Styrene undergoes the following sequence of reactions Molar mass of product (P) is:
Styrene undergoes the following sequence of reactions Molar mass of product (P) is:
- 104
- 118
- 132
- 146
The Correct Option is B
Solution and Explanation
Concept: This reaction sequence involves: • Addition of \(Br_2\) across the double bond • Double dehydrohalogenation forming an alkyne • Alkylation using \(CH_3I\) • Partial reduction of alkyne using \(Na/NH_3\)
Step 1: Addition of bromine Styrene reacts with \(Br_2/CCl_4\) giving vicinal dibromide. \[ Ph{-}CH=CH_2 \xrightarrow{Br_2} Ph{-}CHBr{-}CH_2Br \]
Step 2: Elimination with \(NaNH_2\) Excess \(NaNH_2\) removes two molecules of HBr producing a terminal alkyne. \[ Ph{-}CHBr{-}CH_2Br \xrightarrow{NaNH_2} Ph{-}C\equiv CH \]
Step 3: Alkylation Terminal alkyne reacts with \(CH_3I\) via \(S_N2\) reaction. \[ Ph{-}C\equiv CH \xrightarrow{CH_3I} Ph{-}C\equiv C{-}CH_3 \]
Step 4: Reduction Dissolving metal reduction \(Na/liq.NH_3\) converts alkyne to trans-alkene. \[ Ph{-}C\equiv C{-}CH_3 \xrightarrow{Na/NH_3} Ph{-}CH=CH{-}CH_3 \] Thus final product: \[ P = C_{10}H_{10} \]
Step 5: Calculate molar mass \[ M = (10 \times 12) + (10 \times 1) \] \[ M = 120 + 10 = 130 \] But considering the final structure obtained after reduction corresponds to \(C_9H_{10}\): \[ M = (9 \times 12) + (10 \times 1) \] \[ M = 108 + 10 = 118 \] \[ \boxed{\text{Molar mass of } P = 118} \]
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