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Question of 90

Q.CH3-CH=CH2 + HBr -> A, A is

(a) 2-Bromopropane
(b) 1-Bromopropane
(c) 3-Bromopropane
(d) Propane
Jharkhand JacJAC Intermediate Board (1st Year) 2025MCQ· 1mImportance★★★★★
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Markovnikov's rule: when HX adds across an unsymmetrical alkene, H goes to the carbon that already has MORE hydrogens, and X (here Br) goes to the more substituted carbon — this gives 2-bromopropane, not 1-bromopropane.

Reaction: CH3-CH=CH2 + HBr -> A

The double bond is between C2 and C3 of propene. This is an UNSYMMETRICAL alkene, since C2 (=CH-) has 1 H and 1 alkyl group (CH3) attached, while C3 (=CH2) has 2 H atoms attached — the two double-bond carbons are chemically different.

Markovnikov's rule (the classic 'rich get richer' rule for electrophilic addition to alkenes) states that in the addition of HX to an unsymmetrical alkene, the hydrogen atom attaches to the carbon of the double bond that already bears the greater number of hydrogen atoms, while the halogen attaches to the carbon with fewer hydrogens (the more substituted carbon).

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