According to Markovnikov's rule, the addition of HBr to propene ($\mathrm{CH_3-CH=CH_2}$) gives predominantly:
Apropane
B$1{,}2$-dibromopropane
C$2$-bromopropane
D$1$-bromopropane
Answer & Solution
Correct answer: C. $2$-bromopropane
Markovnikov's rule: the H of HX attaches to the carbon of the double bond bearing more hydrogens, leaving X on the more substituted carbon. Here Br goes to C2 → 2-bromopropane.
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