Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 2: Problem 83

The major product formed in the reaction of \(\mathrm{HBr}\) with \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CH}-\mathrm{CH}=\mathrm{CH}_{2}\) is (a) (b) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CH}-\mathrm{CH}_{2}-\mathrm{CH}_{2} \mathrm{Br}\) (c) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CH}-\mathrm{CHBr}-\mathrm{CH}_{3}\) (d)

Short Answer

Expert verified
a) (CH3)2CH-CHBr-CH2Br b) (CH3)2CBr-CH=CH2 c) (CH3)2CH-CHBr-CH3 d) (CH3)2C(Br)2-CH=CH2 Answer: (c) (CH3)2CH-CHBr-CH3

Step by step solution

01

Identify the type of reaction

In this case, the reaction is taking place between \(\mathrm{HBr}\) (a hydrohalic acid) and an alkene (\(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CH}-\mathrm{CH}=\mathrm{CH}_{2}\)). This kind of reaction is an electrophilic addition reaction.
02

Determine the site of attack of electrophilic addition

According to Markovnikov's rule, when there is an addition of a hydrogen halide (in this case, \(\mathrm{HBr}\)) to an asymmetric alkene, the hydrogen will add to the carbon with the lesser number of hydrogen atoms, so as to create a carbocation with the least steric hindrance. Comparing the two doubly bonded carbons, the first one has only one hydrogen and the second one has two hydrogen atoms. So, the first carbon will bond with the hydrogen, and the bromine will bond with the second carbon.
03

Write the product of the reaction

The product of this reaction, following Markovnikov's rule and considering that the hydrogen will bond with the first carbon and the bromine with the second carbon, is \(\left(\mathrm{CH}_{3}\right)_{2}\mathrm{CH}-\mathrm{CHBr}-\mathrm{CH}_{3}\).
04

Match the reaction product with the given options

Comparing the product obtained in step 3 with the given options, we observe that option (c) is the same as our product. Therefore, the major product formed in the reaction of \(\mathrm{HBr}\) with \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CH}-\mathrm{CH}=\mathrm{CH}_{2}\) is \(\left(\mathrm{CH}_{3}\right)_{2}\mathrm{CH}-\mathrm{CHBr}-\mathrm{CH}_{3}\). The correct answer is (c).

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

Choose the correct statement with regard to 3 -methyl but- 1 -ene (a) A tertiary alcohol is formed by hydration in the presence of an acid. (b) a primary alcohol is formed by oxymercuration reaction. (c) An optically active secondary alcohol is formed by hydroboration- oxidation. (d) The alkene is less reactive than ethene towards hydration under acid catalyzed conditions.

The IUPAC name of the compound is (a) 6-bromo-2-chlorocyclohexene (b) 3-bromo-1-chlorocyclohexene (c) 1 -bromo-3-chlorocyclohexene (d) 2-bromo-6-chlorocyclohexene

This section contains multiple choice questions. Each question has 4 choices (a), (b), (c) and (d), out of which ONLY ONE is correct. A compound with the molecular formula, \(\mathrm{C}_{4} \mathrm{H}_{6}\), does not exist in cis or trans form. It does not produce a precipitate with ammoniacal \(\mathrm{AgNO}_{3}\) solution. The compound as such is optically inactive but a monohalogen derivative can be optically active. The compound is (a) \(\mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_{3}\) (b) \(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{C}=\mathrm{CH}_{2}\) (c) \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{C} \equiv \mathrm{C}-\mathrm{H}\) (d)

Reaction of trans-2-phenyl-1-bromocyclohexane with hot alcoholic \(\mathrm{NaOH}\) gives (a) 3 -phenylcyclohexene (b) 1 -phenylcyclohexene (c) 2 -bromophenyl cyclohexane (d) 2 -phenyl cyclohexanol

\(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{3} \stackrel{\mathrm{Br}_{2} / \mathrm{hv}}{\longrightarrow}(\mathrm{A}) \stackrel{\mathrm{Mg} / \mathrm{Ether}}{\longrightarrow}(\mathrm{B}) \stackrel{\mathrm{D}_{2} \mathrm{O}}{\longrightarrow}(\mathrm{C})\) The major product \(C\) formed is (a) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{D}\) (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{3}\) (c) \(\mathrm{CH}_{3} \mathrm{CHD} \mathrm{CH}_{3}\) (d)
See all solutions

Recommended explanations on Chemistry Textbooks

Chemistry Branches

Read Explanation

Nuclear Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Kinetics

Read Explanation

Chemical Analysis

Read Explanation

Organic Chemistry

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