IIT-JEE 2008 · Paper 1 · Chemistry Q62

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ChemistrySingle correct+4 / -1Question 62 of 69

Passage

Paragraph for Question Nos. 61 to 63

In the following reaction sequence, products I, J and L are formed. K represents a reagent.

Hex-3-ynal $\xrightarrow{\text{1. NaBH}_4;\ \text{2. PBr}_3}$ I $\xrightarrow{\text{1. Mg/ether};\ \text{2. CO}_2;\ \text{3. H}_3\text{O}^+}$ J $\xrightarrow{\textbf{K}}$ $\ce{CH3CH2C#CCH2CH2COCl}$ $\xrightarrow{\text{H}_2,\ \text{Pd/BaSO}_4,\ \text{quinoline}}$ L

(In the original, structures are drawn as skeletal formulae; the options are given here as condensed formulae.)

Hex-3-ynal → (1. NaBH₄, 2. PBr₃) I → (1. Mg/ether, 2. CO₂, 3. H₃O⁺) J → (K) hept-4-ynoyl chloride → (H₂, Pd/BaSO₄, quinoline) L.

The structures of compounds J and K, respectively, are

\ce{CH3CH2C#CCH2CH2COOH}

and

\ce{SOCl2}

\ce{CH3CH2CH=CHCH2CH2COOH}

(trans) and

\ce{SO2Cl2}

\ce{CH3CH2CH=CHCH2CH2COOH}

(cis) and

\ce{SOCl2}

\ce{CH3CH2C#CCH2CH2COOH}

and

\ce{CH3SO2Cl}

Answer and worked solution

Official answer: A

J. The Grignard reagent from I adds to $\ce{CO2}$ ; acid work-up gives hept-4-ynoic acid, $\ce{CH3CH2C#CCH2CH2COOH}$ (one carbon longer).

K. Converting $\ce{-COOH}$ to $\ce{-COCl}$ : thionyl chloride, $\ce{SOCl2}$ . ( $\ce{SO2Cl2}$ and mesyl chloride do not make acid chlorides.)

(A)

Topics:Carboxylic acids Acid derivatives

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IIT-JEE 2008 · Paper 1, Chemistry question 62