Q28 / 180 · AIIMS 2009 · Chemistry

How many geometrical isomers are possible in the following two alkenes ? (i) $\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_3$ (ii) $\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}-\mathrm{Cl}$

  • A
    4 and 4
  • B
    4 and 3
  • C
    3 and 3
  • Correct answer
    D
    3 and 4.

Solution

When the ends of alkene containing $n$ double bonds are different, the number of geometrical isomers is $2^n$. Thus for $\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}-\mathrm{Cl}$. Number of geometrical isomers $=2^2=4$ When the ends of alkene containing $n$ double bonds are same, then the number of geometrical isomers $=2^{n-1}+2^{p-1}$ where $p=\frac{n}{2}$ for even $n$ and $\frac{n+1}{2}$ for odd $n$, thus for $\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_3$ Number of geometrical isomers $=2^{2-1}+2^{\frac{2}{2}-1}=2^1+2^0=2+1=3$.

Hence, the correct answer is option: 'D'.

Relevant Questions