State which of the following can exhibit geometric isomerism: Give reason for your answer

Answer Details

The four compounds shown are:

• (a) \(CH_3-CH{=}CH-CH_3\) (but-2-ene);
• (b) \(CH_3-CH_2-CHBr-CH_2Br\) (1,2-dibromobutane);
• (c) \(HOOC-CH(CH_3)-CH_2-COOH\) (2-methylbutanedioic acid);
• (d) \(HOOC-CH{=}CH-COOH\) (butenedioic acid).

Compounds that exhibit geometric (cis-trans) isomerism: (a) and (d).

Reason. Geometric isomerism requires two conditions to be met together:

1. a carbon-to-carbon double bond (\(C{=}C\)), because the double bond prevents (restricts) free rotation about that bond; and
2. each of the two doubly-bonded carbon atoms must carry two different groups.

In (a) each doubly-bonded carbon carries an \(H\) and a \(CH_3\) group (two different groups), so cis- and trans-but-2-ene exist. In (d) each doubly-bonded carbon carries an \(H\) and a \(-COOH\) group, so the cis form (maleic acid) and the trans form (fumaric acid) exist. Both conditions are satisfied, so (a) and (d) show geometric isomerism.

Compounds (b) and (c) contain only single (C-C) bonds. Single bonds allow free rotation, so the atoms cannot be locked into fixed cis/trans positions; therefore (b) and (c) do not exhibit geometric isomerism.