Q.Alkyl halides are prepared from alcohols by treating with (Two or more than two options may be correct.)

The key idea here is Hindrance By Sulfuric Acid: when $H_2SO_4$ is mixed with KI, the acid oxidises $I^-$ to $I_2$, preventing clean formation of the alkyl iodide.

Reasoning:

1. Option (i) — $\mathrm{HCl + ZnCl_2}$ (Lucas reagent) works for all alcohols to give alkyl chlorides. Correct.
2. Option (ii) — Red P + $\mathrm{Br_2}$ generates $\mathrm{PBr_3}$ in situ, which converts alcohols to alkyl bromides cleanly. Correct. …

To understand why, we need to look at the chemistry of converting an alcohol ($\mathrm{ROH}$) into an alkyl halide ($\mathrm{RX}$). The hydroxyl group is a poor leaving group, so it must be activated—usually by protonation or conversion into a better leaving group—before a halide ion can attack.

Let’s examine each option step by step.

1. Option (i): $\mathrm{HCl + ZnCl_2}$

   This is the Lucas reagent. It works well for tertiary, secondary, and some primary alcohols. The $\mathrm{ZnCl_2}$ acts as a Lewis acid, helping to break the C–O bond after protonation. For example, with a tertiary alcohol:

   $\mathrm{R_3COH + HCl \xrightarrow{ZnCl_2} R_3CCl + H_2O}$

   The reaction proceeds via an $\mathrm{S_N1}$ mechanism for tertiary alcohols, and $\mathrm{S_N2}$ for primary ones (though primary alcohols react slowly). So this is a valid method.

2. Option (ii): Red P + $\mathrm{Br_2}$

   Red phosphorus reacts with bromine to form phosphorus tribromide ($\mathrm{PBr_3}$) in situ:

   $\mathrm{2P + 3Br_2 \rightarrow 2PBr_3}$

   Then $\mathrm{PBr_3}$ reacts with the alcohol:

   $\mathrm{3ROH + PBr_3 \rightarrow 3RBr + H_3PO_3}$

   This is a classic and reliable method for converting alcohols to alkyl bromides. It works well for primary and secondary alcohols. So this is also correct.

3. Option (iii): $\mathrm{H_2SO_4 + KI}$

   Here’s the trap. Sulfuric acid is a strong acid, and potassium iodide provides iodide ions. But $\mathrm{H_2SO_4}$ is also a strong oxidizing agent. Iodide ions ($\mathrm{I^-}$) are easily oxidized to iodine ($\mathrm{I_2}$):

   $\mathrm{2I^- + 2H^+ + H_2SO_4 \rightarrow I_2 + SO_2 + 2H_2O}$ …

Concept: Conversion of Alcohols to Alkyl Halides

This question tests your understanding of nucleophilic substitution reactions where the hydroxyl group ($-OH$) of an alcohol is replaced by a halogen atom ($X = Cl, Br, I$).

Method: Reagent-Specific Analysis

Step 1: Identify the role of each reagent

Step 2: Check each option

• (i) HCl + ZnCl₂ — Correct only for 3° alcohols (but the question doesn't specify alcohol type; generally considered valid for preparation).
• (ii) Red P + Br₂ — Correct for all alcohols. …

Here are the common mistakes students make on this question, along with the correct reasoning and how to avoid each.

The Core Concept

This question tests your knowledge of methods to convert alcohols ($R-OH$) into alkyl halides ($R-X$). The key reagents are:

• Lucas reagent ($HCl + ZnCl_2$) — for $R-Cl$.
• Red phosphorus + halogen ($P/Br_2$ or $P/I_2$) — for $R-Br$ or $R-I$.
• $H_2SO_4 + KI$ — does not give alkyl iodide; it gives iodine ($I_2$) instead.

Mistake 1: Thinking $H_2SO_4 + KI$ works like $HCl$ or $HBr$

The error: Students assume that since $HCl$ reacts with alcohols, $H_2SO_4 + KI$ will also produce alkyl iodide ($R-I$).

Why it’s wrong:

$H_2SO_4$ is a strong oxidising agent. It oxidises $I^-$ (from $KI$) to $I_2$, not to $HI$. The reaction is:

No $HI$ is available to substitute the $OH$ group.

How to avoid:

Remember the redox hierarchy:

• $HCl$ and $HBr$ are not oxidised by $H_2SO_4$ (they can be used directly).
• $HI$ is easily oxidised by $H_2SO_4$ to $I_2$. So never use $H_2SO_4 + KI$ to make alkyl iodides.
• The correct method for $R-I$ is Red P + $I_2$ (option ii with $Br_2$ is for bromides; for iodides, use $I_2$).

Mistake 2: Selecting “All the above” without checking each option

The error: Students see two correct options (i and ii) and assume the third must also be correct, especially when the question says “Two or more may be correct.”

Why it's wrong:

Option iii is chemically incorrect. The question explicitly allows multiple correct answers, but “All the above” is only correct if every individual option is valid.

How to avoid:

• Test each option independently.
  • (i) $HCl + ZnCl_2$ → Lucas reagent → works for $R-Cl$ (especially for tertiary alcohols).
  • (ii) Red P + $Br_2$ → generates $PBr_3$ in situ → works for $R-Br$.
  • (iii) $H_2SO_4 + KI$ → fails (gives $I_2$, not $R-I$).
• Only choose “All the above” if you are 100% sure every sub-option is correct.

Mistake 3: Confusing Red P + $Br_2$ with $PBr_3$ directly

The error: Students think Red P + $Br_2$ is a separate, less effective method, or they forget that it generates $PBr_3$ in situ.

Why it’s wrong:

Red phosphorus reacts with bromine to form phosphorus tribromide ($PBr_3$), which then converts the alcohol to alkyl bromide:

This is a standard, reliable method.

How to avoid: …