Q.Write the equations for the preparation of 1-iodobutane from

Concept: Markovnikov Addition (for the alkene route) — the hydrogen of HX adds to the carbon with more hydrogens, giving the more stable carbocation intermediate.

Step 1 — From 1-butanol (SN2 with HI)

$\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH} + \text{HI} \xrightarrow{\Delta} \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{I} + \text{H}_2\text{O}$

(Protonation of –OH, then iodide displaces water.)

Step 2 — From 1-chlorobutane (Finkelstein reaction)

$\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Cl} + \text{NaI} \xrightarrow{\text{acetone}} \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{I} + \text{NaCl}$

(Iodide is a better nucleophile; NaCl precipitates in acetone, driving equilibrium.)

Step 3 — From but-1-ene (anti-Markovnikov route)

Direct addition of HI would follow Markovnikov's rule — H to the terminal carbon, iodine to the internal one — giving 2-iodobutane, the wrong product. To reach 1-iodobutane, add HBr in the presence of peroxide (anti-Markovnikov, free-radical mechanism — the peroxide effect works for HBr only), then swap Br for I by the Finkelstein reaction:

$\text{CH}_3\text{CH}_2\text{CH}=\text{CH}_2 + \text{HBr} \xrightarrow{\text{peroxide}} \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Br}$ …

1. From 1-butanol ($CH_3CH_2CH_2CH_2OH$)

The hydroxyl is protonated by HI, water leaves, and iodide attacks — a straightforward $S_N2$ substitution on a primary alcohol.

2. From 1-chlorobutane ($CH_3CH_2CH_2CH_2Cl$)

A Finkelstein reaction: sodium iodide in dry acetone exchanges the chlorine for iodine, with the NaCl byproduct precipitating out and driving the exchange forward.

3. From but-1-ene ($CH_3CH_2CH{=}CH_2$)

Direct addition of HI to but-1-ene follows Markovnikov's rule regardless of peroxide, because the peroxide (Kharasch) anti-Markovnikov effect is specific to HBr only — the H–Cl bond is too strong to homolyse under these radical-chain conditions, and with HI the iodine atoms generated preferentially recombine to $I_2$ rather than add to the alkene, so there is no useful anti-Markovnikov pathway for HI at all. Direct HI addition to but-1-ene therefore just gives the Markovnikov product, 2-iodobutane — not the target 1-iodobutane.

The standard route to 1-iodobutane from but-1-ene is two steps, using the halogen the peroxide effect actually works for:

Step 1 — anti-Markovnikov addition of HBr (peroxide effect, genuinely works for HBr):

Step 2 — Finkelstein exchange to swap Br for I: …

Markovnikov Addition — Concept & Method

Method Name: Markovnikov’s Rule (for electrophilic addition to unsymmetrical alkenes)

Concept:

When H–X adds to an unsymmetrical alkene, the hydrogen atom attaches to the carbon with more hydrogen atoms already, and the halogen attaches to the carbon with fewer hydrogen atoms.

Steps to apply:

1. Identify the double bond in the unsymmetrical alkene.
2. Look at the two carbons of the double bond — count the number of H atoms on each.
3. The H⁺ (from H–X) goes to the carbon with more H atoms.
4. The X⁻ goes to the other carbon (more substituted carbon).

Preparation of 1-Iodobutane

(i) From 1-Butanol

Method: Nucleophilic substitution (SN2) using HI

Equation:

Key point:

• 1-Butanol is a primary alcohol → follows SN2 mechanism.
• HI is preferred over other HX because I⁻ is a good nucleophile.

(ii) From 1-Chlorobutane

Method: Finkelstein reaction (Halogen exchange, SN2)

Equation:

Key point:

• Acetone is used as solvent because NaCl is insoluble in acetone, driving the equilibrium forward.
• I⁻ replaces Cl⁻ via SN2.

(iii) From But-1-ene

Method: Anti-Markovnikov HBr addition (peroxide effect), then Finkelstein exchange — a 2-step route, because direct HI addition to but-1-ene follows Markovnikov's rule and gives the secondary iodide (2-iodobutane), not the primary one asked for here.

Step 1 — anti-Markovnikov addition of HBr (peroxide effect):

$\text{CH}_3\text{CH}_2\text{CH}=\text{CH}_2 + \text{HBr} \xrightarrow{(\text{C}_6\text{H}_5\text{CO})_2\text{O}_2} \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Br}$ …

Here are the common mistakes students make when tackling this exact question, along with how to avoid each.

Mistake 1: Using the Wrong Reagent for 1-Butanol → 1-Iodobutane

The Mistake: Students often try to use a simple substitution like $\text{NaI}$ directly with 1-butanol. They forget that the $\text{OH}$ group is a poor leaving group and needs to be converted first.

How to Avoid:

• Remember the principle: You cannot directly substitute an alcohol with $\text{NaI}$ because $\text{OH}^-$ is a strong base and a bad leaving group.
• The correct method: First convert the alcohol to an alkyl halide (like using $\text{P/I}_2$ or $\text{red P + I}_2$), or use a strong acid like $\text{HI}$ (which protonates the $\text{OH}$ to make $\text{H}_2\text{O}$, a good leaving group).
• Correct equation:

(Alternatively: $3\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH} + \text{PI}_3 \rightarrow 3\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{I} + \text{H}_3\text{PO}_3$)

Mistake 2: Forgetting the Finkelstein Reaction Conditions for 1-Chlorobutane

The Mistake: Students write the reaction as $\text{R-Cl} + \text{NaI} \rightarrow \text{R-I} + \text{NaCl}$ but forget that this is an equilibrium that needs to be driven forward.

How to Avoid:

• Know the principle: The Finkelstein reaction works because $\text{NaI}$ is soluble in acetone, but $\text{NaCl}$ (or $\text{NaBr}$) is insoluble in acetone. The precipitate of $\text{NaCl}$ pulls the equilibrium to the right.
• Common error: Writing the solvent as water or ethanol (which would dissolve both salts and stop the reaction).
• Correct equation:

Always mention "acetone" as the solvent and show $\text{NaCl}$ as a precipitate.

Mistake 3: Applying Markovnikov's Rule Backwards for But-1-ene

The Mistake: Students add $\text{HI}$ to but-1-ene and place the iodine on the more substituted carbon (C-2), giving 2-iodobutane instead of 1-iodobutane.

How to Avoid:

• Recall Markovnikov's Rule: The hydrogen adds to the carbon with more hydrogens (less substituted), and the halogen adds to the carbon with fewer hydrogens (more substituted).
• For but-1-ene: $\text{CH}_2=\text{CH}-\text{CH}_2\text{CH}_3$
  • C-1 has 2 H's (more substituted by H)
  • C-2 has 1 H (less substituted by H)
  • So $\text{H}^+$ goes to C-1, $\text{I}^-$ goes to C-2 → 2-iodobutane (wrong product!)
• To get 1-iodobutane, you must use anti-Markovnikov addition — but the peroxide (Kharasch) effect that reverses the regiochemistry only works reliably for HBr, never for HI directly (see Mistake 4). So the anti-Markovnikov step has to be carried out with HBr, not HI.

Mistake 4: Trying to Use HI Directly With Peroxide

The Mistake: Students write "$\text{CH}_2=\text{CH}-\text{CH}_2\text{CH}_3 + \text{HI} \xrightarrow{\text{peroxide}} \text{CH}_2\text{I}-\text{CH}_2-\text{CH}_2\text{CH}_3$" as if the peroxide (anti-Markovnikov) effect works with HI the same way it does with HBr. It doesn't — this equation does not represent a real reaction.

How to Avoid: …