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. Jasperse: If there are two enantiomers, you can draw just one, but you\'ll need to remember that the other forms as well to answer the stereochemistry question.

Answer 1

The product of bromination is shown on the image below

Answer 2

Answer:

The given alkene is in the Z configuration. The bromination will result in the formation of 2,3 Dibromo 3 methyl pentane.

Explanation:

E and Z forms of the isomers are the configuration in which the polarity of groups on different sides of the double bond is different.

When the polar groups are on the opposite side of the double bond, it results in a Z isomer. When the polar groups are on the same side of the double it results in an E isomer.

The given figure is of $\rm CH_3-CH=CH-CH_2-CH_2$

There is the presence of polar groups on the opposite side of the isomer. The molecule posses Z isomer.

The bromination of the molecule will result in the formation of 2,3 dibromo-3 methyl pentane.

The image for the bromination is attached below.

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