Write the equation of the reaction of hydrogen iodide with: (i) 1-propoxypropane (ii) methoxybenzene and (iii)

Exercises · 7.28

Q. Write the equation of the reaction of hydrogen iodide with:

(i) 1-propoxypropane

(ii) methoxybenzene and

(iii) benzyl ethyl ether.

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Concept understanding — Williamson Ether Synthesis

Williamson Ether Synthesis: From Intuition to Mechanism

Imagine you want to build a simple bridge between two carbon chains — an oxygen atom linking them together. That bridge is an ether ( $R-O-R'$ ). The Williamson ether synthesis is the most reliable way to build that bridge in a lab.

The Core Idea

You have two pieces: an alkoxide ion ( $RO^-$ ) and an alkyl halide ( $R'X$ ). The alkoxide is a strong nucleophile — it loves positive charge. The alkyl halide has a carbon attached to a halogen (like Cl, Br, I) that is slightly positive because the halogen pulls electrons away.

When you mix them, the alkoxide attacks that slightly positive carbon, kicks out the halide ion, and forms a new $C-O$ bond. The result? An ether.

R-O^- + R'-X \longrightarrow R-O-R' + X^-

That's the entire reaction in one line. But the devil is in the details — especially which alkyl halide you choose.

The Mechanism (SN2)

This is a classic SN2 reaction — one step, no intermediates. The alkoxide approaches the carbon from the opposite side of the halogen. As the $C-O$ bond forms, the $C-X$ bond breaks. The halide leaves as a stable anion.

Because it's SN2, the reaction is sensitive to steric hindrance. The carbon being attacked must be accessible.

Watch out

If the alkyl halide is tertiary (3°), the reaction will not work via SN2. The bulky carbon blocks the backside attack. Instead, the alkoxide will act as a base and cause elimination (forming an alkene). You'll get no ether.

The Practical Rule

Alkyl halide	Works?	Why
Methyl ( $CH_3X$ )	Yes	Least hindered, fastest SN2
Primary (1°)	Yes	Clean SN2
Secondary (2°)	Sometimes	Works if not too bulky; elimination competes
Tertiary (3°)	No	Elimination dominates
Aryl (e.g., bromobenzene)	No	SN2 impossible on $sp^2$ carbon

Tip

To make an ether like $R-O-R'$ , always use the less hindered alkyl halide and the more hindered alkoxide. For example, to make $CH_3CH_2-O-CH(CH_3)_2$ , use $CH_3CH_2O^-$ (primary alkoxide) + $(CH_3)_2CHBr$ (secondary halide) — not the other way around.

How to Choose the Alkoxide

You can't just buy alkoxide ions in a bottle. You make them by reacting an alcohol with a strong base like sodium hydride ( $NaH$ ) or sodium metal.

ROH + NaH \longrightarrow RO^-Na^+ + H_2

The alkoxide is then used immediately with the alkyl halide.

A Common Exam Trap …

Why this formula?

Williamson Ether Synthesis: Why the Key Principles Hold

The Williamson Ether Synthesis is a classic method to prepare ethers. The core reaction is:

\text{R-O}^- + \text{R'-X} \rightarrow \text{R-O-R'} + \text{X}^-

Where:

$\text{R-O}^-$ is an alkoxide ion (strong nucleophile)
$\text{R'-X}$ is an alkyl halide (electrophile)
$\text{X}^-$ is a halide ion (leaving group)

Let's break down why this works — the reasoning behind the key principles.

1. Why an Alkoxide (Not an Alcohol) is Needed

The Problem with Alcohols

Alcohols ( $\text{R-OH}$ ) are weak nucleophiles. The oxygen has a partial negative charge, but the O–H bond is strong. If you mix an alcohol with an alkyl halide, the reaction is extremely slow or doesn't happen at all.

The Solution: Deprotonation

By treating the alcohol with a strong base (like $\text{NaH}$ , $\text{Na}$ , or $\text{KOH}$ ), you remove the proton:

\text{R-OH} + \text{NaH} \rightarrow \text{R-O}^- \text{Na}^+ + \text{H}_2

The alkoxide ion ( $\text{R-O}^-$ ) has a full negative charge on oxygen. This makes it:

A much stronger nucleophile (higher electron density)
More reactive toward the electrophilic carbon in the alkyl halide

Key takeaway: The alkoxide's full negative charge is what drives the reaction — it's not just about having oxygen, but about having a charged, electron-rich oxygen.

2. Why the Alkyl Halide Must Be Primary (or Methyl)

The Mechanism: $\text{S}_\text{N}2$ is the Only Path

The Williamson synthesis proceeds exclusively via an $\text{S}_\text{N}2$ mechanism (bimolecular nucleophilic substitution). This means:

The nucleophile attacks the carbon from the backside
The leaving group departs from the opposite side
The reaction is concerted (one step, no intermediates)

Why Primary Halides Work Best

In $\text{S}_\text{N}2$ reactions, the rate depends on steric hindrance:

Alkyl Halide Type	Steric Hindrance	$\text{S}_\text{N}2$ Reactivity
Methyl ( $\text{CH}_3\text{X}$ )	Minimal	Very fast
Primary ( $\text{RCH}_2\text{X}$ )	Low	Fast
Secondary ( $\text{R}_2\text{CHX}$ )	Moderate	Slow
Tertiary ( $\text{R}_3\text{CX}$ )	High	Does not occur

Why Tertiary Halides Fail

With a tertiary halide, the bulky alkyl groups block the backside attack. Instead, the alkoxide (a strong base) will eliminate a proton from the halide, forming an alkene:

\text{R-O}^- + \text{R'}_3\text{C-X} \rightarrow \text{R-OH} + \text{alkene} + \text{X}^-

This is an E2 elimination — not the desired ether formation.

Key takeaway: The Williamson synthesis works only when the alkyl halide is primary or methyl because $\text{S}_\text{N}2$ requires an unhindered backside.

3. Why the Leaving Group Must Be Good

The Role of the Halide

The halide ( $\text{X}^-$ ) must be a good leaving group — meaning it can stabilize the negative charge after departure.

Halide	Leaving Group Ability	Reason
$\text{I}^-$	Excellent	Large, polarizable, weak base
$\text{Br}^-$	Good	Moderate size, weak base
$\text{Cl}^-$	Fair	Smaller, stronger base
$\text{F}^-$	Poor	Small, strong base, holds tightly

Why Fluoride Fails

Fluoride is a strong base and a poor leaving group. The C–F bond is very strong, and $\text{F}^-$ does not depart easily. So alkyl fluorides are unreactive in Williamson synthesis.

Key takeaway: The leaving group must be weakly basic and polarizable — iodide and bromide are ideal.

--- …

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The reaction of hydrogen iodide with ethers follows the SN2 mechanism for alkyl ethers and the SN1 mechanism for benzyl/tertiary alkyl ethers. The key is that HI cleaves the C–O bond at the less hindered carbon in SN2, but at the more substituted carbon (via carbocation) in SN1. The products are always an alkyl iodide and an alcohol (or phenol).

The Core Idea: Why HI is Special

Hydrogen iodide is the strongest of the hydrohalic acids (HI > HBr > HCl) and also the best nucleophile (I⁻ is large, polarizable, and a great leaving group). When an ether reacts with HI, the oxygen gets protonated first — this turns the poor leaving group (OR) into an excellent one (OH). Then the iodide ion attacks a carbon, breaking the C–O bond.

The critical question is: Which carbon gets attacked? That depends entirely on the structure of the ether.

Step-by-Step Solutions

1. 1-Propoxypropane (a symmetrical dialkyl ether)

Structure: CH₃–CH₂–CH₂–O–CH₂–CH₂–CH₃

This is a primary–primary ether. Both alkyl groups are straight-chain propyl groups. There is no carbocation stability difference — both carbons are primary.

Mechanism: SN2. The iodide ion attacks the less sterically hindered carbon. But here both carbons are equally hindered. So the reaction cleaves the ether into one molecule of 1-iodopropane and one molecule of propan-1-ol.

The alcohol formed can further react with excess HI to give another molecule of 1-iodopropane, but in equimolar conditions, you stop at the alcohol.

\text{CH}_3\text{CH}_2\text{CH}_2\text{–O–CH}_2\text{CH}_2\text{CH}_3 + \text{HI} \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{–I} + \text{CH}_3\text{CH}_2\text{CH}_2\text{–OH}

Final products: 1-iodopropane and propan-1-ol.

2. Methoxybenzene (anisole)

Structure: C₆H₅–O–CH₃

Here one side is an aryl group (benzene ring) and the other is a methyl group. This is an aryl alkyl ether.

Key insight: The C–O bond to the aromatic ring is very strong — it has partial double-bond character due to resonance of the oxygen lone pair into the ring. An SN2 attack on the aromatic carbon is impossible (it would require breaking aromaticity). An SN1 attack would require forming a phenyl carbocation, which is extremely unstable.

So the only option is SN2 attack on the methyl carbon. The methyl group is primary and unhindered — perfect for SN2.

Watch out

A common mistake is to think the reaction cleaves the O–CH₃ bond on the oxygen side of the methyl. No — the iodide attacks the carbon of the methyl group, displacing the phenoxide ion (which immediately gets protonated to phenol).

\text{C}_6\text{H}_5\text{–O–CH}_3 + \text{HI} \rightarrow \text{C}_6\text{H}_5\text{–OH} + \text{CH}_3\text{–I}

Final products: Phenol and iodomethane.

3. Benzyl ethyl ether

Structure: C₆H₅–CH₂–O–CH₂–CH₃

This is a benzyl–ethyl ether. The benzyl group (C₆H₅–CH₂–) is special because the benzyl carbocation is highly stabilized by resonance with the aromatic ring. …

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Williamson Ether Synthesis — Concept & Method

Method: Nucleophilic Substitution ( $S_N2$ ) of an alkoxide with an alkyl halide

Core Concept

Ethers are formed when an alkoxide ion (strong nucleophile) attacks a primary alkyl halide (electrophile) in an $S_N2$ reaction. The alkoxide comes from deprotonating an alcohol with a strong base like NaH or Na metal.

General Steps

Identify the two alkyl/aryl groups attached to oxygen in the target ether.
Choose the better leaving group — always put the halide on the less hindered carbon (primary > secondary >> tertiary).
Form the alkoxide from the alcohol corresponding to the more hindered side (to avoid elimination).
React the alkoxide with the alkyl halide.

Reaction of Hydrogen Iodide with Ethers

Key concept: HI is a strong acid that cleaves ethers via $S_N2$ or $S_N1$ depending on the alkyl group. The reaction produces an alcohol and an alkyl iodide.

(i) 1-Propoxypropane (di- $n$ -propyl ether)

\text{CH}_3\text{CH}_2\text{CH}_2\text{—O—CH}_2\text{CH}_2\text{CH}_3 + \text{HI} \rightarrow

Step 1: Protonation of oxygen by HI.

Step 2: $S_N2$ attack by $\text{I}^-$ on the less hindered carbon (both are primary, so either can break).

Result: Two products possible — but the major path gives:

\boxed{\text{CH}_3\text{CH}_2\text{CH}_2\text{OH} + \text{CH}_3\text{CH}_2\text{CH}_2\text{I}}

1-propanol and 1-iodopropane

(ii) Methoxybenzene (anisole)

\text{C}_6\text{H}_5\text{—O—CH}_3 + \text{HI} \rightarrow

Step 1: Protonation of oxygen.

Step 2: $S_N2$ attack by $\text{I}^-$ on the methyl carbon (primary, less hindered). The phenyl ring is too hindered for $S_N2$ .

Result:

\boxed{\text{C}_6\text{H}_5\text{OH} + \text{CH}_3\text{I}}

Phenol and iodomethane

(iii) Benzyl ethyl ether …

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🧪 The Core Concept: Ether Cleavage by HI

When an ether reacts with hydrogen iodide (HI), the C–O bond breaks. The alkyl group (not the aryl group) gets converted to an alkyl iodide, and the other fragment becomes an alcohol (or phenol).

The key rule:

The smaller or less hindered alkyl group leaves as the alkyl iodide.

But when one group is aryl (benzene ring), the C–O bond does not break on the aryl side — the alkyl group is always the one that gets iodinated.

✗ Common Mistake #1: Breaking the wrong C–O bond in unsymmetrical ethers

Example: For 1-propoxypropane (a symmetrical ether), students sometimes think both products are the same — but they forget the nucleophilic substitution mechanism.

Correct:

$\text{CH}_3\text{CH}_2\text{CH}_2\text{–O–CH}_2\text{CH}_2\text{CH}_3 + \text{HI} \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{–I} + \text{CH}_3\text{CH}_2\text{CH}_2\text{–OH}$

(Both propyl groups are identical, so only one product set.)
How to avoid:

For symmetrical ethers, the products are one alkyl iodide + one alcohol of the same chain length. Don’t overcomplicate.

✗ Common Mistake #2: Forgetting the aryl–O bond does NOT break

Example: For methoxybenzene (anisole), students often write:

✗ $\text{C}_6\text{H}_5\text{–O–CH}_3 + \text{HI} \rightarrow \text{C}_6\text{H}_5\text{–I} + \text{CH}_3\text{OH}$

(This is wrong — the aryl–O bond is too strong to break under these conditions.)
✓ Correct:

$\text{C}_6\text{H}_5\text{–O–CH}_3 + \text{HI} \rightarrow \text{C}_6\text{H}_5\text{–OH} + \text{CH}_3\text{–I}$

(Phenol + methyl iodide)
How to avoid:

Remember: Aryl–O bonds are resistant to cleavage by HI. The alkyl group (methyl) is the one that gets iodinated. Always check: is one group aromatic? If yes, the alkyl group leaves as iodide.

✗ Common Mistake #3: Misidentifying which alkyl group leaves in benzyl ethyl ether

Example: For benzyl ethyl ether ( $\text{C}_6\text{H}_5\text{CH}_2\text{–O–CH}_2\text{CH}_3$ ), students often think the ethyl group leaves because it’s smaller — but that’s not the rule here.

The correct rule: The group that forms a more stable carbocation (or is less hindered in SN2) gets iodinated.

Benzyl carbocation is very stable (resonance-stabilised), so the benzyl group leaves as benzyl iodide.
✓ Correct: …

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Q.Write the equation of the reaction of hydrogen iodide with:

The Core Idea: Why HI is Special

Step-by-Step Solutions

1. 1-Propoxypropane (a symmetrical dialkyl ether)

2. Methoxybenzene (anisole)

3. Benzyl ethyl ether

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