Which of the following reaction(s) can be used for the preparation of alkyl halides ?
`(i) CH_(3) CH_(2)OH+HCl overset("anhyd."ZnCl_(2))(to)`
`(ii) CH_(3)CH_(2)OH+HCl to `
`(iii) (CH_(3))_(3)C-OH +HCl to `
`(iv) (CH_(3))_(2) CHOH +HCl overset("anhyd"ZnCl_(2))(to)`

Options

A. (a) (IV) only
B. (b) (III) and (IV) only
C. (c) (I), (III) and (IV) only
D. (d) (I) and (II) only

Explanation

To determine which reactions can be used for the preparation of alkyl halides, we will analyze each reaction step by step.

### Step 1: Analyze Reaction (i)
**Reaction:** \( CH_3CH_2OH + HCl \overset{\text{anhyd. } ZnCl_2}{\rightarrow} \)

- **Type of Alcohol:** Ethanol (a primary alcohol).
- **Catalyst:** Anhydrous zinc chloride (ZnCl₂) is used.
- **Mechanism:** The presence of ZnCl₂ facilitates the conversion of the alcohol to an alkyl halide via nucleophilic substitution.
- **Product:** The hydroxyl group (OH) is replaced by a chloride ion (Cl), resulting in the formation of ethyl chloride (\( CH_3CH_2Cl \)).
- **Conclusion:** Alkyl halide can be prepared.

### Step 2: Analyze Reaction (ii)
**Reaction:** \( CH_3CH_2OH + HCl \)

- **Type of Alcohol:** Ethanol (a primary alcohol).
- **Catalyst:** No catalyst is used.
- **Mechanism:** Without a catalyst, the reaction does not proceed effectively. The alcohol does not react with HCl to form an alkyl halide.
- **Product:** No reaction occurs.
- **Conclusion:** Alkyl halide cannot be prepared.

### Step 3: Analyze Reaction (iii)
**Reaction:** \( (CH_3)_3COH + HCl \)

- **Type of Alcohol:** Tertiary alcohol (tert-butanol).
- **Catalyst:** No catalyst is used.
- **Mechanism:** Tertiary alcohols are highly reactive and can undergo nucleophilic substitution via the SN1 mechanism. The hydroxyl group is replaced by a chloride ion.
- **Product:** The product formed is tert-butyl chloride (\( (CH_3)_3CCl \)).
- **Conclusion:** Alkyl halide can be prepared.

### Step 4: Analyze Reaction (iv)
**Reaction:** \( (CH_3)_2CHOH + HCl \overset{\text{anhyd. } ZnCl_2}{\rightarrow} \)

- **Type of Alcohol:** Isopropanol (a secondary alcohol).
- **Catalyst:** Anhydrous zinc chloride (ZnCl₂) is used.
- **Mechanism:** The presence of ZnCl₂ allows the reaction to proceed, facilitating the substitution of the hydroxyl group with a chloride ion.
- **Product:** The product formed is isopropyl chloride (\( (CH_3)_2CHCl \)).
- **Conclusion:** Alkyl halide can be prepared.

### Final Conclusion
From the analysis:
- **Reactions that can prepare alkyl halides:** (i), (iii), and (iv).
- **Reaction that cannot prepare alkyl halides:** (ii).

Thus, the correct answer is **Option C: first, third, and fourth only.**

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