Question:
The number of cis/trans isomers possible for alkenes \(\mathrm{C_5H_{10}}\) and \(\mathrm{C_4H_8}\) respectively is
The number of cis/trans isomers possible for alkenes \(\mathrm{C_5H_{10}}\) and \(\mathrm{C_4H_8}\) respectively is
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Geometrical isomerism is possible only when each carbon of the double bond has two different substituents.
Updated On: Jul 15, 2026
- \(2,\ 4\)
- \(4,\ 4\)
- \(4,\ 2\)
- \(2,\ 2\)
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The Correct Option is D
Solution and Explanation
Step 1: Count cis/trans isomers for \(\mathrm{C_5H_{10}}\).
The alkenes are:
\[
\mathrm{Pent\mbox{-}1\mbox{-}ene}
\]
(no geometrical isomer),
\[
\mathrm{Pent\mbox{-}2\mbox{-}ene}
\]
(gives cis and trans),
\[
\mathrm{2\mbox{-}Methylbut\mbox{-}1\mbox{-}ene}
\]
(no geometrical isomer),
\[
\mathrm{3\mbox{-}Methylbut\mbox{-}1\mbox{-}ene}
\]
(no geometrical isomer),
\[
\mathrm{2\mbox{-}Methylbut\mbox{-}2\mbox{-}ene}
\]
(no geometrical isomer).
Hence,
\[
\boxed{2}
\]
cis/trans isomers are possible.
Step 2: Count cis/trans isomers for \(\mathrm{C_4H_8}\). Among the alkenes, \[ \mathrm{But\mbox{-}2\mbox{-}ene} \] exists as \[ \boxed{\text{cis and trans}.} \] All other alkenes do not exhibit geometrical isomerism. Hence, \[ \boxed{2} \] cis/trans isomers are possible. Therefore, \[ \boxed{2,\ 2} \] Hence, \[ \boxed{(D)} \] is the correct answer.
Step 2: Count cis/trans isomers for \(\mathrm{C_4H_8}\). Among the alkenes, \[ \mathrm{But\mbox{-}2\mbox{-}ene} \] exists as \[ \boxed{\text{cis and trans}.} \] All other alkenes do not exhibit geometrical isomerism. Hence, \[ \boxed{2} \] cis/trans isomers are possible. Therefore, \[ \boxed{2,\ 2} \] Hence, \[ \boxed{(D)} \] is the correct answer.
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