Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 4: Problem 186

Preparation and reactions of alcohols Alcohols can by prepared from alkyl halides, epoxides, alkenes, etc. When treated with acid, \(\mathrm{C}-\mathrm{OH}\) bond is cleaved heterolytically to form a carbocation which could then undergo substitution, elimination or rearrangement depending on the conditions and stability of the carbocation. When 1,2 -dimethyl cyclohexene is treated with cold dilute alkaline \(\mathrm{KMnO}_{4}\), the product obtained is

Short Answer

Expert verified
Based on the given reaction, 1,2-dimethyl cyclohexene reacts with cold dilute alkaline KMnO4. This results in the oxidative cleavage of the alkene (double bond) via a dihydroxylation process. The final product formed is 1,2-dimethylcyclohexane-1,2-diol.

Step by step solution

01

1. Identify reactants and reaction conditions

In this reaction, we have 1,2-dimethyl cyclohexene as the reactant and cold dilute alkaline \(\mathrm{KMnO}_{4}\) as the reagent.
02

2. Find the mechanism of reaction

Cold dilute alkaline \(\mathrm{KMnO}_{4}\) is a strong oxidizing agent that is commonly used for the oxidative cleavage of alkenes (double bonds). In this case, the oxidizing agent reacts with the alkene and breaks the double bond. This process is called "dihydroxylation." The double bond will be cleaved to form two new \(\mathrm{C}-\mathrm{OH}\) bonds on the adjacent carbons where the double bond was initially present.
03

3. Determine the product

In the reaction of 1,2-dimethyl cyclohexene with cold dilute alkaline \(\mathrm{KMnO}_{4}\), dihydroxylation of the double bond occurs. This results in the formation of two new \(\mathrm{C}-\mathrm{OH}\) bonds where the double bond initially was, along with the retention of the two methyl groups present in the parent compound. The final product formed is 1,2-dimethylcyclohexane-1,2-diol.

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.

Start your free trial

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

The correct nucleophilicity order for \(\mathrm{S}_{\mathrm{N}} 2\) reaction in protic solvents is (a) \(\mathrm{CH}_{3} \mathrm{~S}^{-}>\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{~S}^{-}>\mathrm{I}^{-}>\mathrm{CN}^{-}>\mathrm{CH}_{3} \mathrm{O}^{-}>\mathrm{OH}^{-}\) (b) \(\mathrm{CH}_{3} \mathrm{~S}^{-}>\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{~S}^{-}>\mathrm{CN}^{-}>\mathrm{I}^{-}>\mathrm{CH}_{3} \mathrm{O}^{-}>\mathrm{OH}^{-}\) (c) \(\mathrm{CN}^{-}>\mathrm{CH}_{3} \mathrm{~S}^{-}>\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{~S}^{-}>\mathrm{OH}^{-}>\mathrm{CH}_{3} \mathrm{O}^{-}>\mathrm{I}^{-}\) (d) \(\mathrm{OH}^{-}>\mathrm{CH}_{3} \mathrm{O}^{-}>\mathrm{CN}^{-}>\mathrm{CH}_{3} \mathrm{~S}^{-}>\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{~S}^{-}>\mathrm{I}^{-}\)

\(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHBr} \stackrel{\mathrm{Mg} / \mathrm{ether}}{\longrightarrow} \mathrm{A} \stackrel{\mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{CO}-\mathrm{CH}_{3}}{\longrightarrow} \mathrm{B}\) The product \(B\) in the above reaction is (a) 2 -phenyl- 3 -methyl butan- 2 -ol (b) 2 -phenylpentan- 2 -ol (c) 3 -phenyl butan-2-ol (d) propan-2-ol and ethyl benzene

Ethene \(\frac{\mathrm{O}_{2}}{\mathrm{Ag} / \Delta} \mathrm{A}\) dil HCl/\Delta \(\mathrm{B}\). The product \(\mathrm{B}\) in the above reaction is (a) Ethanol (b) Ethane- 1,2 - diol (c) Ethanal (d) epoxyethane

Which of the following is the correct order of dipole moment of methyl halides? (a) \(\mathrm{CH}_{3} \mathrm{~F}>\mathrm{CH}_{3} \mathrm{Cl}>\mathrm{CH}_{3} \mathrm{Br}>\mathrm{CH}_{3} \mathrm{I}\) (b) \(\mathrm{CH}_{3} \mathrm{Cl}>\mathrm{CH}_{3} \mathrm{~F}>\mathrm{CH}_{3} \mathrm{Br}>\mathrm{CH}_{3} \mathrm{I}\) (c) \(\mathrm{CH}_{3} \mathrm{I}>\mathrm{CH}_{3} \mathrm{Br}>\mathrm{CH}_{3} \mathrm{~F}>\mathrm{CH}_{3} \mathrm{I}\) (d) \(\mathrm{CH}_{3} \mathrm{I}>\mathrm{CH}_{3} \mathrm{Br}>\mathrm{CH}_{3} \mathrm{Cl}>\mathrm{CH}_{3} \mathrm{~F}\)

The haloalkane, 2 -iodo-2-methylpentane, can be prepared by treating an alkene (A) with hydroiodic acid. (A) can be (a) 2 -methylpent-2-ene (b) 2 -methylpent-1-ene (c) 2 -methyl-but-2-ene (d) both (a) and (b)
See all solutions

Recommended explanations on Chemistry Textbooks

The Earths Atmosphere

Read Explanation

Chemical Analysis

Read Explanation

Inorganic Chemistry

Read Explanation

Kinetics

Read Explanation

Nuclear Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation
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