Geometric isomerism can exist in

1. Analyze the Question

The question asks us to identify which of the given organic compounds exhibits geometric isomerism (also known as cis-trans isomerism).

2. Explain the Concept

For a molecule to exhibit geometric isomerism, it must satisfy two main conditions:

• Restricted Rotation: There must be a double bond (C=C) or a ring structure that prevents the atoms from rotating freely.
• Different Groups: Each carbon atom involved in the double bond must be attached to two different groups.

If one of the carbons in the double bond is attached to two identical groups (e.g., two hydrogen atoms), geometric isomerism is impossible because swapping those identical groups does not create a new spatial arrangement.

3. Show the Solution

Let’s examine each option:

• A. Hex-3-ene ($CH_3CH_2CH=CHCH_2CH_3$): The double bond is between carbon 3 and carbon 4. • Carbon 3 is attached to a hydrogen atom and an ethyl group ($-CH_2CH_3$). • Carbon 4 is also attached to a hydrogen atom and an ethyl group ($-CH_2CH_3$). Since both carbons have two different groups attached, this molecule can exist as cis-hex-3-ene or trans-hex-3-ene. This satisfies the condition.

• B. Hexane ($CH_3CH_2CH_2CH_2CH_2CH_3$): This is an alkane with only single bonds. There is free rotation around all carbon-carbon bonds, so it cannot exhibit geometric isomerism.

• C. Prop-1-ene ($CH_2=CHCH_3$): The double bond is at the end (carbon 1). Carbon 1 is attached to two identical hydrogen atoms. Because of these identical atoms, it cannot show geometric isomerism.

• D. 3-methylbut-1-ene ($CH_2=CHCH(CH_3)_2$): Again, the double bond is at the end (carbon 1). Carbon 1 is attached to two identical hydrogen atoms, which prevents geometric isomerism.

4. Identify the Answer

Based on the analysis, hex-3-ene is the only compound where the groups attached to the double-bonded carbons are distinct enough to allow for cis and trans configurations.

5. Exam Tip

The "Terminal Double Bond" Rule: In JAMB Chemistry, if you see an alkene with the double bond at the very beginning of the chain (like prop-1-ene or but-1-ene), it cannot exhibit geometric isomerism because the terminal carbon will always have two hydrogen atoms attached to it. Look for double bonds located towards the middle of the carbon chain!