JEE Challenger
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ChemHardAdv 2018 • Paper 1

Identification of Compound S from Carboxylic Acid P Reaction Sequence

Comprehension Passage

PARAGRAPH "A"

An organic acid P\mathbf{P} (C11H12O2\text{C}_{11}\text{H}_{12}\text{O}_2) can easily be oxidized to a dibasic acid which reacts with ethylene glycol to form the polymer dacron. Upon ozonolysis, P\mathbf{P} yields an aliphatic ketone as one of the products. P\mathbf{P} undergoes the following reaction sequences to furnish R\mathbf{R} via Q\mathbf{Q}. The compound P\mathbf{P} also undergoes another sequence of reactions to produce S\mathbf{S}.

S←4) CHCl3, KOH, Δ5) H2/Pd-C1) H2/Pd-C2) NH3/Δ3) Br2/NaOHP→3) MeMgBr, CdCl24) NaBH41) H2/Pd-C2) SOCl2Q→3) CO2 (dry ice)4) H3O+1) HCl2) Mg/Et2OR\begin{array}{ccc} \mathbf{S} & \xleftarrow[\begin{array}{l} \text{4) }\text{CHCl}_3,\,\text{KOH},\,\Delta \\ \text{5) }\text{H}_2/\text{Pd-C} \end{array}]{\begin{array}{l} \text{1) }\text{H}_2/\text{Pd-C} \\ \text{2) }\text{NH}_3/\Delta \\ \text{3) }\text{Br}_2/\text{NaOH} \end{array}} & \mathbf{P} \xrightarrow[\begin{array}{l} \text{3) }\text{MeMgBr},\,\text{CdCl}_2 \\ \text{4) }\text{NaBH}_4 \end{array}]{\begin{array}{l} \text{1) }\text{H}_2/\text{Pd-C} \\ \text{2) }\text{SOCl}_2 \end{array}} & \mathbf{Q} \xrightarrow[\begin{array}{l} \text{3) }\text{CO}_2\text{ (dry ice)} \\ \text{4) }\text{H}_3\text{O}^+ \end{array}]{\begin{array}{l} \text{1) }\text{HCl} \\ \text{2) }\text{Mg/Et}_2\text{O} \end{array}} & \mathbf{R} \end{array}

(There are two questions based on PARAGRAPH "A", the question given below is one of them)

Determine the structure of the compound S\mathbf{S}:

Options

A
Option A
B
Option B
Correct
C
Option C
D
Option D

Topics & Concepts

AminesCarboxylic Acids and DerivativesAminesOxidation and Reduction (Organic)Polymers

Step-by-Step Solution

To determine the structure of compound S\mathbf{S}, let us analyze the given information step-by-step:

1. Determination of the Structure of Compound P\mathbf{P} (C11H12O2\text{C}_{11}\text{H}_{12}\text{O}_2)

  1. Degree of Unsaturation (DU): DU=11+1−122=6\text{DU} = 11 + 1 - \frac{12}{2} = 6
  2. Oxidation product: Compound P\mathbf{P} is oxidized easily to a dibasic acid that condenses with ethylene glycol to give Dacron (poly(ethylene terephthalate)). Therefore, the dibasic acid is terephthalic acid (benzene-1,4-dicarboxylic acid), indicating that P\mathbf{P} is a 1,4-disubstituted (para-substituted) benzene derivative.
  3. Ozonolysis product: Upon ozonolysis, P\mathbf{P} yields an aliphatic ketone, which means the side chain has a terminal branched alkene unit, −CH=C(CH3)2-\text{CH}=\text{C}(\text{CH}_3)_2, yielding acetone (CH3COCH3\text{CH}_3\text{COCH}_3).
  4. Structure of P\mathbf{P}: Combining the carboxylic acid group (−COOH-\text{COOH}), the benzene ring (−C6H4−-\text{C}_6\text{H}_4-), and the alkene group (−CH=C(CH3)2-\text{CH}=\text{C}(\text{CH}_3)_2): P=HOOC−C6H4−CH=C(CH3)2(4-(2-methylprop-1-en-1-yl)benzoic acid)\mathbf{P} = \text{HOOC}-\text{C}_6\text{H}_4-\text{CH}=\text{C}(\text{CH}_3)_2 \quad (\text{4-(2-methylprop-1-en-1-yl)benzoic acid})

2. Reaction Sequence to Form Compound S\mathbf{S}

The conversion of P\mathbf{P} to S\mathbf{S} proceeds through the following steps:

  1. Step 1: Catalytic Hydrogenation (H2/Pd-C\text{H}_2/\text{Pd-C}) The aliphatic C=C\text{C}=\text{C} double bond is reduced to give 4-isobutylbenzoic acid: HOOC−C6H4−CH=C(CH3)2→H2/Pd-CHOOC−C6H4−CH2−CH(CH3)2\text{HOOC}-\text{C}_6\text{H}_4-\text{CH}=\text{C}(\text{CH}_3)_2 \xrightarrow{\text{H}_2/\text{Pd-C}} \text{HOOC}-\text{C}_6\text{H}_4-\text{CH}_2-\text{CH}(\text{CH}_3)_2

  2. Step 2: Amidation (NH3,Δ\text{NH}_3, \Delta) The carboxylic acid group is converted into a primary amide: HOOC−C6H4−CH2−CH(CH3)2→NH3,ΔH2NOC−C6H4−CH2−CH(CH3)2\text{HOOC}-\text{C}_6\text{H}_4-\text{CH}_2-\text{CH}(\text{CH}_3)_2 \xrightarrow{\text{NH}_3, \Delta} \text{H}_2\text{NOC}-\text{C}_6\text{H}_4-\text{CH}_2-\text{CH}(\text{CH}_3)_2

  3. Step 3: Hoffmann Bromamide Degradation (Br2/NaOH\text{Br}_2/\text{NaOH}) The amide is degraded to a primary aromatic amine: H2NOC−C6H4−CH2−CH(CH3)2→Br2/NaOHH2N−C6H4−CH2−CH(CH3)2(4-isobutylaniline)\text{H}_2\text{NOC}-\text{C}_6\text{H}_4-\text{CH}_2-\text{CH}(\text{CH}_3)_2 \xrightarrow{\text{Br}_2/\text{NaOH}} \text{H}_2\text{N}-\text{C}_6\text{H}_4-\text{CH}_2-\text{CH}(\text{CH}_3)_2 \quad (\text{4-isobutylaniline})

  4. Step 4: Carbylamine Reaction (CHCl3,KOH,Δ\text{CHCl}_3, \text{KOH}, \Delta) The primary amine reacts to form an isocyanide (isonitrile): H2N−C6H4−CH2−CH(CH3)2→CHCl3,KOH,ΔC≡N−C6H4−CH2−CH(CH3)2\text{H}_2\text{N}-\text{C}_6\text{H}_4-\text{CH}_2-\text{CH}(\text{CH}_3)_2 \xrightarrow{\text{CHCl}_3, \text{KOH}, \Delta} \text{C}\equiv\text{N}-\text{C}_6\text{H}_4-\text{CH}_2-\text{CH}(\text{CH}_3)_2

  5. Step 5: Reduction of Isocyanide (H2/Pd-C\text{H}_2/\text{Pd-C}) Catalytic hydrogenation reduces the isocyanide group (−N≡C-\text{N}\equiv\text{C}) to a secondary methylamino group (−NH−CH3-\text{NH}-\text{CH}_3): C≡N−C6H4−CH2−CH(CH3)2→H2/Pd-CCH3−NH−C6H4−CH2−CH(CH3)2\text{C}\equiv\text{N}-\text{C}_6\text{H}_4-\text{CH}_2-\text{CH}(\text{CH}_3)_2 \xrightarrow{\text{H}_2/\text{Pd-C}} \text{CH}_3-\text{NH}-\text{C}_6\text{H}_4-\text{CH}_2-\text{CH}(\text{CH}_3)_2

Thus, compound S\mathbf{S} is NN-methyl-4-isobutylaniline, which corresponds to structure (B).

Correct Option: (B)

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