Question

Which of the following compounds will show geometrical isomerism?  I. $(CH_{3})_{2}C=CH~C_{2}H_{5}$  II. $C_{6}H_{5}CH=CH~C_{2}H_{5}$  III. $C_{6}H_{5}CH=CH_{2}$  IV. $CH_{3}CH=C(Cl)CH_{3}$  The correct answer is:

• A. I & II only
• B. II & III only
• C. I & III only
• D. II & IV only

Hint:

Check both alkene carbons individually. If you spot two identical groups on the same carbon (like $=CH_{2}$ or $=C(CH_{3})_{2}$), rule it out immediately!

Answer 1

The Correct Option is D

Step 1: Recall the rule.
An alkene shows geometrical isomerism only if each carbon of the double bond carries two different groups. If a carbon has two identical groups, it cannot show this isomerism.

Step 2: Test compound I.
Compound I is $(CH_3)_2C=CH\,C_2H_5$. One double-bond carbon has two identical methyl groups, so it has no geometrical isomerism.

Step 3: Test compound II.
Compound II is $C_6H_5CH=CH\,C_2H_5$. One carbon has H and phenyl, the other has H and ethyl. Both carbons carry different groups, so it shows geometrical isomerism.

Step 4: Test compound III.
Compound III is $C_6H_5CH=CH_2$. The end carbon has two identical hydrogens, so it has no geometrical isomerism.

Step 5: Test compound IV.
Compound IV is $CH_3CH=C(Cl)CH_3$. One carbon has H and methyl, the other has Cl and methyl. Both carry different groups, so it shows geometrical isomerism.

Step 6: Pick the answer.
Only compounds II and IV satisfy the rule. \[ \boxed{\text{II and IV only}} \]