The compound A , B and C in the reaction sequence
`CH_(3) - CH_(2) OH overset(PBr_(3))to A overset(Alc. KOH) to B overset(Br_(2))to C` are given by the set :

Options

A. Ethyl bromide, `CH_(3) CH_(2) OH , CH_(3) CHBr_(2)`
B. `C_(2)H_(5) Br, CH -= CH , CH_(2) = CHBr`
C. `C_(2) H_(5) Br, CH_(2) = CH_(2) , CH_(2)Br- CH_(2)Br`
D. `C_(2) H_(5)Br, CH_(3)CH_(2)OH, BrCH_(2)- CH_(2)Br`

Answer

To solve the problem, we will identify the compounds A, B, and C in the given reaction sequence step by step.

### Step 1: Identify Compound A
The starting compound is ethyl alcohol, which is represented as \( CH_3CH_2OH \).

When ethyl alcohol reacts with phosphorus tribromide (\( PBr_3 \)), the hydroxyl group (\( -OH \)) is replaced by a bromine atom (\( -Br \)). This reaction converts the alcohol into an alkyl bromide.

**Reaction:**
\[ CH_3CH_2OH + PBr_3 \rightarrow CH_3CH_2Br + HBr + H_3PO_3 \]

Thus, compound A is:
\[ A = CH_3CH_2Br \] (ethyl bromide)

### Step 2: Identify Compound B
Next, we react compound A (ethyl bromide) with alcoholic potassium hydroxide (\( Alc. KOH \)). In this reaction, an elimination reaction occurs where a bromine atom is removed along with a hydrogen atom from the adjacent carbon, resulting in the formation of a double bond.

**Reaction:**
\[ CH_3CH_2Br + Alc. KOH \rightarrow CH_2=CH_2 + KBr + H_2O \]

Thus, compound B is:
\[ B = CH_2=CH_2 \] (ethylene or ethene)

### Step 3: Identify Compound C
Finally, compound B (ethylene) reacts with bromine (\( Br_2 \)). In this reaction, the double bond in ethylene is broken, and two bromine atoms add across the double bond, resulting in the formation of 1,2-dibromoethane.

**Reaction:**
\[ CH_2=CH_2 + Br_2 \rightarrow CH_2Br-CH_2Br \]

Thus, compound C is:
\[ C = CH_2Br-CH_2Br \] (1,2-dibromoethane)

### Summary of Compounds
- Compound A: \( CH_3CH_2Br \) (ethyl bromide)
- Compound B: \( CH_2=CH_2 \) (ethylene)
- Compound C: \( CH_2Br-CH_2Br \) (1,2-dibromoethane)