider rearrangements in carbocation reactions. - Not identifying the correct leaving group or active nucleophile/electrophile. #### Solved JEE Main Example **Q1:** In an $\text{S}_{\text{N}}1$ reaction, what is the role of the solvent? **A1:** The solvent in an $\text{S}_{\text{N}}1$ reaction is typically a polar protic solvent (e.g., water, ethanol). Its role is to: 1. **Solvate and stabilize the carbocation intermediate:** The polar solvent molecules surround the positively charged carbocation, dispersing the charge and making it more stable, thereby facilitating its formation. 2. **Help in the departure of the leaving group:** The solvent also solvates the departing leaving group (anion), aiding its dissociation from the substrate. #### Practice Questions 1. Draw the curved arrows for the reaction of $\text{CH}_3\text{Cl}$ with $\text{OH}^-$ to form $\text{CH}_3\text{OH}$ and $\text{Cl}^-$. What type of reaction is this? 2. What is the difference between a transition state and a reaction intermediate? 3. Which of the following is a good leaving group: $\text{OH}^-$, $\text{NH}_2^-$, $\text{H}_2\text{O}$? Why? 4. Explain why the rate of an $\text{S}_{\text{N}}1$ reaction is independent of the concentration of the nucleophile. 5. In the electrophilic addition of $\text{HBr}$ to propene, which species attacks the double bond first? **Answers:** 1. $\text{OH}^- \curvearrowright \text{CH}_3-\text{Cl} \curvearrowleft \text{Cl}^-$. This is an $\text{S}_{\text{N}}2$ (nucleophilic substitution bimolecular) reaction. 2. A **transition state** is a fleeting, high-energy state where bonds are partially forming and breaking; it cannot be isolated. A **reaction intermediate** is a relatively stable (though reactive) species that exists for a measurable time and can sometimes be isolated or detected. It lies in an energy valley between two transition states. 3. $\text{H}_2\text{O}$ (water) is a good leaving group. Good leaving groups are weak bases. $\text{OH}^-$ and $\text{NH}_2^-$ are strong bases, hence poor leaving groups. $\text{H}_2\text{O}$ is the conjugate acid of $\text{OH}^-$, making it a very weak base and thus a good leaving group. 4. The rate-determining step of an $\text{S}_{\text{N}}1$ reaction involves only the unimolecular ionization of the substrate to form a carbocation. The nucleophile attacks the carbocation in a subsequent, faster step, so its concentration does not affect the overall rate. 5. $\text{H}^+$ (proton) from $\text{HBr}$. It acts as an electrophile and adds to the double bond to form a more stable secondary carbocation. ### General Organic Chemistry (GOC) - Isomerism #### Definition - **Isomers:** Compounds that have the same molecular formula but different structural or spatial arrangements of atoms. #### 1. Structural Isomerism (Constitutional Isomerism) - **Definition:** Isomers with the same molecular formula but different connectivity of atoms. - **Types:** a. **Chain Isomerism (Skeletal Isomerism):** Difference in the arrangement of the carbon skeleton. - Example: *n*-butane ($\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_3$) and isobutane ($\text{CH}_3\text{CH}(\text{CH}_3)\text{CH}_3$). b. **Position Isomerism:** Difference in the position of the functional group, substituent, or multiple bond on the same carbon skeleton. - Example: 1-chloropropane ($\text{CH}_3\text{CH}_2\text{CH}_2\text{Cl}$) and 2-chloropropane ($\text{CH}_3\text{CH}(\text{Cl})\text{CH}_3$). c. **Functional Group Isomerism:** Different functional groups. - Example: Ethanol ($\text{CH}_3\text{CH}_2\text{OH}$) and Dimethyl ether ($\text{CH}_3\text{OCH}_3$). (Alcohol and Ether) - Example: Propanal ($\text{CH}_3\text{CH}_2\text{CHO}$) and Propanone ($\text{CH}_3\text{COCH}_3$). (Aldehyde and Ketone) - Example: Carboxylic acid and Ester ($\text{CH}_3\text{COOH}$ and $\text{HCOOCH}_3$). d. **Metamerism:** Different alkyl groups attached to the same polyvalent functional group. - Example: Diethyl ether ($\text{CH}_3\text{CH}_2\text{OCH}_2\text{CH}_3$) and Methyl propyl ether ($\text{CH}_3\text{OCH}_2\text{CH}_2\text{CH}_3$). - Example: Pentan-3-one ($\text{CH}_3\text{CH}_2\text{COCH}_2\text{CH}_3$) and Methyl propyl ketone ($\text{CH}_3\text{COCH}_2\text{CH}_2\text{CH}_3$). e. **Tautomerism:** A special type of functional group isomerism where isomers exist in dynamic equilibrium, differing in the position of a proton and a double bond. - **Keto-enol Tautomerism:** Most common type. The keto form is generally more stable than the enol form, except when the enol is stabilized by aromaticity, intramolecular H-bonding, or extensive conjugation (e.g., 1,3-dicarbonyl compounds). - **Mechanism:** Involves proton transfer and electron rearrangement. Catalyzed by acids or bases. $$\text{R}-\text{CH}_2-\text{C}(=\text{O})-\text{R}' \rightleftharpoons \text{R}-\text{CH}=\text{C}(\text{OH})-\text{R}'$$ #### 2. Stereoisomerism - **Definition:** Isomers with the same molecular formula and connectivity but different spatial arrangements of atoms. - **Types:** a. **Conformational Isomerism (Rotational Isomerism):** Different spatial arrangements that arise from rotation around single bonds. - Not true isomers as they interconvert rapidly at room temperature. - **Ethane:** Staggered (more stable) and Eclipsed (less stable). - **Butane:** Anti (most stable), Gauche, Eclipsed, Fully Eclipsed (least stable). - **Cyclohexane:** Chair (most stable), Boat, Twist-boat. Chair form has axial and equatorial positions. Larger groups prefer equatorial position to minimize 1,3-diaxial interactions. b. **Configurational Isomerism:** Isomers that cannot be interconverted by simple rotation around single bonds (bond breaking and reforming is required). - **Geometrical Isomerism (cis-trans Isomerism):** - Occurs in compounds with restricted rotation around a double bond or in cyclic structures. - Requires two different groups on each carbon of the double bond (or on adjacent carbons in a ring). - **cis-isomer:** Same groups on the same side. - **trans-isomer:** Same groups on opposite sides. - **E/Z Notation:** For more complex cases. E (entgegen = opposite), Z (zusammen = together). Priority assigned using Cahn-Ingold-Prelog rules. - **Properties:** Differ in melting point, boiling point, dipole moment, stability (trans generally more stable due to less steric hindrance). - **Optical Isomerism (Enantiomerism/Diastereomerism):** - Occurs in compounds that are chiral. - **Chiral Center:** A carbon atom bonded to four different groups (asymmetric carbon). - **Chiral Molecule:** A molecule that is non-superimposable on its mirror image. - **Enantiomers:** Stereoisomers that are non-superimposable mirror images of each other. - Have identical physical properties (MP, BP, density, refractive index) except for their interaction with plane-polarized light (rotate it in opposite directions). - Have identical chemical properties except when reacting with other chiral molecules. - **Diastereomers:** Stereoisomers that are not mirror images of each other. - Have different physical and chemical properties. - **Meso Compounds:** Achiral compounds that contain chiral centers. They have an internal plane of symmetry. - **Racemic Mixture:** An equimolar mixture of two enantiomers. Optically inactive due to external compensation. - **Specific Rotation:** $[\alpha]_{\text{D}}^\text{T} = \frac{\alpha}{l \times c}$ (where $\alpha$ = observed rotation, $l$ = path length, $c$ = concentration). - **R/S Configuration:** Absolute configuration assigned using Cahn-Ingold-Prelog rules. 1. Assign priorities (higher atomic number = higher priority). 2. Orient molecule so lowest priority group (4) is pointing away. 3. Trace path from 1 -> 2 -> 3. - Clockwise = R (Rectus) - Counter-clockwise = S (Sinister) #### Memory Tricks & Common Mistakes - **Isomerism Hierarchy:** Molecular Formula -> (Different Connectivity?) -> Structural Isomers. If same connectivity -> Stereoisomers. - **Stereoisomers:** (Interconvert by rotation?) -> Conformational. If not -> Configurational. - **Configurational:** (Mirror Images?) -> Enantiomers. If not -> Diastereomers. - **Chiral vs. Achiral:** If a molecule has a plane of symmetry, it's achiral (even if it has chiral centers - meso compounds). - **Common Mistake:** Confusing cis/trans with E/Z. E/Z is more general. - **Common Mistake:** Assuming all molecules with chiral centers are chiral. Meso compounds are the exception. #### Solved JEE Main Example **Q1:** How many structural isomers are possible for $\text{C}_4\text{H}_{10}$? **A1:** Two structural isomers are possible for $\text{C}_4\text{H}_{10}$. **Explanation:** 1. *n*-Butane: $\text{CH}_3-\text{CH}_2-\text{CH}_2-\text{CH}_3$ (straight chain) 2. Isobutane (2-methylpropane): $\text{CH}_3-\text{CH}(\text{CH}_3)-\text{CH}_3$ (branched chain) #### Practice Questions 1. Draw all possible structural isomers for $\text{C}_3\text{H}_6\text{O}$ containing a carbonyl group. 2. Identify the type of isomerism between 1-butene and 2-butene. 3. How many chiral centers are present in 2,3-dibromobutane? Draw its stereoisomers. 4. Explain why 2-butanol is optically active but 2-methyl-2-butanol is not. 5. Assign R/S configuration to (2R,3S)-2,3-dibromobutane. (No need to draw, just understand the concept) **Answers:** 1. Three structural isomers: - Propanal ($\text{CH}_3\text{CH}_2\text{CHO}$) - aldehyde - Propanone ($\text{CH}_3\text{COCH}_3$) - ketone - Allyl alcohol ($\text{CH}_2=\text{CH}-\text{CH}_2\text{OH}$) - alcohol (contains C=C and -OH) 2. Position isomerism (the double bond is at a different position). 3. Two chiral centers (C2 and C3). - Stereoisomers: (2R,3R), (2S,3S) - enantiomers; (2R,3S) - meso compound (achiral due to plane of symmetry). Total 3 stereoisomers. 4. 2-butanol ($\text{CH}_3\text{CH}(\text{OH})\text{CH}_2\text{CH}_3$) has a chiral carbon (C2, bonded to H, OH, $\text{CH}_3$, $\text{CH}_2\text{CH}_3$), so it exists as enantiomers and is optically active. 2-methyl-2-butanol ($\text{CH}_3\text{C}(\text{OH})(\text{CH}_3)\text{CH}_2\text{CH}_3$) has no chiral carbon (C2 is bonded to two $\text{CH}_3$ groups), so it is achiral and optically inactive. 5. This question asks to understand the concept without drawing. (2R,3S)-2,3-dibromobutane is a meso compound. The 'R' at C2 implies a clockwise priority sequence for groups 1,2,3 when 4 is away. The 'S' at C3 implies a counter-clockwise priority sequence for groups 1,2,3 when 4 is away. ### Hydrocarbons - Alkanes #### Structure and Nomenclature - **General Formula:** $\text{C}_n\text{H}_{2n+2}$ - **Hybridization:** All carbon atoms are $\text{sp}^3$ hybridized, forming tetrahedral geometry. - **Bonding:** Only C-C and C-H single bonds. - **Nomenclature (IUPAC):** 1. Find the longest continuous carbon chain (parent chain). 2. Number the parent chain such that substituents get the lowest possible numbers. 3. Name substituents alphabetically. 4. Use prefixes (di-, tri-, tetra-) for multiple identical substituents. 5. Example: 2,2,4-trimethylpentane. #### Preparation of Alkanes 1. **Hydrogenation of Unsaturated Hydrocarbons:** - Alkenes/Alkynes $\xrightarrow{\text{H}_2/\text{Ni, Pt, or Pd}}$ Alkanes - $\text{R}-\text{CH}=\text{CH}-\text{R}' + \text{H}_2 \xrightarrow{\text{Ni, Pt, Pd}} \text{R}-\text{CH}_2-\text{CH}_2-\text{R}'$ 2. **Wurtz Reaction:** (For symmetrical alkanes, coupling of alkyl halides) - $2\text{R}-\text{X} + 2\text{Na} \xrightarrow{\text{Dry Ether}} \text{R}-\text{R} + 2\text{NaX}$ - **Limitations:** Poor yield for unsymmetrical alkanes (mixture of products). 3. **Corey-House Synthesis (Gilman Reagent):** (For symmetrical and unsymmetrical alkanes, better than Wurtz) - $\text{R}-\text{X} + 2\text{Li} \rightarrow \text{R}-\text{Li} + \text{LiX}$ - $2\text{R}-\text{Li} + \text{CuI} \rightarrow (\text{R})_2\text{CuLi}$ (Lithium dialkylcuprate, Gilman Reagent) - $(\text{R})_2\text{CuLi} + \text{R}'-\text{X} \rightarrow \text{R}-\text{R}' + \text{R}-\text{Cu} + \text{LiX}$ 4. **Decarboxylation of Carboxylic Acids:** - $\text{R}-\text{COONa} + \text{NaOH} \xrightarrow{\text{CaO}, \Delta} \text{R}-\text{H} + \text{Na}_2\text{CO}_3$ - Product alkane has one carbon less than the parent carboxylic acid. 5. **Kolbe's Electrolytic Method:** (For symmetrical alkanes) - $2\text{R}-\text{COONa} \xrightarrow{\text{Electrolysis}} \text{R}-\text{R} + 2\text{CO}_2 + 2\text{NaOH} + \text{H}_2$ - **Mechanism:** Free radical process. 6. **Reduction of Alkyl Halides:** - $\text{R}-\text{X} \xrightarrow{\text{Zn/HCl or LiAlH}_4 \text{ or Red P/HI}} \text{R}-\text{H}$ #### Physical Properties - Nonpolar, insoluble in water, soluble in organic solvents. - Boiling points increase with increasing molecular weight (due to increased van der Waals forces). - Branching decreases boiling point (less surface area for interaction). - Melting points generally increase with molecular weight, but highly symmetrical molecules have higher MP. #### Chemical Properties (Reactions) 1. **Combustion:** - $\text{C}_n\text{H}_{2n+2} + (\frac{3n+1}{2})\text{O}_2 \xrightarrow{\Delta} n\text{CO}_2 + (n+1)\text{H}_2\text{O} + \text{Heat}$ - Complete combustion gives $\text{CO}_2$ and $\text{H}_2\text{O}$. Incomplete combustion gives CO or C (soot). 2. **Halogenation (Free Radical Substitution):** - $\text{R}-\text{H} + \text{X}_2 \xrightarrow{h\nu \text{ or } \Delta} \text{R}-\text{X} + \text{HX}$ ($\text{X}_2 = \text{Cl}_2 \text{ or } \text{Br}_2$) - **Mechanism:** - **Initiation:** $\text{X}-\text{X} \xrightarrow{h\nu} 2\text{X}^\cdot$ (Homolytic fission) - **Propagation:** - $\text{X}^\cdot + \text{R}-\text{H} \rightarrow \text{H}-\text{X} + \text{R}^\cdot$ - $\text{R}^\cdot + \text{X}-\text{X} \rightarrow \text{R}-\text{X} + \text{X}^\cdot$ - **Termination:** Radicals combine (e.g., $\text{R}^\cdot + \text{R}^\cdot \rightarrow \text{R}-\text{R}$, $\text{X}^\cdot + \text{X}^\cdot \rightarrow \text{X}_2$, $\text{R}^\cdot + \text{X}^\cdot \rightarrow \text{R}-\text{X}$) - **Reactivity of H:** $3^\circ > 2^\circ > 1^\circ$ (due to stability of free radicals). - **Reactivity of Halogens:** $\text{F}_2 > \text{Cl}_2 > \text{Br}_2 > \text{I}_2$. Fluorination is explosive, iodination is reversible. - **Selectivity:** Bromination is more selective than chlorination (due to less reactive and more stable $\text{Br}^\cdot$). 3. **Nitration (vapour phase):** - $\text{R}-\text{H} + \text{HNO}_3 \xrightarrow{\text{400-500}^\circ\text{C}} \text{R}-\text{NO}_2 + \text{H}_2\text{O}$ 4. **Sulphonation (vapour phase):** - $\text{R}-\text{H} + \text{H}_2\text{SO}_4 \xrightarrow{\text{400}^\circ\text{C}} \text{R}-\text{SO}_3\text{H} + \text{H}_2\text{O}$ 5. **Isomerization:** - Straight chain alkanes $\xrightarrow{\text{Anhydrous AlCl}_3/\text{HCl}, \Delta}$ Branched alkanes. 6. **Aromatization:** - *n*-Hexane $\xrightarrow{\text{Cr}_2\text{O}_3/\text{Al}_2\text{O}_3, \text{500}^\circ\text{C}, \text{10-20 atm}}$ Benzene + $4\text{H}_2$ (for $\text{C}_6$ and higher, forms aromatic compounds). 7. **Cracking/Pyrolysis:** (Thermal decomposition) - Large alkanes $\xrightarrow{\text{Heat}}$ Smaller alkanes, alkenes, $\text{H}_2$. #### Memory Tricks & Common Mistakes - **Wurtz vs. Corey-House:** Wurtz for symmetrical, Corey-House for unsymmetrical. - **Halogenation Selectivity:** Bromination is "pickier" than chlorination. - **Common Mistake:** Forgetting that free radical halogenation leads to a mixture of products if multiple types of hydrogens are present. #### Solved JEE Main Example **Q1:** What is the major product when *n*-pentane is subjected to free radical chlorination? **A1:** The major product will be 2-chloropentane. **Explanation:** *n*-Pentane: $\text{CH}_3-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{CH}_3$ - C1 and C5 are primary (6 H's total) - C2 and C4 are secondary (4 H's total) - C3 is secondary (2 H's total) The order of reactivity for free radical halogenation is $3^\circ > 2^\circ > 1^\circ$. Although there are more primary hydrogens, the secondary hydrogens are more reactive. Between C2/C4 and C3, C2/C4 are equivalent and statistically more abundant than C3. However, the stability of the intermediate free radical determines the product. A secondary radical is more stable than a primary radical. Thus, attack at C2 (or C4) will be favored over C1 (or C5). The relative reactivity of $2^\circ \text{H} : 1^\circ \text{H}$ is approximately 3.8:1 for chlorination. Number of $2^\circ$ H's at C2/C4 = 4. Number of $1^\circ$ H's at C1/C5 = 6. So, $\text{Ratio} = (4 \times 3.8) : (6 \times 1) = 15.2 : 6$. So, substitution at C2/C4 is favored. Therefore, 2-chloropentane ($\text{CH}_3-\text{CHCl}-\text{CH}_2-\text{CH}_2-\text{CH}_3$) is the major product. #### Practice Questions 1. Complete the reaction: $\text{CH}_3\text{CH}_2\text{COONa} + \text{NaOH} \xrightarrow{\text{CaO}, \Delta}$ 2. Write the initiation and one propagation step for the free radical bromination of propane. 3. Why is Wurtz reaction not suitable for preparing unsymmetrical alkanes? 4. Predict the main product of the reaction of 2,2-dimethylpropane with $\text{Cl}_2$ in the presence of $\text{UV}$ light. 5. Which alkane would give benzene upon aromatization? **Answers:** 1. $\text{CH}_3\text{CH}_2\text{COONa} + \text{NaOH} \xrightarrow{\text{CaO}, \Delta} \text{CH}_3\text{CH}_3 + \text{Na}_2\text{CO}_3$ (Ethane) 2. **Initiation:** $\text{Br}-\text{Br} \xrightarrow{h\nu} 2\text{Br}^\cdot$ **Propagation:** $\text{Br}^\cdot + \text{CH}_3\text{CH}_2\text{CH}_3 \rightarrow \text{HBr} + \text{CH}_3\text{CH}^\cdot\text{CH}_3$ (or $\text{CH}_3\text{CH}_2\text{CH}_2^\cdot$) 3. If two different alkyl halides are used ($\text{R}-\text{X}$ and $\text{R}'-\text{X}$), a mixture of three alkanes ($\text{R}-\text{R}$, $\text{R}'-\text{R}'$, and $\text{R}-\text{R}'$) will be formed, making separation difficult and yield of desired unsymmetrical alkane low. 4. 2,2-dimethylpropane (neopentane) has only primary hydrogens. All 12 hydrogens are equivalent. So, only one product is formed: 1-chloro-2,2-dimethylpropane. 5. *n*-Hexane (or cyclohexane, which is an isomer of *n*-hexane). ### Hydrocarbons - Alkenes #### Structure and Nomenclature - **General Formula:** $\text{C}_n\text{H}_{2n}$ (for monounsaturated acyclic alkenes) - **Hybridization:** Carbon atoms involved in the double bond are $\text{sp}^2$ hybridized, trigonal planar geometry. - **Bonding:** Contains at least one C=C double bond (one $\sigma$ and one $\pi$ bond). - **Nomenclature (IUPAC):** 1. Longest carbon chain containing the double bond is the parent chain. 2. Number the chain to give the double bond carbons the lowest possible numbers. 3. Suffix is "-ene". Indicate position of double bond. 4. Substituents named as usual. 5. Cis/trans or E/Z notation for geometrical isomers. #### Preparation of Alkenes 1. **Dehydration of Alcohols:** - $\text{R}-\text{CH}_2-\text{CH}_2-\text{OH} \xrightarrow{\text{Conc. H}_2\text{SO}_4/\Delta \text{ or } \text{Al}_2\text{O}_3/\Delta} \text{R}-\text{CH}=\text{CH}_2 + \text{H}_2\text{O}$ - **Mechanism:** $\text{E}1$ (for $2^\circ, 3^\circ$ alcohols) or $\text{E}2$ (for $1^\circ$ alcohols). Carbocation intermediate in E1. - **Reactivity:** $3^\circ > 2^\circ > 1^\circ$. - **Saytzeff's Rule:** In elimination reactions, the major product is the most substituted alkene (more stable). 2. **Dehydrohalogenation of Alkyl Halides:** - $\text{R}-\text{CH}_2-\text{CH}_2-\text{X} \xrightarrow{\text{Alc. KOH}, \Delta} \text{R}-\text{CH}=\text{CH}_2 + \text{HX}$ - **Mechanism:** E2 reaction (concerted). - **Reactivity:** $3^\circ > 2^\circ > 1^\circ$. Reactivity of halide: $\text{I} > \text{Br} > \text{Cl}$. - **Saytzeff's Rule:** Followed for major product. 3. **Dehalogenation of Vicinal Dihalides:** - $\text{R}-\text{CH}(\text{X})-\text{CH}(\text{X})-\text{R}' \xrightarrow{\text{Zn dust/Alcohol}, \Delta} \text{R}-\text{CH}=\text{CH}-\text{R}' + \text{ZnX}_2$ 4. **Partial Hydrogenation of Alkynes:** - $\text{R}-\text{C}\equiv\text{C}-\text{R}' + \text{H}_2 \xrightarrow{\text{Pd/BaSO}_4 \text{ (Lindlar's catalyst) or Na/liq. NH}_3} \text{R}-\text{CH}=\text{CH}-\text{R}'$ - **Lindlar's catalyst:** Gives *cis*-alkene. - **Na/liq. $\text{NH}_3$ (Birch Reduction):** Gives *trans*-alkene. 5. **Wittig Reaction:** (Forms alkene from carbonyl compound) - $\text{R}_2\text{C}=\text{O} + \text{Ph}_3\text{P}=\text{CR}'_2 \rightarrow \text{R}_2\text{C}=\text{CR}'_2 + \text{Ph}_3\text{P}=\text{O}$ #### Physical Properties - Nonpolar, insoluble in water, soluble in organic solvents. - Boiling points increase with molecular weight. - Branching decreases boiling point. - *Cis*-alkenes generally have higher boiling points and dipole moments than *trans*-alkenes due to dipole moments not cancelling out. #### Chemical Properties (Reactions) - Electrophilic Addition The $\pi$-bond is an electron-rich region, so alkenes undergo electrophilic addition reactions. 1. **Addition of Hydrogen (Hydrogenation):** - $\text{R}-\text{CH}=\text{CH}_2 + \text{H}_2 \xrightarrow{\text{Ni, Pt, or Pd}} \text{R}-\text{CH}_2-\text{CH}_3$ - *Syn*-addition (both H's add from the same side). 2. **Addition of Halogens (Halogenation):** - $\text{R}-\text{CH}=\text{CH}_2 + \text{X}_2 \xrightarrow{\text{CCl}_4} \text{R}-\text{CH}(\text{X})-\text{CH}_2\text{X}$ (Vicinal dihalide) - **Mechanism:** Forms a cyclic halonium ion intermediate. *Anti*-addition (halogens add from opposite sides). - Test for unsaturation (decolorizes $\text{Br}_2$ water). 3. **Addition of Hydrogen Halides (HX):** - $\text{R}-\text{CH}=\text{CH}_2 + \text{HX} \rightarrow \text{R}-\text{CH}(\text{X})-\text{CH}_3$ - **Markovnikov's Rule:** The negative part of the adding reagent adds to the carbon atom of the double bond that has fewer hydrogen atoms. (The hydrogen adds to the carbon with more hydrogens). - **Mechanism:** Carbocation intermediate. Rearrangements possible. - **Anti-Markovnikov Addition (Peroxide Effect/Kharasch Effect):** Only for HBr in the presence of peroxides. - $\text{R}-\text{CH}=\text{CH}_2 + \text{HBr} \xrightarrow{\text{Peroxides}} \text{R}-\text{CH}_2-\text{CH}_2\text{Br}$ - **Mechanism:** Free radical addition. 4. **Addition of Water (Hydration):** - $\text{R}-\text{CH}=\text{CH}_2 + \text{H}_2\text{O} \xrightarrow{\text{H}_2\text{SO}_4} \text{R}-\text{CH}(\text{OH})-\text{CH}_3$ (Alcohol) - Follows Markovnikov's rule. Carbocation intermediate, rearrangements possible. - Alternative methods: Hydroboration-oxidation (anti-Markovnikov addition of H and OH, *syn*-addition), Oxymercuration-demercuration (Markovnikov addition of H and OH, *anti*-addition). 5. **Oxidation Reactions:** - **Baeyer's Reagent (Cold, Dilute, Alkaline $\text{KMnO}_4$):** - $\text{R}-\text{CH}=\text{CH}-\text{R}' \xrightarrow{\text{Cold, dil. alk. KMnO}_4} \text{R}-\text{CH}(\text{OH})-\text{CH}(\text{OH})-\text{R}'$ (Vicinal diol/Glycol) - *Syn*-dihydroxylation. Test for unsaturation (purple $\text{KMnO}_4$ decolorizes). - **Hot, Acidic $\text{KMnO}_4$ or $\text{K}_2\text{Cr}_2\text{O}_7$ or $\text{O}_3$ (Ozonolysis):** Cleaves the double bond. - **Ozonolysis:** - $\text{R}_2\text{C}=\text{CR}'_2 \xrightarrow{1. \text{O}_3, \text{CCl}_4; 2. \text{Zn/H}_2\text{O} \text{ or } (\text{CH}_3)_2\text{S}} \text{R}_2\text{C}=\text{O} + \text{R}'_2\text{C}=\text{O}$ (Aldehydes/Ketones) - **Reductive work-up (Zn/$\text{H}_2\text{O}$ or DMS):** Aldehydes are formed from $\text{R}-\text{CH}=$. - **Oxidative work-up ($\text{H}_2\text{O}_2$):** Aldehydes are further oxidized to carboxylic acids. - **Strong Oxidation (Hot $\text{KMnO}_4$):** - $\text{R}-\text{CH}=\text{CH}-\text{R}' \xrightarrow{\text{Hot KMnO}_4} \text{R}-\text{COOH} + \text{R}'-\text{COOH}$ (Carboxylic acids) - Terminal alkene ($\text{R}-\text{CH}=\text{CH}_2$) gives $\text{R}-\text{COOH} + \text{CO}_2 + \text{H}_2\text{O}$. - Internal alkene with tertiary carbon ($\text{R}_2\text{C}=\text{CR}'-\text{R}''$) gives ketone and carboxylic acid. - Terminal carbon with two H's ($\text{CH}_2=$) gives $\text{CO}_2$. 6. **Polymerization:** - *n* ($\text{CH}_2=\text{CH}_2$) $\xrightarrow{\text{High Temp/Pressure, Catalyst}}$ $(-\text{CH}_2-\text{CH}_2-)_n$ (Polyethylene) #### Memory Tricks & Common Mistakes - **Markovnikov's Rule:** "Rich get richer" (H goes to the carbon with more H's). - **Anti-Markovnikov:** Only HBr with peroxides (Free radical mechanism). - **Ozonolysis:** "Cut the double bond in half, add oxygen to each side." Reductive work-up preserves aldehydes, oxidative converts to acids. - **Saytzeff's Rule:** "More substituted alkene is major product." - **Common Mistake:** Forgetting carbocation rearrangements in Markovnikov additions (hydration, HX addition). - **Common Mistake:** Confusing *syn*-addition and *anti*-addition. #### Solved JEE Main Example **Q1:** What are the products when 2-methylbut-2-ene reacts with $\text{O}_3$ followed by $\text{Zn/H}_2\text{O}$? **A1:** The products are propanone (acetone) and ethanal (acetaldehyde). **Explanation:** 2-methylbut-2-ene: $\text{CH}_3-\text{CH}=\text{C}(\text{CH}_3)_2$ Ozonolysis cleaves the double bond and adds oxygen to each carbon. $\text{CH}_3-\text{CH}=\text{C}(\text{CH}_3)_2 \xrightarrow{1. \text{O}_3; 2. \text{Zn/H}_2\text{O}} \text{CH}_3-\text{CHO} + \text{CH}_3-\text{CO}-\text{CH}_3$ (Ethanal) + (Propanone) #### Practice Questions 1. Predict the major product when 3-methylbut-1-ene reacts with HBr. 2. Suggest reagents to convert propene to propan-1-ol. 3. What is the stereochemistry of the dihydroxylation of *cis*-but-2-ene using Baeyer's reagent? 4. Identify the alkene that would give only propanone on ozonolysis followed by reductive work-up. 5. Distinguish between *cis* and *trans* isomers of but-2-ene based on their physical properties. **Answers:** 1. 3-methylbut-1-ene ($\text{CH}_3-\text{CH}(\text{CH}_3)-\text{CH}=\text{CH}_2$) Addition of HBr follows Markovnikov's rule. The initial carbocation formed at C2 ($\text{CH}_3-\text{CH}(\text{CH}_3)-\text{CH}^+-\text{CH}_3$) is secondary. A 1,2-hydride shift can occur to form a more stable tertiary carbocation at C3 ($\text{CH}_3-\text{C}^+(\text{CH}_3)-\text{CH}_2-\text{CH}_3$). Thus, the major product is 2-bromo-2-methylbutane. 2. Hydroboration-oxidation: $\text{CH}_3-\text{CH}=\text{CH}_2 \xrightarrow{1. \text{BH}_3/\text{THF}; 2. \text{H}_2\text{O}_2/\text{OH}^-} \text{CH}_3-\text{CH}_2-\text{CH}_2-\text{OH}$ (Propan-1-ol) 3. Baeyer's reagent causes *syn*-dihydroxylation. When *cis*-but-2-ene undergoes *syn*-dihydroxylation, it yields a *meso* compound (2,3-butanediol). 4. 2,3-dimethylbut-2-ene ($\text{CH}_3-\text{C}(\text{CH}_3)=\text{C}(\text{CH}_3)-\text{CH}_3$). Ozonolysis would cleave the double bond to give two molecules of propanone ($\text{CH}_3\text{COCH}_3$). 5. *cis*-But-2-ene has a net dipole moment because the methyl groups are on the same side, causing a slight charge separation. *trans*-But-2-ene has zero dipole moment because the dipole moments of the C-CH3 bonds cancel out. Therefore, *cis*-but-2-ene has a higher boiling point than *trans*-but-2-ene. *trans*-isomer is generally more stable due to less steric hindrance. ### Hydrocarbons - Alkynes #### Structure and Nomenclature - **General Formula:** $\text{C}_n\text{H}_{2n-2}$ (for monounsaturated acyclic alkynes) - **Hybridization:** Carbon atoms involved in the triple bond are $\text{sp}$ hybridized, linear geometry. - **Bonding:** Contains at least one $\text{C}\equiv\text{C}$ triple bond (one $\sigma$ and two $\pi$ bonds). - **Nomenclature (IUPAC):** 1. Longest carbon chain containing the triple bond is the parent chain. 2. Number the chain to give the triple bond carbons the lowest possible numbers. 3. Suffix is "-yne". Indicate position of triple bond. 4. Substituents named as usual. 5. Terminal alkynes have acidic hydrogens. #### Preparation of Alkynes 1. **Dehydrohalogenation of Vicinal or Geminal Dihalides:** - $\text{R}-\text{CH}(\text{X})-\text{CH}_2\text{X} \xrightarrow{\text{Alc. KOH}, \Delta} \text{R}-\text{CH}=\text{CHX} \xrightarrow{\text{NaNH}_2 \text{ (strong base)}} \text{R}-\text{C}\equiv\text{CH}$ (Terminal alkyne) - $\text{R}-\text{CX}_2-\text{CH}_3 \xrightarrow{\text{Alc. KOH}, \Delta} \text{R}-\text{C}(\text{X})=\text{CH}_2 \xrightarrow{\text{NaNH}_2 \text{ (strong base)}} \text{R}-\text{C}\equiv\text{CH}$ (Terminal alkyne) - Requires a strong base like $\text{NaNH}_2$ (sodamide) for the second elimination to form the triple bond, especially for terminal alkynes. 2. **From Tetrahalides:** - $\text{R}-\text{CX}_2-\text{CX}_2-\text{R}' \xrightarrow{\text{Zn dust}, \Delta} \text{R}-\text{C}\equiv\text{C}-\text{R}'$ 3. **Addition of $\text{CH}_3\text{I}$ to Sodium Acetylide:** (Chain lengthening) - $\text{CH}\equiv\text{C}^-\text{Na}^+ + \text{CH}_3\text{I} \rightarrow \text{CH}\equiv\text{C}-\text{CH}_3 + \text{NaI}$ - Only for terminal alkynes. #### Physical Properties - Nonpolar, insoluble in water, soluble in organic solvents. - Boiling points increase with molecular weight. - Terminal alkynes are slightly polar. #### Chemical Properties (Reactions) 1. **Acidity of Terminal Alkynes:** - Due to $\text{sp}$ hybridization, the C-H bond in terminal alkynes is polarized, and the hydrogen atom is acidic. - $\text{R}-\text{C}\equiv\text{C}-\text{H} + \text{Na} \rightarrow \text{R}-\text{C}\equiv\text{C}^-\text{Na}^+ + \frac{1}{2}\text{H}_2$ - $\text{R}-\text{C}\equiv\text{C}-\text{H} + \text{NaNH}_2 \rightarrow \text{R}-\text{C}\equiv\text{C}^-\text{Na}^+ + \text{NH}_3$ - Form acetylides with heavy metal ions (e.g., $\text{AgNO}_3$, $\text{CuCl}$), which are used as a test for terminal alkynes. - $\text{R}-\text{C}\equiv\text{C}-\text{H} + \text{AgNO}_3 \xrightarrow{\text{NH}_4\text{OH}} \text{R}-\text{C}\equiv\text{C}-\text{Ag}\downarrow$ (White ppt) - $\text{R}-\text{C}\equiv\text{C}-\text{H} + \text{CuCl} \xrightarrow{\text{NH}_4\text{OH}} \text{R}-\text{C}\equiv\text{C}-\text{Cu}\downarrow$ (Red ppt) 2. **Hydrogenation:** - Full hydrogenation: $\text{R}-\text{C}\equiv\text{C}-\text{R}' + 2\text{H}_2 \xrightarrow{\text{Ni, Pt, or Pd}} \text{R}-\text{CH}_2-\text{CH}_2-\text{R}'$ (Alkane) - Partial hydrogenation: - $\text{R}-\text{C}\equiv\text{C}-\text{R}' + \text{H}_2 \xrightarrow{\text{Pd/BaSO}_4 \text{ (Lindlar's catalyst)}} \text{R}-\text{CH}=\text{CH}-\text{R}'$ (*cis*-alkene) - $\text{R}-\text{C}\equiv\text{C}-\text{R}' + \text{H}_2 \xrightarrow{\text{Na/liq. NH}_3 \text{ (Birch Reduction)}} \text{R}-\text{CH}=\text{CH}-\text{R}'$ (*trans*-alkene) 3. **Addition of Halogens:** - $\text{R}-\text{C}\equiv\text{C}-\text{R}' + \text{X}_2 \xrightarrow{\text{CCl}_4} \text{R}-\text{CX}=\text{CX}-\text{R}'$ (Dihaloalkene) - $\text{R}-\text{CX}=\text{CX}-\text{R}' + \text{X}_2 \xrightarrow{\text{CCl}_4} \text{R}-\text{CX}_2-\text{CX}_2-\text{R}'$ (Tetrahaloalkane) - Decolorizes $\text{Br}_2$ water. 4. **Addition of Hydrogen Halides (HX):** - $\text{R}-\text{C}\equiv\text{C}-\text{H} + \text{HX} \rightarrow \text{R}-\text{C}(\text{X})=\text{CH}_2$ (Vinyl halide) - $\text{R}-\text{C}(\text{X})=\text{CH}_2 + \text{HX} \rightarrow \text{R}-\text{CX}_2-\text{CH}_3$ (Geminal dihalide) - Follows Markovnikov's Rule. 5. **Addition of Water (Hydration):** - $\text{R}-\text{C}\equiv\text{C}-\text{H} + \text{H}_2\text{O} \xrightarrow{\text{HgSO}_4/\text{H}_2\text{SO}_4} [\text{R}-\text{C}(\text{OH})=\text{CH}_2] \rightleftharpoons \text{R}-\text{CO}-\text{CH}_3$ (Ketone, via enol-keto tautomerism) - For acetylene ($\text{HC}\equiv\text{CH}$), it gives acetaldehyde ($\text{CH}_3\text{CHO}$). - Follows Markovnikov's Rule. 6. **Oxidation (Ozonolysis):** - $\text{R}-\text{C}\equiv\text{C}-\text{R}' \xrightarrow{1. \text{O}_3; 2. \text{H}_2\text{O}} \text{R}-\text{COOH} + \text{R}'-\text{COOH}$ (Carboxylic acids) - Terminal alkynes ($\text{R}-\text{C}\equiv\text{C}-\text{H}$) give $\text{R}-\text{COOH} + \text{CO}_2 + \text{H}_2\text{O}$. 7. **Polymerization:** - Linear polymerization: Acetylene $\xrightarrow{\text{CuCl}} \text{Vinylacetylene} \xrightarrow{\text{CuCl}} \text{Divinylacetylene}$ - Cyclic polymerization: $3\text{HC}\equiv\text{CH} \xrightarrow{\text{Red hot Fe tube, } 873\text{K}} \text{Benzene}$ #### Memory Tricks & Common Mistakes - **Terminal Alkyne Acidity:** Remember the test with $\text{AgNO}_3$ and $\text{CuCl}$. - **Partial Hydrogenation:** Lindlar gives *cis*, Birch gives *trans*. - **Hydration:** Always leads to a carbonyl compound (aldehyde from acetylene, ketone from other alkynes) via enol tautomerization. - **Common Mistake:** Forgetting the two-step Markovnikov addition for HX or $\text{X}_2$ to alkynes. #### Solved JEE Main Example **Q1:** What is the product when propyne reacts with excess HBr? **A1:** The product is 2,2-dibromopropane. **Explanation:** Propyne ($\text{CH}_3-\text{C}\equiv\text{CH}$) First addition of HBr: Following Markovnikov's rule, H adds to C1 and Br to C2. $\text{CH}_3-\text{C}\equiv\text{CH} + \text{HBr} \rightarrow \text{CH}_3-\text{C}(\text{Br})=\text{CH}_2$ (2-bromopropene) Second addition of HBr: Again, following Markovnikov's rule, H adds to C1 (which now has one H) and Br to C2 (which is now more substituted). $\text{CH}_3-\text{C}(\text{Br})=\text{CH}_2 + \text{HBr} \rightarrow \text{CH}_3-\text{CBr}_2-\text{CH}_3$ (2,2-dibromopropane, a geminal dihalide) #### Practice Questions 1. How can you distinguish between propyne and propene using a chemical test? (Give two tests) 2. What reagents are needed to convert 1,2-dibromopropane to propyne? 3. Predict the product when but-1-yne is hydrated in the presence of $\text{HgSO}_4/\text{H}_2\text{SO}_4$. 4. Why are terminal alkynes acidic while alkanes and alkenes are not? 5. What happens when acetylene is passed through a red hot iron tube? **Answers:** 1. **Test 1 (Tollen's Reagent/Ammoniacal $\text{AgNO}_3$):** Propyne (terminal alkyne) will react to form a white precipitate of silver propynylide. Propene (alkene) will not react. **Test 2 (Ammoniacal $\text{CuCl}$):** Propyne will form a red precipitate of cuprous propynylide. Propene will not react. 2. 1,2-dibromopropane ($\text{CH}_3-\text{CHBr}-\text{CH}_2\text{Br}$) is a vicinal dihalide. $\text{CH}_3-\text{CHBr}-\text{CH}_2\text{Br} \xrightarrow{\text{Alc. KOH}, \Delta} \text{CH}_3-\text{CH}=\text{CHBr} \xrightarrow{\text{NaNH}_2} \text{CH}_3-\text{C}\equiv\text{CH}$ (Propyne) 3. But-1-yne ($\text{CH}_3-\text{CH}_2-\text{C}\equiv\text{CH}$) upon hydration follows Markovnikov's rule, forming an enol intermediate which tautomerizes to a ketone. $\text{CH}_3-\text{CH}_2-\text{C}\equiv\text{CH} \xrightarrow{\text{HgSO}_4/\text{H}_2\text{SO}_4} [\text{CH}_3-\text{CH}_2-\text{C}(\text{OH})=\text{CH}_2] \rightleftharpoons \text{CH}_3-\text{CH}_2-\text{CO}-\text{CH}_3$ (Butan-2-one) 4. Terminal alkynes have an $\text{sp}$ hybridized carbon atom directly bonded to hydrogen. The $\text{sp}$ hybrid orbital has 50% s-character, meaning the electrons in the C-H bond are held closer to the nucleus and are more electronegative. This makes the C-H bond more polar and the hydrogen more acidic, allowing it to be removed as a proton. Alkanes ($\text{sp}^3$) and alkenes ($\text{sp}^2$) have less s-character and are not acidic. 5. Acetylene undergoes cyclic polymerization to form benzene. $3\text{HC}\equiv\text{CH} \xrightarrow{\text{Red hot Fe tube, } 873\text{K}} \text{C}_6\text{H}_6$ (Benzene) ### Hydrocarbons - Aromatic Hydrocarbons #### Definition - Aromatic compounds are cyclic, planar molecules with a conjugated system of $\pi$-electrons that exhibit enhanced stability due to delocalization. Benzene is the simplest aromatic hydrocarbon. #### Aromaticity and Huckel Rule - **Conditions for Aromaticity (Hückel's Rule):** 1. **Cyclic:** The molecule must be cyclic. 2. **Planar:** All atoms in the ring must be $\text{sp}^2$ or $\text{sp}$ hybridized to allow for continuous overlap of p-orbitals (essential for delocalization). 3. **Conjugated:** The ring must have a continuous system of p-orbitals (no $\text{sp}^3$ carbons interrupting the conjugation). 4. **($4n+2$) $\pi$-electrons:** The cyclic conjugated system must contain ($4n+2$) $\pi$-electrons, where *n* is an integer (0, 1, 2, 3, ...). - *n*=0: 2 $\pi$-electrons (e.g., cyclopropenyl cation) - *n*=1: 6 $\pi$-electrons (e.g., benzene, pyridine, pyrrole, furan, thiophene) - *n*=2: 10 $\pi$-electrons (e.g., naphthalene) - **Anti-aromatic Compounds:** Cyclic, planar, conjugated systems with $4n$ $\pi$-electrons. They are highly unstable. (e.g., cyclobutadiene, cyclooctatetraene is non-planar to avoid anti-aromaticity). - **Non-aromatic Compounds:** Compounds that do not meet all criteria for aromaticity or anti-aromaticity (e.g., cyclooctatetraene is non-planar, making it non-aromatic). #### Structure of Benzene - $\text{C}_6\text{H}_6$. All carbons are $\text{sp}^2$ hybridized. - Planar hexagonal structure. All C-C bond lengths are identical (139 pm, intermediate between single and double bond). - Each carbon has an unhybridized p-orbital perpendicular to the ring, which overlap to form a continuous $\pi$-electron cloud above and below the ring. This delocalization is responsible for its high stability (resonance energy). #### Preparation of Benzene 1. **Cyclic Polymerization of Ethyne (Acetylene):** - $3\text{HC}\equiv\text{CH} \xrightarrow{\text{Red hot Fe tube, } 873\text{K}} \text{C}_6\text{H}_6$ 2. **Decarboxylation of Benzoic Acid:** - $\text{C}_6\text{H}_5\text{COONa} + \text{NaOH} \xrightarrow{\text{CaO}, \Delta} \text{C}_6\text{H}_6 + \text{Na}_2\text{CO}_3$ 3. **Reduction of Phenol:** - $\text{C}_6\text{H}_5\text{OH} + \text{Zn} \xrightarrow{\Delta} \text{C}_6\text{H}_6 + \text{ZnO}$ 4. **From Chlorobenzene:** - $\text{C}_6\text{H}_5\text{Cl} + 2\text{H} \xrightarrow{\text{Ni/Al alloy + NaOH}} \text{C}_6\text{H}_6 + \text{HCl}$ (Catalytic reduction) 5. **Aromatization of *n*-Hexane:** - $\text{CH}_3(\text{CH}_2)_4\text{CH}_3 \xrightarrow{\text{Cr}_2\text{O}_3/\text{Al}_2\text{O}_3, \text{500}^\circ\text{C}, \text{10-20 atm}} \text{C}_6\text{H}_6 + 4\text{H}_2$ #### Chemical Properties - Electrophilic Aromatic Substitution (EAS) - Benzene undergoes EAS reactions due to its electron-rich $\pi$-cloud. It does not undergo addition reactions easily (to preserve aromaticity). - **General Mechanism:** 1. **Generation of Electrophile ($\text{E}^+$):** Reagent reacts to form a strong electrophile. 2. **Attack of Electrophile on Benzene:** Benzene acts as a nucleophile, attacking the electrophile. This forms a resonance-stabilized carbocation intermediate (arenium ion or $\sigma$-complex). Aromaticity is temporarily lost. 3. **Loss of Proton:** A base abstracts a proton from the carbon bearing the electrophile, restoring aromaticity. 1. **Nitration:** - $\text{C}_6\text{H}_6 + \text{HNO}_3 \xrightarrow{\text{Conc. H}_2\text{SO}_4, \Delta} \text{C}_6\text{H}_5\text{NO}_2 + \text{H}_2\text{O}$ (Nitrobenzene) - **Electrophile:** $\text{NO}_2^+$ (Nitronium ion), generated from $\text{HNO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{NO}_2^+ + \text{HSO}_4^- + \text{H}_2\text{O}$. 2. **Halogenation:** - $\text{C}_6\text{H}_6 + \text{X}_2 \xrightarrow{\text{FeX}_3 \text{ (Lewis acid)}} \text{C}_6\text{H}_5\text{X} + \text{HX}$ ($\text{X}_2 = \text{Cl}_2 \text{ or } \text{Br}_2$) - **Electrophile:** $\text{X}^+$ (or $\text{X}^\delta+-\text{X}-\text{FeX}_3^\delta-$ complex), generated from $\text{X}_2 + \text{FeX}_3 \rightarrow \text{X}^+[\text{FeX}_4^-]$. 3. **Sulfonation:** - $\text{C}_6\text{H}_6 + \text{Conc. H}_2\text{SO}_4 \xrightarrow{\Delta} \text{C}_6\text{H}_5\text{SO}_3\text{H} + \text{H}_2\text{O}$ (Benzenesulfonic acid) - **Electrophile:** $\text{SO}_3$ (sulfur trioxide), generated from $2\text{H}_2\text{SO}_4 \rightleftharpoons \text{H}_3\text{O}^+ + \text{HSO}_4^- + \text{SO}_3$. - Reversible reaction. 4. **Friedel-Crafts Alkylation:** - $\text{C}_6\text{H}_6 + \text{R}-\text{X} \xrightarrow{\text{Anhydrous AlCl}_3} \text{C}_6\text{H}_5-\text{R} + \text{HX}$ (Alkylbenzene) - **Electrophile:** $\text{R}^+$ (carbocation), generated from $\text{R}-\text{X} + \text{AlCl}_3 \rightarrow \text{R}^+[\text{AlCl}_4^-]$. - **Limitations:** - Polyalkylation (product is more reactive than reactant). - Rearrangements of carbocations (e.g., $1^\circ$ alkyl halides can rearrange to $2^\circ$ or $3^\circ$ carbocations). - Cannot be used with strongly deactivating groups on benzene (e.g., $-\text{NO}_2$). - Cannot be used with aryl or vinyl halides. 5. **Friedel-Crafts Acylation:** - $\text{C}_6\text{H}_6 + \text{R}-\text{CO}-\text{Cl} \xrightarrow{\text{Anhydrous AlCl}_3} \text{C}_6\text{H}_5-\text{CO}-\text{R} + \text{HCl}$ (Acylbenzene/Ketone) - **Electrophile:** $\text{R}-\text{CO}^+$ (acylium ion), resonance stabilized. - **Advantages over alkylation:** No polyacylation (acyl group is deactivating), no rearrangements. #### Directing Effects of Substituents in EAS - When a substituted benzene undergoes EAS, the existing substituent directs the incoming electrophile to specific positions (*ortho*, *meta*, or *para*) and also affects the reaction rate. - **Activating Groups (o,p-directing):** Electron-donating groups (EDG) increase electron density of the ring, especially at *ortho* and *para* positions, making the ring more reactive towards electrophiles. - Examples: $-\text{NH}_2$, $-\text{NHR}$, $-\text{NR}_2$, $-\text{OH}$, $-\text{OR}$, $-\text{NHCOCH}_3$, $-\text{C}_6\text{H}_5$, $-\text{R}$, $-\text{CH}=\text{CH}_2$. - **Halogens (-F, -Cl, -Br, -I):** Deactivating but *o,p*-directing (due to strong -I effect and weaker +R effect). The +R effect directs, while the -I effect deactivates. - **Deactivating Groups (m-directing):** Electron-withdrawing groups (EWG) decrease electron density of the ring, especially at *ortho* and *para* positions, making the ring less reactive. They direct electrophiles to *meta* position. - Examples: $-\text{NO}_2$, $-\text{CN}$, $-\text{CHO}$, $-\text{COOH}$, $-\text{COOR}$, $-\text{SO}_3\text{H}$, $-\text{COR}$, $-\text{NR}_3^+$. #### Other Reactions of Benzene 1. **Addition Reactions (under harsh conditions):** - Hydrogenation: $\text{C}_6\text{H}_6 + 3\text{H}_2 \xrightarrow{\text{Ni, Pt, or Pd}, \text{High Temp/Pressure}} \text{C}_6\text{H}_{12}$ (Cyclohexane) - Chlorination: $\text{C}_6\text{H}_6 + 3\text{Cl}_2 \xrightarrow{h\nu} \text{C}_6\text{H}_6\text{Cl}_6$ (Benzene Hexachloride, BHC or Lindane) 2. **Oxidation (Combustion):** Burns with a sooty flame. #### Memory Tricks & Common Mistakes - **Aromaticity:** Cyclic, Planar, Conjugated, $4n+2 \pi$-electrons. - **EAS Mechanism:** Electrophile generation -> Arenium ion -> Proton loss. - **Friedel-Crafts Alkylation limitations:** Polyalkylation, rearrangements, deactivated rings. - **Directing Effects:** Activators are *o,p*-directing, Deactivators are *m*-directing (except halogens). - **Common Mistake:** Confusing activating/deactivating with directing effects. Halogens are deactivating but *o,p*-directing. #### Solved JEE Main Example **Q1:** Predict the major product when toluene undergoes nitration. **A1:** The major products will be *o*-nitrotoluene and *p*-nitrotoluene. **Explanation:** Toluene is methylbenzene. The methyl group ($-\text{CH}_3$) is an electron-donating group (+I effect and hyperconjugation). It is an activating group and *ortho*,*para*-director. Therefore, the incoming nitronium ion ($\text{NO}_2^+$) will predominantly attack the *ortho* and *para* positions. *Para* product is usually favored due to less steric hindrance. #### Practice Questions 1. Explain why cyclobutadiene is anti-aromatic. 2. What is the electrophile in Friedel-Crafts acylation, and why does it not undergo rearrangements? 3. Arrange the following compounds in increasing order of reactivity towards electrophilic substitution: Benzene, Toluene, Nitrobenzene. 4. Draw the resonance structures of the arenium ion formed during the nitration of benzene. 5. Predict the major product when bromobenzene undergoes Friedel-Crafts alkylation with $\text{CH}_3\text{Cl}$ in presence of $\text{AlCl}_3$. **Answers:** 1. Cyclobutadiene is cyclic, planar, and fully conjugated. However, it has 4 $\pi$-electrons ($4n$ where $n=1$). According to Hückel's rule, systems with $4n$ $\pi$-electrons are anti-aromatic and highly unstable. 2. The electrophile in Friedel-Crafts acylation is the acylium ion ($\text{R}-\text{CO}^+$). It is resonance-stabilized (e.g., $\text{R}-\text{C}^+=\text{O} \leftrightarrow \text{R}-\text{C}\equiv\text{O}^+$), which prevents it from undergoing rearrangements. 3. Nitrobenzene ### Haloalkanes and Haloarenes #### Haloalkanes (Alkyl Halides, $\text{R}-\text{X}$) - **Nomenclature:** IUPAC (Haloalkanes) or Common (Alkyl halides). - **Nature of C-X Bond:** Polar, with carbon being partially positive and halogen partially negative. This polarity makes them susceptible to nucleophilic attack. #### Preparation of Haloalkanes 1. **From Alcohols:** - $\text{R}-\text{OH} + \text{HX} \xrightarrow{\text{ZnCl}_2 \text{ (Lucas Reagent for HCl)}} \text{R}-\text{X} + \text{H}_2\text{O}$ - $\text{R}-\text{OH} + \text{PCl}_5 \rightarrow \text{R}-\text{Cl} + \text{POCl}_3 + \text{HCl}$ - $\text{R}-\text{OH} + \text{PCl}_3 \rightarrow 3\text{R}-\text{Cl} + \text{H}_3\text{PO}_3$ - $\text{R}-\text{OH} + \text{SOCl}_2 \xrightarrow{\text{Pyridine}} \text{R}-\text{Cl} + \text{SO}_2 \uparrow + \text{HCl} \uparrow$ (Darzens process, best for purity as byproducts are gases). - $\text{R}-\text{OH} + \text{Red P}/\text{X}_2 \rightarrow \text{R}-\text{X}$ (for $\text{Br}_2$, $\text{I}_2$) 2. **From Alkanes (Free Radical Halogenation):** - $\text{R}-\text{H} + \text{X}_2 \xrightarrow{h\nu \text{ or } \Delta} \text{R}-\text{X} + \text{HX}$ (Mixture of isomers, poor for specific products). 3. **From Alkenes:** - **Addition of HX:** $\text{R}-\text{CH}=\text{CH}_2 + \text{HX} \rightarrow \text{R}-\text{CH}(\text{X})-\text{CH}_3$ (Markovnikov's rule). - Anti-Markovnikov with HBr/peroxides. - **Addition of $\text{X}_2$:** $\text{R}-\text{CH}=\text{CH}_2 + \text{X}_2 \xrightarrow{\text{CCl}_4} \text{R}-\text{CH}(\text{X})-\text{CH}_2\text{X}$ (Vicinal dihalide). 4. **Halogen Exchange Reactions:** - **Finkelstein Reaction:** $\text{R}-\text{X} + \text{NaI} \xrightarrow{\text{Acetone}, \Delta} \text{R}-\text{I} + \text{NaX}\downarrow$ (for preparing alkyl iodides, $\text{NaX}$ precipitates). - **Swarts Reaction:** $\text{R}-\text{Br}/\text{Cl} + \text{AgF}/\text{Hg}_2\text{F}_2/\text{CoF}_2/\text{SbF}_3 \rightarrow \text{R}-\text{F} + \text{AgBr}/\text{Cl}$ (for preparing alkyl fluorides). #### Chemical Properties of Haloalkanes **A. Nucleophilic Substitution Reactions ($\text{S}_{\text{N}}1$ and $\text{S}_{\text{N}}2$)** - **$\text{S}_{\text{N}}2$ (Bimolecular Nucleophilic Substitution):** - **Mechanism:** Concerted, one-step reaction. Nucleophile attacks from the backside, simultaneously forming a new bond and breaking the C-X bond through a single transition state. - **Stereochemistry:** Inversion of configuration (Walden inversion). - **Rate:** Rate $= k[\text{R}-\text{X}][\text{Nu}^-]$. - **Reactivity of Alkyl Halides:** $\text{CH}_3\text{X} > 1^\circ > 2^\circ > 3^\circ$ (Steric hindrance). - **Reactivity of Leaving Group:** $\text{I}^- > \text{Br}^- > \text{Cl}^- > \text{F}^-$. (Weaker base = better leaving group). - **Nucleophile:** Strong, unhindered nucleophiles favored. - **Solvent:** Polar aprotic solvents (e.g., DMSO, acetone, DMF) increase rate by not solvating nucleophile. - **$\text{S}_{\text{N}}1$ (Unimolecular Nucleophilic Substitution):** - **Mechanism:** Two-step reaction. 1. Slow, rate-determining step: Ionization of alkyl halide to form a carbocation (planar intermediate) and a leaving group. 2. Fast step: Nucleophile attacks the carbocation. - **Stereochemistry:** Racemization (attack from both sides of planar carbocation), though often slight preference for inversion. - **Rate:** Rate $= k[\text{R}-\text{X}]$. - **Reactivity of Alkyl Halides:** $3^\circ > 2^\circ > 1^\circ > \text{CH}_3\text{X}$ (Carbocation stability). - **Reactivity of Leaving Group:** $\text{I}^- > \text{Br}^- > \text{Cl}^- > \text{F}^-$. - **Nucleophile:** Weak nucleophiles favored (solvent often acts as nucleophile). - **Solvent:** Polar protic solvents (e.g., $\text{H}_2\text{O}$, alcohols) stabilize carbocation. **B. Elimination Reactions (E1 and E2)** - **Dehydrohalogenation:** Elimination of HX to form an alkene. - **$\text{E}2$ (Bimolecular Elimination):** - **Mechanism:** Concerted, one-step. Strong base abstracts a $\beta$-hydrogen, while the leaving group departs, forming a double bond. - **Stereochemistry:** *Anti*-periplanar geometry preferred for $\text{H}$ and $\text{X}$. - **Rate:** Rate $= k[\text{R}-\text{X}][\text{Base}]$. - **Reactivity of Alkyl Halides:** $3^\circ > 2^\circ > 1^\circ$. - **Reactivity of Leaving Group:** $\text{I}^- > \text{Br}^- > \text{Cl}^- > \text{F}^-$. - **Base:** Strong, bulky bases (e.g., *tert*-BuOK) favor E2. - **Product:** Follows Saytzeff's Rule (most substituted alkene is major) unless a bulky base is used (then Hofmann product - less substituted alkene). - **$\text{E}1$ (Unimolecular Elimination):** - **Mechanism:** Two-step. 1. Slow, rate-determining step: Formation of carbocation. 2. Fast step: Base abstracts $\beta$-hydrogen from carbocation. - **Rate:** Rate $= k[\text{R}-\text{X}]$. - **Reactivity of Alkyl Halides:** $3^\circ > 2^\circ$. (Carbocation stability). - **Product:** Follows Saytzeff's Rule. - **Competition between $\text{S}_{\text{N}}1/\text{E}1$ and $\text{S}_{\text{N}}2/\text{E}2$:** - **Substrate:** $1^\circ$ favors $\text{S}_{\text{N}}2$. $3^\circ$ favors $\text{S}_{\text{N}}1/\text{E}1/\text{E}2$. $2^\circ$ is mixed. - **Nucleophile/Base:** Strong Nu/Base favors $\text{S}_{\text{N}}2/\text{E}2$. Weak Nu/Base favors $\text{S}_{\text{N}}1/\text{E}1$. Bulky base favors E2. - **Solvent:** Polar protic favors $\text{S}_{\text{N}}1/\text{E}1$. Polar aprotic favors $\text{S}_{\text{N}}2$. - **Temperature:** High temperature favors elimination (E). **C. Other Reactions:** 1. **Reaction with Metals:** - **Grignard Reagent:** $\text{R}-\text{X} + \text{Mg} \xrightarrow{\text{Dry Ether}} \text{R}-\text{MgX}$ (Organomagnesium halide). - **Wurtz Reaction:** $2\text{R}-\text{X} + 2\text{Na} \xrightarrow{\text{Dry Ether}} \text{R}-\text{R} + 2\text{NaX}$. - **Corey-House Reaction:** For unsymmetrical alkanes. 2. **Reduction:** $\text{R}-\text{X} \xrightarrow{\text{Zn/HCl or LiAlH}_4} \text{R}-\text{H}$. #### Haloarenes (Aryl Halides, $\text{Ar}-\text{X}$) - Halogen directly attached to an aromatic ring. - **Nature of C-X bond:** - **Resonance:** Lone pair on halogen is delocalized into the benzene ring, giving partial double bond character to C-X bond. This makes the bond shorter and stronger than in haloalkanes. - **$\text{sp}^2$ Hybridization:** Carbon of the C-X bond is $\text{sp}^2$ hybridized, which is more electronegative than $\text{sp}^3$ carbon, making the bond shorter and stronger. - These factors make haloarenes less reactive towards nucleophilic substitution than haloalkanes. #### Preparation of Haloarenes 1. **Halogenation of Benzene (EAS):** - $\text{C}_6\text{H}_6 + \text{X}_2 \xrightarrow{\text{FeX}_3} \text{C}_6\text{H}_5\text{X} + \text{HX}$ 2. **Sandmeyer Reaction:** (From diazonium salts) - $\text{Ar}-\text{N}_2^+\text{Cl}^- \xrightarrow{\text{CuCl/HCl or CuBr/HBr}} \text{Ar}-\text{Cl}/\text{Ar}-\text{Br} + \text{N}_2$ 3. **Gattermann Reaction:** (Similar to Sandmeyer, uses Cu powder instead of $\text{Cu}_2\text{X}_2$) - $\text{Ar}-\text{N}_2^+\text{Cl}^- \xrightarrow{\text{Cu powder/HCl or HBr}} \text{Ar}-\text{Cl}/\text{Ar}-\text{Br} + \text{N}_2$ 4. **Balz-Schiemann Reaction:** (For fluorobenzene) - $\text{Ar}-\text{N}_2^+\text{Cl}^- \xrightarrow{1. \text{HBF}_4; 2. \Delta} \text{Ar}-\text{F} + \text{BF}_3 + \text{N}_2$ #### Chemical Properties of Haloarenes 1. **Nucleophilic Substitution Reactions:** - Generally unreactive towards $\text{S}_{\text{N}}1$ and $\text{S}_{\text{N}}2$ due to: - Partial double bond character of C-X bond. - $\text{sp}^2$ hybridized carbon of C-X bond (more stable C-X bond). - Instability of phenyl carbocation (if $\text{S}_{\text{N}}1$). - Repulsion between nucleophile and electron-rich $\pi$-cloud (if $\text{S}_{\text{N}}2$). - **Exceptions (Activated Nucleophilic Aromatic Substitution):** - Presence of strong electron-withdrawing groups (EWG) at *ortho* and *para* positions (e.g., $-\text{NO}_2$) activates the ring towards nucleophilic substitution. - Example: Chlorobenzene $\xrightarrow{\text{NaOH}, 623\text{K}, 300\text{atm}} \text{Phenol}$ (Dow's process, harsh conditions). - *o, p*-Nitrophenol can be prepared under much milder conditions. - **Benzyne Mechanism:** Reactions with very strong bases (e.g., $\text{NaNH}_2$) can proceed via a benzyne intermediate. 2. **Electrophilic Aromatic Substitution (EAS):** - Halogens are deactivating but *ortho, para*-directing. - Example: Chlorination of chlorobenzene gives *o*-dichlorobenzene and *p*-dichlorobenzene. 3. **Reaction with Metals:** - **Wurtz-Fittig Reaction:** $\text{Ar}-\text{X} + \text{R}-\text{X} + 2\text{Na} \xrightarrow{\text{Dry Ether}} \text{Ar}-\text{R} + 2\text{NaX}$ (Alkylarene). - **Fittig Reaction:** $2\text{Ar}-\text{X} + 2\text{Na} \xrightarrow{\text{Dry Ether}} \text{Ar}-\text{Ar} + 2\text{NaX}$ (Biaryl). - **Ullmann Reaction:** $2\text{Ar}-\text{I} \xrightarrow{\text{Cu}, \Delta} \text{Ar}-\text{Ar} + \text{CuI}_2$. - **Grignard Reagent:** $\text{Ar}-\text{X} + \text{Mg} \xrightarrow{\text{Dry Ether}} \text{Ar}-\text{MgX}$. #### Memory Tricks & Common Mistakes - **Haloalkane Reactivity:** $3^\circ$ for $\text{S}_{\text{N}}1/\text{E}1$ (carbocation stability), $1^\circ$ for $\text{S}_{\text{N}}2$ (steric hindrance). - **Haloarene Reactivity:** Less reactive for $\text{S}_{\text{N}}$ due to C-X bond strength and $\pi$-electron repulsion. - **Common Mistake:** Forgetting that halogens are *deactivating* but *o,p*-directing in EAS. - **Common Mistake:** Confusing different named reactions for coupling (Wurtz, Fittig, Wurtz-Fittig). #### Solved JEE Main Example **Q1:** Predict the major product when 1-bromobutane reacts with ethanolic KOH. **A1:** The major product is but-1-ene. **Explanation:** 1-bromobutane is a primary alkyl halide. Ethanolic KOH is a strong base and favors E2 elimination. Since it's a primary halide with only one $\beta$-carbon, it can only form but-1-ene. If it were a secondary or tertiary halide, Saytzeff's rule would apply for the major product. $\text{CH}_3-\text{CH}_2-\text{CH}_2-\text{CH}_2\text{Br} \xrightarrow{\text{Alc. KOH}} \text{CH}_3-\text{CH}_2-\text{CH}=\text{CH}_2 + \text{KBr} + \text{H}_2\text{O}$ #### Practice Questions 1. Explain why chlorobenzene is less reactive than chlorocyclohexane towards nucleophilic substitution. 2. What is the product when 2-bromopropane reacts with $\text{KCN}$ in an aqueous alcoholic solution? 3. Give the reagents for the conversion of aniline to bromobenzene. 4. Why is $\text{SOCl}_2$ preferred for converting alcohols to alkyl chlorides? 5. Predict the products when *p*-chlorotoluene is subjected to nitration. **Answers:** 1. Chlorobenzene is less reactive than chlorocyclohexane towards nucleophilic substitution due to: - Resonance stabilization of the C-Cl bond in chlorobenzene (partial double bond character). - $\text{sp}^2$ hybridized carbon of C-Cl bond in chlorobenzene (stronger bond). - Repulsion between the nucleophile and the electron-rich $\pi$-electron cloud of the benzene ring. 2. 2-bromopropane is a secondary alkyl halide. $\text{KCN}$ is a strong nucleophile. It will undergo $\text{S}_{\text{N}}2$ reaction (predominantly) to form 2-cyanopropane (isobutyl cyanide). $\text{CH}_3-\text{CHBr}-\text{CH}_3 + \text{KCN} \rightarrow \text{CH}_3-\text{CH}(\text{CN})-\text{CH}_3 + \text{KBr}$ 3. Aniline $\xrightarrow{\text{NaNO}_2/\text{HCl}, 0-5^\circ\text{C}} \text{Benzenediazonium chloride}$ Benzenediazonium chloride $\xrightarrow{\text{CuBr/HBr}} \text{Bromobenzene}$ (Sandmeyer reaction) 4. $\text{SOCl}_2$ (thionyl chloride) is preferred because the byproducts ($\text{SO}_2$ and $\text{HCl}$) are gases that escape easily, leaving behind a pure alkyl chloride. 5. *p*-Chlorotoluene has two substituents: -Cl (deactivating, *o,p*-directing) and $-\text{CH}_3$ (activating, *o,p*-directing). The activating effect of $-\text{CH}_3$ will dominate. The $-\text{CH}_3$ group directs to *ortho* (C2, C6) and *para* (C4). Since C4 is already occupied by -Cl, the nitronium ion will attack C2 and C6. The -Cl group directs to *ortho* (C2, C6) and *para* (C4). Therefore, the incoming $-\text{NO}_2$ group will predominantly attack the positions *ortho* to the methyl group and *ortho* to the chlorine atom (which are the same positions, C2 and C6). The major products will be 4-chloro-2-nitrotoluene and 4-chloro-3-nitrotoluene (due to attack at C2 and C6 relative to methyl, and C3 and C5 relative to chlorine). Since methyl is activating, it dominates. So, 4-chloro-2-nitrotoluene and 4-chloro-6-nitrotoluene (same as 2-chloro-4-nitrotoluene). The most favorable positions are *ortho* to $-\text{CH}_3$ and *meta* to $-\text{Cl}$. Thus, 4-chloro-2-nitrotoluene and 4-chloro-6-nitrotoluene will be formed. ### Alcohols, Phenols and Ethers #### Alcohols ($\text{R}-\text{OH}$) - **Nomenclature:** Suffix "-ol". Number chain to give -OH lowest number. - **Classification:** Primary ($1^\circ$), Secondary ($2^\circ$), Tertiary ($3^\circ$) based on the number of alkyl groups attached to the carbon bearing the -OH group. #### Preparation of Alcohols 1. **From Alkenes:** - **Hydration:** $\text{R}-\text{CH}=\text{CH}_2 + \text{H}_2\text{O} \xrightarrow{\text{H}^+} \text{R}-\text{CH}(\text{OH})-\text{CH}_3$ (Markovnikov, rearrangements possible). - **Hydroboration-Oxidation:** $\text{R}-\text{CH}=\text{CH}_2 \xrightarrow{1. \text{BH}_3/\text{THF}; 2. \text{H}_2\text{O}_2/\text{OH}^-} \text{R}-\text{CH}_2-\text{CH}_2-\text{OH}$ (Anti-Markovnikov, *syn*-addition). - **Oxymercuration-Demercuration:** $\text{R}-\text{CH}=\text{CH}_2 \xrightarrow{1. \text{Hg(OAc)}_2/\text{H}_2\text{O}; 2. \text{NaBH}_4} \text{R}-\text{CH}(\text{OH})-\text{CH}_3$ (Markovnikov, no rearrangements). 2. **From Carbonyl Compounds (Reduction):** - **Aldehydes:** $\text{RCHO} \xrightarrow{\text{LiAlH}_4 \text{ or } \text{NaBH}_4 \text{ or } \text{H}_2/\text{Ni, Pt, Pd}} \text{RCH}_2\text{OH}$ ($1^\circ$ alcohol). - **Ketones:** $\text{RCOR}' \xrightarrow{\text{LiAlH}_4 \text{ or } \text{NaBH}_4 \text{ or } \text{H}_2/\text{Ni, Pt, Pd}} \text{RCH}(\text{OH})\text{R}'$ ($2^\circ$ alcohol). - **Carboxylic Acids/Esters:** $\text{RCOOH/RCOOR}' \xrightarrow{\text{LiAlH}_4} \text{RCH}_2\text{OH}$ ($1^\circ$ alcohol, $\text{NaBH}_4$ cannot reduce acids/esters). 3. **From Grignard Reagents ($\text{R}-\text{MgX}$):** - **Formaldehyde:** $\text{HCHO} + \text{RMgX} \rightarrow \text{RCH}_2\text{OMgX} \xrightarrow{\text{H}_2\text{O}/\text{H}^+} \text{RCH}_2\text{OH}$ ($1^\circ$ alcohol). - **Other Aldehydes:** $\text{R}'\text{CHO} + \text{RMgX} \rightarrow \text{RR}'\text{CHOMgX} \xrightarrow{\text{H}_2\text{O}/\text{H}^+} \text{RR}'\text{CHOH}$ ($2^\circ$ alcohol). - **Ketones:** $\text{R}'\text{COR}'' + \text{RMgX} \rightarrow \text{RR}'\text{R}''\text{COMgX} \xrightarrow{\text{H}_2\text{O}/\text{H}^+} \text{RR}'\text{R}''\text{COH}$ ($3^\circ$ alcohol). - **Esters:** React with 2 moles of Grignard reagent to form $3^\circ$ alcohols (except formate esters give $2^\circ$ alcohols). 4. **From Primary Amines (via Diazotization):** - $\text{RCH}_2\text{NH}_2 \xrightarrow{\text{NaNO}_2/\text{HCl}} [\text{RCH}_2\text{N}_2^+] \xrightarrow{\text{H}_2\text{O}} \text{RCH}_2\text{OH} + \text{N}_2 \uparrow$ (Rearrangements are common). #### Physical Properties of Alcohols - Higher boiling points than corresponding alkanes, alkyl halides, or ethers due to hydrogen bonding. - Lower alcohols (up to 3 carbons) are miscible with water due to hydrogen bonding with water. Miscibility decreases with increasing carbon chain length. #### Chemical Properties of Alcohols **A. Reactions involving C-O bond cleavage (Reactivity order: $3^\circ > 2^\circ > 1^\circ$)** 1. **Reaction with Hydrogen Halides (HX):** - $\text{R}-\text{OH} + \text{HX} \xrightarrow{\text{ZnCl}_2 \text{ (for HCl)}} \text{R}-\text{X} + \text{H}_2\text{O}$ - **Lucas Test:** Distinguishes $1^\circ, 2^\circ, 3^\circ$ alcohols using $\text{HCl}/\text{ZnCl}_2$. - $3^\circ$: Immediate turbidity. - $2^\circ$: Turbidity within 5-10 minutes. - $1^\circ$: No turbidity at room temperature. 2. **Reaction with $\text{PCl}_5$, $\text{PCl}_3$, $\text{SOCl}_2$:** Forms alkyl halides. 3. **Dehydration:** - $\text{R}-\text{CH}_2-\text{CH}_2-\text{OH} \xrightarrow{\text{Conc. H}_2\text{SO}_4/\Delta \text{ or } \text{Al}_2\text{O}_3/\Delta} \text{R}-\text{CH}=\text{CH}_2 + \text{H}_2\text{O}$ (Alkene, follows Saytzeff's rule). - **Mechanism:** $\text{E}1$ (for $2^\circ, 3^\circ$) or $\text{E}2$ (for $1^\circ$). Rearrangements possible. **B. Reactions involving O-H bond cleavage (Acidity)** 1. **Reaction with Active Metals:** - $2\text{R}-\text{OH} + 2\text{Na} \rightarrow 2\text{R}-\text{O}^-\text{Na}^+ + \text{H}_2$ (Forms sodium alkoxide). 2. **Esterification:** - $\text{R}-\text{OH} + \text{R}'-\text{COOH} \rightleftharpoons \text{R}'-\text{COOR} + \text{H}_2\text{O}$ (Acid catalyzed, reversible). - $\text{R}-\text{OH} + \text{R}'-\text{COCl} \rightarrow \text{R}'-\text{COOR} + \text{HCl}$ (Acyl chloride, irreversible). - $\text{R}-\text{OH} + (\text{R}'\text{CO})_2\text{O} \rightarrow \text{R}'-\text{COOR} + \text{R}'-\text{COOH}$ (Acid anhydride). 3. **Reaction with Carboxylic Acids (Esterification)** **C. Oxidation Reactions** 1. **$1^\circ$ Alcohols:** - $\text{RCH}_2\text{OH} \xrightarrow{\text{PCC} \text{ (Pyridinium chlorochromate, mild)}} \text{RCHO}$ (Aldehyde). - $\text{RCH}_2\text{OH} \xrightarrow{\text{KMnO}_4/\text{H}^+ \text{ or } \text{K}_2\text{Cr}_2\text{O}_7/\text{H}^+ \text{ (strong)}} \text{RCOOH}$ (Carboxylic acid). 2. **$2^\circ$ Alcohols:** - $\text{RCH}(\text{OH})\text{R}' \xrightarrow{\text{PCC} \text{ or } \text{KMnO}_4/\text{H}^+ \text{ or } \text{K}_2\text{Cr}_2\text{O}_7/\text{H}^+} \text{RCOR}'$ (Ketone). 3. **$3^\circ$ Alcohols:** - Resistant to oxidation under mild conditions. Under strong conditions, dehydration to alkene followed by oxidation. 4. **Catalytic Dehydrogenation (Cu at 573K):** - $1^\circ$: Alcohol $\rightarrow$ Aldehyde. - $2^\circ$: Alcohol $\rightarrow$ Ketone. - $3^\circ$: Alcohol $\rightarrow$ Alkene (dehydration). #### Phenols ($\text{Ar}-\text{OH}$) - **Definition:** Hydroxyl group directly attached to a benzene ring. - **Acidity:** More acidic than alcohols due to resonance stabilization of the phenoxide ion (negative charge delocalized into the ring). Weaker than carboxylic acids. - **Effect of Substituents on Acidity:** - EWG (*o,p*-directing) increase acidity (e.g., nitrophenols). - EDG (*o,p*-directing) decrease acidity (e.g., cresols). #### Preparation of Phenols 1. **From Haloarenes (Dow's Process):** - Chlorobenzene $\xrightarrow{\text{NaOH}, 623\text{K}, 300\text{atm}} \text{Sodium phenoxide} \xrightarrow{\text{H}^+} \text{Phenol}$. 2. **From Benzenediazonium Salt:** - $\text{Ar}-\text{N}_2^+\text{Cl}^- \xrightarrow{\text{H}_2\text{O}, \Delta} \text{Ar}-\text{OH} + \text{N}_2 + \text{HCl}$. 3. **From Cumene (Isopropylbenzene):** - Cumene $\xrightarrow{1. \text{O}_2; 2. \text{H}^+} \text{Phenol} + \text{Acetone}$ (Important industrial method). 4. **From Benzenesulfonic Acid:** - $\text{C}_6\text{H}_5\text{SO}_3\text{Na} + \text{NaOH} \xrightarrow{\Delta} \text{Sodium phenoxide} \xrightarrow{\text{H}^+} \text{Phenol}$. #### Chemical Properties of Phenols 1. **Acidity:** - Reacts with $\text{NaOH}$ to form sodium phenoxide. - Does not react with $\text{NaHCO}_3$ (weaker acid than carbonic acid). 2. **Electrophilic Aromatic Substitution (EAS):** - -OH group is strongly activating and *ortho, para*-directing. - **Halogenation (Bromination):** - Phenol $\xrightarrow{\text{Br}_2/\text{H}_2\text{O}} 2,4,6-\text{Tribromophenol}$ (White ppt, very reactive). - Phenol $\xrightarrow{\text{Br}_2/\text{CS}_2 \text{ or } \text{CHCl}_3} \text{o-Bromophenol} + \text{p-Bromophenol}$ (Milder conditions). - **Nitration:** - Phenol $\xrightarrow{\text{Dil. HNO}_3} \text{o-Nitrophenol} + \text{p-Nitrophenol}$. - Phenol $\xrightarrow{\text{Conc. HNO}_3} 2,4,6-\text{Trinitrophenol}$ (Picric acid, explosive). - **Sulfonation:** - Phenol $\xrightarrow{\text{Conc. H}_2\text{SO}_4, \text{low temp}} \text{o-Phenolsulfonic acid}$. - Phenol $\xrightarrow{\text{Conc. H}_2\text{SO}_4, \text{high temp}} \text{p-Phenolsulfonic acid}$. - **Kolbe's Reaction:** Phenol $\xrightarrow{1. \text{NaOH}; 2. \text{CO}_2, 400\text{K}, 4-7 \text{atm}; 3. \text{H}^+} \text{Salicylic acid}$. - **Reimer-Tiemann Reaction:** Phenol $\xrightarrow{1. \text{CHCl}_3/\text{NaOH}; 2. \text{H}^+} \text{Salicylaldehyde}$. 3. **Reaction with Zinc Dust:** Phenol $\xrightarrow{\text{Zn dust}, \Delta} \text{Benzene}$. 4. **Oxidation:** - Phenol $\xrightarrow{\text{K}_2\text{Cr}_2\text{O}_7/\text{H}^+} \text{Benzoquinone}$. 5. **Coupling Reaction (Azo Dye Formation):** With diazonium salts. #### Ethers ($\text{R}-\text{O}-\text{R}'$) - **Nomenclature:** Alkoxyalkane (IUPAC), Dialkyl ether (Common). - **Structure:** C-O-C bond angle is tetrahedral-like. Oxygen is $\text{sp}^3$ hybridized. #### Preparation of Ethers 1. **Dehydration of Alcohols:** - $2\text{R}-\text{OH} \xrightarrow{\text{Conc. H}_2\text{SO}_4, 413\text{K}} \text{R}-\text{O}-\text{R} + \text{H}_2\text{O}$ (For symmetrical ethers, $1^\circ$ alcohols). - **Mechanism:** $\text{S}_{\text{N}}2$ (for $1^\circ$) or $\text{E}1$ (for $2^\circ, 3^\circ$, but E1/E2 often compete to give alkenes). 2. **Williamson Synthesis (Best method for unsymmetrical ethers):** - $\text{R}-\text{X} + \text{R}'-\text{O}^-\text{Na}^+ \rightarrow \text{R}-\text{O}-\text{R}' + \text{NaX}$ - **Mechanism:** $\text{S}_{\text{N}}2$ reaction. - **Key:** Primary alkyl halide ($\text{R}-\text{X}$) must be used to minimize elimination (E2) side reactions. Alkoxide ($\text{R}'-\text{O}^-\text{Na}^+$) can be $1^\circ, 2^\circ, 3^\circ$. - Example: $\text{CH}_3\text{Br} + (\text{CH}_3)_3\text{CO}^-\text{Na}^+ \rightarrow \text{CH}_3-\text{O}-\text{C}(\text{CH}_3)_3$ (*tert*-butyl methyl ether). - If a $3^\circ$ alkyl halide is used with a strong nucleophile/base, elimination (E2) will be the major product. #### Physical Properties of Ethers - Lower boiling points than alcohols (no H-bonding). - Higher boiling points than alkanes of comparable molecular weight (due to dipole-dipole interactions). - Slightly soluble in water (due to H-bonding with water). #### Chemical Properties of Ethers 1. **Cleavage by Hot Concentrated HI/HBr:** - $\text{R}-\text{O}-\text{R}' + \text{HX} \xrightarrow{\Delta} \text{R}-\text{X} + \text{R}'-\text{OH}$ - $\text{R}-\text{OH} + \text{HX} \rightarrow \text{R}-\text{X} + \text{H}_2\text{O}$ (If excess HX is used, alcohol is converted to halide). - **Mechanism:** $\text{S}_{\text{N}}1$ for $3^\circ$ alkyl groups, $\text{S}_{\text{N}}2$ for $1^\circ$ alkyl groups. - Example: $\text{CH}_3-\text{O}-\text{C}(\text{CH}_3)_3 + \text{HI} \rightarrow \text{CH}_3\text{I} + (\text{CH}_3)_3\text{COH}$ (Here, $\text{S}_{\text{N}}2$ on methyl, $\text{S}_{\text{N}}1$ on *tert*-butyl. Due to carbocation stability, *tert*-butyl alcohol is formed). - For aryl alkyl ethers (e.g., anisole), the alkyl-oxygen bond is cleaved. The phenyl-oxygen bond has partial double bond character due to resonance and is stronger. - $\text{C}_6\text{H}_5-\text{O}-\text{CH}_3 + \text{HI} \rightarrow \text{C}_6\text{H}_5-\text{OH} + \text{CH}_3\text{I}$. 2. **Electrophilic Substitution (for aromatic ethers, e.g., Anisole):** - $-\text{OR}$ group is strongly activating and *ortho, para*-directing. - Similar to phenol, but less reactive than phenol (due to less electron-donating ability of -OR compared to $-\text{O}^-$). 3. **Formation of Peroxides:** Ethers react with atmospheric oxygen to form peroxides, which are explosive. #### Memory Tricks & Common Mistakes - **Alcohol Acidity:** Phenol > Water > Alcohol. - **Lucas Test:** $3^\circ$ (fastest) to $1^\circ$ (slowest). - **Williamson Synthesis:** Use $1^\circ$ alkyl halide and any alkoxide for desired ether. If $3^\circ$ alkyl halide, elimination predominates. - **Ether Cleavage (HI):** Look for carbocation stability or steric hindrance for bond breaking. Aryl-O bond is rarely broken. - **Common Mistake:** Confusing products of alcohol oxidation (aldehyde/acid from $1^\circ$, ketone from $2^\circ$). #### Solved JEE Main Example **Q1:** How will you prepare *tert*-butyl methyl ether using Williamson synthesis? **A1:** To prepare *tert*-butyl methyl ether, we need to use a primary alkyl halide and a *tert*-butoxide. **Correct reaction:** $\text{CH}_3-\text{Br} + (\text{CH}_3)_3\text{CO}^-\text{Na}^+ \rightarrow \text{CH}_3-\text{O}-\text{C}(\text{CH}_3)_3 + \text{NaBr}$ **Explanation:** If we tried to use *tert*-butyl bromide and sodium methoxide, the *tert*-butyl bromide, being a $3^\circ$ alkyl halide, would undergo E2 elimination with the strong base methoxide, giving 2-methylpropene as the major product. $(\text{CH}_3)_3\text{C}-\text{Br} + \text{CH}_3\text{O}^-\text{Na}^+ \rightarrow (\text{CH}_3)_2\text{C}=\text{CH}_2 + \text{CH}_3\text{OH} + \text{NaBr}$ (Elimination product) #### Practice Questions 1. Give the products of oxidation of propan-1-ol and propan-2-ol with chromic acid ($\text{H}_2\text{CrO}_4$). 2. Explain why phenols are acidic but do not react with sodium bicarbonate solution. 3. Suggest a synthesis for 1-phenyl-1-ethanol from benzaldehyde using a Grignard reagent. 4. What happens when anisole reacts with hot concentrated HI? 5. Which alcohol will give an immediate turbidity with Lucas reagent? Write the reaction. **Answers:** 1. Propan-1-ol ($\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}$) is a $1^\circ$ alcohol. With strong oxidizing agent like chromic acid, it will be oxidized to propanoic acid ($\text{CH}_3\text{CH}_2\text{COOH}$). Propan-2-ol ($\text{CH}_3\text{CH}(\text{OH})\text{CH}_3$) is a $2^\circ$ alcohol. With chromic acid, it will be oxidized to propanone ($\text{CH}_3\text{COCH}_3$). 2. Phenols are acidic because the phenoxide ion (conjugate base) is resonance-stabilized by delocalization of the negative charge into the benzene ring. This stabilization makes the removal of $\text{H}^+$ easier. However, phenols are weaker acids than carbonic acid ($\text{H}_2\text{CO}_3$), so they cannot protonate bicarbonate ($\text{HCO}_3^-$) to form $\text{CO}_2$. 3. Benzaldehyde ($\text{C}_6\text{H}_5\text{CHO}$) is an aldehyde. To get 1-phenyl-1-ethanol ($\text{C}_6\text{H}_5-\text{CH}(\text{OH})-\text{CH}_3$), we need to add a methyl group. $\text{C}_6\text{H}_5\text{CHO} + \text{CH}_3\text{MgBr} \rightarrow \text{C}_6\text{H}_5-\text{CH}(\text{OMgBr})-\text{CH}_3 \xrightarrow{\text{H}_2\text{O}/\text{H}^+} \text{C}_6\text{H}_5-\text{CH}(\text{OH})-\text{CH}_3$ 4. Anisole ($\text{C}_6\text{H}_5-\text{O}-\text{CH}_3$) is an aryl methyl ether. The alkyl-oxygen bond will be cleaved. The phenyl-oxygen bond has partial double bond character due to resonance and is not cleaved. $\text{C}_6\text{H}_5-\text{O}-\text{CH}_3 + \text{HI} \xrightarrow{\Delta} \text{C}_6\text{H}_5-\text{OH} + \text{CH}_3\text{I}$ (Phenol and Iodomethane). 5. A tertiary alcohol will give an immediate turbidity with Lucas reagent ($\text{HCl}/\text{ZnCl}_2$). For example, 2-methylpropan-2-ol (t-butyl alcohol). $(\text{CH}_3)_3\text{C}-\text{OH} + \text{HCl} \xrightarrow{\text{ZnCl}_2} (\text{CH}_3)_3\text{C}-\text{Cl} + \text{H}_2\text{O}$ ### Aldehydes and Ketones #### Structure and Nomenclature - **Carbonyl Group:** Contains a $\text{C}=\text{O}$ group. Carbon is $\text{sp}^2$ hybridized, trigonal planar geometry. - **Polarity:** The $\text{C}=\text{O}$ bond is highly polar due to oxygen's electronegativity ($\text{C}^\delta+=\text{O}^\delta-$). This makes the carbonyl carbon an electrophilic center. - **Aldehydes ($\text{RCHO}$):** At least one hydrogen atom attached to the carbonyl carbon. Suffix "-al". - **Ketones ($\text{RCOR}'$):** Two alkyl or aryl groups attached to the carbonyl carbon. Suffix "-one". #### Preparation of Aldehydes and Ketones **A. Preparation of Aldehydes** 1. **Oxidation of $1^\circ$ Alcohols:** - $\text{RCH}_2\text{OH} \xrightarrow{\text{PCC} \text{ or } \text{CrO}_3/\text{anhydrous CrO}_3} \text{RCHO}$ (Mild oxidizing agents). 2. **Dehydrogenation of $1^\circ$ Alcohols:** - $\text{RCH}_2\text{OH} \xrightarrow{\text{Cu}, 573\text{K}} \text{RCHO}$. 3. **From Acyl Chlorides (Rosenmund Reduction):** - $\text{RCOCl} + \text{H}_2 \xrightarrow{\text{Pd/BaSO}_4} \text{RCHO} + \text{HCl}$ - $\text{BaSO}_4$ acts as a poison to prevent further reduction to alcohol. 4. **From Nitriles (Stephen Reaction):** - $\text{RCN} \xrightarrow{1. \text{SnCl}_2/\text{HCl}; 2. \text{H}_3\text{O}^+} \text{RCHO}$ - $\text{RCN} \xrightarrow{1. \text{DIBAL-H} \text{ (Diisobutylaluminium hydride)}; 2. \text{H}_2\text{O}} \text{RCHO}$ (Also works for esters). 5. **From Esters (DIBAL-H):** - $\text{RCOOR}' \xrightarrow{1. \text{DIBAL-H}, -78^\circ\text{C}; 2. \text{H}_2\text{O}} \text{RCHO}$. 6. **From Hydrocarbons (Toluene to Benzaldehyde):** - **Etard Reaction:** Toluene $\xrightarrow{1. \text{CrO}_2\text{Cl}_2/\text{CS}_2; 2. \text{H}_3\text{O}^+} \text{Benzaldehyde}$. - **Gattermann-Koch Reaction:** Benzene $\xrightarrow{\text{CO/HCl}, \text{Anhydrous AlCl}_3/\text{CuCl}} \text{Benzaldehyde}$. - **Side-chain Chlorination:** Toluene $\xrightarrow{\text{Cl}_2/h\nu} \text{Benzal chloride} \xrightarrow{\text{H}_2\text{O}, \Delta} \text{Benzaldehyde}$. **B. Preparation of Ketones** 1. **Oxidation of $2^\circ$ Alcohols:** - $\text{RCH}(\text{OH})\text{R}' \xrightarrow{\text{PCC} \text{ or } \text{K}_2\text{Cr}_2\text{O}_7/\text{H}^+ \text{ or } \text{CrO}_3} \text{RCOR}'$. 2. **Dehydrogenation of $2^\circ$ Alcohols:** - $\text{RCH}(\text{OH})\text{R}' \xrightarrow{\text{Cu}, 573\text{K}} \text{RCOR}'$. 3. **From Nitriles (with Grignard Reagent):** - $\text{RCN} + \text{R}'\text{MgX} \xrightarrow{1. \text{Ether}; 2. \text{H}_3\text{O}^+} \text{RCOR}'$. 4. **From Acyl Chlorides (with Dialkylcadmium):** - $2\text{RCOCl} + (\text{R}')_2\text{Cd} \rightarrow 2\text{RCOR}' + \text{CdCl}_2$. 5. **Friedel-Crafts Acylation:** - Benzene $\xrightarrow{\text{RCOCl}/\text{Anhydrous AlCl}_3} \text{ArCOR}$ (Aromatic ketone). 6. **From Alkynes (Hydration):** - $\text{R}-\text{C}\equiv\text{CH} \xrightarrow{\text{HgSO}_4/\text{H}_2\text{SO}_4} \text{RCOR}'$ (Except acetylene which gives aldehyde). 7. **From Cumene:** - Cumene $\xrightarrow{1. \text{O}_2; 2. \text{H}^+} \text{Phenol} + \text{Acetone}$ (Acetone is a ketone). #### Physical Properties - Boiling points are higher than hydrocarbons and ethers of comparable molecular weight (due to dipole-dipole interactions). - Lower boiling points than alcohols (no H-bonding). - Lower members (up to 4 carbons) are miscible with water due to H-bonding with water. #### Chemical Properties - General Reactions (Nucleophilic Addition) - Aldehydes are generally more reactive than ketones towards nucleophilic addition due to: - **Steric Hindrance:** Ketones have two alkyl groups, more hindered. Aldehydes have at least one small H-atom. - **Electronic Effect:** Alkyl groups are EDG (+I effect), stabilizing the partial positive charge on carbonyl carbon. Ketones have two EDG, aldehydes have one (or none for formaldehyde), making aldehydes more electrophilic. 1. **Addition of HCN (Cyanohydrin Formation):** - $\text{RCHO/RCOR}' + \text{HCN} \xrightarrow{\text{Base catalyst}} \text{RCH}(\text{OH})\text{CN} / \text{RC}(\text{OH})(\text{CN})\text{R}'$ (Cyanohydrins). 2. **Addition of Sodium Bisulfite ($\text{NaHSO}_3$):** - Forms crystalline bisulfite addition products (useful for separation/purification of aldehydes/ketones). 3. **Addition of Grignard Reagents:** (See alcohol preparation) - Aldehydes $\rightarrow 1^\circ/2^\circ$ alcohols. Ketones $\rightarrow 3^\circ$ alcohols. 4. **Addition of Alcohols (Acetal/Ketal Formation):** - $\text{RCHO} + \text{R}'\text{OH} \xrightarrow{\text{H}^+} \text{RCH}(\text{OR}')_2$ (Acetal). - $\text{RCOR}' + \text{R}''\text{OH} \xrightarrow{\text{H}^+} \text{RC}(\text{OR}'')_2\text{R}'$ (Ketal). - Cyclic acetals/ketals are formed with diols (e.g., ethylene glycol) and are used as protecting groups for carbonyls. 5. **Addition of Ammonia Derivatives (Condensation Reactions):** - $\text{C}=\text{O} + \text{H}_2\text{N}-\text{Z} \rightarrow \text{C}=\text{N}-\text{Z} + \text{H}_2\text{O}$ (Imine, Oxime, Hydrazone, Semicarbazone etc.) - $\text{Z} = -\text{OH}$ (hydroxylamine) $\rightarrow$ Oxime - $\text{Z} = -\text{NH}_2$ (hydrazine) $\rightarrow$ Hydrazone - $\text{Z} = -\text{NHC}_6\text{H}_5$ (phenylhydrazine) $\rightarrow$ Phenylhydrazone - $\text{Z} = -\text{NHCONH}_2$ (semicarbazide) $\rightarrow$ Semicarbazone - $\text{Z} = -\text{NHC}_6\text{H}_3(\text{NO}_2)_2$ (2,4-DNP) $\rightarrow$ 2,4-Dinitrophenylhydrazone (yellow/orange/red ppt, test for carbonyls). #### Chemical Properties - Specific Reactions 1. **Reduction Reactions:** - **Reduction to Alcohols:** $\text{RCHO/RCOR}' \xrightarrow{\text{LiAlH}_4 \text{ or } \text{NaBH}_4 \text{ or } \text{H}_2/\text{Ni, Pt, Pd}} \text{Alcohols}$ (See alcohol preparation). - **Reduction to Hydrocarbons:** - **Clemmensen Reduction:** $\text{RCHO/RCOR}' \xrightarrow{\text{Zn-Hg/Conc. HCl}} \text{RCH}_2\text{R}'$ (Alkane). - **Wolff-Kishner Reduction:** $\text{RCHO/RCOR}' \xrightarrow{1. \text{NH}_2\text{NH}_2; 2. \text{KOH/Ethylene Glycol}, \Delta} \text{RCH}_2\text{R}'$ (Alkane). 2. **Oxidation Reactions:** - **Aldehydes are easily oxidized to carboxylic acids:** - $\text{RCHO} \xrightarrow{\text{KMnO}_4/\text{H}^+ \text{ or } \text{K}_2\text{Cr}_2\text{O}_7/\text{H}^+ \text{ or } \text{HNO}_3 \text{ or } \text{Tollen's or Fehling's}} \text{RCOOH}$. - **Ketones are resistant to oxidation:** Require strong conditions, C-C bond cleavage occurs to give mixture of acids. - **Tollen's Test (Silver Mirror Test):** $\text{RCHO} + 2[\text{Ag(NH}_3)_2]^+\text{OH}^- \rightarrow \text{RCOO}^- + 2\text{Ag}\downarrow + \text{H}_2\text{O} + 4\text{NH}_3$. (Positive for aldehydes, negative for ketones). - **Fehling's Test:** $\text{RCHO} + 2\text{Cu}^{2+} + 5\text{OH}^- \rightarrow \text{RCOO}^- + \text{Cu}_2\text{O}\downarrow (\text{red ppt}) + 3\text{H}_2\text{O}$. (Positive for aliphatic aldehydes, negative for ketones and aromatic aldehydes). - **Benedict's Test:** Similar to Fehling's. 3. **Reactions involving $\alpha$-hydrogens:** - **Aldol Condensation:** Carbonyl compounds with at least one $\alpha$-hydrogen react in presence of dilute base (or acid) to form $\beta$-hydroxy aldehydes/ketones, which often dehydrate to $\alpha,\beta$-unsaturated carbonyl compounds. - **Mechanism:** Enolate formation, nucleophilic attack on another carbonyl, dehydration. - **Cross Aldol Condensation:** Between two different carbonyl compounds. If both have $\alpha$-hydrogens, mixture of products. If one has no $\alpha$-hydrogens (e.g., formaldehyde, benzaldehyde), it acts as electrophile. - **Cannizzaro Reaction:** Aldehydes with NO $\alpha$-hydrogen (e.g., formaldehyde, benzaldehyde) undergo disproportionation in concentrated base. One molecule is oxidized to carboxylic acid (salt) and another is reduced to alcohol. - $\text{2HCHO} \xrightarrow{\text{Conc. NaOH}} \text{CH}_3\text{OH} + \text{HCOONa}$. - **Cross Cannizzaro:** Between two different aldehydes, one with no $\alpha$-H. - **Haloform Reaction (Iodoform Test):** Compounds with $\text{CH}_3\text{CO}-$ group or $\text{CH}_3\text{CH}(\text{OH})-$ group react with $\text{X}_2/\text{NaOH}$ to give $\text{CHX}_3$ (haloform, e.g., $\text{CHI}_3$ is yellow ppt). - Positive for acetaldehyde, methyl ketones, ethanol, $2^\circ$ alcohols with a methyl group at C2. 4. **Other Named Reactions:** - **Perkin Reaction:** Aromatic aldehyde + Anhydride + Sodium salt of acid $\rightarrow$ $\alpha,\beta$-unsaturated carboxylic acid. - **Knoevenagel Condensation:** Carbonyl compound + active methylene compound $\rightarrow$ $\alpha,\beta$-unsaturated compound. - **Benzoin Condensation:** Aromatic aldehydes with $\text{KCN}$ catalyst. - **Wittig Reaction:** Aldehyde/Ketone + Phosphorane $\rightarrow$ Alkene. #### Memory Tricks & Common Mistakes - **Reactivity:** Aldehydes > Ketones (Steric and Electronic). - **Oxidation:** Aldehydes oxidize easily; Ketones are resistant. Tollen's/Fehling's are specific for aldehydes. - **$\alpha$-hydrogens:** Aldol if present, Cannizzaro if absent. - **Haloform:** Check for $\text{CH}_3\text{CO}-$ or $\text{CH}_3\text{CH}(\text{OH})-$ group. - **Reduction to alkane:** Clemmensen (acidic), Wolff-Kishner (basic). - **Common Mistake:** Confusing products of cross-aldol or cross-cannizzaro reactions. #### Solved JEE Main Example **Q1:** An organic compound (A) with molecular formula $\text{C}_2\text{H}_4\text{O}$ gives a positive Tollen's test and forms a yellow precipitate with $\text{I}_2/\text{NaOH}$. Identify (A) and write the reactions. **A1:** The compound (A) is ethanal (acetaldehyde). **Explanation:** - Molecular formula $\text{C}_2\text{H}_4\text{O}$ suggests an aldehyde or ketone. - Positive Tollen's test indicates it's an aldehyde. - Forms yellow precipitate with $\text{I}_2/\text{NaOH}$ (iodoform test) indicates presence of $\text{CH}_3\text{CO}-$ group. Only acetaldehyde ($\text{CH}_3\text{CHO}$) fits these criteria among $\text{C}_2\text{H}_4\text{O}$ isomers (ethylene oxide is an ether, vinyl alcohol is an enol). **Reactions:** 1. **Tollen's Test:** $\text{CH}_3\text{CHO} + 2[\text{Ag(NH}_3)_2]^+\text{OH}^- \rightarrow \text{CH}_3\text{COO}^- + 2\text{Ag}\downarrow (\text{silver mirror}) + \text{H}_2\text{O} + 4\text{NH}_3$ 2. **Iodoform Test:** $\text{CH}_3\text{CHO} + 3\text{I}_2 + 4\text{NaOH} \rightarrow \text{CHI}_3\downarrow (\text{yellow ppt}) + \text{HCOONa} + 3\text{NaI} + 3\text{H}_2\text{O}$ #### Practice Questions 1. How will you convert propanal to propan-1-ol and propan-2-ol? 2. An unknown compound (X) with molecular formula $\text{C}_3\text{H}_6\text{O}$ does not give Tollen's test but responds to iodoform test. Identify (X) and write the reactions. 3. Write the products of aldol condensation of propanal. 4. Why does benzaldehyde undergo Cannizzaro reaction but not aldol condensation? 5. Suggest reagents to convert acetophenone to ethylbenzene. **Answers:** 1. **Propanal to Propan-1-ol (Reduction):** $\text{CH}_3\text{CH}_2\text{CHO} \xrightarrow{\text{LiAlH}_4 \text{ or } \text{NaBH}_4 \text{ or } \text{H}_2/\text{Ni}} \text{CH}_3\text{CH}_2\text{CH}_2\text{OH}$ **Propanal to Propan-2-ol (via Grignard):** $\text{CH}_3\text{CH}_2\text{CHO} + \text{CH}_3\text{MgBr} \rightarrow \text{CH}_3\text{CH}_2-\text{CH}(\text{OMgBr})-\text{CH}_3 \xrightarrow{\text{H}_2\text{O}/\text{H}^+} \text{CH}_3\text{CH}_2-\text{CH}(\text{OH})-\text{CH}_3$ 2. (X) is propanone (acetone). - No Tollen's test: It's a ketone. - Positive iodoform test: It has a $\text{CH}_3\text{CO}-$ group. Propanone is $\text{CH}_3\text{COCH}_3$. **Iodoform Test:** $\text{CH}_3\text{COCH}_3 + 3\text{I}_2 + 4\text{NaOH} \rightarrow \text{CHI}_3\downarrow (\text{yellow ppt}) + \text{CH}_3\text{COONa} + 3\text{NaI} + 3\text{H}_2\text{O}$ 3. Propanal has $\alpha$-hydrogens. In aldol condensation, two molecules react. $2\text{CH}_3\text{CH}_2\text{CHO} \xrightarrow{\text{Dil. NaOH}} \text{CH}_3\text{CH}_2-\text{CH}(\text{OH})-\text{CH}(\text{CH}_3)-\text{CHO}$ (3-hydroxy-2-methylpentanal) This can dehydrate to $\text{CH}_3\text{CH}_2-\text{CH}=\text{C}(\text{CH}_3)-\text{CHO}$ (2-methylpent-2-enal). 4. Benzaldehyde ($\text{C}_6\text{H}_5\text{CHO}$) does not have any $\alpha$-hydrogen atoms (the carbon adjacent to the carbonyl group is part of the benzene ring and has no hydrogens). Therefore, it cannot form an enolate ion, which is a key intermediate in aldol condensation. It undergoes Cannizzaro reaction because it lacks $\alpha$-hydrogens and can undergo hydride transfer in concentrated base. 5. Acetophenone ($\text{C}_6\text{H}_5\text{COCH}_3$) to Ethylbenzene ($\text{C}_6\text{H}_5\text{CH}_2\text{CH}_3$). This is a reduction of the carbonyl group to a methylene group. **Clemmensen Reduction:** $\text{C}_6\text{H}_5\text{COCH}_3 \xrightarrow{\text{Zn-Hg/Conc. HCl}} \text{C}_6\text{H}_5\text{CH}_2\text{CH}_3$ **Wolff-Kishner Reduction:** $\text{C}_6\text{H}_5\text{COCH}_3 \xrightarrow{1. \text{NH}_2\text{NH}_2; 2. \text{KOH/Ethylene Glycol}, \Delta} \text{C}_6\text{H}_5\text{CH}_2\text{CH}_3$ ### Carboxylic Acids and their Derivatives #### Carboxylic Acids ($\text{RCOOH}$) - **Functional Group:** Contains a carboxyl group $(-\text{COOH})$. - **Structure:** Carbonyl carbon is $\text{sp}^2$ hybridized. Planar geometry around carbonyl. - **Acidity:** Carboxylic acids are more acidic than alcohols and phenols but generally weaker than mineral acids. This is due to the resonance stabilization of the carboxylate anion ($\text{RCOO}^-$). - **Factors affecting acidity:** - **EWG (-I, -R):** Increase acidity by stabilizing the carboxylate anion. - **EDG (+I, +R):** Decrease acidity by destabilizing the carboxylate anion. - **Hybridization:** $\text{sp}$ > $\text{sp}^2$ > $\text{sp}^3$ (e.g., $\text{HC}\equiv\text{CCOOH}$ is more acidic than $\text{CH}_2=\text{CHCOOH}$). - **Nomenclature:** Suffix "-oic acid". #### Preparation of Carboxylic Acids 1. **From Primary Alcohols/Aldehydes (Oxidation):** - $\text{RCH}_2\text{OH} \xrightarrow{\text{KMnO}_4/\text{H}^+ \text{ or } \text{K}_2\text{Cr}_2\text{O}_7/\text{H}^+ \text{ or } \text{HNO}_3} \text{RCOOH}$. - $\text{RCHO} \xrightarrow{\text{Mild oxidizing agents (Tollen's/Fehling's) or strong oxidizing agents}} \text{RCOOH}$. 2. **From Nitriles and Amides (Hydrolysis):** - $\text{RCN} \xrightarrow{\text{H}_3\text{O}^+ \text{ or } \text{OH}^-, \Delta} \text{RCOOH}$. (via amide intermediate). - $\text{RCONH}_2 \xrightarrow{\text{H}_3\text{O}^+ \text{ or } \text{OH}^-, \Delta} \text{RCOOH}$. 3. **From Grignard Reagents:** - $\text{RMgX} + \text{CO}_2 \xrightarrow{1. \text{Dry Ether}; 2. \text{H}_3\text{O}^+} \text{RCOOH}$. (Increases carbon chain by one). 4. **From Alkylbenzenes (Side-chain Oxidation):** - Alkylbenzene (e.g., Toluene) $\xrightarrow{\text{KMnO}_4/\text{KOH}, \Delta; \text{then } \text{H}^+} \text{Benzoic acid}$. - Regardless of length of alkyl chain, if it has at least one benzylic hydrogen, it's oxidized to $-\text{COOH}$. 5. **From Acyl Halides/Anhydrides (Hydrolysis):** - $\text{RCOCl} \xrightarrow{\text{H}_2\text{O}} \text{RCOOH}$. - $(\text{RCO})_2\text{O} \xrightarrow{\text{H}_2\text{O}} \text{RCOOH}$. #### Physical Properties - Higher boiling points than alcohols, aldehydes, ketones, ethers of comparable molecular weight due to strong intermolecular hydrogen bonding (form dimers). - Lower members are miscible with water. #### Chemical Properties of Carboxylic Acids 1. **Acidity (Reactions involving O-H bond cleavage):** - React with active metals (Na, K, Mg, Zn) to form salts and $\text{H}_2$. - React with bases (NaOH, KOH) to form salts and water. - React with carbonates ($\text{Na}_2\text{CO}_3$) and bicarbonates ($\text{NaHCO}_3$) to form salts, water, and $\text{CO}_2$ (distinguishes from phenols). 2. **Reactions involving -COOH group (replacement of -OH):** - **Formation of Acyl Chlorides:** $\text{RCOOH} + \text{PCl}_5/\text{PCl}_3/\text{SOCl}_2 \rightarrow \text{RCOCl}$. - **Formation of Anhydrides:** $2\text{RCOOH} \xrightarrow{\text{P}_2\text{O}_5, \Delta} (\text{RCO})_2\text{O} + \text{H}_2\text{O}$. - **Esterification:** $\text{RCOOH} + \text{R}'\text{OH} \rightleftharpoons \text{RCOOR}' + \text{H}_2\text{O}$ (Acid catalyzed, reversible). - **Formation of Amides:** $\text{RCOOH} + \text{NH}_3 \xrightarrow{\Delta} \text{RCOONH}_4 \xrightarrow{\Delta} \text{RCONH}_2 + \text{H}_2\text{O}$. 3. **Reduction:** - $\text{RCOOH} \xrightarrow{\text{LiAlH}_4/\text{H}_2\text{O}} \text{RCH}_2\text{OH}$ ($1^\circ$ alcohol). (Only $\text{LiAlH}_4$, $\text{NaBH}_4$ cannot reduce acids). 4. **Decarboxylation:** - $\text{RCOONa} + \text{NaOH} \xrightarrow{\text{CaO}, \Delta} \text{R}-\text{H} + \text{Na}_2\text{CO}_3$. - **Kolbe's Electrolytic Reaction:** $2\text{RCOONa} \xrightarrow{\text{Electrolysis}} \text{R}-\text{R} + 2\text{CO}_2 + 2\text{NaOH} + \text{H}_2$. - $\beta$-keto acids readily decarboxylate on heating. 5. **Hell-Volhard-Zelinsky (HVZ) Reaction:** - $\text{R}-\text{CH}_2-\text{COOH} \xrightarrow{\text{X}_2/\text{Red P}} \text{R}-\text{CH}(\text{X})-\text{COOH}$ ($\alpha$-halo carboxylic acid). - Only for acids with $\alpha$-hydrogens. #### Carboxylic Acid Derivatives - **Acyl Chlorides ($\text{RCOCl}$):** Most reactive derivative. - **Anhydrides ($(\text{RCO})_2\text{O}$):** Second most reactive. - **Esters ($\text{RCOOR}'$):** Less reactive than anhydrides. - **Amides ($\text{RCONH}_2$):** Least reactive derivative. - **Order of Reactivity towards Nucleophilic Acyl Substitution:** Acyl Chloride > Anhydride > Ester > Amide. #### Reactions of Derivatives 1. **Hydrolysis:** All derivatives hydrolyze to carboxylic acids. - $\text{RCOCl} \xrightarrow{\text{H}_2\text{O}} \text{RCOOH}$. - $(\text{RCO})_2\text{O} \xrightarrow{\text{H}_2\text{O}} \text{RCOOH}$. - $\text{RCOOR}' \xrightarrow{\text{H}_2\text{O}/\text{H}^+ \text{ or } \text{OH}^-} \text{RCOOH} + \text{R}'\text{OH}$ (Saponification with base). - $\text{RCONH}_2 \xrightarrow{\text{H}_3\text{O}^+ \text{ or } \text{OH}^-, \Delta} \text{RCOOH} + \text{NH}_3$. 2. **Reaction with Alcohols (Transesterification for esters):** - Acyl chloride + Alcohol $\rightarrow$ Ester. - Anhydride + Alcohol $\rightarrow$ Ester + Carboxylic acid. - Ester + Alcohol $\rightleftharpoons$ New Ester + New Alcohol. 3. **Reaction with Ammonia/Amines:** - Acyl chloride/Anhydride + $\text{NH}_3/\text{RNH}_2 \rightarrow$ Amide. 4. **Reduction:** - **Esters:** $\text{RCOOR}' \xrightarrow{\text{LiAlH}_4} \text{RCH}_2\text{OH} + \text{R}'\text{OH}$. (Can be partially reduced to aldehydes using DIBAL-H). - **Amides:** $\text{RCONH}_2 \xrightarrow{\text{LiAlH}_4} \text{RCH}_2\text{NH}_2$ ($1^\circ$ amine). - **Nitriles:** $\text{RCN} \xrightarrow{\text{LiAlH}_4} \text{RCH}_2\text{NH}_2$ ($1^\circ$ amine). #### Memory Tricks & Common Mistakes - **Acidity Order:** Carboxylic acid > Phenol > Alcohol. - **Carboxylic Acid Tests:** Reacts with $\text{NaHCO}_3$ (releases $\text{CO}_2$). - **HVZ:** Halogenation at $\alpha$-carbon. - **Derivative Reactivity:** Acyl Chloride > Anhydride > Ester > Amide. Relates to leaving group ability. - **Common Mistake:** Forgetting that $\text{NaBH}_4$ cannot reduce carboxylic acids or esters. $\text{LiAlH}_4$ is a stronger reducing agent. - **Common Mistake:** Confusing the products of hydrolysis of nitriles (acid) vs. reduction of nitriles (amine). #### Solved JEE Main Example **Q1:** How will you convert ethanoic acid to methanamine? **A1:** This requires reducing the carbon chain by one carbon atom and converting the carboxyl group to an amine. This can be done using the Hoffmann bromamide degradation reaction. **Steps:** 1. **Convert ethanoic acid to ethanamide:** $\text{CH}_3\text{COOH} \xrightarrow{\text{NH}_3, \Delta} \text{CH}_3\text{CONH}_2 + \text{H}_2\text{O}$ 2. **Hoffmann bromamide degradation:** $\text{CH}_3\text{CONH}_2 + \text{Br}_2 + 4\text{NaOH} \rightarrow \text{CH}_3\text{NH}_2 + \text{Na}_2\text{CO}_3 + 2\text{NaBr} + 2\text{H}_2\text{O}$ (Methanamine) #### Practice Questions 1. Arrange the following compounds in increasing order of acidity: Formic acid, Acetic acid, Benzoic acid. 2. Suggest a method to prepare propanoic acid from an alkyl halide with one less carbon atom. 3. What are the products when ethyl acetate is hydrolyzed in acidic medium? 4. Write the reaction for the conversion of propanoic acid to 2-bromopropanoic acid. 5. Explain why amides are the least reactive among carboxylic acid derivatives towards nucleophilic acyl substitution. **Answers:** 1. Acetic acid ### Amines #### Definition and Classification - **Definition:** Organic compounds derived from ammonia ($\text{NH}_3$) by replacing one or more hydrogen atoms with alkyl or aryl groups. - **Classification:** - **Primary (1°):** One alkyl/aryl group attached to nitrogen ($\text{RNH}_2, \text{ArNH}_2$). - **Secondary (2°):** Two alkyl/aryl groups attached to nitrogen ($\text{R}_2\text{NH}, \text{Ar}_2\text{NH}$). - **Tertiary (3°):** Three alkyl/aryl groups attached to nitrogen ($\text{R}_3\text{N}, \text{Ar}_3\text{N}$). - **Quaternary Ammonium Salts:** $\text{R}_4\text{N}^+\text{X}^-$. - **Nomenclature:** Suffix "-amine" for primary amines. For secondary/tertiary, use "N-alkyl...". #### Structure of Amines - Nitrogen atom in amines is $\text{sp}^3$ hybridized. - Pyramidal geometry (like ammonia) due to the presence of a lone pair of electrons. - The lone pair is responsible for their basicity and nucleophilicity. - Chiral amines (with three different groups + lone pair) rapidly interconvert (pyramidal inversion), so they are usually not optically active at room temperature. #### Preparation of Amines 1. **Reduction of Nitro Compounds:** (For aromatic amines) - $\text{Ar}-\text{NO}_2 \xrightarrow{\text{Fe/HCl} \text{ or } \text{Sn/HCl} \text{ or } \text{H}_2/\text{Pd}} \text{Ar}-\text{NH}_2$. 2. **Ammonolysis of Alkyl Halides:** - $\text{R}-\text{X} + \text{NH}_3 \rightarrow \text{R}-\text{NH}_2 + \text{HX}$. - **Limitation:** Gives a mixture of $1^\circ, 2^\circ, 3^\circ$ amines and quaternary ammonium salts due to overalkylation. 3. **Gabriel Phthalimide Synthesis:** (For $1^\circ$ aliphatic amines, avoids polyalkylation) - Phthalimide $\xrightarrow{1. \text{KOH}; 2. \text{R}-\text{X}; 3. \text{H}_2\text{O}/\text{H}^+ \text{ or } \text{N}_2\text{H}_4} \text{R}-\text{NH}_2$. - Only for $1^\circ$ amines. Aromatic $1^\circ$ amines cannot be prepared as aryl halides do not undergo nucleophilic substitution with phthalimide anion. 4. **Hoffmann Bromamide Degradation Reaction:** (Decreases carbon chain by one carbon) - $\text{RCONH}_2 + \text{Br}_2 + 4\text{NaOH} \rightarrow \text{RNH}_2 + \text{Na}_2\text{CO}_3 + 2\text{NaBr} + 2\text{H}_2\text{O}$. - Converts primary amides to primary amines with one carbon less. 5. **Reduction of Nitriles:** - $\text{RCN} \xrightarrow{\text{LiAlH}_4 \text{ or } \text{H}_2/\text{Ni}} \text{RCH}_2\text{NH}_2$ ($1^\circ$ amine). 6. **Reduction of Amides:** - $\text{RCONH}_2 \xrightarrow{\text{LiAlH}_4} \text{RCH}_2\text{NH}_2$ ($1^\circ$ amine). 7. **Reductive Amination of Aldehydes/Ketones:** - $\text{RCHO/RCOR}' \xrightarrow{1. \text{NH}_3; 2. \text{H}_2/\text{Ni or NaBH}_3\text{CN}} \text{RCH}_2\text{NH}_2 \text{ or } \text{R}_2\text{CHNH}_2$. #### Physical Properties - $1^\circ$ and $2^\circ$ amines form intermolecular hydrogen bonds (higher boiling points than alkanes). - $3^\circ$ amines do not have H attached to N, so no intermolecular H-bonding (lower boiling points than $1^\circ/2^\circ$). - Lower members are soluble in water (H-bonding with water). Solubility decreases with increasing alkyl chain. - Lower aliphatic amines have fishy smell. Aniline is odorless. #### Chemical Properties (Basicity) - Amines are basic due to the lone pair on nitrogen. They are Lewis bases. - **Basicity Order (Gas Phase):** $3^\circ > 2^\circ > 1^\circ > \text{NH}_3$. (Due to +I effect of alkyl groups stabilizing the conjugate acid). - **Basicity Order (Aqueous Solution):** Complex due to solvation and steric hindrance. - **For Methyl Amines:** $2^\circ > 1^\circ > 3^\circ > \text{NH}_3$. (e.g., $(\text{CH}_3)_2\text{NH} > \text{CH}_3\text{NH}_2 > (\text{CH}_3)_3\text{N} > \text{NH}_3$) - **For Ethyl Amines:** $2^\circ > 3^\circ > 1^\circ > \text{NH}_3$. (e.g., $(\text{C}_2\text{H}_5)_2\text{NH} > (\text{C}_2\text{H}_5)_3\text{N} > \text{C}_2\text{H}_5\text{NH}_2 > \text{NH}_3$) - **Aromatic Amines (e.g., Aniline):** Much weaker bases than aliphatic amines due to resonance delocalization of the lone pair of electrons on nitrogen into the benzene ring, making it less available for protonation. - **Effect of Substituents on Basicity:** - EDG (e.g., $-\text{CH}_3$, $-\text{OCH}_3$) increase basicity. - EWG (e.g., $-\text{NO}_2$, $-\text{CN}$, $-\text{COOH}$) decrease basicity. #### Other Chemical Properties 1. **Alkylation:** Amines react with alkyl halides to form higher amines and quaternary ammonium salts. - $\text{RNH}_2 \xrightarrow{\text{R}'\text{X}} \text{RNHR}' \xrightarrow{\text{R}'\text{X}} \text{RR}'_2\text{N} \xrightarrow{\text{R}'\text{X}} \text{RR}'_3\text{N}^+\text{X}^-$. 2. **Acylation:** Amines react with acyl chlorides, anhydrides, or esters to form amides. - $\text{RNH}_2 + \text{R}'\text{COCl} \xrightarrow{\text{Pyridine}} \text{RNHCOR}' + \text{HCl}$. - $1^\circ$ and $2^\circ$ amines react, $3^\circ$ amines do not (no H on N). 3. **Carbylamine Reaction (Isocyanide Test):** - $1^\circ$ amine + $\text{CHCl}_3 + 3\text{KOH} \xrightarrow{\Delta} \text{R}-\text{NC} (\text{isocyanide, foul smell}) + 3\text{KCl} + 3\text{H}_2\text{O}$. - Specific test for $1^\circ$ amines (aliphatic and aromatic). 4. **Reaction with Nitrous Acid ($\text{HNO}_2$, generated from $\text{NaNO}_2/\text{HCl}$):** - **$1^\circ$ Aliphatic Amines:** $\text{RNH}_2 \xrightarrow{\text{HNO}_2, 0-5^\circ\text{C}} [\text{R}-\text{N}_2^+\text{Cl}^-] \xrightarrow{\text{H}_2\text{O}} \text{ROH} + \text{N}_2 \uparrow + \text{HCl}$. (Nitrogen gas evolved, used to estimate $1^\circ$ aliphatic amines). - **$1^\circ$ Aromatic Amines:** $\text{ArNH}_2 \xrightarrow{\text{NaNO}_2/\text{HCl}, 0-5^\circ\text{C}} \text{Ar}-\text{N}_2^+\text{Cl}^-$ (Benzenediazonium chloride, stable at low temperature). Used for Sandmeyer/Gattermann reactions, coupling reactions. - **$2^\circ$ Amines (aliphatic and aromatic):** $\text{R}_2\text{NH} \xrightarrow{\text{HNO}_2} \text{R}_2\text{N}-\text{NO}$ (N-Nitrosamine, yellow oily compound). - **$3^\circ$ Aliphatic Amines:** React to form a salt and an alcohol. - **$3^\circ$ Aromatic Amines:** React with nitrous acid to form *p*-nitroso-N,N-dialkylaniline. 5. **Hinsberg Test:** (Distinguishes $1^\circ, 2^\circ, 3^\circ$ amines) - Reagent: Benzenesulfonyl chloride ($\text{C}_6\text{H}_5\text{SO}_2\text{Cl}$). - **$1^\circ$ Amine:** Forms N-alkylbenzenesulfonamide, which is soluble in $\text{KOH}$ (due to acidic H on N). Acidification regenerates insoluble sulfonamide. - **$2^\circ$ Amine:** Forms N,N-dialkylbenzenesulfonamide, which is insoluble in $\text{KOH}$ (no acidic H on N). - **$3^\circ$ Amine:** Does not react with benzenesulfonyl chloride. Remains insoluble (or dissolves if amine is very basic and forms salt). 6. **Electrophilic Substitution in Aromatic Amines (e.g., Aniline):** - $-\text{NH}_2$ group is strongly activating and *ortho, para*-directing. - **Bromination:** Aniline $\xrightarrow{\text{Br}_2/\text{H}_2\text{O}} 2,4,6-\text{Tribromoaniline}$ (White ppt). - To get monosubstituted product, amino group must be protected by acylation (e.g., with acetic anhydride to form acetanilide). - Aniline $\xrightarrow{(\text{CH}_3\text{CO})_2\text{O}} \text{Acetanilide} \xrightarrow{\text{Br}_2/\text{CH}_3\text{COOH}} \text{p-Bromoacetanilide} \xrightarrow{\text{H}^+/\text{H}_2\text{O}} \text{p-Bromoaniline}$. #### Memory Tricks & Common Mistakes - **Basicity:** Gas phase ($3^\circ > 2^\circ > 1^\circ$). Aqueous (methyl: $2^\circ > 1^\circ > 3^\circ$; ethyl: $2^\circ > 3^\circ > 1^\circ$). Aromatic amines are weak. - **Named Reactions:** Gabriel (1° aliphatic), Hoffmann (1° with C-loss). - **Tests:** Carbylamine ($1^\circ$), Hinsberg ($1^\circ, 2^\circ, 3^\circ$), Nitrous acid ($1^\circ, 2^\circ, 3^\circ$ different products). - **Aromatic Amines:** $-\text{NH}_2$ is highly activating; protect it for monosubstitution. - **Common Mistake:** Confusing the aqueous basicity order of methyl vs. ethyl amines. #### Solved JEE Main Example **Q1:** What is the product when aniline reacts with bromine water? How can you obtain *p*-bromoaniline from aniline? **A1:** 1. When aniline reacts with bromine water, it gives 2,4,6-tribromoaniline as a white precipitate. $\text{C}_6\text{H}_5\text{NH}_2 + 3\text{Br}_2(\text{aq}) \rightarrow \text{C}_6\text{H}_2(\text{Br})_3\text{NH}_2 + 3\text{HBr}$ 2. To obtain *p*-bromoaniline from aniline, the amino group needs to be protected to moderate its activating effect and prevent polysubstitution. **Steps:** a. **Acylation of Aniline:** $\text{C}_6\text{H}_5\text{NH}_2 + (\text{CH}_3\text{CO})_2\text{O} \xrightarrow{\text{Pyridine}} \text{C}_6\text{H}_5\text{NHCOCH}_3 + \text{CH}_3\text{COOH}$ (Acetanilide) b. **Bromination of Acetanilide:** $\text{C}_6\text{H}_5\text{NHCOCH}_3 \xrightarrow{\text{Br}_2/\text{CH}_3\text{COOH}} \text{p-Br}-\text{C}_6\text{H}_4-\text{NHCOCH}_3 + \text{o-Br}-\text{C}_6\text{H}_4-\text{NHCOCH}_3$ (Major product is *p*-bromoacetanilide due to steric hindrance at *ortho* positions). c. **Hydrolysis of *p*-Bromoacetanilide:** $\text{p-Br}-\text{C}_6\text{H}_4-\text{NHCOCH}_3 \xrightarrow{\text{H}^+ \text{ or } \text{OH}^-/\text{H}_2\text{O}} \text{p-Br}-\text{C}_6\text{H}_4-\text{NH}_2 + \text{CH}_3\text{COOH}$ (*p*-Bromoaniline) #### Practice Questions 1. Explain why aniline is a weaker base than methylamine. 2. What is the product when ethylamine reacts with nitrous acid? 3. Write the reaction for the Hoffmann bromamide degradation of propanamide. 4. How can you distinguish between ethylamine and diethylamine using Hinsberg test? 5. Suggest a synthesis for benzylamine from benzonitrile. **Answers:** 1. In aniline, the lone pair of electrons on the nitrogen atom is delocalized into the benzene ring through resonance. This delocalization makes the lone pair less available for protonation, reducing its basicity compared to methylamine, where the lone pair is localized and made more available by the +I effect of the methyl group. 2. Ethylamine is a $1^\circ$ aliphatic amine. It reacts with nitrous acid to form an unstable diazonium salt, which immediately decomposes to form ethanol, nitrogen gas, and HCl. $\text{CH}_3\text{CH}_2\text{NH}_2 + \text{HNO}_2 \xrightarrow{0-5^\circ\text{C}} [\text{CH}_3\text{CH}_2\text{N}_2^+\text{Cl}^-] \xrightarrow{\text{H}_2\text{O}} \text{CH}_3\text{CH}_2\text{OH} + \text{N}_2 \uparrow + \text{HCl}$ 3. Propanamide ($\text{CH}_3\text{CH}_2\text{CONH}_2$) will be converted to ethylamine ($\text{CH}_3\text{CH}_2\text{NH}_2$). $\text{CH}_3\text{CH}_2\text{CONH}_2 + \text{Br}_2 + 4\text{NaOH} \rightarrow \text{CH}_3\text{CH}_2\text{NH}_2 + \text{Na}_2\text{CO}_3 + 2\text{NaBr} + 2\text{H}_2\text{O}$ 4. **Ethylamine (1° amine):** Reacts with benzenesulfonyl chloride to form N-ethylbenzenesulfonamide. This product is soluble in aqueous KOH solution because the hydrogen attached to nitrogen is acidic. Upon acidification, the sulfonamide precipitates. **Diethylamine (2° amine):** Reacts with benzenesulfonyl chloride to form N,N-diethylbenzenesulfonamide. This product is insoluble in aqueous KOH solution because there is no acidic hydrogen attached to nitrogen. 5. Benzonitrile ($\text{C}_6\text{H}_5\text{CN}$) can be reduced to benzylamine ($\text{C}_6\text{H}_5\text{CH}_2\text{NH}_2$) using lithium aluminium hydride or catalytic hydrogenation. $\text{C}_6\text{H}_5\text{CN} \xrightarrow{\text{LiAlH}_4 \text{ or } \text{H}_2/\text{Ni}} \text{C}_6\text{H}_5\text{CH}_2\text{NH}_2$ ### Biomolecules (JEE Main level) #### Carbohydrates - **Definition:** Polyhydroxy aldehydes or ketones, or compounds which produce these upon hydrolysis. - **Classification:** 1. **Monosaccharides:** Simplest carbohydrates, cannot be hydrolyzed further. (e.g., Glucose, Fructose, Ribose). - **Glucose (Aldohexose):** $\text{C}_6\text{H}_{12}\text{O}_6$. Aldehyde group and five hydroxyl groups. Exists in open-chain and cyclic (pyranose) forms. Forms two anomers ($\alpha$-D-Glucose and $\beta$-D-Glucose). Reducing sugar (has free aldehyde/ketone group or hemiacetal/hemiketal). - **Fructose (Ketohexose):** $\text{C}_6\text{H}_{12}\text{O}_6$. Ketone group and five hydroxyl groups. Exists in open-chain and cyclic (furanose) forms. Reducing sugar. 2. **Oligosaccharides:** Yield 2-10 monosaccharide units on hydrolysis. - **Disaccharides:** Yield two monosaccharide units. - **Sucrose:** Glucose + Fructose. Non-reducing sugar (glycosidic bond involves both anomeric carbons). Hydrolysis gives invert sugar. - **Maltose:** Glucose + Glucose. Reducing sugar. - **Lactose:** Glucose + Galactose. Reducing sugar. 3. **Polysaccharides:** High molecular weight polymers yielding many monosaccharide units on hydrolysis. - **Starch:** Main storage polysaccharide for plants. Polymer of $\alpha$-D-Glucose. Consists of Amylose (linear, water-soluble) and Amylopectin (branched, water-insoluble). - **Cellulose:** Structural polysaccharide in plants. Polymer of $\beta$-D-Glucose. Humans cannot digest cellulose. - **Glycogen:** Animal starch, storage polysaccharide in animals. Highly branched polymer of $\alpha$-D-Glucose. - **Reactions of Glucose:** - With HI/Red P: *n*-Hexane (confirms straight chain). - With $\text{Br}_2$ water: Gluconic acid (confirms aldehyde group). - With hydroxylamine: Oxime (confirms carbonyl group). - With HCN: Cyanohydrin (confirms carbonyl group). - With acetic anhydride: Pentaacetate (confirms 5 -OH groups). - With Tollen's/Fehling's reagent: Positive (reducing sugar). #### Proteins - **Definition:** Polymers of $\alpha$-amino acids linked by peptide bonds. - **Amino Acids:** Contain both an amino ($\text{-NH}_2$) group and a carboxyl ($\text{-COOH}$) group, attached to the same carbon atom ($\alpha$-carbon). - **Zwitterionic form:** In aqueous solution, amino acids exist as dipolar ions. - **Isoelectric point (pI):** pH at which amino acid has zero net charge. - **Peptide Bond:** Formed by condensation reaction between the carboxyl group of one amino acid and the amino group of another, eliminating water. $$\text{R}_1\text{CH}(\text{NH}_2)\text{COOH} + \text{R}_2\text{CH}(\text{NH}_2)\text{COOH} \rightarrow \text{R}_1\text{CH}(\text{NH}_2)\text{CO}-\text{NH}-\text{CH}(\text{R}_2)\text{COOH} + \text{H}_2\text{O}$$ - **Classification of Proteins (Structure):** 1. **Primary Structure:** Sequence of amino acids in a polypeptide chain. 2. **Secondary Structure:** Local spatial arrangement of polypeptide backbone. (e.g., $\alpha$-helix, $\beta$-pleated sheet) due to H-bonding. 3. **Tertiary Structure:** Overall three-dimensional folding of the polypeptide chain (due to H-bonding, disulfide bridges, ionic interactions, van der Waals forces). 4. **Quaternary Structure:** Arrangement of multiple polypeptide subunits. - **Denaturation of Proteins:** Loss of $2^\circ, 3^\circ, 4^\circ$ structures (e.g., by heat, acid/base, heavy metal salts) leading to loss of biological activity. Primary structure remains intact. #### Nucleic Acids - **Definition:** Biopolymers responsible for genetic information storage and transfer. - **Monomers:** Nucleotides. - **Nucleotide:** Consists of a pentose sugar (ribose in RNA, deoxyribose in DNA), a nitrogenous base, and a phosphate group. - **Nitrogenous Bases:** - **Purines:** Adenine (A), Guanine (G). - **Pyrimidines:** Cytosine (C), Thymine (T, in DNA), Uracil (U, in RNA). - **DNA (Deoxyribonucleic Acid):** - Double helix structure (Watson-Crick model). - Sugar: Deoxyribose. - Bases: A, G, C, T. - Base pairing: A with T (2 H-bonds), G with C (3 H-bonds). - Function: Stores genetic information. - **RNA (Ribonucleic Acid):** - Single-stranded. - Sugar: Ribose. - Bases: A, G, C, U. - Function: Involved in protein synthesis (mRNA, tRNA, rRNA). #### Vitamins - **Definition:** Organic compounds required in small amounts for normal growth and metabolic functions, not synthesized by the body (or in sufficient amounts). - **Classification:** - **Fat-soluble:** A, D, E, K. Stored in adipose tissue and liver. - **Water-soluble:** B-complex vitamins, C. Must be supplied regularly, excreted in urine. - **Deficiency Diseases (Examples):** - Vitamin A: Xerophthalmia, Night Blindness. - Vitamin C: Scurvy. - Vitamin D: Rickets, Osteomalacia. - Vitamin B1 (Thiamine): Beriberi. - Vitamin B2 (Riboflavin): Cheilosis. - Vitamin B6 (Pyridoxine): Convulsions. - Vitamin B12 (Cyanocobalamin): Pernicious anemia. - Vitamin K: Impaired blood clotting. #### Hormones - **Definition:** Chemical messengers synthesized by endocrine glands, secreted directly into bloodstream, act on target cells/organs. - **Types:** Steroid hormones (e.g., estrogen, testosterone), peptide hormones (e.g., insulin), amine hormones (e.g., adrenaline). - **Function:** Regulate various physiological processes. #### Enzymes - **Definition:** Biological catalysts, mostly proteins, that accelerate biochemical reactions without being consumed. - **Specificity:** Highly specific for their substrates. - **Mechanism:** Lock and key model, induced fit model. - **Optimum pH and Temperature:** Activity is maximal within a narrow range. #### Memory Tricks & Common Mistakes - **Carbohydrates:** Reducing sugars have free anomeric carbon (hemiacetal/ketal). Sucrose is non-reducing. - **Proteins:** Primary structure is sequence, not lost in denaturation. - **Nucleic Acids:** DNA (T, deoxyribose, double helix), RNA (U, ribose, single strand). A-T (2H), G-C (3H). - **Vitamins:** Fat-soluble (ADEK), Water-soluble (B, C). - **Common Mistake:** Confusing the structure/function of different biomolecules. #### Solved JEE Main Example **Q1:** What is the difference between a nucleoside and a nucleotide? **A1:** - **Nucleoside:** Formed by the attachment of a nitrogenous base to a pentose sugar (ribose or deoxyribose). It lacks the phosphate group. Examples: Adenosine, Guanosine. - **Nucleotide:** Formed when a phosphate group is attached to the hydroxyl group of the sugar in a nucleoside. It is the monomer unit of DNA and RNA. Examples: Adenosine monophosphate (AMP), Deoxyadenosine triphosphate (dATP). #### Practice Questions 1. Draw the open-chain structure of D-glucose. 2. What type of glycosidic linkage is present in sucrose? Why is it a non-reducing sugar? 3. Name the forces responsible for the secondary structure of proteins. 4. If a double-stranded DNA contains 20% Adenine, what is the percentage of Guanine? 5. What is denaturation of proteins? Give an example. **Answers:** 1. (Drawing not possible here, but describe) D-Glucose is an aldohexose. Its open-chain structure is a six-carbon chain with an aldehyde group at C1, and hydroxyl groups at C2, C3, C4, C5, and C6. C2, C3, C4, C5 are chiral centers. For D-glucose, -OH at C5 is on the right. CHO | CHOH (R at C2) | CHOH (S at C3) | CHOH (R at C4) | CHOH (R at C5) | CH2OH 2. Sucrose contains an $\alpha$, $\beta$-glycosidic linkage between C1 of $\alpha$-D-glucose and C2 of $\beta$-D-fructose. It is a non-reducing sugar because the anomeric carbons of both glucose and fructose are involved in the glycosidic bond, meaning neither sugar can open up to its free aldehyde or ketone form. 3. The secondary structure of proteins (like $\alpha$-helix and $\beta$-pleated sheet) is primarily stabilized by **hydrogen bonding** between the carbonyl oxygen of one peptide bond and the amide hydrogen of another peptide bond within the polypeptide backbone. 4. According to Chargaff's rules for DNA, the amount of Adenine (A) is equal to the amount of Thymine (T), and the amount of Guanine (G) is equal to the amount of Cytosine (C). If Adenine (A) = 20%, then Thymine (T) = 20%. Total A + T = 20% + 20% = 40%. Remaining percentage for G + C = 100% - 40% = 60%. Since G = C, then Guanine (G) = 60% / 2 = 30%. So, the percentage of Guanine is 30%. 5. Denaturation of proteins is the process by which a protein loses its specific three-dimensional (secondary, tertiary, and quaternary) structure due to disruption of non-covalent interactions (like hydrogen bonds, ionic bonds, disulfide bridges, van der Waals forces) that maintain its native conformation. This often leads to loss of its biological activity. The primary structure (amino acid sequence) remains intact. **Example:** Coagulation of egg white when boiled (due to heat) or curdling of milk (due to acid, changing pH). ### Polymers #### Definition - **Polymers:** Large macromolecules formed by the repetitive bonding of small molecules called monomers. - **Monomers:** The repeating structural units from which polymers are formed. - **Polymerization:** The process of forming polymers from monomers. #### Classification of Polymers 1. **Based on Source:** - **Natural Polymers:** Found in nature (e.g., starch, cellulose, proteins, natural rubber). - **Synthetic Polymers:** Man-made (e.g., polyethylene, nylon, PVC, teflon). - **Semi-synthetic Polymers:** Derived from natural polymers by chemical modifications (e.g., cellulose acetate, rayon). 2. **Based on Structure:** - **Linear Polymers:** Long straight chains (e.g., HDPE, PVC). - **Branched-chain Polymers:** Linear chains with some branches (e.g., LDPE). - **Cross-linked (Network) Polymers:** Monomers are polyfunctional and form a 3D network structure (e.g., bakelite, melamine-formaldehyde resin). 3. **Based on Mode of Polymerization:** - **Addition Polymers:** Formed by the direct addition of monomers without the elimination of any small molecules. Monomers usually contain double or triple bonds. (e.g., polyethylene, polypropylene, PVC, teflon, natural rubber). - **Mechanism:** Free radical, cationic, or anionic polymerization. - **Condensation Polymers:** Formed by the combination of monomers with the elimination of small molecules like $\text{H}_2\text{O}$, $\text{HCl}$, $\text{NH}_3$, etc. Monomers usually have two functional groups. (e.g., nylon 6,6, terylene, bakelite). 4. **Based on Molecular Forces:** - **Elastomers:** Rubber-like solids with elastic properties (e.g., natural rubber, buna-S, buna-N). Weakest intermolecular forces. - **Fibers:** Thread-like, strong, high tensile strength (e.g., nylon 6,6, terylene). Strong H-bonding or dipole-dipole forces. - **Thermoplastics:** Soften on heating, harden on cooling (reversible). Can be reshaped (e.g., polyethylene, PVC, polystyrene). Intermolecular forces intermediate between elastomers and fibers. - **Thermosetting Polymers:** Undergo extensive cross-linking on heating and become hard, infusible. Cannot be reshaped (e.g., bakelite, melamine-formaldehyde resin). Strong covalent bonds formed during curing. #### Important Polymers (Monomers and Uses) 1. **Polyethylene:** - **Monomer:** Ethene ($\text{CH}_2=\text{CH}_2$). - **Types:** HDPE (High Density Polyethylene, linear, for buckets, bottles), LDPE (Low Density Polyethylene, branched, for plastic bags, films). - **Mode:** Addition. 2. **Polypropylene:** - **Monomer:** Propene ($\text{CH}_2=\text{CHCH}_3$). - **Mode:** Addition. - **Uses:** Ropes, carpets, plastic containers. 3. **Polyvinyl Chloride (PVC):** - **Monomer:** Vinyl chloride ($\text{CH}_2=\text{CHCl}$). - **Mode:** Addition. - **Uses:** Pipes, flooring, raincoats. 4. **Polytetrafluoroethene (Teflon):** - **Monomer:** Tetrafluoroethene ($\text{CF}_2=\text{CF}_2$). - **Mode:** Addition. - **Uses:** Non-stick cookware coatings, gaskets. 5. **Polystyrene:** - **Monomer:** Styrene ($\text{C}_6\text{H}_5-\text{CH}=\text{CH}_2$). - **Mode:** Addition. - **Uses:** Packaging, insulation (thermocol). 6. **Polyacrylonitrile (PAN, Orlon):** - **Monomer:** Acrylonitrile ($\text{CH}_2=\text{CHCN}$). - **Mode:** Addition. - **Uses:** Acrylic fibers, artificial wool. 7. **Natural Rubber:** - **Monomer:** Isoprene (2-methylbuta-1,3-diene). - **Structure:** *cis*-polyisoprene. - **Vulcanization:** Heating rubber with sulfur to improve properties (elasticity, strength, resistance to chemicals). Forms cross-links. 8. **Buna-S Rubber:** - **Monomers:** Buta-1,3-diene + Styrene. - **Mode:** Addition (Copolymerization). - **Uses:** Tyres, footwear. 9. **Buna-N Rubber:** - **Monomers:** Buta-1,3-diene + Acrylonitrile. - **Mode:** Addition (Copolymerization). - **Uses:** Oil seals, tank linings. 10. **Nylon 6,6:** - **Monomers:** Hexamethylenediamine ($\text{H}_2\text{N}-(\text{CH}_2)_6-\text{NH}_2$) + Adipic acid ($\text{HOOC}-(\text{CH}_2)_4-\text{COOH}$). - **Mode:** Condensation. - **Bond:** Amide linkage. - **Uses:** Fibers, fabrics, bristles. 11. **Nylon 6:** - **Monomer:** Caprolactam. - **Mode:** Condensation (ring-opening polymerization). - **Uses:** Tyres, fabrics. 12. **Polyesters (e.g., Terylene/Dacron):** - **Monomers:** Ethylene glycol ($\text{HO}-(\text{CH}_2)_2-\text{OH}$) + Terephthalic acid ($\text{HOOC}-\text{C}_6\text{H}_4-\text{COOH}$). - **Mode:** Condensation. - **Bond:** Ester linkage. - **Uses:** Fabrics, safety belts. 13. **Bakelite:** - **Monomers:** Phenol + Formaldehyde. - **Mode:** Condensation (Thermosetting plastic). - **Uses:** Electrical switches, handles of utensils. 14. **Melamine-Formaldehyde Resin:** - **Monomers:** Melamine + Formaldehyde. - **Mode:** Condensation (Thermosetting plastic). - **Uses:** Unbreakable crockery. #### Biodegradable Polymers - Polymers that can be decomposed by microorganisms. - **PHBV (Poly-$\beta$-hydroxybutyrate-co-$\beta$-hydroxyvalerate):** - **Monomers:** 3-Hydroxybutanoic acid + 3-Hydroxypentanoic acid. - **Uses:** Orthopaedic devices, drug capsules. - **Nylon 2-Nylon 6:** - **Monomers:** Glycine ($\text{H}_2\text{N}-\text{CH}_2-\text{COOH}$) + Aminocaproic acid ($\text{H}_2\text{N}-(\text{CH}_2)_5-\text{COOH}$). #### Memory Tricks & Common Mistakes - **Addition vs. Condensation:** Addition involves unsaturated monomers, no byproduct. Condensation involves bifunctional monomers, small byproduct. - **Thermoplastics vs. Thermosetting:** Thermoplastics (linear/branched, meltable), Thermosetting (cross-linked, non-meltable). - **Rubber:** Isoprene is natural rubber. Buna-S, Buna-N are synthetic. Vulcanization improves rubber properties. - **Nylon:** Nylon 6,6 (diamine + diacid). Nylon 6 (caprolactam). - **Common Mistake:** Confusing monomers for different polymers, especially between similar-sounding ones. #### Solved JEE Main Example **Q1:** What is the monomer of natural rubber? What is the function of vulcanization of rubber? **A1:** - The monomer of natural rubber is **isoprene** (2-methylbuta-1,3-diene). - **Vulcanization** is the process of heating natural rubber with sulfur (typically 3-10%). This process introduces sulfur cross-links between the polymer chains. The function of vulcanization is to: 1. Improve the physical properties of rubber, such as **elasticity, tensile strength, and hardness**. 2. Increase its **resistance to abrasion and chemical attack**. 3. Reduce its **plasticity and tackiness**. 4. Make it **less susceptible to temperature changes** (less brittle in cold, less sticky in hot). #### Practice Questions 1. Classify polyethylene based on its mode of polymerization. Name its monomer. 2. What are the monomers of Nylon 6,6? What type of linkage holds them together? 3. Distinguish between thermoplastics and thermosetting polymers with an example of each. 4. Give the monomer of Teflon and its main use. 5. Why is Buna-N preferred over natural rubber for making oil seals? **Answers:** 1. Polyethylene is an **addition polymer**. Its monomer is **ethene** ($\text{CH}_2=\text{CH}_2$). 2. The monomers of Nylon 6,6 are **hexamethylenediamine** ($\text{H}_2\text{N}-(\text{CH}_2)_6-\text{NH}_2$) and **adipic acid** ($\text{HOOC}-(\text{CH}_2)_4-\text{COOH}$). They are held together by **amide linkages** ($\text{-CO-NH-}$). 3. **Thermoplastics:** Polymers that soften on heating and can be molded into various shapes repeatedly. They have linear or branched structures with relatively weak intermolecular forces. **Example:** Polyethylene, PVC. **Thermosetting Polymers:** Polymers that undergo extensive cross-linking upon heating, becoming hard, rigid, and infusible. They cannot be reshaped once molded. **Example:** Bakelite, Melamine-formaldehyde resin. 4. The monomer of Teflon is **tetrafluoroethene** ($\text{CF}_2=\text{CF}_2$). Its main use is in making **non-stick coatings for cookware** and for chemical-resistant gaskets and seals. 5. Buna-N (Nitrile rubber) is a copolymer of buta-1,3-diene and acrylonitrile. The presence of polar nitrile groups ($-\text{CN}$) makes it resistant to the action of oils, gasoline, and other organic solvents. Natural rubber, being nonpolar, swells and degrades in such solvents. Therefore, Buna-N is preferred for making oil seals and tank linings. ### Practical Organic Chemistry #### Purification Methods 1. **Crystallization:** - **Principle:** Based on the difference in solubility of the organic compound and its impurities in a suitable solvent. The compound is sparingly soluble at room temperature but appreciably soluble at higher temperatures. Impurities are either highly soluble or insoluble. - **Procedure:** Dissolve crude solid in minimum hot solvent, filter hot (to remove insoluble impurities), cool slowly (compound crystallizes out), filter cold, wash with cold solvent, and dry. - **Criteria for solvent:** Should not react with compound, dissolve compound sparingly at room temp and readily at boiling temp, dissolve impurities readily or not at all. 2. **Sublimation:** - **Principle:** Used for solids that directly change from solid to gas phase on heating without passing through liquid phase, and then back to solid on cooling. Impurities should not sublime. - **Examples:** Naphthalene, Anthracene, Benzoic acid, Camphor. 3. **Distillation:** - **Principle:** Used for liquids based on difference in their boiling points. - **Simple Distillation:** For liquids boiling below $200^\circ\text{C}$ and having a difference of at least $25^\circ\text{C}$ in their boiling points. - **Fractional Distillation:** For liquids having boiling points close to each other (difference less than $25^\circ\text{C}$). Uses a fractionating column. - **Distillation Under Reduced Pressure (Vacuum Distillation):** For liquids that decompose below their normal boiling points. Boiling point is lowered by reducing external pressure. - **Steam Distillation:** For compounds that are steam volatile (volatile in steam), immiscible with water, and decompose at their normal boiling points. (e.g., Aniline, Nitrobenzene, essential oils). 4. **Differential Extraction:** - **Principle:** Used for separating organic compounds from aqueous solutions. The organic compound is more soluble in an immiscible organic solvent than in water. - **Procedure:** Shake aqueous solution with organic solvent, two layers form, separate using separating funnel. Repeat multiple times. 5. **Chromatography:** - **Principle:** Based on the difference in adsorption or partition of compounds between a stationary phase and a mobile phase. - **Adsorption Chromatography:** - **Column Chromatography:** Stationary phase (solid adsorbent like $\text{Al}_2\text{O}_3$ or $\text{SiO}_2$) packed in a column. Mobile phase (liquid solvent) moves through. - **Thin Layer Chromatography (TLC):** Adsorbent coated on a glass plate. Sample applied, solvent moves up by capillary action. $\text{R}_{\text{f}} \text{ value} = \frac{\text{distance travelled by substance}}{\text{distance travelled by solvent}}$. - **Partition Chromatography:** - **Paper Chromatography:** Stationary phase (water adsorbed on paper). Mobile phase (organic solvent). #### Qualitative Analysis of Organic Compounds **1. Detection of Elements:** - **Lassaigne's Test (Sodium Fusion Test):** To detect N, S, Halogens. Organic compound is fused with sodium metal to convert these elements into ionic forms ($\text{NaCN}$, $\text{Na}_2\text{S}$, $\text{NaX}$) which can be tested. - **Nitrogen:** $\text{NaCN} + \text{FeSO}_4 \rightarrow \text{Na}_4[\text{Fe(CN)}_6]$ (Sodium ferrocyanide). Then $\text{Fe}^{3+} \rightarrow \text{Prussian blue}$ color. - **Sulfur:** $\text{Na}_2\text{S} + \text{Na}_2[\text{Fe(CN)}_5\text{NO}] \rightarrow \text{Na}_4[\text{Fe(CN)}_5\text{NOS}]$ (Violet color with Sodium nitroprusside). Or $\text{Na}_2\text{S} + \text{Pb(OAc)}_2 \rightarrow \text{PbS}\downarrow$ (Black ppt). - **Halogens:** $\text{NaX} + \text{AgNO}_3 \rightarrow \text{AgX}\downarrow$. - $\text{AgCl}$: White ppt, soluble in $\text{NH}_4\text{OH}$. - $\text{AgBr}$: Pale yellow ppt, sparingly soluble in $\text{NH}_4\text{OH}$. - $\text{AgI}$: Yellow ppt, insoluble in $\text{NH}_4\text{OH}$. - **Carbon and Hydrogen:** Heated with $\text{CuO}$. C $\rightarrow \text{CO}_2$ (turns limewater milky). H $\rightarrow \text{H}_2\text{O}$ (turns anhydrous $\text{CuSO}_4$ blue). **2. Detection of Functional Groups:** - **Unsaturation ($\text{C}=\text{C}, \text{C}\equiv\text{C}$):** - **Baeyer's Test:** Decolorizes cold, dilute, alkaline $\text{KMnO}_4$ (purple to colorless). - **Bromine Water Test:** Decolorizes $\text{Br}_2$ water (reddish-brown to colorless). - **Alcohols:** - **Lucas Test:** $1^\circ, 2^\circ, 3^\circ$ alcohols with $\text{HCl}/\text{ZnCl}_2$. - **Ceric Ammonium Nitrate Test:** Red color with alcohols and phenols. - **Sodium metal:** Evolution of $\text{H}_2$ gas. - **Phenols:** - **Ferric Chloride Test:** Violet/blue/green color with neutral $\text{FeCl}_3$. - **Bromine water:** White precipitate (2,4,6-tribromophenol). - **Aldehydes and Ketones:** - **2,4-Dinitrophenylhydrazine (2,4-DNP) Test:** Yellow/orange/red precipitate (Confirms carbonyl group). - **Tollen's Test:** Silver mirror (for aldehydes). - **Fehling's/Benedict's Test:** Red precipitate of $\text{Cu}_2\text{O}$ (for aliphatic aldehydes). - **Iodoform Test:** Yellow precipitate of $\text{CHI}_3$ (for $\text{CH}_3\text{CO}-$ or $\text{CH}_3\text{CH}(\text{OH})-$ groups). - **Carboxylic Acids:** - **Litmus Test:** Turns blue litmus red. - **Sodium Bicarbonate Test:** Effervescence of $\text{CO}_2$ gas. - **Amines:** - **Litmus Test:** Turns red litmus blue (basic). - **Carbylamine Test:** Foul-smelling isocyanide (for $1^\circ$ amines). - **Hinsberg Test:** Distinguishes $1^\circ, 2^\circ, 3^\circ$ amines. #### Important Reagents and their Uses - **Grignard reagent ($\text{RMgX}$):** Forms alcohols from carbonyls, forms carboxylic acids from $\text{CO}_2$. - **$\text{LiAlH}_4$ (Lithium Aluminium Hydride):** Strong reducing agent, reduces aldehydes, ketones, carboxylic acids, esters, amides, nitriles. - **$\text{NaBH}_4$ (Sodium Borohydride):** Milder reducing agent, reduces aldehydes and ketones. Does not reduce carboxylic acids, esters, amides. - **PCC (Pyridinium Chlorochromate):** Mild oxidizing agent, converts $1^\circ$ alcohols to aldehydes, $2^\circ$ alcohols to ketones. - **$\text{KMnO}_4$ (Potassium Permanganate):** Strong oxidizing agent. - Cold, dilute, alkaline: Baeyer's reagent, dihydroxylation of alkenes. - Hot, acidic: Cleaves double/triple bonds, oxidizes $1^\circ$ alcohols to acids, alkylbenzenes to benzoic acid. - **$\text{CrO}_3$ (Chromium Trioxide) / Jones Reagent ($\text{CrO}_3/\text{H}_2\text{SO}_4$):** Strong oxidizing agent, oxidizes $1^\circ$ alcohols to acids, $2^\circ$ alcohols to ketones. - **$\text{Zn-Hg/HCl}$ (Clemmensen Reagent):** Reduces carbonyl group to methylene group. (Acidic conditions). - **$\text{NH}_2\text{NH}_2/\text{KOH}$ (Wolff-Kishner Reagent):** Reduces carbonyl group to methylene group. (Basic conditions). - **$\text{Pd/BaSO}_4$ (Lindlar's Catalyst):** Partially hydrogenates alkynes to *cis*-alkenes. - **$\text{Na/liq. NH}_3$ (Birch Reduction):** Partially hydrogenates alkynes to *trans*-alkenes. - **$\text{DIBAL-H}$ (Diisobutylaluminium Hydride):** Reduces esters/nitriles to aldehydes at low temperature. - **$\text{NaNO}_2/\text{HCl}$ (Nitrous Acid):** Diazotization of $1^\circ$ amines. - **$\text{FeCl}_3$ (Ferric Chloride):** Test for phenols. - **Tollen's Reagent:** Test for aldehydes. - **Fehling's/Benedict's Reagent:** Test for aliphatic aldehydes. - **$\text{I}_2/\text{NaOH}$ (Iodoform Reagent):** Test for $\text{CH}_3\text{CO}-$ or $\text{CH}_3\text{CH}(\text{OH})-$ groups. - **$\text{CHCl}_3/\text{KOH}$ (Carbylamine Reagent):** Test for $1^\circ$ amines. - **Benzenesulfonyl chloride (Hinsberg Reagent):** Distinguishes $1^\circ, 2^\circ, 3^\circ$ amines. #### Memory Tricks & Common Mistakes - **Lassaigne's:** Remember $\text{NaCN}$ (N), $\text{Na}_2\text{S}$ (S), $\text{NaX}$ (Halogens). - **Tests for Carbonyl:** 2,4-DNP (general), Tollen's (aldehyde), Fehling's (aliphatic aldehyde), Iodoform (methyl ketone/alcohol). - **Reductions:** $\text{LiAlH}_4$ (strong, reduces almost all functional groups to alcohols or amines), $\text{NaBH}_4$ (mild, only aldehydes/ketones). Clemmensen/Wolff-Kishner (carbonyl to alkane). - **Common Mistake:** Confusing the conditions (acidic/basic) for Clemmensen/Wolff-Kishner reductions. - **Common Mistake:** Forgetting the specific reagents for *cis* vs *trans* alkene formation from alkynes. #### Solved JEE Main Example **Q1:** An organic compound (A) gives a positive Lassaigne's test for nitrogen and forms a red precipitate with ammoniacal cuprous chloride. Compound (A) on hydration with $\text{HgSO}_4/\text{H}_2\text{SO}_4$ gives an aldehyde. Identify (A). **A1:** The compound (A) is propyne. **Explanation:** - **Positive Lassaigne's test for nitrogen:** This information seems irrelevant or a distractor for this specific question, as the other clues point to a hydrocarbon. (Unless the compound is a nitrile or amine, but the hydration reaction points otherwise). Let's assume the question implies nitrogen is present in some other part of the molecule or this is a general test question. Let's focus on the other clues. - **Forms a red precipitate with ammoniacal cuprous chloride:** This is a specific test for **terminal alkynes**. So, (A) must be a terminal alkyne. - **On hydration with $\text{HgSO}_4/\text{H}_2\text{SO}_4$ gives an aldehyde:** Hydration of alkynes generally gives ketones (Markovnikov's rule) via enol tautomerization, *except for acetylene* which gives acetaldehyde. If the product is an aldehyde, it means either (A) is acetylene, or it's a terminal alkyne that forms a primary enol. - If (A) is acetylene ($\text{HC}\equiv\text{CH}$), hydration gives acetaldehyde ($\text{CH}_3\text{CHO}$). - If (A) is propyne ($\text{CH}_3-\text{C}\equiv\text{CH}$), hydration gives propanone ($\text{CH}_3\text{COCH}_3$). - The question says "an aldehyde", not specifically acetaldehyde. However, if the hydration gives an aldehyde, it usually implies that the initial enol formed must be a primary enol, which only happens with acetylene. **Re-evaluating the "aldehyde" part:** For a terminal alkyne, Markovnikov's rule applies, placing -OH on the more substituted carbon. $\text{R}-\text{C}\equiv\text{CH} \xrightarrow{\text{H}_2\text{O}/\text{HgSO}_4} [\text{R}-\text{C}(\text{OH})=\text{CH}_2] \rightleftharpoons \text{R}-\text{CO}-\text{CH}_3$ (Ketone) So, if the product is an aldehyde, the only possible terminal alkyne is **acetylene** ($\text{HC}\equiv\text{CH}$). Let's recheck the question: "forms a red precipitate with ammoniacal cuprous chloride." Acetylene also gives a red precipitate with ammoniacal cuprous chloride (cuprous acetylide). So, compound (A) is **acetylene** ($\text{HC}\equiv\text{CH}$). *(Note: The initial interpretation of "an aldehyde" as possibly meaning any aldehyde led to propyne, but then the hydration product would be a ketone. This highlights careful reading of "an aldehyde" vs "a ketone" and the specific case of acetylene).* #### Practice Questions 1. How will you distinguish between phenol and benzoic acid using a chemical test? Write the reaction. 2. An organic compound (X) gives a yellow precipitate with $\text{I}_2/\text{NaOH}$ and forms an orange-red precipitate with 2,4-DNP. It does not reduce Tollen's reagent. Identify (X) and draw its structure. 3. Name a suitable method for the purification of a mixture of aniline and nitrobenzene. 4. Describe Lassaigne's test for the detection of sulfur in an organic compound. 5. What is the $\text{R}_{\text{f}}$ value in Thin Layer Chromatography? How is it determined? **Answers:** 1. **Benzoic acid** will react with sodium bicarbonate ($\text{NaHCO}_3$) to produce effervescence of carbon dioxide gas, as it is a stronger acid than carbonic acid. **Phenol** is a weaker acid than carbonic acid and will not react with $\text{NaHCO}_3$. $\text{C}_6\text{H}_5\text{COOH} + \text{NaHCO}_3 \rightarrow \text{C}_6\text{H}_5\text{COONa} + \text{H}_2\text{O} + \text{CO}_2 \uparrow$ 2. (X) gives yellow precipitate with $\text{I}_2/\text{NaOH}$ (iodoform test) $\rightarrow$ Contains $\text{CH}_3\text{CO}-$ group. (X) forms orange-red precipitate with 2,4-DNP $\rightarrow$ Contains carbonyl group ($\text{C}=\text{O}$). (X) does not reduce Tollen's reagent $\rightarrow$ It is a ketone, not an aldehyde. Combining these, (X) is a methyl ketone. The simplest methyl ketone is **Propanone (Acetone)**. Structure: $\text{CH}_3-\text{CO}-\text{CH}_3$. 3. Aniline and nitrobenzene are both immiscible with water and have different boiling points (Aniline b.p. $184^\circ\text{C}$, Nitrobenzene b.p. $210^\circ\text{C}$). They can be separated by **steam distillation**. Aniline is steam volatile while nitrobenzene is not to the same extent. Or, by **fractional distillation** since their boiling points are sufficiently different ($>25^\circ\text{C}$). Alternatively, Aniline is basic and can react with dilute HCl to form anilinium chloride (water-soluble salt), while nitrobenzene is neutral and insoluble in water. So, **differential extraction** using dilute HCl can separate them. The aqueous layer containing anilinium chloride can then be basified to regenerate aniline. 4. **Lassaigne's Test for Sulfur:** **Procedure:** A small piece of dried organic compound is fused with a clean piece of sodium metal in a fusion tube. The red-hot tube is plunged into distilled water, and the contents are boiled and filtered. This filtrate is called Lassaigne's extract or Sodium fusion extract. **Test:** To a portion of Lassaigne's extract, add a few drops of sodium nitroprusside solution ($\text{Na}_2[\text{Fe(CN)}_5\text{NO}]$). **Observation:** If sulfur is present, a **deep violet/purple color** is observed. **Reaction:** $\text{Na}_2\text{S} + \text{Na}_2[\text{Fe(CN)}_5\text{NO}] \rightarrow \text{Na}_4[\text{Fe(CN)}_5\text{NOS}]$ (Sodium thionitroprusside, violet). Alternatively, add lead acetate solution. A black precipitate of lead sulfide indicates sulfur. $\text{Na}_2\text{S} + \text{Pb(OAc)}_2 \rightarrow \text{PbS}\downarrow (\text{black}) + 2\text{NaOAc}$. 5. The $\text{R}_{\text{f}}$ (retardation factor) value in Thin Layer Chromatography is the ratio of the distance travelled by the solute (compound) to the distance travelled by the solvent front, both measured from the starting line. $$\text{R}_{\text{f}} = \frac{\text{Distance travelled by solute from baseline}}{\text{Distance travelled by solvent front from baseline}}$$ It is a characteristic constant for a given compound under specific chromatographic conditions (adsorbent, solvent, temperature). It is determined by applying the sample to the TLC plate, developing the chromatogram in a solvent, marking the solvent front, and then measuring the distances.