Which of the following compounds will exhibit geometrical isomerism?

First, recall the basic condition for geometrical (cis-trans) isomerism in alkenes. We state the rule:

For an alkene $$R^1R^2C=CR^3R^4$$ to show geometrical isomerism, each of the two doubly‐bonded carbons must be attached to two different groups. Because rotation about the $$C=C$$ bond is restricted, different spatial arrangements (cis and trans or the more general $$E$$/$$Z$$) become possible only when the four substituents $$R^1,\,R^2,\,R^3,\,R^4$$ are not all identical in pairs on either carbon.

Now we examine every option one by one, drawing or visualising the structure of the compound and checking the substituents on each doubly-bonded carbon.

Option A : 1,1-Diphenyl-1-propane

The word “propane’’ tells us there is no double bond at all; the compound is saturated. Without a $$C=C$$ bond (or a suitably constrained ring), geometrical isomerism cannot arise. So Option A fails the very first requirement.

Option B : 1-Phenyl-2-butene

Let us write its carbon skeleton. Numbering starts from the end nearer the double bond:

$$\text{Ph-CH}_2-CH=CH-CH_3$$

• The double bond is between C-2 and C-3.
• On C-2 we have the groups $$\text{Ph-CH}_2-$$ and $$H$$, clearly two different substituents.
• On C-3 we have $$CH_3$$ and $$H$$, again two different substituents.

Since both doubly-bonded carbons possess two different groups, the alkene fulfils the stated rule. Rotation is locked, so we can arrange similar groups on the same side (cis) or on opposite sides (trans). Therefore Option B does exhibit geometrical isomerism.

Option C : 3-Phenyl-1-butene

Its structure is

$$CH_2=CH-CH_2-CH_2-\text{Ph}$$

The double bond is now at the very end, between C-1 and C-2.

• Carbon C-1 carries two identical substituents, namely $$H$$ and $$H$$ (it is $$CH_2=$$).
Because the two groups on C-1 are identical, the required difference is missing; no cis-trans pair can be generated. Hence Option C does not show geometrical isomerism.

Option D : 2-Phenyl-1-butene

The formula becomes

$$CH_2=CH-CH(\text{Ph})-CH_3$$

Again the double bond is terminal (between C-1 and C-2).

• Carbon C-1 still has two identical $$H$$ atoms.
Thus the same objection raised for Option C applies here as well. Option D cannot display geometrical isomerism.

Among the four possibilities, only Option B satisfies the necessary condition on both double-bonded carbons. All the others fail because either there is no double bond or one of the double-bonded carbons bears two identical substituents.

Hence, the correct answer is Option 2.