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Draw structures for the following compounds:a.

Chapter 5, Problem 52P

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QUESTION:

Problem 52P

Draw structures for the following compounds:

a. (2E,4E)-1-chloro-3-methyl-2,4-hexadiene

b. (3Z,5E)-4-methyl-3,5-nonadiene

c. (3Z,5Z)-4,5-dimethyl-3,5-nonadiene

d. (3E,5E)-2,5-dibromo-3,5-octadiene

Questions & Answers

QUESTION:

Problem 52P

Draw structures for the following compounds:

a. (2E,4E)-1-chloro-3-methyl-2,4-hexadiene

b. (3Z,5E)-4-methyl-3,5-nonadiene

c. (3Z,5Z)-4,5-dimethyl-3,5-nonadiene

d. (3E,5E)-2,5-dibromo-3,5-octadiene

ANSWER:

Solution:

Here we have to draw the structure of the following molecules.

Isomer: -The compounds of same molecular formula with different structure and properties is called as isomer.

E and Z-isomer:-

When the higher priority groups are present on the opposite sides of the double bond then it is said to be E-isomer. E comes from the German letter i.e “entgegen” which means the opposite. The Higher priority groups are on the same sides of the double bond than the isomer is said to be Z-isomer. Z comes from the German word “Zusammen” which means together.

Step 1

a. (2E,4E)-1-chloro-3-methyl-2,4-hexadiene

In order to show the isomers, 1st we have to draw the structure of the given molecule, then mark all the substituents depending on the priority. If we will place higher priority groups on the opposite side of the double bonds then it is said to be E and when the higher priority groups are present on the same side of the double bond then it is said to be Z-isomer.

Thus the structure of (2E,4E)-1-chloro-3-methyl-2,4-hexadiene is,

       (2E-1-chloro-3-methyl-2,4-hexadiene)           (4E-1-chloro-3-methyl-2,4-hexadiene)

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