Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 2: Problem 7

What are the products formed when buta-1, 3 -diene is treated with HBr in \(1: 1\) molar proportion? Explain the formation of the products.

Short Answer

Expert verified
Answer: The major product is 3-bromo-1-butene.

Step by step solution

01

Identify the reaction mechanism

Buta-1,3-diene is a conjugated diene, which means the double bonds are separated by a single bond. Conjugated dienes commonly undergo electrophilic addition reactions due to the presence of multiple pi bonds. In this case, the electrophile is HBr, and the reaction is an electrophilic addition to the diene.
02

Consider regioselectivity to predict the possible products

Conjugated dienes exhibit regioselectivity when reacting with electrophiles, which means the electrophile can add to different carbons of the diene, yielding different products. In this case, there are two possible products: 3-bromo-1-butene and 1-bromo-2-butene.
03

Evaluate the stability of the carbocations formed during the reaction

The stability of carbocations is crucial in determining the major product in the electrophilic addition of HBr to buta-1,3-diene. If the carbocation is more stable, it is more likely to form from the initial attack of the hydrogen to the pi bond. For 3-bromo-1-butene, the intermediate carbocation would be on the second carbon, which is an allylic carbocation. Allylic carbocations are resonance stabilized and thus more stable than non-allylic carbocations. For 1-bromo-2-butene, the intermediate carbocation would be on the third carbon, which is not an allylic carbocation, and less stable than the carbocation formed in the other pathway.
04

Identify the major product

Based on the regioselectivity and the stability of the intermediate carbocations, the major product of the reaction between buta-1,3-diene and HBr in a 1:1 molar proportion is 3-bromo-1-butene, which forms through a more stable allylic carbocation.

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 total number of chlorinated products given by ethane is (a) 6 (b) 7 (c) 8 (d) 9

Question contains Statement-1 and Statement-2 and has the following choices (a), (b), (c) and (d), out of which ONLY ONE is correct. (a) Statement- 1 is True, Statement- 2 is True; Statement- 2 is a correct explanation for Statement-1 (b) Statement- 1 is True, Statement- 2 is True; Statement- 2 is NOT a correct explanation for Statement-1 (c) Statement- 1 is True, Statement- 2 is False (d) Statement-1 is False, Statement- 2 is True Statement 1 Alkynes are more reactive than alkenes in electrophilic addition reactions. and Statement 2 Electrophilic addition to both alkenes and alkynes involve a three membered cyclic intermediate and the bromonium ion type intermediate is less stable for an alkyne than for an alkene.

Alkynes are less reactive than alkenes towards electrophilic reagents. But the alkyne chemistry is useful for organic synthesis, in the form of hydration of the triple bond, formation of metalacetylides, selective reduction of alkynes and few other related reactions.

\(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_{3}\) product The product formed is (a) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{CO} \mathrm{CH}_{3}\) (b) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO} \mathrm{CH}_{2} \mathrm{CH}_{3}\) (c) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO} \mathrm{CO} \mathrm{CH}_{3}\) (d)

Directions: This section contains 1 paragraph. Based upon the paragraph, 3 multiple choice questions have to be answered. Each question has 4 choices (a), (b), (c) and (d), out of which ONLY ONE is correct. Passage I Alkenes undergo a variety of reactions like hydrogenation, halogenation, oxidation, epoxidation and so on. The double bond of the alkenes acts as electron donor. An unsaturated hydrocarbon gave, on treatment with alkaline \(\mathrm{KMnO}_{4}\) under hot conditions, one mole of adipic acid. The unsaturated hydrocarbon could be (a) \(\mathrm{H}_{3} \mathrm{C}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{3}\) (b) \(\mathrm{H}_{3} \mathrm{C}-\mathrm{CH}_{2}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_{2}-\mathrm{CH}_{3}\)
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Reactions

Read Explanation

Kinetics

Read Explanation

Physical Chemistry

Read Explanation

Making Measurements

Read Explanation

Chemistry Branches

Read Explanation

The Earths Atmosphere

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