Q.Why is sulphuric acid not used during the reaction of alcohols with KI?

Concept: Reagent Incompatibility — $H_2SO_4$ is an oxidising acid. The conversion of an alcohol to an alkyl iodide needs the acid to generate HI from KI (and to activate the $-OH$ group) — so the acid must not destroy the iodide it liberates.

Reasoning:

1. In the reaction mixture, the acid first converts KI to the corresponding halogen acid:

2. But HI is a strong reducing agent, and concentrated $H_2SO_4$ is a strong oxidising agent — so the HI formed is immediately oxidised to iodine instead of reacting with the alcohol:

3. With the iodide destroyed as $I_2$ (violet vapours), no alkyl iodide can form. A non-oxidising acid such as phosphoric acid ($H_3PO_4$) is used instead.

The Core Idea: Why the Acid Matters

When you want to convert an alcohol ($\ce{ROH}$) into an alkyl iodide ($\ce{RI}$) using potassium iodide ($\ce{KI}$), the acid has two jobs: it liberates HI from the ionic salt KI, and it protonates the alcohol's hydroxyl group. Protonation turns the poor leaving group ($\ce{-OH}$) into a good one ($\ce{-OH2+}$), which can then be displaced by iodide.

The problem is that HI (and iodide in general) is a strong reducing agent — it is very easily oxidised. If the acid you add is itself a strong oxidising agent, like concentrated sulphuric acid, it destroys the very reagent it has just produced.

Step-by-Step Breakdown

1. The intended reaction (with a non-oxidising acid)

The ideal pathway is:

This works beautifully with a non-oxidising acid like phosphoric acid ($\ce{H3PO4}$). The iodide survives and acts as the nucleophile.

2. What happens with $\ce{H2SO4}$ — two steps, the second one fatal

Step (a): acid–base. Sulphuric acid first converts KI into the corresponding halogen acid:

Step (b): redox. Concentrated sulphuric acid is a powerful oxidising agent, and HI is a strong reducing agent. The HI formed in step (a) is immediately oxidised. The half-reactions are:

• Oxidation: $\ce{2I- -> I2 + 2e-}$
• Reduction: $\ce{H2SO4 + 2H+ + 2e- -> SO2 + 2H2O}$

Overall:

You get violet iodine vapour ($\ce{I2}$) and the choking gas sulphur dioxide ($\ce{SO2}$). The iodide is gone, so no substitution can occur.

3. The correct choice: Phosphoric acid ($\ce{H3PO4}$)

Phosphoric acid is a strong enough acid to generate HI from KI and to protonate the alcohol, but it is not an oxidising agent — it cannot accept electrons from HI. The iodide therefore stays intact and successfully performs the nucleophilic substitution.

The Final Answer

Method: Reagent Incompatibility Analysis (Hindrance by Sulfuric Acid)

Concept Explanation

The reaction of alcohols with KI (potassium iodide) is meant to produce alkyl iodides. However, sulfuric acid ($H_2SO_4$) cannot be used as the acid catalyst here because it creates a dangerous side reaction.

Steps of the Method

1. Identify the intended reaction In the acidic mixture, KI first gives HI, and the HI then converts the alcohol to the alkyl iodide (shown here with the non-oxidising acid that actually works):

2. Recognize the role of sulfuric acid

   $H_2SO_4$ is a strong acid that provides $H^+$ ions to protonate the alcohol, making it a better leaving group.

3. Check for redox incompatibility

   Sulfuric acid ($H_2SO_4$) is a strong oxidizing agent. Potassium iodide (KI) contains $\text{I}^-$ ions, which are easily oxidized.

4. Write the side reaction

   With $H_2SO_4$, the HI formed in the first (acid–base) step is immediately oxidised to iodine ($I_2$) instead of participating in the substitution:

(The second step is a redox reaction — $H_2SO_4$ is reduced to $SO_2$, and the iodide in HI is oxidised to $I_2$)

5. Observe the consequence
   • No alkyl iodide is formed.
   • Instead, violet iodine vapors appear (a visible sign of the side reaction).
   • The reaction mixture turns brown/black due to iodine.

Final Answer

Sulfuric acid is not used because it first converts KI to HI and then oxidises that HI to iodine ($I_2$) instead of allowing the substitution reaction to proceed. The correct acid to use is phosphoric acid ($H_3PO_4$), which is non-oxidizing.

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

The Core Concept (Why You Need This)

The question tests your understanding of competing reactions. You are not just asked to state a fact; you are asked to explain why a common reagent (sulfuric acid) fails in a specific context.

The reaction in question is the preparation of alkyl iodides (R–I) from alcohols:

The problem is that sulfuric acid is a strong oxidizing agent, and hydrogen iodide (HI) is a strong reducing agent. They react with each other before HI can react with the alcohol.

Mistake #1: Saying "Sulfuric Acid Decomposes KI"

The Mistake: Students often write: "H₂SO₄ decomposes KI to give I₂." This is incomplete and misses the key chemical step.

Why it's wrong: Sulfuric acid does not directly attack KI. The actual sequence is:

1. Step 1 (Acid-Base): H₂SO₄ reacts with KI to produce HI (hydrogen iodide).

2. Step 2 (Redox): The freshly formed HI is then oxidized by the concentrated H₂SO₄.

How to Avoid: Always write the two-step mechanism in your answer. Show that HI is first generated, then destroyed. This proves you understand the sequence of reactions, not just a vague "decomposition."

Mistake #2: Confusing the Oxidizing Agent

The Mistake: Students say: "H₂SO₄ oxidizes KI to I₂."

Why it's wrong: H₂SO₄ does not oxidize KI directly. It oxidizes HI (which is a much stronger reducing agent than I⁻). The oxidation potential of HI is higher than that of I⁻.

How to Avoid: Memorize the reducing power order: $\text{HI} > \text{HBr} > \text{HCl}$. HI is so strong a reducing agent that even concentrated H₂SO₄ can oxidize it. In your answer, explicitly state: "The HI produced is a strong reducing agent and is oxidized by H₂SO₄."

Mistake #3: Forgetting the Product (I₂) is a Side-Reaction

The Mistake: Students stop at "I₂ is formed" and don't explain why this is a problem for the desired reaction.

Why it's wrong: The question asks why H₂SO₄ is not used. The answer must connect the side reaction to the failure to form the alkyl iodide.

How to Avoid: Complete the logic chain:

• Goal: We want HI to react with the alcohol ($R\text{–OH} + \text{HI} \rightarrow R\text{–I} + H_2O$).
• Problem: The HI is consumed by H₂SO₄ before it can reach the alcohol.
• Result: No alkyl iodide is formed; instead, you get iodine (I₂) and sulfur dioxide (SO₂).

Correct final statement: "Therefore, the desired product (alkyl iodide) is not obtained, and the reaction mixture becomes contaminated with iodine."

Mistake #4: Not Mentioning the Alternative Reagent

The Mistake: Students answer the "why" but don't suggest the correct method.

Why it's wrong: In exams, this question is often followed by or paired with: "What should be used instead?" Not mentioning the alternative shows incomplete understanding.

How to Avoid: Always add the correct alternative at the end of your answer:

• Use phosphoric acid ($H_3PO_4$) instead of H₂SO₄.
• Why? H₃PO₄ is a non-oxidizing acid. It provides the acidic medium to generate HI but does not oxidize it.

Summary Table: Mistakes vs. Fixes

Final Tip for Exams: Write your answer in this order:

1. Goal: To make R–I from R–OH + KI.
2. Problem: H₂SO₄ reacts with KI to form HI, but then oxidizes HI to I₂.
3. Result: No R–I is formed.
4. Solution: Use H₃PO₄ (non-oxidizing).