The mechanism for the epimerization of cis -declone to trans -declone, and the chair conformational structure that show trans -declone is more stable than cis -declone are to be drawn. Concept introduction: Keto and enol forms of carbonyl compounds are called constitutional isomers. Interconvertible keto and enol forms are called tautomers. Racemization takes place at the α hydrogen in the presence of acids or bases. Cis – trans isomers are known as configurational isomers. If two groups are present on the same side of the carbon double bond, the isomer is a cis -isomer. If two groups are present on the opposite side of the carbon double bond, the isomer is a trans -isomer. Epimerization is the process by which less stable epimers are converted to more stable epimers.

Question

Want to see the full answer?

Answer 1

Question

Chapter 18, Problem 3PP

Interpretation Introduction

Interpretation:

The mechanism for the epimerization of cis-declone to trans-declone, and the chair conformational structure that show trans-declone is more stable than cis-declone are to be drawn.

Concept introduction:

Keto and enol forms of carbonyl compounds are called constitutional isomers. Interconvertible keto and enol forms are called tautomers.

Racemization takes place at the α hydrogen in the presence of acids or bases.

Cis–trans isomers are known as configurational isomers. If two groups are present on the same side of the carbon double bond, the isomer is a cis-isomer. If two groups are present on the opposite side of the carbon double bond, the isomer is a trans-isomer.

Epimerization is the process by which less stable epimers are converted to more stable epimers.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Students have asked these similar questions

In some nucleophilic substitutions under SN1 conditions, complete racemization does not occur and a small excess of one enantiomer is present. For example, treatment of optically pure 1-bromo-1-phenylpropane with water forms 1-phenyl-1-propanol. (a) Calculate how much of each enantiomer is present using the given optical rotation data. (b) Which product predominates-the product of inversion or the product of retention of configuration? (c) Suggest an explanation for this phenomenon. H Br он H20 1-bromo-1-phenylpropane 1-phenyl-1-propanol observed [a) = +5.0 optically pure S isomer, [a] =-48 %3D

In some nucleophilic substitutions under SN1 conditions, complete racemization does not occur and a small excess of one enantiomer is present. For example, treatment of optically pure 1-bromo-1-phenylpropane with water forms 1-phenylpropan-1-ol. (a) Calculate how much of each enantiomer is present using the given optical rotation data. (b) Which product predominates—the product of inversion or the product of retention of configuration? (c) Suggest an explanation for this phenomenon.

In some nucleophilic substitutions under SN1 conditions, complete racemization does not occur and a small excess of one enantiomer is present. For example, treatment of optically pure 1-bromo-1-phenylpropane with water forms 1- phenylpropan-1-ol. (a) Calculate how much of each enantiomer is present using the given optical rotation data. (b) Whichproduct predominates—the product of inversion or the product of retention of conguration? (c) Suggest an explanation for this phenomenon.