Question 100

HBr reacts with $$CH_2 = CH - OCH_3$$ under anhydrous conditions at room temperature to give

Solution

Protonation (Electrophilic Attack):
The reaction proceeds via electrophilic addition across the carbon-carbon double bond.
The $$H^{+}$$ ion from $$HBr$$ adds to the terminal $$CH_{2}$$ carbon following Markovnikov's rule, producing a highly stable carbocation.
$$CH_{2}=CH-OCH_{3}+H^{+}$$ $$\longrightarrow CH_{3}-\overset{+}{C}H-OCH_{3}$$

Resonance Stabilization:
The resulting intermediate carbocation is exceptionally stable because the adjacent oxygen atom can donate its lone pair of electrons through resonance $$(+M$$ effect):
$$CH_{3}-\overset{+}{C}H-{O}CH_{3}$$ $$\longleftrightarrow CH_{3}-CH=\overset{+}{O}CH_{3}$$

Nucleophilic Attack:
The bromide ion $$(Br^{-}$$) then attacks the resonance-stabilized carbocation center to form the final addition product:
$$CH_{3}-\overset{+}{C}H-OCH_{3}+Br^{-}\longrightarrow \mathbf{CH}_{\mathbf{3}}\mathbf{-CH(Br)-OCH}_{\mathbf{3}}$$

Hence, Option D is correct.