Question

Which of the following compound shows E-Z configuration ?
(A) 2-Chloro-3-Bromo-pentane
(B) Formic acid
(C) 2-Chlorobutane
(D) 1-Chloro-1-bromo-but-1-ene​

Answer 1

Explanation:

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OPTION:-(D)

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Answer 2

Answer:

Option (D) 1-bromo-1-chlorobut-1-ene is the correct option.

Explanation:

We all are familiar with the terms cis and trans as it is related to alkenes. When we come across an alkene having different atoms or functional groups on both sides of the C=C, we cannot label it with either trans or cis. Here, E-Z configuration comes to play here. On each C of the C=C, we check the groups or atoms attached, and rank them according to Cahn-Ingold-Prelog (CIP) priority rule. If both the priority groups/atoms are on the same side it is Z, and if they are on the opposite it is E of the double bond.

In 1-bromo-1-chloro-but-1-ene, (please see the image attached) the groups/atoms attached to both ends of C=C are different. One C is attached to bromine and chlorine while the other C is attached to H and ethyl group. In the former C, the priority is given to bromine and in the latter ethyl group is given the priority. Both the priority groups are on  the opposite side so it is (E)-1-bromo-1-chlorobut-1-ene.

Conclusion:

1-bromo-1-chlorobut-1-ene shows E-Z configuration.