Identify (Z) in the series.
`C_(3)H_(7)OH underset(443K)overset("conc."H_(2)SO_(4))(rarr)(X)overset(Br_(2))(rarr)(Y) underset("alc. KOH")overset("Excess of ")(rarr)(Z)`

Options

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

Answer

To identify compound Z in the given reaction series, we will break down the steps involved in the transformations of the starting material, C₃H₇OH (which is propanol), through to the final product.

### Step 1: Identify the starting material
The starting compound is C₃H₇OH, which is propanol (specifically, 1-propanol or n-propanol).

### Step 2: Reaction with concentrated H₂SO₄ at 443 K
When propanol reacts with concentrated sulfuric acid at elevated temperatures (443 K), it undergoes dehydration to form an alkene. The dehydration of propanol leads to the formation of propene (C₃H₆), which is the compound X.

**Reaction:**
\[ \text{C}_3\text{H}_7\text{OH} \xrightarrow{\text{conc. H}_2\text{SO}_4, 443 \text{K}} \text{C}_3\text{H}_6 \text{ (propene)} \]

### Step 3: Reaction with Br₂
Next, propene (X) reacts with bromine (Br₂) in a non-polar solvent like carbon tetrachloride (CCl₄). This is an electrophilic addition reaction where bromine adds across the double bond of propene, resulting in the formation of 1,2-dibromopropane (Y).

**Reaction:**
\[ \text{C}_3\text{H}_6 + \text{Br}_2 \xrightarrow{\text{CCl}_4} \text{C}_3\text{H}_6\text{Br}_2 \text{ (1,2-dibromopropane)} \]

### Step 4: Reaction with excess alcoholic KOH
The final step involves the reaction of 1,2-dibromopropane (Y) with excess alcoholic KOH. In this case, KOH acts as a strong base, promoting elimination reactions. The dibromopropane will undergo dehydrohalogenation to form an alkyne. Specifically, it will eliminate both bromine atoms and form propyne (Z).

**Reaction:**
\[ \text{C}_3\text{H}_6\text{Br}_2 + \text{alc. KOH} \rightarrow \text{C}_3\text{H}_4 \text{ (propyne)} \]

### Conclusion
Thus, the final compound Z is propyne, which can be represented as:
\[ \text{Z} = \text{C}_3\text{H}_4 \text{ (propyne)} \]

### Summary of the Steps:
1. **Starting Material:** C₃H₇OH (propanol)
2. **Step 1:** Dehydration to form propene (C₃H₆)
3. **Step 2:** Electrophilic addition with Br₂ to form 1,2-dibromopropane (C₃H₆Br₂)
4. **Step 3:** Dehydrohalogenation with alcoholic KOH to form propyne (C₃H₄)