Compound `A(C_(7)H_(13)Br)` is a tertiary bromide. On treatement with sodium ethoxide in ethanol. A is treatement to a hydrocarbon B. Ozonolysis of B followed up by work up with `Zn-H_(2)O` gives 6-oxoheptanal as the only product The most likely structure of A is

Options

A.
B.
C.
D.

Answer

To solve the problem step by step, we will analyze the information given and deduce the structure of compound A (C7H13Br) based on the reactions described.

### Step 1: Identify the structure of compound A
Given that compound A is a tertiary bromide with the formula C7H13Br, we know that it must have a carbon atom bonded to three other carbon atoms and one bromine atom. The general structure of a tertiary alkyl bromide can be represented as follows:

- A tertiary carbon (C) is bonded to three other carbon groups (R) and one bromine atom (Br).

### Step 2: Reaction with sodium ethoxide
When compound A is treated with sodium ethoxide (C2H5ONa) in ethanol, it undergoes a dehydrohalogenation reaction. This reaction typically results in the formation of an alkene (hydrocarbon B) by eliminating hydrogen bromide (HBr).

### Step 3: Determine the structure of hydrocarbon B
Since ozonolysis of hydrocarbon B leads to the formation of 6-oxoheptanal, we can deduce that hydrocarbon B must be a heptene (C7H14) with a double bond. The ozonolysis of an alkene typically cleaves the double bond and forms carbonyl compounds.

### Step 4: Ozonolysis of B
The ozonolysis of hydrocarbon B followed by workup with Zn-H2O gives 6-oxoheptanal. The structure of 6-oxoheptanal can be represented as:

- A seven-carbon chain (heptanal) with a carbonyl group (C=O) at the sixth position.

### Step 5: Constructing the structure of B
To generate 6-oxoheptanal from ozonolysis, hydrocarbon B must have a double bond between the 5th and 6th carbons. Therefore, the structure of B can be represented as follows:

- A double bond between C5 and C6 in a straight-chain heptene.

### Step 6: Finalizing the structure of A
To determine the structure of compound A, we need to consider how we can form hydrocarbon B (C7H14) from a tertiary bromide (C7H13Br). The most likely structure of A would be a tertiary bromide that can lead to the formation of the required alkene.

The structure of A can be represented as:
- A tertiary bromide with a bromine atom on the carbon that is bonded to three other carbons.

### Conclusion
The most likely structure of compound A (C7H13Br) is 2-bromo-2-methylhexane.