There are six isomeric alkenes (A, B, C, D, E, and F) that require 1 mol of `H_(2)` per mole of alkene for hydrogenation and give the same product (G) on hydrogenation. G is an alkane having the lowest molecular mass and is optically active. Write the structure of compounds from A to G.

Explanation

To solve the problem, we need to identify six isomeric alkenes (A, B, C, D, E, and F) that can be hydrogenated to produce the same alkane (G), which is optically active and has the lowest molecular mass.

### Step-by-Step Solution:

1. **Identify the Alkane (G)**:
- Since G is the lowest molecular mass alkane and is optically active, it must have a chiral center. The simplest alkane that meets these criteria is 2-methylpropane (or isobutane), which has the formula C4H10. However, to be optically active, we need a compound with a chiral carbon.
- The simplest optically active alkane is 2-pentanol (C5H12), which can be derived from 2-pentene (C5H10) upon hydrogenation.

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