JEE Advanced 2014 · Paper 2 · Chemistry Q32

ChemistrySingle correct+3 / -1Question 32 of 60

Passage

Paragraph For Question 31 and 32

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.

The correct statement with respect to product Y is

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

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

It gives a positive iodoform test and is a functional isomer of X.

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

Answer and worked solution

Official answer: C

Scheme 2 (N is but-1-yne).

Two equivalents of $\ce{NaNH2}$ : the acetylide $\ce{CH3CH2C#C^-}$ forms, and the extra base deprotonates the bromohydrin's OH.
The acetylide attacks the $\ce{CH2}$ carbon (directly, or by opening the epoxide formed from the bromohydrin): $\ce{CH3CH2C#C-CH2-CH(O^-)CH3}$ .
Mild acid gives hept-4-yn-2-ol.
$\ce{H2}$ , Pd/C reduces the triple bond fully: heptan-2-ol.
$\ce{CrO3}$ oxidises the secondary alcohol: Y = heptan-2-one, $\ce{CH3CO(CH2)4CH3}$ , $\ce{C7H14O}$ .

Judging the options.

Y is a methyl ketone, so it gives a positive iodoform test; ketones do not reduce Tollens' reagent.
Y and X are both $\ce{C7H14O}$ with different functional groups (ketone versus ether + alkene): functional isomers. Y has no C=C, so it cannot be a geometrical isomer.

(C)

Topics:Oxidation and reduction Alkynes Structural isomerism

JEE Advanced 2014 · Paper 2, Chemistry question 32