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. 1. $\mathrm{NaNH}_{2}$ (excess) 2. $\mathrm{NaNH}_{2}$ (2 equivalent) The correct statement with respect to product Y is

(A) It gives a positive Tollens test and is a functional isomer of X.

(B) It gives a positive Tollens test and is a geometrical isomer of X.

(D) It gives a positive iodoform test and is a geometrical isomer of X.

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

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 correct statement with respect to product Y is

(A)It gives a positive Tollens test and is a functional isomer of X.
(B)It gives a positive Tollens test and is a geometrical isomer of X.
(C)It gives a positive iodoform test and is a functional isomer of X.✓ Correct
(D)It gives a positive iodoform test and is a geometrical isomer of X.

Answer: (C)

Solution

Product Y is formed from Scheme 2. Following the transformations, Y is a ketone with a

CH_{3} CO -

group (methyl ketone structure). Methyl ketones give a positive iodoform test when treated with

I_{2} / NaOH

.

Product X from Scheme 1 is an ether, while Y is a ketone. They have the same molecular formula but different functional groups, making them functional isomers.

Y does not give a positive Tollens test because ketones (except α-hydroxy ketones) do not reduce Tollens' reagent. Only aldehydes give a positive Tollens test.

Therefore, Y gives a positive iodoform test and is a functional isomer of X.

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

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

Solution

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