Question:

In the following structure, the double bonds are marked as I, II, III and IV

Geometrical isomerism is not possible at site (s) :

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Check each C=C for two different substituents on both carbons; the gem-dimethyl end fails.
Updated On: Nov 11, 2026
  • III
  • I
  • I and III
  • III and IV
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The Correct Option is B

Solution and Explanation

Geometrical isomerism at a C=C needs two different groups on each doubly bonded carbon.
Double bond I: the terminal carbon carries two identical \(\mathrm{CH_3}\) groups, so no geometrical isomerism.
Double bond II: one carbon has \(\mathrm{CH_3}\) and a chain, the other has H and a \(\mathrm{CH_2}\) chain; all different, so it is possible.
Double bond III: one carbon has \(\mathrm{CH_3}\) and a \(\mathrm{CH_2}\) chain; the other has \(\mathrm{CH_3}\) and the carbon of bond IV: different, so it is possible.
Double bond IV: one carbon has \(\mathrm{CH_3}\) and the carbon of bond III; the other has \(\mathrm{CH_3}\) and \(\mathrm{C_2H_5}\): different, so it is possible.
Only site I lacks geometrical isomerism.
Answer: (2) I
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