Q.How can you obtain iodoethane from ethanol when no other iodine containing reagent except NaI is available in the laboratory?

Concept: Make a good leaving group first, then let iodide swap in (Finkelstein) — ethanol's –OH cannot be displaced by I⁻ directly, and conc. H₂SO₄ must never be used to liberate HI from NaI (it oxidises the iodide to I₂ — the same doctrine as "why H₂SO₄ is not used with KI to make alkyl iodides").

Reasoning (the NCERT route):

1. Convert ethanol to chloroethane with HCl gas in the presence of anhydrous $ZnCl_2$: $C_2H_5OH + HCl \xrightarrow{ZnCl_2} C_2H_5Cl + H_2O$.
2. Heat the chloroethane with NaI in dry acetone (Finkelstein reaction): $C_2H_5Cl + NaI \xrightarrow{\text{dry acetone}} C_2H_5I + NaCl\downarrow$ — NaCl is insoluble in acetone and precipitates, driving the halide exchange forward. …

You have ethanol ($\text{CH}_3\text{CH}_2\text{OH}$) and only NaI as an iodine source. Two dead ends frame the problem:

• Directly mixing ethanol with NaI does nothing — hydroxide is a terrible leaving group, so I⁻ cannot displace the –OH.
• Liberating HI from NaI with conc. H₂SO₄ fails — concentrated sulphuric acid is an oxidising acid: it oxidises I⁻ (and any HI formed) to I₂, destroying the very nucleophile you need. This is the same doctrine this chapter teaches for why H₂SO₄ is never paired with KI in making alkyl iodides.

So the working strategy is: first give the carbon a good leaving group using reagents that don't involve iodine, then let iodide swap in.

Here's the step-by-step route:

1. Convert ethanol to chloroethane. Pass dry HCl gas into ethanol in the presence of anhydrous $\text{ZnCl}_2$ (the Lucas-type reagent). The Lewis-acidic ZnCl₂ coordinates to the –OH oxygen, converting it into a good leaving group so that chloride can substitute:

2. Exchange Cl for I — the Finkelstein reaction. Reflux the chloroethane with NaI in dry acetone:

This halide exchange works beautifully for two reasons: iodide is an excellent nucleophile in a polar aprotic solvent like acetone (it isn't smothered by hydrogen bonding), and — the real driving force — NaCl is insoluble in acetone, so it precipitates out and pulls the equilibrium continuously toward iodoethane.

3. Isolate the product. Iodoethane is a dense liquid (b.p. ≈ 72 °C) and is distilled from the mixture. …

Concept: Nucleophilic Substitution — Converting a Poor Leaving Group into a Good One, then a Finkelstein Halide Exchange

Method: Alcohol → alkyl chloride (HCl/anhydrous ZnCl₂), then chloride → iodide with NaI in dry acetone (Finkelstein reaction) — the route the NCERT Exemplar's own hint gives:

Why this works (the "why" before the "how")

Ethanol's $-\text{OH}$ is a poor leaving group — I⁻ cannot displace it directly. And the tempting shortcut of liberating HI from NaI with conc. $\text{H}_2\text{SO}_4$ fails: concentrated sulphuric acid oxidises I⁻/HI to I₂ (this is exactly why this chapter teaches that H₂SO₄ is not used with KI for making alkyl iodides). So we first install a halogen that is a good leaving group (–Cl), then exchange it for iodine.

Step-by-step method

1. Step 1 — Ethanol to chloroethane Pass dry HCl gas into ethanol in the presence of anhydrous $\text{ZnCl}_2$ (Lewis-acid catalyst; this HCl/ZnCl₂ mixture is the Lucas-type reagent):

The ZnCl₂ coordinates to the –OH oxygen, turning it into a good leaving group so chloride can substitute.

2. Step 2 — Finkelstein exchange (Cl → I) Reflux the chloroethane with NaI in dry acetone:

The driving force: NaCl is insoluble in acetone and precipitates out, pulling the equilibrium toward iodoethane (I⁻ is also an excellent nucleophile in this aprotic solvent).

3. Step 3 — Isolation …

Common Mistakes: Converting Ethanol to Iodoethane Using NaI

This question tests your understanding of nucleophilic substitution and the limitations of NaI as an iodine source. The route to have in mind throughout (it is the NCERT Exemplar's own hint) is:

Mistake 1: Trying a Direct SN2 Reaction

The error:

Students write:

$C_2H_5OH + NaI \rightarrow C_2H_5I + NaOH$

Why it's wrong:

• $\ce{OH-}$ is a poor leaving group (a strong base — the conjugate acid has pKa ≈ 16).
• $\ce{I-}$ cannot displace it under any normal lab conditions.

How to avoid:

Remember: the –OH must first be converted into a good leaving group — here, into –Cl using HCl/anhydrous ZnCl₂ (or into $-\overset{+}{O}H_2$ by in-situ protonation with HI from a non-oxidising acid). Only then can iodide take its place.

Mistake 2: Proposing HI from NaI + Conc. H₂SO₄

The error:

NaI + conc. H₂SO₄ → HI (then add to ethanol)

Why it is wrong:

• Conc. H₂SO₄ is an oxidising acid: it oxidises I⁻ (and the HI formed) to I₂ instead of cleanly liberating HI.
• You get purple iodine vapour contaminating the mixture, not a working iodinating agent — this is exactly the same doctrine as "why is sulphuric acid not used during the reaction of alcohols with KI."

How to avoid:

Keep iodide away from oxidising acids entirely. Either follow the main route (make chloroethane first, then Finkelstein), or — if generating HI in situ — use the non-oxidising ~95% H₃PO₄:

NaI + H₃PO₄ → HI + NaH₂PO₄, then C₂H₅OH + HI → C₂H₅I + H₂O.

Mistake 3: Getting the Finkelstein Conditions Wrong

The error:

Writing:

$C_2H_5Cl + NaI \rightarrow C_2H_5I + NaCl$ (in water)

Why it's wrong:

• In water, $\ce{I-}$ is heavily solvated and much less nucleophilic.
• More importantly, the equilibrium is only pulled forward in acetone, where the NaCl byproduct is insoluble and precipitates out.

How to avoid:

Always specify dry acetone (and reflux) for the Finkelstein exchange:

$C_2H_5Cl + NaI \xrightarrow[\text{reflux}]{\text{dry acetone}} C_2H_5I + NaCl \downarrow$

Mistake 4: Skipping the First Step's Conditions

The error:

Writing "ethanol + HCl → chloroethane" with no catalyst.

Why it is wrong:

HCl alone reacts sluggishly with a primary alcohol — the –OH is not a good enough leaving group. The anhydrous ZnCl₂ (Lewis acid) is what activates the C–O bond by coordinating to the oxygen.

How to avoid: …