Which of the following compounds shows geometrical isomerism?

First, let us recall the fundamental requirement for the existence of geometrical (cis-trans) isomerism in an alkene. In an alkene $$\mathrm{R^1R^2C = CR^3R^4}$$ each of the two doubly-bonded carbon atoms must be attached to two different groups. Symbolically:

$$\text{For C}_\alpha:\; R^1 \neq R^2 \quad\text{and}\quad \text{For C}_\beta:\; R^3 \neq R^4$$

If either carbon carries two identical substituents, interchange of positions of the groups around the $$\pi$$-bond no longer produces a distinct compound, and geometrical isomerism becomes impossible. Now we apply this criterion to every option one by one, writing out the full structure each time.

Option A : 2-methylpent-2-ene

The parent pent-2-ene is $$\mathrm{CH_3-CH = CH-CH_2-CH_3}$$. Introducing the 2-methyl group on the second carbon gives

$$\mathrm{CH_3-C(CH_3)=CH-CH_2-CH_3}$$

At the doubly-bonded carbon on the left (the second carbon) the attached groups are $$\mathrm{CH_3}$$ and another $$\mathrm{CH_3}$$ (identical) while the other carbon of the double bond has $$\mathrm{H}$$ and $$\mathrm{CH_2-CH_2-CH_3}$$. Because the first carbon of the $$\mathrm{C=C}$$ bears two identical $$\mathrm{CH_3}$$ groups, we have

$$R^1 = R^2 = \mathrm{CH_3}$$

which violates $$R^1 \neq R^2$$, so no geometrical isomerism is possible here.

Option B : 4-methylpent-2-ene

First write pent-2-ene again: $$\mathrm{CH_3-CH = CH-CH_2-CH_3}$$. Now place a methyl group on carbon 4 (counting from the left end):

$$\mathrm{CH_3-CH = CH-CH(CH_3)-CH_3}$$

Let us examine the two double-bonded carbons separately.

• For the left carbon of the double bond (carbon 2) the two single-bonded substituents are

$$R^1 = \mathrm{CH_3}\quad\text{and}\quad R^2 = \mathrm{H}$$

Clearly, $$R^1 \neq R^2$$.

• For the right carbon of the double bond (carbon 3) the two single-bonded substituents are

$$R^3 = \mathrm{H}\quad\text{and}\quad R^4 = \mathrm{CH(CH_3)-CH_3}$$

Again, $$R^3 \neq R^4$$.

Since both carbons meet the inequality condition, two distinct spatial arrangements (cis and trans, or Z and E) are possible. Therefore geometrical isomerism is exhibited by 4-methylpent-2-ene.

Option C : 4-methylpent-1-ene

Its structure is $$\mathrm{CH_2 = CH-CH_2-CH(CH_3)-CH_3}$$. The leftmost carbon of the double bond is $$\mathrm{CH_2}$$ and hence has two identical hydrogens:

$$R^1 = R^2 = \mathrm{H}$$

Because $$R^1 = R^2$$, the necessary condition fails and geometrical isomerism is not possible.

Option D : 2-methylpent-1-ene

The structure is $$\mathrm{CH_2 = C(CH_3)-CH_2-CH_2-CH_3}$$. The terminal $$\mathrm{CH_2}$$ of the double bond again carries two identical hydrogens:

$$R^1 = R^2 = \mathrm{H}$$

With identical groups on one carbon, geometrical isomerism is not feasible.

Among all the given compounds, only 4-methylpent-2-ene satisfies the requirement that each carbon of the $$\mathrm{C=C}$$ bond bears two different substituents. Hence it alone can exist as a pair of geometrical isomers.

Hence, the correct answer is Option B.