JEE Main 2026 — Apr 4, Shift 2 — Chemistry solution

JEE Main2026Apr 4, Shift 2Chemistry

Compound (X) is subjected to the sequence of reactions as shown above. Molar mass of the major product (Y) formed is ______ g mol $^{- 1}$ .

(Given molar mass in g mol $^{- 1}$ C: 12, H: 1, O: 16)

A
90
B
118
C
160
D
125

Solution

Trace the transformations step-by-step.

(i) Br $_{2}$ /CHCl $_{3}$ : anti-addition across the vinyl C=C $\to$ Ph–CHBr–CH $_{2}$ Br (vicinal dibromide).

(ii) NaNH $_{2}$ (excess): double dehydrohalogenation (two successive E2 eliminations) gives the terminal alkyne Ph–C $\equiv$ CH, which the excess base deprotonates to the acetylide Ph–C $\equiv$ C $^{-}$ .

(iii) CH $_{3}$ I: the acetylide attacks methyl iodide (S $_{N}$ 2) $\to$ Ph–C $\equiv$ C–CH $_{3}$ (1-phenylpropyne).

(iv) Na / NH $_{3}$ (l): dissolving-metal reduction selectively gives the trans-alkene $\to$ (E)-Ph–CH=CH–CH $_{3}$ , molecular formula C $_{9}$ H $_{10}$ .

Molar mass of C $_{9}$ H $_{10}$ $= 9 \times 12 + 10 \times 1 = 108 + 10 = 118$ g/mol.

← Back to full paper

Practice more Chemistry

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →