What is Z in the following sequence of reactions?
`Z overset(PCl_(3))toX overset(alc.KOH)to Y underset((ii)H_(2)O boil)overset((i)Conc.H_(2)SO_(4))toZ`

Options

A. `H_(2)C=CH-CH_(2)-OH`
B. `CH_(3)CHOHCH_(3)`
C. `(CH_(3)CH_(2))COOH`
D. `CH_(3)CH=CH_(2)`

Answer

To determine what Z is in the given sequence of reactions, let's analyze each step systematically.

### Step 1: Identify the starting compound Z
We need to identify the starting compound Z, which is transformed through a series of reactions and eventually returns to the same compound Z.

### Step 2: Reaction with PCl3
The first reaction involves Z being treated with PCl3. This suggests that Z is likely an alcohol since PCl3 is commonly used to convert alcohols into alkyl chlorides.

Let’s assume Z is isopropyl alcohol (CH3-CHOH-CH3). When isopropyl alcohol reacts with PCl3, it forms isopropyl chloride (CH3-CHCl-CH3).

### Step 3: Reaction with alcoholic KOH
Next, the product (isopropyl chloride) undergoes a reaction with alcoholic KOH. This reaction typically leads to an elimination reaction, where a hydrogen atom and a chlorine atom are removed, resulting in the formation of an alkene.

For isopropyl chloride, the elimination reaction would yield propene (CH3-CH=CH2).

### Step 4: Reaction with concentrated H2SO4
The propene then reacts with concentrated H2SO4. In this step, the alkene undergoes protonation to form a carbocation. The more stable secondary carbocation (CH3-CH+-CH3) is formed.

### Step 5: Reaction with water
After the formation of the carbocation, water acts as a nucleophile and attacks the positively charged carbon, leading to the formation of an oxonium ion (CH3-CHOH-CH3+).

### Step 6: Deprotonation
Finally, deprotonation occurs, where the oxonium ion loses a proton to regenerate isopropyl alcohol (Z).

### Conclusion
After analyzing the sequence of reactions, we can conclude that the compound Z is isopropyl alcohol (CH3-CHOH-CH3).

### Final Answer
Z = Isopropyl alcohol (CH3-CHOH-CH3)

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