Hydrocarbon 'A' reacts with `KMnO_4` under neutral conditions at room temperature to form only a diketone,
`CH_3CH_2underset(O)underset(||)C-underset(O)underset(||)C"CH"_2CH_2CH_3`. The hydrocarbon is

Options

A. `CH_3C-="CCH"_2CH_2CH_2CH_3`
B. `HC-="CCH"_2CH_2CH_2CH_2CH_3`
C. `CH_3C-=underset(CH_3)underset(|)C"CHCH"_2CH_3`
D. `CH_3CH_2C-="CCH"_2CH_2CH_3`

Answer

To solve the problem, we need to identify the hydrocarbon 'A' that reacts with `KMnO4` under neutral conditions to produce the diketone `CH3CH2C(=O)C(=O)CH2CH2CH3`. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Reaction
`KMnO4` is a strong oxidizing agent that can oxidize alkynes to diketones under neutral conditions. The presence of a triple bond in the hydrocarbon will lead to the formation of a diketone.

**Hint:** Remember that alkynes (triple bonds) are oxidized to diketones by `KMnO4`.

### Step 2: Analyze the Diketone Structure
The given diketone is `CH3CH2C(=O)C(=O)CH2CH2CH3`. This indicates that there are two carbonyl (C=O) groups located on a carbon chain. The structure suggests that the diketone is formed from an alkyne with two terminal groups.

**Hint:** Look for the carbon backbone of the diketone to deduce the original hydrocarbon.

### Step 3: Identify the Hydrocarbon
To find the hydrocarbon 'A', we can work backwards from the diketone. The diketone has two carbonyl groups, which means the original hydrocarbon must have been an alkyne with a structure that allows for the formation of these two carbonyls.

1. The diketone has a total of 8 carbon atoms (C8).
2. The hydrocarbon must have been an alkyne with a total of 8 carbon atoms as well, specifically a terminal alkyne.

The structure of the hydrocarbon 'A' can be deduced as follows:
- The diketone has the structure `CH3-CH2-C(=O)-C(=O)-CH2-CH2-CH3`, which indicates that the alkyne must have been `CH3-CH2-C≡C-CH2-CH2-CH3`.

**Hint:** The hydrocarbon must have a triple bond that corresponds to the two carbonyl groups in the diketone.

### Step 4: Conclusion
The hydrocarbon 'A' that reacts with `KMnO4` to form the diketone is `CH3-CH2-C≡C-CH2-CH2-CH3`, which is 1-heptyne.

**Final Answer:** The hydrocarbon 'A' is 1-heptyne.