i. `A + 2Na overset ("Dryehter") rarr Me_(2)CH - CJMe_(2)`
ii. `A + 2Na overset ("Dryether") rarr Me_(3)C - Cme_(3)`
iii. One mole of `CH_(3)Br` are reacted with 2 mole of sodium metal in dry ether. The productus thus obtained are:
a. Ethane + 2NaBr
b. Ethane + Propane + 2 NaBr
c. Ethane + Propane + Butane + 2 NaBr
d. Ethane + Propane + Butane + Ethene + 2 NaBr

Explanation

To solve the given question, we will analyze each part step by step.

### Step 1: Identify Compound A in the First Reaction
The first reaction is:
\[ A + 2Na \overset{\text{Dry Ether}}{\rightarrow} \text{Me}_2\text{CH} - \text{CHMe}_2 \]

In this reaction, we observe that the product is a symmetrical alkane, which indicates that the reactant A must also be a symmetrical alkyl halide. The product can be identified as isopropyl bromide (or any similar halide), which can be represented as:
\[ \text{Me}_2\text{CH}X \]
...