Organic Chemistry
Organic Chemistry
12th Edition
ISBN: 9781118875766
Author: T. W. Graham Solomons, Craig B. Fryhle, Scott A. Snyder
Publisher: WILEY
Question
Book Icon
Chapter FRP, Problem 15P
Interpretation Introduction

Interpretation:

The statement, “The products formed by the hydrolysis of enantiomers of 3-chloro-1-butene are optically inactive”, is to be explained.

Concept Introduction:

Enantiomers are stereoisomers of a compound which are non-superimposable mirror images of each other. Enantiomers are optically active.

Electrophiles are electron deficient species which has positive or partially positive charge. Lewis acids are electrophiles which accept electron pair.

Nucleophiles are electron rich species which has negative or partially negative charge. Lewis bases are nucleophiles which donate electron pair.

Free radical is an atom, molecule or ion that has unpaired electrons which makes it highly chemically reactive.

Substitution reaction: A reaction in which one of the hydrogen atoms of a hydrocarbon or a functional group is substituted by any other functional group is called substitution reaction.

Elimination reaction: A reaction in which two substituent groups are detached and a double bond is formed is called elimination reaction.

Addition reaction: It is the reaction in which unsaturated bonds are converted to saturated molecules by the addition of molecules.

Carbocation is a molecule having a carbon atom bearing three bonds and a positive formal charge.

Carbocation are generally unstable because they do not have eight electrons to satisfy the octet rule.

The order of stability of carbocation is such that the tertiary carbocation is the most stable whereas the primary carbocation is the least stable, and secondary carbocation lies between primary and tertiary carbocations.

If primary carbocation is obtained in product, it rearranges itself to secondary or tertiary carbocation to form more stable product.

If secondary carbocation is obtained in product, it rearranges itself to tertiary carbocation to form more stable product.

The stability of carbocation:

3ocarbocation>2ocarbocation>1ocarbocation>methylcarbocation

Blurred answer
Students have asked these similar questions
a: Is the product chiral? _____ b: Assuming that the reaction takes place with equal likelihood from both Re and Si faces of the carbonyl group, is the product optically active? ____
Does bromochloromethane contain a plane of symmetry? is it a chiral?
Using what you know about the conformational energetics of substituted cyclohexanes, predict which of the two decalin isomers is more stable. Explain your reasoning.

Chapter FRP Solutions

Organic Chemistry

Knowledge Booster
Background pattern image
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Organic Chemistry
Chemistry
ISBN:9781305080485
Author:John E. McMurry
Publisher:Cengage Learning
Text book image
Macroscale and Microscale Organic Experiments
Chemistry
ISBN:9781305577190
Author:Kenneth L. Williamson, Katherine M. Masters
Publisher:Brooks Cole
Text book image
Organic Chemistry
Chemistry
ISBN:9781305580350
Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. Foote
Publisher:Cengage Learning