Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 1: Problem 111

Statement 1 All olefinic compounds exhibit geometric isomerism. and Statement 2 Geometric isomerism in olefins arises due to restricted or hindered rotation between \(\mathrm{C}=\mathrm{C}\) bonds.

Short Answer

Expert verified
Answer: The restricted rotation around C=C bonds in olefinic compounds forces the different groups attached to the double bond to have fixed relative positions. This leads to the formation of geometric isomers, specifically E and Z isomers (or cis and trans isomers).

Step by step solution

01

Defining Olefinic Compounds

Olefinic compounds are hydrocarbons that contain at least one carbon-carbon double bond (\(\mathrm{C}=\mathrm{C}\)). These compounds are also known as alkenes.
02

Defining Geometric Isomerism

Geometric isomerism, also known as cis-trans or E-Z isomerism, is a type of stereoisomerism where isomers have the same molecular formula and sequence of bonded atoms (constitution), but differ in the three-dimensional orientations of their atoms in space.
03

Understanding Restricted Rotation

In single bonds (like \(\mathrm{C}-\mathrm{C}\)), rotation is relatively free and molecules can change their conformation without much energy. In contrast, double bonds (like \(\mathrm{C}=\mathrm{C}\)) have restricted rotation due to the presence of the second (pi) bond. This restricted rotation creates a barrier for the atoms to change their relative positions, which can lead to geometric isomers.
04

Relating Geometric Isomerism to Olefins

For olefinic compounds, geometric isomerism arises due to the restricted rotation around the \(\mathrm{C}=\mathrm{C}\) bond. Both statements are correct. The presence of a double bond forces different groups attached to the double bond to have fixed relative positions, which can lead to the formation of E and Z isomers (or cis and trans isomers). In summary, geometric isomerism in olefins arises due to the hindered rotation between \(\mathrm{C}=\mathrm{C}\) bonds.

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 most stable free radical among the following is (a) C=CC(C)C (b) c1ccccc1 (c) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}\) (d) \(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{2}-\mathrm{CH}_{2}\).

The correct IUPAC name of \(\mathrm{CH}_{3}-\mathrm{O}-\mathrm{CH}_{2}-\mathrm{O}-\mathrm{CH}_{2}-\mathrm{O}-\mathrm{CH}_{3}\) is (a) \(1,3,5\) -Trioxabutane (b) \(2,4,6\) -Trioxabutane (c) \(2,4,6\) -Trioxaheptane (d) Dimethoxydimethyl ether

The molecules which have dipole moment is/are (a) \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\) (b) 2, 2-dimethylpropane (c) Cis-3-hexene (d) p-chloronitrobenzene

Homolytic bond fission is observed in (a) Syn addition of bromine to ethylene (b) trans addition of bromine to butene (c) halogenation of alkanes (d) nitration of benzene

IUPAC name of the compound (a) Trans-3,4-dimethylhex-3-ene (b) Cis- 3,4 -dimethylhex-3-ene (c) Cis- 3,4 -dimethylhex-4-ene (d) Cis-2,3-diethylbut-2-ene
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Making Measurements

Read Explanation

Physical Chemistry

Read Explanation

Kinetics

Read Explanation

Chemical Reactions

Read Explanation

Chemical Analysis

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