JEE Chemistry 2026 Question with Solution

Given: Four compounds are listed: n-propyl chloride, iso-propyl chloride, sec-butyl chloride, and neo-pentyl chloride.

Find: The percentage of carbon in the compound that exhibits optical isomerism.

A compound shows optical isomerism when it contains a chiral carbon bonded to four different groups.

From the given compounds:

• n-propyl chloride: $CH_3CH_2CH_2Cl$ has no chiral carbon.
• iso-propyl chloride: $(CH_3)_2CHCl$ has no chiral carbon.
• sec-butyl chloride: $CH_3CH(Cl)CH_2CH_3$ has a carbon attached to $H$, $Cl$, $CH_3$, and $C_2H_5$, so it is chiral.
• neo-pentyl chloride: $(CH_3)_3CCH_2Cl$ has no chiral carbon.

Therefore, the required compound is sec-butyl chloride with molecular formula $C_4H_9Cl$.

Now calculate the mass percentage of carbon:

$$\%C = \frac{\text{Mass of Carbon atoms}}{\text{Molar mass of the compound}} \times 100$$

Mass of carbon:

$$4 \times 12 = 48$$

Molar mass of $C_4H_9Cl$:

$$48 + 9 + 35.5 = 92.5$$

So,

$$\%C = \frac{48}{92.5} \times 100 \approx 51.89\%$$

Rounding to the nearest integer:

$$52\%$$

Therefore, the correct option is D.

Given: The four compounds are compared for optical isomerism.

Find: Which one is optically active and then its carbon percentage.

The key idea is that optical isomerism requires a carbon attached to four different substituents.

In sec-butyl chloride, the relevant carbon is bonded to:

• $H$
• $Cl$
• $CH_3$
• $CH_2CH_3$

Since all four groups are different, this carbon is chiral. The other listed chlorides do not contain such a carbon.

Hence the compound is sec-butyl chloride, $C_4H_9Cl$.

Now compute carbon percentage using atomic masses $C = 12$, $H = 1$, and $Cl = 35.5$:

$$\text{Mass of carbon} = 4 \times 12 = 48$$ $$\text{Total molar mass} = (4 \times 12) + (9 \times 1) + 35.5 = 92.5$$ $$\%C = \frac{48}{92.5} \times 100 \approx 51.89$$

Thus the percentage of carbon is approximately $52$, so the correct option is D.