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:
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:
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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!
Updated On: Jun 4, 2026
- I & II only
- II & III only
- I & III only
- II & IV only
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The Correct Option is D
Solution and Explanation
Step 1: Concept
Geometrical isomerism ($cis/trans$ or $E/Z$) occurs in alkenes when rotation around the carbon-carbon double bond is restricted, and each carbon atom of the double bond is bonded to two completely different groups.
Step 2: Meaning
For a molecule of the type $abC=Ccd$ to exhibit geometrical isomerism, $a \neq b$ and $c \neq d$. If either carbon holds two identical attachments, geometrical isomerism becomes impossible.
Step 3: Analysis
Let's test each compound against this criteria: Compound I: $(CH_{3})_{2}C=CH(C_{2}H_{5})$. One double-bonded carbon carries two identical methyl ($-CH_{3}$) groups. No geometrical isomerism. Compound II: $(C_{6}H_{5})CH=CH(C_{2}H_{5})$. Carbon-1 has $-H$ and $-C_{6}H_{5}$ ($a \neq b$). Carbon-2 has $-H$ and $-C_{2}H_{5}$ ($c \neq d$). It shows geometrical isomerism. Compound III: $C_{6}H_{5}CH=CH_{2}$. The terminal carbon carries two identical hydrogen atoms. No geometrical isomerism. Compound IV: $(CH_{3})CH=C(Cl)CH_{3}$. Carbon-1 has $-H$ and $-CH_{3}$ ($a \neq b$). Carbon-2 has $-Cl$ and $-CH_{3}$ ($c \neq d$). It shows geometrical isomerism.
Step 4: Conclusion
Compounds II and IV satisfy the necessary rules to exhibit geometrical isomerism.
Final Answer: (D)
Geometrical isomerism ($cis/trans$ or $E/Z$) occurs in alkenes when rotation around the carbon-carbon double bond is restricted, and each carbon atom of the double bond is bonded to two completely different groups.
Step 2: Meaning
For a molecule of the type $abC=Ccd$ to exhibit geometrical isomerism, $a \neq b$ and $c \neq d$. If either carbon holds two identical attachments, geometrical isomerism becomes impossible.
Step 3: Analysis
Let's test each compound against this criteria: Compound I: $(CH_{3})_{2}C=CH(C_{2}H_{5})$. One double-bonded carbon carries two identical methyl ($-CH_{3}$) groups. No geometrical isomerism. Compound II: $(C_{6}H_{5})CH=CH(C_{2}H_{5})$. Carbon-1 has $-H$ and $-C_{6}H_{5}$ ($a \neq b$). Carbon-2 has $-H$ and $-C_{2}H_{5}$ ($c \neq d$). It shows geometrical isomerism. Compound III: $C_{6}H_{5}CH=CH_{2}$. The terminal carbon carries two identical hydrogen atoms. No geometrical isomerism. Compound IV: $(CH_{3})CH=C(Cl)CH_{3}$. Carbon-1 has $-H$ and $-CH_{3}$ ($a \neq b$). Carbon-2 has $-Cl$ and $-CH_{3}$ ($c \neq d$). It shows geometrical isomerism.
Step 4: Conclusion
Compounds II and IV satisfy the necessary rules to exhibit geometrical isomerism.
Final Answer: (D)
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