an Organic compound `(A) (C_(4)H_(6))` forms a precipitate with Tollens and Fehling's reagents. `(A)` has an isomer `(B)`. `(B)` reacts with `1 mol` of `Br_(2)` to form `1,4`-dibromo-2-butene. `(A)` and `(B)` are:

Options

A. a.
B. b.
C. c.
D. d.

Explanation

To solve the problem, we need to identify the organic compounds (A) and (B) based on the given information. Let's break down the solution step by step.

### Step 1: Determine the Degree of Unsaturation
The molecular formula of compound (A) is C₄H₆. We can calculate the degree of unsaturation (DU) using the formula:

\[ \text{Degree of Unsaturation} = \frac{2C + 2 - H}{2} \]

Where C is the number of carbon atoms and H is the number of hydrogen atoms.

Substituting the values:

\[ \text{DU} = \frac{2(4) + 2 - 6}{2} = \frac{8 + 2 - 6}{2} = \frac{4}{2} = 2 \]

### Step 2: Analyze the Degree of Unsaturation
A degree of unsaturation of 2 indicates that the compound could have either:
- One triple bond (alkyne) or
- Two double bonds (alkenes).

### Step 3: Identify Compound (A)
Since compound (A) forms a precipitate with Tollens' and Fehling's reagents, it suggests that (A) is likely a terminal alkyne. Terminal alkynes can reduce these reagents.

A suitable structure for compound (A) is 1-butyne (or methylacetylene), represented as:

\[ \text{CH}_3\text{C} \equiv \text{C}\text{H} \]

### Step 4: Confirm the Reaction with Tollens' and Fehling's Reagents
When 1-butyne reacts with Tollens' reagent (ammonical silver nitrate), it forms a white precipitate of silver butylenide (AgC≡C-CH₃). Similarly, it will react with Fehling's solution to produce a red precipitate of copper(I) oxide.

### Step 5: Identify Compound (B)
Compound (B) is an isomer of (A) and has the same molecular formula (C₄H₆). It reacts with 1 mole of Br₂ to form 1,4-dibromo-2-butene.

The structure of compound (B) can be deduced as follows:

\[ \text{C} \text{C} \text{C} \text{C} \]
with two double bonds, specifically 1,3-butadiene:

\[ \text{CH}_2=\text{CH}-\text{CH}=\text{CH}_2 \]

### Step 6: Confirm the Reaction with Bromine
When 1,3-butadiene reacts with bromine (Br₂), it can add across the double bonds to yield 1,4-dibromo-2-butene:

\[ \text{Br}-\text{CH}_2-\text{C}=\text{C}-\text{Br}-\text{CH}_2 \]

### Conclusion
Thus, the compounds are:
- Compound (A): 1-butyne (C₄H₆)
- Compound (B): 1,3-butadiene (C₄H₆)