See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Concept: When allyl alcohol (H2C=CH-CH2-OH) reacts with excess HBr, two successive reactions occur. Step 1: The hydroxyl group of the allylic alcohol reacts with HBr. The -OH is protonated and leaves as water, generating an allylic carbocation (H2C=CH-CH2+ <-> +CH2-CH=CH2). Bromide attacks to give allyl bromide: H2C=CH-CH2-Br. This is the product with 1 equivalent of HBr. Step 2: Since HBr is in excess, the allyl bromide (H2C=CH-CH2-Br) now undergoes electrophilic addition of a second HBr across the double bond. H+ adds to the terminal carbon (CH2=) following Markovnikov's rule, placing the positive charge on the more substituted internal carbon: CH3-CH(+)-CH2-Br. Bromide then attacks this carbocation to give: CH3-CH(Br)-CH2-Br, which is 1,2-dibromopropane. Why other options fail: - Option (b) H2C=CH-CH2-Br: This is only the intermediate product after reaction with 1 equivalent of HBr; with excess HBr, the double bond also reacts. - Option (c) CH3-CH(Br)-CH2-OH: This would require HBr to add across the double bond without reacting with the -OH, which is not favored; moreover, excess HBr would convert -OH to -Br as well. - Option (d) CH3-CH(OH)-CH2-OH: This is a diol and there is no mechanism to produce it from HBr addition. The major product with excess HBr is CH3-CH(Br)-CH2-Br (1,2-dibromopropane). Therefore, the correct answer is A.