Question

Characterize the following alkene as having the E or Z configuration.

Draw the product(s) of bromination of this compound, including all expected stereoisomers (if any).

Use wedge-and-dash bonds to designate the stereochemistry at any chirality centers, and make sure to draw an explicit hydrogen if a chirality center has one.

Answer 1

Concepts and reason

Alkenes involve in electrophilic addition relations. Where, alkene acts as nucleophile and electrophile attacks on, one of the carbon of a double bond. Draw the major organic product for the reaction between an alkene and bromine.

Fundamentals

If two high priority groups are on the same side of the double bond, then the compound is called as Z-alkene. On the other hand, if the two highest priority groups on the opposite side, then the compound is called as E-alkene.

Alkenes react with dibromine ${\rm{(B}}{{\rm{r}}_{\rm{2}}})$ to form 1,2-dibromo substituted compound through formation of a bromonium intermediate. Z-alkenes react with dibromine ${\rm{(B}}{{\rm{r}}_{\rm{2}}})$ gives enantiomers as products. E-alkene reacts with dibromine ${\rm{(B}}{{\rm{r}}_{\rm{2}}})$ gives meso compound.

The configuration at double bond is Z.

The reaction is,

Stereochemistry to bromonium intermediate can be applied is as follows:

In (Z)-3-methylpent-2-ene, the groups away from the observer are shown with hashed bond, and groups towards the observer are shown with dashed bond.

The formation of products is as follows:

Ans:

Thus, the configuration of the alkene is Z.

Thus, the major organic products are,

The products of the reaction are characterized as, (R, R), (S, S) and racemic.