JEE Main 2026 · 4 Apr · Shift 2 · Chemistry Q64

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ChemistrySingle correct+4 / -1Question 64 of 75

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)

Styrene (X) → (i) Br₂/CHCl₃; (ii) NaNH₂ excess; (iii) CH₃I; (iv) H₂, Na/NH₃(l) → major product (Y).

118

160

125

Answer and worked solution

Official answer: B

Sequence.

$\ce{Br2}$ gives $\ce{PhCHBrCH2Br}$ .
Excess $\ce{NaNH2}$ gives the phenylacetylide.
$\ce{CH3I}$ gives $\ce{PhC#CCH3}$ .
Na/NH₃ reduces it to trans-1-phenylpropene, $\ce{C9H10}$ .

Molar mass $108+10=118$ g mol⁻¹.

(B)

Topics:Alkynes Alkenes

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JEE Main 2026 · 4 Apr · Shift 2, Chemistry question 64