Hydrocarbon (A) `C_(6)H_(10)` on treatment with `H_(2),Ni, H_(2)//`Lindlar's catalyst or `Na//`liq. `NH_(3)` forms three different reduction products (B), (C), (D) respectively. (A) does not form any salt with ammonical `AgNO_(3)` solution, but (E) forms salt on heating with `NaNH_(2)` in an inert solvent. Compound (E) reacts with `CH_(3)`I to give (F). Compound (D) on oxidative ozonolysis gives n-butanoic acid along with other product.
If (E) is reacted with acetaldehyde followed by acidification, product is

Options

A. acid
B. ketone
C. ether
D. alcohol

Answer

To solve the problem step by step, we will analyze the information given about the hydrocarbon (A) and its reactions to identify the products formed.

### Step 1: Identify Compound A
The hydrocarbon (A) has the formula `C6H10`. Given that it does not form a salt with ammonical `AgNO3`, it indicates that (A) is a non-terminal alkyne. A possible structure for (A) could be 2-heptyne or 3-heptyne. For simplicity, let's consider it as 2-heptyne:
\[ \text{A: } CH_3CH_2C \equiv CCH_3 \]

### Step 2: Reduction of Compound A
When (A) is treated with hydrogen in the presence of nickel (Ni), it undergoes catalytic hydrogenation to form an alkane (B):
\[ \text{B: } CH_3CH_2CH_2CH_2CH_2CH_3 \]
This is hexane.

### Step 3: Reduction with Lindlar's Catalyst
When (A) is treated with hydrogen in the presence of Lindlar's catalyst, it selectively reduces to a cis-alkene (C):
\[ \text{C: } CH_3CH_2C=CCH_3 \]
This is cis-2-hexene.

### Step 4: Reduction with Sodium in Liquid Ammonia
When (A) is treated with sodium in liquid ammonia, it undergoes Birch reduction to form a trans-alkene (D):
\[ \text{D: } CH_3CH_2CH_2CH=CHCH_3 \]
This is trans-2-hexene.

### Step 5: Formation of Compound E
When (A) is heated with sodium amide (NaNH2) in an inert solvent, it forms a sodium alkanide (E):
\[ \text{E: } CH_3CH_2CH_2C \equiv CNa \]

### Step 6: Reaction of Compound E with Methyl Iodide
Compound (E) reacts with methyl iodide (CH3I) to give (F):
\[ \text{F: } CH_3CH_2CH_2C \equiv CH_3 \]
This is 1-heptyne.

### Step 7: Ozonolysis of Compound D
When compound (D) undergoes oxidative ozonolysis, it produces n-butanoic acid and another product (acetic acid):
\[ \text{Products: } CH_3CH_2CH_2COOH + CH_3COOH \]

### Step 8: Reaction of Compound E with Acetaldehyde
When compound (E) is reacted with acetaldehyde (CH3CHO) followed by acidification, it forms an alcohol:
1. The sodium alkanide (E) adds to acetaldehyde to form an alkoxide.
2. Upon acidification, the alkoxide is converted to an alcohol.

The final product is:
\[ \text{Product: } CH_3CH_2CH_2C(OH)(CH_3) \]
This is 2-heptanol.

### Final Answer
The product formed when compound (E) is reacted with acetaldehyde followed by acidification is **2-heptanol**.

---