Question

Which of the following will not show geometrical isomerism?

Options

• \[\begin{array}{cc}
  \phantom{}\ce{F}\phantom{........}\ce{H}\phantom{}\\
  \phantom{}\backslash\phantom{......}/\phantom{}\\
  \phantom{}\ce{C = C}\\
  \phantom{}/\phantom{......}\backslash\phantom{}\\
  \phantom{}\ce{Cl}\phantom{........}\ce{D}\phantom{}\\
  \end{array}\]

• \[\begin{array}{cc}
  \phantom{}\ce{F}\phantom{........}\ce{F}\phantom{}\\
  \phantom{}\backslash\phantom{......}/\phantom{}\\
  \phantom{}\ce{C = C}\\
  \phantom{}/\phantom{......}\backslash\phantom{}\\
  \phantom{}\ce{Cl}\phantom{........}\ce{Cl}\phantom{}\\
  \end{array}\]

• \[\begin{array}{cc}
  \phantom{}\ce{H3C}\phantom{........}\ce{C2H5}\phantom{}\\
  \phantom{}\backslash\phantom{......}/\phantom{}\\
  \phantom{}\ce{C = C}\\
  \phantom{}/\phantom{......}\backslash\phantom{}\\
  \phantom{}\ce{H5C2}\phantom{........}\ce{CH3}\phantom{}\\
  \end{array}\]

• \[\begin{array}{cc}
  \phantom{}\ce{H3C}\phantom{........}\ce{CH3}\phantom{}\\
  \phantom{}\backslash\phantom{......}/\phantom{}\\
  \phantom{}\ce{C = C}\\
  \phantom{}/\phantom{......}\backslash\phantom{}\\
  \phantom{}\ce{H3C}\phantom{........}\ce{C2H5}\phantom{}\\
  \end{array}\]

Answer 1

\[\begin{array}{cc}
\phantom{}\ce{H3C}\phantom{........}\ce{CH3}\phantom{}\\
\phantom{}\backslash\phantom{......}/\phantom{}\\
\phantom{}\ce{C = C}\\
\phantom{}/\phantom{......}\backslash\phantom{}\\
\phantom{}\ce{H3C}\phantom{........}\ce{C2H5}\phantom{}\\
\end{array}\]

Explanation:

Will not show geometrical isomerism due to presence of same group on one carbon atom of the double bond.