1. Use the molecular formula to find unsaturation

Given compound $P$ has formula $\mathrm{C_6H_{12}O}$.

Degree of unsaturation:

So, $P$ has one degree of unsaturation.

2. Interpret the 2,4-DNP test

A positive 2,4-dinitrophenylhydrazine test indicates the presence of a carbonyl group ($>C=O$), so $P$ is either an aldehyde or a ketone.

3. Interpret the Tollens test

$P$ gives a negative Tollens test, so it is not an aldehyde. Therefore, $P$ must be a ketone.

4. Use optical activity condition

$P$ is stated to be optically active, so it must contain at least one chiral carbon.

Thus we need a compound with:

• formula $\mathrm{C_6H_{12}O}$
• a ketone functional group
• at least one chiral center

5. Find a suitable ketone isomer

Acyclic saturated ketones with formula $\mathrm{C_6H_{12}O}$ are possible. We need one having a stereogenic center.

A suitable structure is:

This is 3-methyl-2-pentanone.

Check chirality at the carbon $\mathrm{CH(CH_3)}$: It is attached to four different groups:

• $\mathrm{H}$
• $\mathrm{CH_3}$
• $\mathrm{CH_2CH_3}$
• $\mathrm{COCH_3}$

Hence this carbon is chiral, so the compound is optically active.

6. Conclusion

Therefore, $P$ must be 3-methyl-2-pentanone, which corresponds to option B.

7. Comparison with stored answer

Stored correct answer is B. My derived answer is also B, so they agree.