Compound (A) ${C_6}{H_{12}}O$ is optically active. Compound (A) gives negative Tollens test and positive test with 2-4-di-nitro phenyl hydrazine. Identify A.
A.$C{H_3} - \left( {C = O} \right) - C{H_2} - C\left( { - C{H_3}} \right)H - C{H_3}$
B.$C{H_3} - \left( {C = O} \right) - \left( {C - C{H_2} - C{H_3}} \right)H - C{H_3}$
C.$H - \left( {C = O} \right) - C{H_2} - \left( {C - C{H_2} - C{H_3}} \right)H - C{H_3}$
D.$C{H_3} - \left( {C = O} \right) - C{H_2} - C{H_2} - C{H_2} - C{H_3}$

Hint: Optical activity is the ability of a material to move a light beam's polarization axis. The amplitude of optical action is represented as a number, called the specific rotation, which corresponds to a relation that relates the angle at which a plane rotates, to the sample duration and the sample density (or the frequency whether it is present in the plane's sample) (in a plane-polarised light, the vibration of an electric field is limited to a single planet). As the specific rotation relies on the temperature and the light's wavelength, both amounts must be observed.

Complete answer:
As we already know that the Tollens test is given by aldehydes $\left( { - CHO} \right)$. And the same test is never given by ketones $\left[ { - \left( {C = O} \right) - } \right]$.
2,4 DNP (2-4-di-nitro phenyl hydrazine) test will always be given by aldehydes as well as ketones.
Which means that all the substrates surrounding the four sides of the carbon atom should be different. This would mean that the compound is optically active.
Since in option A, the fourth carbon atom is not optically active due to two substrates being the same, A is not the answer.
In option B, ketone is present (which means that it will test negative for tollens and positive for 2,4 DNP), as well as the third carbon atom in the chain is optically active as well, it is the correct option.
In option C, aldehyde is present (which means that it will test positive for tollens and positive for 2,4 DNP), but the third carbon atom in the chain is not optically active, it is not the correct option.
In option D, ketone is present (which means that it will test negative for tollens and positive for 2,4 DNP), but no carbon atom in the chain is optically active, it is not the correct option.
Hence, from the above data collected.

It is clear that option B is the correct option.

Note:
A positive value is given to a rotation when it is clockwise, negative if counter clockwise, with respect to the observer facing the light. A material that has a definite positive rotation is considered a dextrorotatory and the prefix d or (+) is indicated; one with a similar negative rotation is levorotatory and the prefix l or-) (is applied to it.