JEE Advanced 2021 · Paper 1 · Chemistry Q1

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ChemistrySingle correct+3 / -1Question 20 of 57

The major product formed in the following reaction is

Undeca-1,7-diyne (HC≡C-(CH2)4-C≡C-CH2CH2CH3, skeletal) treated with NaNH2, then Na in liquid NH3.

(E)-Undeca-1,7-diene (structure shown)

(E)-Undeca-1,7-diene, drawn as a skeletal zigzag with a terminal C=C and a trans C7=C8 double bond.

(E)-Undec-7-en-1-yne (structure shown)

(E)-Undec-7-en-1-yne: the terminal C≡C kept, the internal triple bond reduced to a trans C=C.

(Z)-Undec-7-en-1-yne (structure shown)

(Z)-Undec-7-en-1-yne: the terminal C≡C kept, the internal triple bond reduced to a cis C=C.

(Z)-Undeca-1,7-diene (structure shown)

(Z)-Undeca-1,7-diene: terminal C=C and a cis C7=C8 double bond.

Answer and worked solution

Official answer: B

Step 1. $\ce{NaNH2}$ deprotonates the terminal alkyne, giving the acetylide $\ce{RC#C^- Na+}$ .

Step 2. Na in liquid $\ce{NH3}$ reduces the internal alkyne to the trans alkene. The acetylide anion is not reduced, and the terminal C≡C is restored on work-up.

Product. (E)-Undec-7-en-1-yne. (B)

Topics:Alkynes

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JEE Advanced 2021 · Paper 1, Chemistry question 1