An ether, `(A)` having molecular formula, `C_(6)H_(14)O`, when treated with excess of `HI` produced two alkyl iodides which on hydrolysis yield compounds `(B)` and `(C)`. Oxidation of `(B)` gives an acid `(D)`, whereas oxidation of `(C)` results in the formation of a mixed ketone, `(E)`. Give graphic representation of `(A)` to `(E)`.

Explanation

To solve the problem step by step, let's analyze the information provided and derive the structures of compounds (A) to (E).

### Step 1: Identify Compound (A)
We know that compound (A) is an ether with the molecular formula \( C_6H_{14}O \). A common structure for an ether with this formula could be:

\[ \text{Compound (A): } CH_3CH_2OCH_2CH_2CH_3 \]

This structure can also be represented as:
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