Question

treatment of 2 - methylpropene with which of the following reaction conditions results in an anti - markovnikov addition product?
bh₃ - thf, followed by alkaline h₂o₂
dilute h₂so₄ and heat
aqueous hg(oac)₂, followed by alkaline nabh₄
dry gaseous hbr with peroxides present

Explanation:

Step1: Analyze \(BH_3\)-THF, followed by alkaline \(H_2O_2\)

Hydroboration - oxidation of alkenes (e.g., 2 - methylpropene) follows anti - Markovnikov rule. In hydroboration (\(BH_3\)-THF step), the boron atom adds to the less - substituted carbon of the double bond. Then, in the oxidation step (alkaline \(H_2O_2\)), the boron is replaced by a hydroxyl group, resulting in an anti - Markovnikov alcohol product.

Step2: Analyze dilute \(H_2SO_4\) and heat

Acid - catalyzed hydration (dilute \(H_2SO_4\) and heat) of alkenes follows Markovnikov rule. The carbocation intermediate formed during the reaction leads to the addition of the hydroxyl group to the more - substituted carbon.

Step3: Analyze aqueous \(Hg(OAc)_2\), followed by alkaline \(NaBH_4\)

Oxymercuration - demercuration (aqueous \(Hg(OAc)_2\), followed by alkaline \(NaBH_4\)) of alkenes follows Markovnikov rule. The mercury - containing intermediate and the reaction mechanism favor the addition of the hydroxyl group to the more - substituted carbon.

Step4: Analyze dry gaseous \(HBr\) with peroxides present

Addition of \(HBr\) to alkenes in the presence of peroxides follows anti - Markovnikov rule. A radical mechanism is involved. The bromine radical adds to the less - substituted carbon of the double bond first, leading to an anti - Markovnikov alkyl bromide product.

Answer:

A. \(BH_3\)-THF, followed by alkaline \(H_2O_2\); D. dry gaseous \(HBr\) with peroxides present