Compound (X) with molecular formula `C_(3)H_(8)O` is treated with acidified potassium dichromate to form a product (Y) with molecular formula `C_(3)H_(6)O`. (Y) does not form a shining silver mirror on warming with ammoniacal `AgNO_(3)`. (Y) when treated with an aqueous solution of `NH_(2)CONHNH_(2).HCl` and sodium acetate, gives a product (Z). The structure of (Z) is

Options

A. `CH_(3)CH_(2)CH=N NHCONH_(2)`
B. `(CH_(3))_(2)C=N NHCONH_(2)`
C. `(CH_(3))_(2)C=NCONHNH_(2)`
D. `CH_(3)CH_(2)CH=NCONHNH_(2)`

Explanation

To solve the problem step by step, let's analyze the information provided:

### Step 1: Identify Compound (X)
The molecular formula of compound (X) is C₃H₈O. The unsaturation factor (UF) can be calculated using the formula:
\[ \text{UF} = \frac{(2C + 2 + N - H - X)}{2} \]
For C₃H₈O:
- C = 3
- H = 8
- O = 1
- N = 0

Substituting the values:
\[ \text{UF} = \frac{(2(3) + 2 + 0 - 8 - 0)}{2} = \frac{(6 + 2 - 8)}{2} = \frac{0}{2} = 0 \]

Since the unsaturation factor is 0, compound (X) is likely an alcohol or ether.

**Hint for Step 1:** The unsaturation factor helps determine the type of compound based on its molecular formula.

### Step 2: Reaction with Acidified Potassium Dichromate
When compound (X) is treated with acidified potassium dichromate (K₂Cr₂O₇), it gets oxidized to form compound (Y) with the molecular formula C₃H₆O. The unsaturation factor for C₃H₆O is:
\[ \text{UF} = \frac{(2(3) + 2 - 6)}{2} = \frac{(6 + 2 - 6)}{2} = \frac{2}{2} = 1 \]

Since the unsaturation factor is 1, compound (Y) could be either an aldehyde or a ketone.

**Hint for Step 2:** Oxidation can change the functional group of a compound, leading to different types of compounds.

### Step 3: Determine the Type of Compound (Y)
The problem states that (Y) does not form a shining silver mirror on warming with ammoniacal AgNO₃ (Tollens' reagent). This indicates that (Y) is a ketone, as aldehydes would give a positive Tollens' test.

Possible structures for (Y) could be:
1. Propan-2-one (Acetone): CH₃C(=O)CH₃
2. Propanal: CH₃CH₂C(=O)H

Since (Y) is a ketone, we conclude:
\[ Y = \text{Propan-2-one (Acetone)} \]

**Hint for Step 3:** The Tollens' test is a key indicator for distinguishing between aldehydes and ketones.

### Step 4: Reaction of Compound (Y) with Hydrazine
Compound (Y) (Propan-2-one) is then treated with an aqueous solution of hydrazine (NH₂CONHNH₂·HCl) and sodium acetate. This reaction is known as the formation of a hydrazone.

The reaction can be summarized as follows:
\[ \text{Propan-2-one} + \text{Hydrazine} \rightarrow \text{Hydrazone} \]

The structure of the product (Z) formed will be:
\[ Z = CH₃C(=NNH₂)CH₃ \]

This is the structure of the hydrazone derived from propan-2-one.

**Hint for Step 4:** The reaction with hydrazine typically leads to the formation of a hydrazone, which is characterized by the presence of a C=N bond.

### Final Answer
The structure of product (Z) is:
\[ Z = CH₃C(=NNH₂)CH₃ \]