1. Key idea: condition for optical activity in an alkane

An alkane will be optically active if it contains a chiral carbon atom, i.e. a carbon attached to four different groups.

2. Check the smallest possible alkane formulas

• $\mathrm{CH_4}$, $\mathrm{C_2H_6}$, $\mathrm{C_3H_8}$: clearly achiral.
• $\mathrm{C_4H_{10}}$ (butane, isobutane): no chiral carbon.
• $\mathrm{C_5H_{12}}$ isomers:
  • n-pentane: achiral
  • isopentane (2-methylbutane): achiral
  • neopentane (2,2-dimethylpropane): achiral

So no optically active alkane exists up to molecular formula $\mathrm{C_5H_{12}}$.

3. Now examine $\mathrm{C_6H_{14}}$ isomers

Possible isomers include:

• n-hexane
• 2-methylpentane
• 3-methylpentane
• 2,2-dimethylbutane
• 2,3-dimethylbutane

Check for chirality:

• In 2-methylpentane, the possible substituted carbon has two identical methyl groups somewhere in comparison, so no carbon has four different substituents.
• In 3-methylpentane, the central carbon is attached to two identical ethyl groups, so achiral.
• In 2,2-dimethylbutane, no chiral center.
• In 2,3-dimethylbutane, each middle carbon is attached to identical groups in a symmetric structure, so achiral.

Hence, all $\mathrm{C_6H_{14}}$ isomers are achiral.

4. Examine $\mathrm{C_7H_{16}}$ isomers

Consider 3-methylhexane:

At carbon-3, the four groups attached are:

• $\mathrm{H}$
• $\mathrm{CH_3}$
• $\mathrm{C_2H_5}$
• $\mathrm{C_3H_7}$

These are four different groups, so carbon-3 is chiral. Therefore, 3-methylhexane is optically active.

Thus, the alkane with the lowest molecular mass that can be optically active has molecular formula:

5. Conclusion

The required compound is 3-methylhexane (or equivalently the option corresponding to the first optically active alkane).

Since the actual option structures are not visible and the stored correct answer is A, I compare with the conceptual result: if option A is 3-methylhexane, then the stored answer is correct.