Molar Mass of Major Product in Multistep Reaction of Alkyne

To determine the molar mass of the major product $(Y)$, we analyze each step of the reaction sequence starting from compound $(X)$, which is styrene ($\text{C}_6\text{H}_5\text{-CH=CH}_2$):

1. Step (i): Addition of Bromine ($\text{Br}_2 / \text{CHCl}_3$) Styrene undergoes electrophilic addition of bromine across its alkene double bond to form 1,2-dibromo-1-phenylethane: $\text{C}_6\text{H}_5\text{-CH=CH}_2 + \text{Br}_2 \xrightarrow{\text{CHCl}_3} \text{C}_6\text{H}_5\text{-CH(Br)-CH}_2\text{Br}$

2. Step (ii): Double Dehydrohalogenation ($\text{NaNH}_2 \text{ excess}$) Treatment of the vicinal dibromide with excess sodium amide ($\text{NaNH}_2$) causes two consecutive $\text{E2}$ eliminations to yield phenylacetylene. Since excess $\text{NaNH}_2$ is present, the acidic hydrogen of the terminal alkyne is deprotonated to form sodium phenylacetylide: $\text{C}_6\text{H}_5\text{-CH(Br)-CH}_2\text{Br} \xrightarrow{\text{excess NaNH}_2} \text{C}_6\text{H}_5\text{-C}\equiv\text{C}^- \text{Na}^+$

3. Step (iii): Alkylation ($\text{CH}_3\text{I}$) The acetylide ion acts as a strong nucleophile and reacts with iodomethane via an $\text{S}_\text{N}2$ pathway to yield an internal alkyne, 1-phenylprop-1-yne: $\text{C}_6\text{H}_5\text{-C}\equiv\text{C}^- \text{Na}^+ + \text{CH}_3\text{I} \longrightarrow \text{C}_6\text{H}_5\text{-C}\equiv\text{C-CH}_3 + \text{NaI}$

4. Step (iv): Birch Reduction of Alkyne ($\text{Na/NH}_3 (l)$) Reduction of internal alkynes using sodium in liquid ammonia selectively converts the alkyne into a trans-alkene. Thus, 1-phenylprop-1-yne is reduced to (E)-1-phenylprop-1-ene: $\text{C}_6\text{H}_5\text{-C}\equiv\text{C-CH}_3 \xrightarrow{\text{Na/NH}_3(l)} \text{C}_6\text{H}_5\text{-CH=CH-CH}_3 \quad \text{Product }(Y)$

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Calculation of Molar Mass of Product $(Y)$:

• Molecular formula of $(Y)$: $\text{C}_9\text{H}_{10}$

  • Benzene ring: $\text{C}_6\text{H}_5$
  • Propenyl group: $-\text{CH}=\text{CH}-\text{CH}_3$

• Molar Mass: $\text{Molar Mass} = (9 \times 12) + (10 \times 1) = 108 + 10 = 118 \text{ g mol}^{-1}$

Therefore, the molar mass of the major product $(Y)$ is $118 \text{ g mol}^{-1}$, which corresponds to Option B.