Question

Observe the following set of reactions
2-Methylpropene $\xrightarrow{\text{HBr}, (\text{CH}_3\text{COO})_2}$ Y
2-Methylpropene $\xrightarrow{\text{HBr}}$ X
Correct statement regarding X and Y is

• A. Both X and Y undergo nucleophilic substitution by $\text{S}_\text{N}1$ mechanism
• B. Both X and Y undergo nucleophilic substitution by $\text{S}_\text{N}2$ mechanism
• C. X undergoes nucleophilic substitution by $\text{S}_\text{N}1$ and Y by $\text{S}_\text{N}2$ mechanism
• D. X undergoes nucleophilic substitution by $\text{S}_\text{N}2$ and Y by $\text{S}_\text{N}1$ mechanism

Hint:

HBr addition without peroxide is Markovnikov ($\text{S}_\text{N}1$-like via carbocation). Peroxide triggers anti-Markovnikov ($\text{S}_\text{N}2$-like). Draw carbocation/radical intermediates to confirm.

Answer 1

The Correct Option is D

1. 2-Methylpropene ((CH$_3$)$_2$C=CH$_2$) reacts with HBr via electrophilic addition, not nucleophilic substitution, but the question likely refers to the mechanism of Br$^-$ addition.
2. Without peroxide: HBr adds via Markovnikov rule, forming (CH$_3$)$_3$CBr (X) via carbocation (tertiary, stable), so $\text{S}_\text{N}1$-like addition.
3. With peroxide ((CH$_3$COO)$_2$): Free radical mechanism gives anti-Markovnikov product (CH$_3$)$_2$CHCH$_2$Br (Y) via primary radical, $\text{S}_\text{N}2$-like addition.
4. Thus, X (Markovnikov) is $\text{S}_\text{N}1$, Y (anti-Markovnikov) is $\text{S}_\text{N}2$, so the answer is (4).