Identification of A and B in the Reaction:
- A: The optically active alcohol A is most likely a chiral compound containing a stereocenter.
- B: Compound B, which is formed after the hydrogenation of A, is not optically active. This suggests that the chiral center in A has been removed during the reaction.

Explanation:
- When an optically active compound undergoes hydrogenation, the chiral center is often reduced to a non-chiral center. This results in the loss of optical activity in the product.
- In this case, alcohol A absorbs two moles of H2 per mole of A during hydrogenation. This indicates that A contains a double bond that is being reduced to a saturated compound in B.
- The molecular formula of A is C6H10O, which suggests that A is a cyclic compound with either a ketone or aldehyde functional group.
- The presence of a double bond in A, along with its optical activity, points towards a compound like 2-methylcyclopentanone or a similar chiral cyclic ketone/aldehyde.
- Upon hydrogenation, the double bond in A is reduced to a single bond, leading to compound B, which is not optically active. This loss of chirality confirms the removal of the chiral center present in A.
Therefore, based on the given information, it can be inferred that compound A is an optically active cyclic ketone/aldehyde, while compound B is the product obtained after the hydrogenation of A, resulting in the loss of optical activity.