(i) $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{MgBr}$ $\xrightarrow[\text { (ii) } \mathrm{H}_{3} \mathrm{O}^{+}]{\text {(1.0 equivalent), dry Ether }} \underset{\text { Major Product }}{\mathrm{X}}$ The structure of $X$ is:

JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Haloalkanes and Haloarenes

(i)

C_{6} H_{5} MgBr

(1.0 equivalent), dry Ether (ii) H_{3} O^{+} Major Product X

The structure of

X

is:

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

Answer: (B)

Solution

A single equivalent of phenylmagnesium bromide adds across the carbon-nitrogen triple bond of the nitrile to give a magnesium ketimine salt, and the acidic work-up hydrolyses that imine to a carbonyl group, so the cyano group becomes a benzoyl group. A Grignard reagent does not cleave an ether, so the methoxy-bearing side chain on the neighbouring ring position comes through untouched, and with only one equivalent there is nothing left to attack the ketone further. The product is phenyl 2-(1-methoxyethyl)phenyl ketone.

JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Single correct option

Practise more: all JEE Main PYQs on Haloalkanes and Haloarenes →

JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Haloalkanes and Haloarenes

Solution

More JEE Main questions on Haloalkanes and Haloarenes