Which of the following compounds shows geometrical isomerism?

Step 1: Identifying Geometrical Isomerism in Alkenes Geometrical isomerism ($cis/trans$ or $E/Z$) requires each carbon of the double bond to be attached to two different groups.

• 2-methylpent-2-ene: $\ce{(CH3)2C=CHCH2CH3}$ (One carbon has two methyls - No GI)
• 4-methylpent-1-ene: $\ce{CH2=CHCH2CH(CH3)2}$ (Terminal bond - No GI)
• 2-methylpent-1-ene: $\ce{CH2=C(CH3)CH2CH2CH3}$ (Terminal bond - No GI)
• 4-methylpent-2-ene: $\ce{CH3CH=CHCH(CH3)2}$

Step 2: Final Evaluation In 4-methylpent-2-ene, the C2 is attached to H and Me, and C3 is attached to H and Isopropyl. Both possess two different groups.

$\boxed{\text{Answer: (b) 4-methylpent-2-ene}}$