JEE Advanced · 2014 · 25 May 2014 · Paper 2 · Hydrocarbons

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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.

$NaNH_{2}$ (excess)

$NaNH_{2}$ (2 equivalent)

The product X is

(A)(see structure)
✓ Correct
(B)(see structure)
(C)(see structure)
(D)(see structure)

Answer: (A)

Solution

In Scheme 1, the alkyne M is treated with excess

NaNH_{2}

(a strong base), which deprotonates the terminal alkyne. The resulting acetylide anion then undergoes nucleophilic substitution with an alkyl halide, followed by reduction with

Na/NH_{3}

(Birch reduction conditions) to give a trans-alkene. The final step involves reaction with an alcohol or ether-forming reagent to give product X, which is the cis-alkene ether shown in option (A).

JEE Advanced · 2014 · 25 May 2014 · Paper 2 · Single correct option

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