Answer to Question #190634 in Organic Chemistry for Kaygee

Question #190634

What stereochemisyry would you expect for the alkene obtained by E2 elimination of(1R,2R-1,2-dibromo--1,2-diphenylethane?Draw the Newman structure.

2.Draw the resonance structures for the products of all nitrogen in a compound below and determine which will protonate first(6marks)

H2N-C=NH-NH-CH3


3.How do I identify the most acidic proton in molecules using electronic factors to explain their stabilization conjugated bases?


More especially in molecules having a Carbon bonded to a carbon and 2 Oxygens??

4.From the reaction below,only one product will form.Propose a mechanism and use your knowledge of intermediate stability to predict the structure of the expected product.

Reaction:

Cyclohexene-CH=CH-CH3 reacting in the presence of HBr/H2O2


1
Expert's answer
2021-05-08T23:36:13-0400

1.

E2

 stereochemistry

In E2

 eliminations, the β-hydrogen must be antiperiplanar to the leaving group.

www.chem.ucalgary.ca

1,2-Dibromo-1,2-diphenylethane

The structure of 1,2-dibromo-1,2-diphenylethane is

images-a.chemnet.com

E2 elimination of meso-1,2-dibromo-1,2-diphenylethane

The Newman projection of the meso isomer is

Meso

We know that the structure is meso because the order of groups H → Ph → Br

 is the same direction (clockwise) on each carbon atom. There must be an internal mirror plane.

In the transition state, the Ph

 on the back carbon is gauche to Ph

 and Br

 on the front carbon.

This gives a large activation energy and make for a slow reaction.

The product will have the two phenyl groups on the same side, giving an E

-configuration.

meso


(from chemwiki.ucdavis.edu)

E2 elimination of (±)-1,2-dibromo-1,2-diphenylethane

The Newman projection for one of the enantiomers is

Racemic

We know that the structure is chiral because the order of groups H → Br → Ph

 is clockwise on one carbon atom and counterclockwise on the other. There can be no internal mirror plane.

In the transition state, the Br

 on the back carbon is gauche to Ph

 and Br

 on the front carbon.

The activation energy will be less, so this will be a faster reaction.

The product will have the two phenyl groups on opposite sides, giving a Z

-configuration.



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