The cis-trans isomerism is a form of stereoisomerism also called geometrical isomerism, which deals with spatial arrangement of atoms. Concept introduction: There are two requirements for cis-trans isomerism: Rotation must be restricted in the molecule i.e. presence of = bond is necessary. There must be two non-identical groups on each doubly bonded carbon atom, the following statements hold true in cis-trans isomerism: Alkenes with a C=CH 2 unit do not exist as cis-trans isomers. Alkenes with a C=CR 2 unit, where the two R groups are the same, do not exist as cis-trans isomers. Alkenes of the type R–CH=CH–R can exist as cis and trans isomers; cis if the two R groups are on the same side of the carbon-to-carbon double bond, and trans if the two R groups are on opposite sides of the carbon-to-carbon double bond. Alkenes of the type R 1 –CH=CH–R 2 can exist as cis and trans isomers; cis if the two R groups are on the same side of the carbon-to-carbon double bond, and trans if the two R groups are on opposite sides of the carbon-to-carbon double bond.

Question

Want to see the full answer?

Answer 1

Question

Chapter 12.6, Problem 12.37QAP

Interpretation Introduction

Interpretation:

The cis-trans isomerism is a form of stereoisomerism also called geometrical isomerism, which deals with spatial arrangement of atoms.

Concept introduction:

There are two requirements for cis-trans isomerism:

Rotation must be restricted in the molecule i.e. presence of = bond is necessary.

There must be two non-identical groups on each doubly bonded carbon atom, the following statements hold true in cis-trans isomerism:

Alkenes with a C=CH₂ unit do not exist as cis-trans isomers.

Alkenes with a C=CR₂ unit, where the two R groups are the same, do not exist as cis-trans isomers.

Alkenes of the type R–CH=CH–R can exist as cis and trans isomers; cis if the two R groups are on the same side of the carbon-to-carbon double bond, and trans if the two R groups are on opposite sides of the carbon-to-carbon double bond.

Alkenes of the type R₁–CH=CH–R₂ can exist as cis and trans isomers; cis if the two R groups are on the same side of the carbon-to-carbon double bond, and trans if the two R groups are on opposite sides of the carbon-to-carbon double bond.

Interpretation Introduction

To determine:

Among of the following will show cis-trans isomerism-

a.

b.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Students have asked these similar questions

Hi!! Please provide a solution that is handwritten. Ensure all figures, reaction mechanisms (with arrows and lone pairs please!!), and structures are clearly drawn to illustrate the synthesis of the product as per the standards of a third year organic chemistry course. ****the solution must include all steps, mechanisms, and intermediate structures as required. Please hand-draw the mechanisms and structures to support your explanation. Don’t give me AI-generated diagrams or text-based explanations, no wordy explanations on how to draw the structures I need help with the exact mechanism hand drawn by you!!! I am reposting this—ensure all parts of the question are straightforward and clear or please let another expert handle it thanks!!

. (11pts total) Consider the arrows pointing at three different carbon-carbon bonds in the molecule depicted below. Bond B 2°C. +2°C. < cleavage Bond A • CH3 + 26. t cleavage 2°C• +3°C• Bond C Cleavage CH3 ZC '2°C. 26. E Strongest 3°C. 2C. Gund Largest BDE weakest bond In that molecule a. (2pts) Which bond between A-C is weakest? Which is strongest? Place answers in appropriate boxes. Weakest C bond Produces A Weakest Bond Most Strongest Bond Stable radical Strongest Gund produces least stable radicals b. (4pts) Consider the relative stability of all cleavage products that form when bonds A, B, AND C are homolytically cleaved/broken. Hint: cleavage products of bonds A, B, and C are all carbon radicals. i. Which ONE cleavage product is the most stable? A condensed or bond line representation is fine. 人 8°C. formed in bound C cleavage ii. Which ONE cleavage product is the least stable? A condensed or bond line representation is fine. methyl radical •CH3 formed in bund A Cleavage