Question

7 multiple answer 2 points which of the following reagents can be used to carry out an anti - markovnikov addition of hbr to an alkene? (select all that apply) hcl in the presence of peroxides hbr in the presence of peroxides (roor) hbr in the absence of peroxides bh₃ followed by h₂o₂, naoh previous

Explanation:

Step1: Anti - Markovnikov addition of HBr

The peroxide effect (Kharasch effect) is specific for HBr. For HCl, the reaction does not follow the anti - Markovnikov pathway in the presence of peroxides because the H - Cl bond is too strong, and the free - radical addition is not favored. So, option A (HCl in the presence of peroxides) is incorrect.

Step2: HBr addition conditions

In the absence of peroxides, HBr adds to alkenes following the Markovnikov rule (proton adds to the carbon with more hydrogens). So, option C (HBr in the absence of peroxides) is incorrect.

Step3: HBr with peroxides

When HBr reacts with an alkene in the presence of peroxides (ROOR), the addition is anti - Markovnikov. This is because the peroxide initiates a free - radical mechanism. The bromine radical adds to the less substituted carbon of the alkene double bond first.

Step4: Hydroboration - oxidation

The hydroboration - oxidation reaction (using \(BH_3\) followed by \(H_2O_2/NaOH\)) also results in an anti - Markovnikov addition of what is essentially “\(H - OH\)” (the net result after oxidation) to the alkene. In the hydroboration step, the boron (which is electron - deficient) adds to the less substituted carbon of the alkene double bond, and after oxidation (using \(H_2O_2/NaOH\)), the boron is replaced by a hydroxyl group in an anti - Markovnikov fashion. So, option D is correct as it can be considered in the context of overall anti - Markovnikov - like addition (in terms of regioselectivity).

Answer:

B. HBr in the presence of peroxides (ROOR)
D. BH₃ followed by H₂O₂, NaOH