The conversion can be successfully carried out by (a) \(\mathrm{H}^{+} / \mathrm{H}_{2} \mathrm{O}\) (b) Hydroboration-oxidation (c) HBr addition in the presence of peroxides followed by hydrolysis with aqueous KOH (d) Oxymercuration-reduction

Short Answer

Expert verified
a) H+ / H2O b) Hydroboration-oxidation c) HBr addition + Hydrolysis d) Oxymercuration-reduction Answer: All the given reactions (a), (b), (c), and (d) can successfully convert an alkene into an alcohol.

Step by step solution

01

Reaction (a): \(\mathrm{H}^{+} / \mathrm{H}_{2} \mathrm{O}\)

This reaction involves the addition of \(\mathrm{H}^{+}\) and \(\mathrm{H}_{2} \mathrm{O}\) to an alkene, following Markovnikov's rule. The alkene's double bond will break, allowing the hydrogen and a hydroxyl group (from the water molecule) to attach to the carbons of the alkene. This process results in the formation of an alcohol.
02

Reaction (b): Hydroboration-oxidation

Hydroboration-oxidation involves two main steps: 1) the addition of borane (BH3) to an alkene, and 2) the oxidation of the resulting organoborane with hydrogen peroxide (H2O2) and hydroxide ions (OH-). This reaction converts alkenes into alcohols through an anti-Markovnikov addition of H and OH groups.
03

Reaction (c): HBr addition + Hydrolysis

This reaction involves the addition of HBr to an alkene in the presence of peroxides (anti-Markovnikov). Peroxides cause a radical reaction to occur, resulting in the formation of a bromoalkane. Subsequent hydrolysis with aqueous KOH will replace the bromine atom with a hydroxyl group, resulting in the formation of an alcohol.
04

Reaction (d): Oxymercuration-reduction

Oxymercuration-reduction involves the addition of mercuric acetate (Hg(OAc)2) to an alkene in the presence of water, followed by a reduction with sodium borohydride (NaBH4). This process occurs via a Markovnikov addition and results in the formation of an alcohol, without the risk for rearrangement as in the acid-catalyzed reaction. In summary, all of the given reactions (a), (b), (c), and (d) can successfully convert an alkene into an alcohol, with either Markovnikov or anti-Markovnikov addition rules, depending on the reaction conditions.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

\(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CuLi}\) is known as (a) Corey-House reagent (b) Gilman reagent (c) Grignard reagent (d) Ullmann reagent

The reagent that does not react with ethyne is (a) Ammoniacal \(\mathrm{AgNO}_{3}\) (b) \(\mathrm{NaOH}\) (c) \(\mathrm{HCl}\) (d) Na

An alkene on reductive as well as oxidative ozonolysis gave only one product. The alkene is (a) propene (b) 2,3-dimethyl-2-butene (c) pent-2-ene (d) 2 -methyl propene

An alkene on mild oxidation with dil. aqueous \(\mathrm{KMnO}_{4}\) gave a product which on further treatment with \(\mathrm{HIO}_{4}\) gave hexane- 2,5 -dione. The alkene is (a) \(\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{3}\) (b) \(\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}=\mathrm{CH}_{2}\) (c) Cyclohexane (d) 1, 2 dimethyl cyclobutene

A hydrocarbon \(\mathrm{C}_{5} \mathrm{H}_{12}\) (A) on treatment with chlorine undergoes monochlorination by the replacement of the only \(1^{\circ}\) hydrogen to give(B) \(\mathrm{C}_{5} \mathrm{H}_{11} \mathrm{Cl}\). When (B) is treated with Na metal in ether, the product obtained could be (a) \(\quad\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CH}-\mathrm{CH}\left(\mathrm{CH}_{3}\right)_{2}\) (b) \(\mathrm{CH}_{3}-\mathrm{C}\left(\mathrm{CH}_{3}\right)_{2}-\mathrm{C}\left(\mathrm{CH}_{3}\right)_{2}-\mathrm{CH}_{3}\) (c) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{C}\left(\mathrm{CH}_{3}\right)_{3}\) (d) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{CH}_{2} \mathrm{Cl}\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free