Identify Z in the series
`CH_(3)CH_(2)CH_(2)OHunderset(160-180^(@)C)overset(conc.H_(2)SO_(4))rarr(X)overset(Br_(2))rarr(Y)underset(NaNH_(2)//Delta)overset(alc,KOH//Delta)rarr(Z)`

Options

A. `CH_(3)-underset(underset(OH)|)(CH)-underset(underset(OH)|)(CH_(2))`
B. `CH_(3)CH_(2)CH(OH)_(2)`
C. `CH_(3)-underset(underset(OH)|)(C)=CH_(2)`
D. `CH_(3)-C=CH`

Explanation

To identify the compound Z in the given reaction series, we will analyze each step of the transformation starting from propanol (CH₃CH₂CH₂OH).

### Step 1: Dehydration of Propanol
The first step involves heating propanol (CH₃CH₂CH₂OH) with concentrated sulfuric acid (H₂SO₄) at 160-180°C. This process leads to dehydration, where a water molecule is removed, resulting in the formation of an alkene.

**Product after Step 1:**
- The dehydration of propanol yields propene (CH₃CH=CH₂).

### Step 2: Electrophilic Addition with Br₂
Next, the alkene (propene) is treated with bromine (Br₂). This reaction is an electrophilic addition reaction, where bromine adds across the double bond.

**Product after Step 2:**
- The product of this reaction is 1,2-dibromopropane (CH₃CHBrCH₂Br).

### Step 3: Dehydrohalogenation with Alcoholic KOH
The 1,2-dibromopropane is then treated with alcoholic potassium hydroxide (alc. KOH) under heating conditions. This step involves dehydrohalogenation, where hydrogen and bromine are removed, resulting in the formation of another alkene.

**Product after Step 3:**
- The product formed is 1-bromo-1-propene (CH₃C=CHBr).

### Step 4: Reaction with NaNH₂
Finally, the compound (1-bromo-1-propene) is treated with sodium amide (NaNH₂), which is a strong base. This step leads to the elimination of the bromine atom and the formation of a triple bond.

**Product after Step 4:**
- The final product is propyne (CH≡C-CH₃).

### Final Answer
Thus, the compound Z in the series is propyne (CH≡C-CH₃).

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