Question

Geometrical isomerism is exhibited by (1) 2-chloro but-2-ene (2) but-2-ene (3) 3-methyl pent-2-ene (4) 2-methyl but-2-ene

• A. 1, 2 and 3 are correct.
• B. 1 and 2 are correct.
• C. 2 and 4 are correct.
• D. 1 and 3 are correct.

Answer 1

Answer: (A)

The main conditions for exhibiting geometrical isomerism are (i) presence of double bond. (ii) presence of different groups on double bonded C-atom. (iii) presence of atleast one similar group on adjacent double bonded carbon atoms. $\underset{2-chlorobut-2-ene}{H_{3}C-\underset{\begin{array}{c}|\\ H\end{array}}{C}=\stackrel{\begin{array}{c}Cl\\ |\end{array}}{C}-C{H}_3,}$ $\underset{butene-2}{H_{3}C-\stackrel{\begin{array}{c}H\\ |\end{array}}{C}=\stackrel{\begin{array}{c}H\\ |\end{array}}{C}-C{H}_3}$ $\underset{3-methylpent-2-ene}{H_{3}C-C{H}_2--\stackrel{\begin{array}{c}C{H}_3\\ |\end{array}}{\mathrm{CH}}=CH-C{H}_3}$ $\underset{2-methylbut-2-ene}{H_{3}C-CH=\stackrel{\begin{array}{c}C{H}_3\\ |\end{array}}{C}=CH-C{H}_3}$ (2-methyl but-2-ene does not show geometrical isomerism because same $(-C{H}_3)$ group are present on double bonded carbon).
$\therefore$ Only 2-chloro but-2-ene, butene-2 and 3 methyl pent-2-ene will exhibit geometrical isomerism.