Question

The number of geometrical isomers possible for the compound, CH$_3$CH = CH - CH = CH$_2$ is:

• A. 2
• B. 3
• C. 4
• D. 6

Hint:

For cis-trans isomerism, each carbon in the double bond must have two different substituents.

Answer 1

The Correct Option is A

Step 1: Identifying the Double Bonds
The given compound has two double bonds.
The key factor in determining geometrical isomerism is whether each double bond has two different substituents.
Step 2: Checking for Geometrical Isomerism
The double bond at position \( {C}_2 - {C}_3 \) has two different groups, allowing cis-trans isomerism.
The double bond at \( {C}_4 - {C}_5 \) has identical hydrogen atoms, preventing cis-trans isomerism.
Step 3: Number of Isomers
Only one double bond exhibits cis-trans isomerism.
Therefore, only two geometrical isomers exist.
Thus, the correct answer is (A).

Answer 2

Step 1: Analyze the compound structure
The given compound is CH₃–CH=CH–CH=CH₂.
This is a conjugated diene with two double bonds: one between C2=C3 and another between C4=C5.
Carbon numbering: CH₃–(C1)–(C2)=C3–C4=C5–H

Step 2: Identify centers of geometrical (cis-trans) isomerism
Geometrical isomerism is possible around a double bond when each carbon of the double bond has two different groups attached.

- Double bond C2=C3:
- C2 is attached to CH₃ and H (different)
- C3 is attached to C4 and C2 (different groups)
⇒ This double bond can show geometrical isomerism.

- Double bond C4=C5:
- C4 is connected to C3 and H (different)
- C5 is connected to H and no other alkyl group (it’s terminal: CH=CH₂)
⇒ One side has two hydrogens ⇒ No geometrical isomerism possible here.

Step 3: Count possible geometrical isomers
Only one of the two double bonds (C2=C3) can exhibit cis-trans isomerism.
Thus, the total number of geometrical isomers possible = 2 (cis and trans forms).

Final Answer: 2