A compound X declourises `Br_(2)` water and reacts slowly with conc `H_(2)SO_(4)` to give an addition product. X reacts with HBr to form Y. Y reacts with NaOH to form Z. on oxidation Z gives hexan-3-one. X, Y and Z in the reactions are
`underset("Addition product") underset(darr "conc. "H_(2)SO_(4))(X) overset(Br_(2)" water")to"Decolourisation"`
`X overset(HBr) to Y overset(NaOH) to Z overset([O])to CH_(3)CH_(2)underset("Hexan-3-one")(CH_(2)-underset(O)underset(||)(C)-CH_(2)CH_(3))`

Options

A. `X=CH_(3)CH_(2)CH=CHCH_(3),Y=CH_(3)CH_(2)CH(Br)CH(Br)CH_(2)CH_(3),Z=CH_(3)CH_(2)CH_(3)`
B. `X=CH_(3)CH=CHCH_(3),Y=CH_(3)CH(Br)CH(Br)CH_(3),Z=CH_(3)CH_(2)CH_(2)OH`
C. `X=CH_(3)CH_(2)CH=CHCH_(2)CH_(3),Y=CH_(3)CH_(2)-underset(Br)underset(|)(C)H-CH_(2)CH_(2)CH_(3),Z=CH_(3)CH_(2)CH_(2)-underset(OH)underset(|)(C)H-CH_(2)CH_(3)`
D. `X=CH_(3)CH_(2)CH_(2)CH=CHCH_(3),Y=CH_(3)CH_(2)CH_(2)CH_(2)CH_(2)CH_(2)Br, Z=CH_(3)CH_(2)CH_(2)CH_(2)OH`

Explanation

To solve the problem step by step, let's analyze the information given in the question and the reactions involved.

### Step 1: Identify Compound X
The question states that compound X decolorizes bromine water, indicating that it is an unsaturated compound (most likely an alkene).

**Hint:** Unsaturated compounds (like alkenes) can react with bromine, leading to decolorization.

### Step 2: Reaction of X with HBr
When compound X (an alkene) reacts with HBr, it undergoes an addition reaction according to Markovnikov's rule. This means that the hydrogen atom from HBr will attach to the carbon with the most hydrogen atoms, while the bromine will attach to the other carbon.

**Hint:** Remember Markovnikov's rule: in the addition of HX to alkenes, the hydrogen atom attaches to the carbon with more hydrogen atoms.

### Step 3: Identify Compound Y
The product of the reaction of X with HBr is compound Y, which is a bromoalkane.

**Hint:** The product formed from the addition of HBr to an alkene is a haloalkane.

### Step 4: Reaction of Y with NaOH
Compound Y then reacts with NaOH. This is a nucleophilic substitution reaction where the bromine atom is replaced by a hydroxyl group (OH). The product formed is compound Z, which is an alcohol.

**Hint:** Nucleophilic substitution reactions involve the replacement of a leaving group (like Br) with a nucleophile (like OH).

### Step 5: Identify Compound Z
The product Z is a secondary alcohol because it is formed from a bromoalkane where the carbon attached to the OH group is bonded to two other carbons.

**Hint:** Secondary alcohols can be oxidized to form ketones.

### Step 6: Oxidation of Z
When compound Z is oxidized, it produces a ketone. The question states that Z gives hexan-3-one upon oxidation.

**Hint:** The structure of Z should allow for oxidation to yield a ketone, specifically hexan-3-one.

### Step 7: Determine the Structure of Compounds
Now, let's summarize the structures:
- **X**: An alkene (C6H12) that can decolorize bromine water.
- **Y**: A bromoalkane formed from X.
- **Z**: A secondary alcohol that can be oxidized to hexan-3-one.

### Step 8: Identify the Correct Option
From the analysis, we can conclude that:
- **X** is 1-hexene (CH3CH2CH=CHCH2CH3).
- **Y** is 2-bromohexane (CH3CH2CHBrCH2CH2CH3).
- **Z** is 3-hexanol (CH3CH2C(OH)CH2CH2CH3).
- Upon oxidation, Z gives hexan-3-one (CH3CH2C(=O)CH2CH2CH3).

After checking the options provided in the question, option number 3 matches all the criteria.

### Final Answer
The correct option is **Option 3**.

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