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← IIT-JEE 2008 · Paper 1
ChemistrySingle correct+4 / -1Question 61 of 69

IIT-JEE 2008 · Paper 1, Chemistry question 61

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 →1. NaBH4; 2. PBr3\xrightarrow{\text{1. NaBH}_4;\ \text{2. PBr}_3} I →1. Mg/ether; 2. CO2; 3. H3O+\xrightarrow{\text{1. Mg/ether};\ \text{2. CO}_2;\ \text{3. H}_3\text{O}^+} J →K\xrightarrow{\textbf{K}} CHX3CHX2C≡CCHX2CHX2COCl\ce{CH3CH2C#CCH2CH2COCl} →H2, Pd/BaSO4, quinoline\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 structure of the product I is

Choose one option
Answer and worked solution

Official answer: D

Step 1. NaBHX4\ce{NaBH4} reduces the aldehyde (it does not touch the C≡C): hex-3-yn-1-ol, CHX3CHX2C≡CCHX2CHX2OH\ce{CH3CH2C#CCH2CH2OH}.

Step 2. PBrX3\ce{PBr3} converts the primary alcohol to the bromide: I = CHX3CHX2C≡CCHX2CHX2Br\ce{CH3CH2C#CCH2CH2Br}.

(D)

Topics:SN1 and SN2Oxidation and reduction
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