JEE Advanced 2014 · Paper 2 · Chemistry Q31 · Dronacharya Exams

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ChemistrySingle correct+3 / -1Question 31 of 60

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Schemes 1 and 2 describe sequential transformation of alkynes M and N. Consider only the major products formed in each step for both the schemes.

Scheme-1: M = HO-CH₂CH₂-C≡C-H treated with 1. NaNH₂ (excess); 2. CH₃CH₂I (1 equivalent); 3. CH₃I (1 equivalent); 4. H₂, Lindlar's catalyst gives X. Scheme-2: N = but-1-yne, CH₃CH₂-C≡C-H, treated with 1. NaNH₂ (2 equivalent); 2. BrCH₂CH(OH)CH₃; 3. H₃O⁺ (mild); 4. H₂, Pd/C; 5. CrO₃ gives Y.

The product X is

A

(Z)-1-Methoxyhex-3-ene, CH₃O-CH₂CH₂-CH=CH-CH₂CH₃ with both H on the same side (cis).

B

(E)-1-Methoxyhex-3-ene, CH₃O-CH₂CH₂-CH=CH-CH₂CH₃ with the H atoms on opposite sides (trans).

C

(Z)-1-Ethoxypent-3-ene, CH₃CH₂O-CH₂CH₂-CH=CH-CH₃ with both H on the same side (cis).

D

(E)-1-Ethoxypent-3-ene, CH₃CH₂O-CH₂CH₂-CH=CH-CH₃ with the H atoms on opposite sides (trans).

Answer and worked solution

Official answer: A

Scheme 1.

Excess $\ce{NaNH2}$ removes both acidic protons, giving the dianion $\ce{^-O-CH2CH2-C#C^-}$ .
One equivalent of ethyl iodide alkylates the more nucleophilic site, the acetylide carbon: $\ce{^-OCH2CH2C#C-CH2CH3}$ .
Methyl iodide then alkylates the alkoxide: $\ce{CH3O-CH2CH2-C#C-CH2CH3}$ .
Lindlar's catalyst gives syn addition of $\ce{H2}$ : the cis alkene.

X is (Z)-1-methoxyhex-3-ene, $\ce{CH3OCH2CH2CH=CHCH2CH3}$ , $\ce{C7H14O}$ .

(A)

Topics:Alkynes Ethers Geometrical isomerism

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