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Chemistry Question on Organic Chemistry- Some Basic Principles and Techniques

How many geometrical isomers are possible in the following two alkenes ? (i) \quad CH3−CH=CH−CH=CH−CH3CH_3 - CH = CH - CH = CH - CH_3 (ii) \quad CH3−CH=CH−CH=CH−ClCH_3 - CH = CH - CH = CH - Cl

A

4 and 4

B

4 and 3

C

3 and 3

D

3 and 4

Answer

3 and 4

Explanation

Solution

When the ends of alkene containing n double bonds are different, the number of geometrical isomers is 2n2^n. Thus for CH3−CH=CH−CH=CH−ClCH_3 - CH = CH - CH = CH - Cl. Number of geometrical isomers =22=4= 2^2 = 4 When the ends of alkene containing nn double bonds are same, then the number of geometrical isomers =2n−1+2p−1= 2^{n-1} + 2^{p-1} where p=n2p=\frac{n}{2} for even nn and n+12\frac{n+1}{2} for odd nn, thus for CH3−CH=CH−CH=CH−CH3CH_3 - CH = CH - CH = CH - CH_3 Number of geometrical isomers =22−1+222−1=21+20=2+1=3.=2^{2-1}+2^{\frac{2}{2}-1}=2^{1}+2^{0}=2+1=3.