Preparation of ethers by acid dehydration of secondary or tertiary alcohols is not a suitable method. Give rea

Exercises · 7.27

Q. Preparation of ethers by acid dehydration of secondary or tertiary alcohols is not a suitable method. Give reason.

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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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Acid-catalysed dehydration of secondary and tertiary alcohols leads predominantly to alkenes via elimination (E1 mechanism), not to ethers via substitution (SN1), because the carbocation intermediate favours loss of a proton over attack by a second alcohol molecule. This makes the Williamson ether synthesis the preferred route for unsymmetrical ethers.

Why acid dehydration fails for secondary/tertiary alcohols

The acid-catalysed dehydration of alcohols to form ethers works well only for primary alcohols. The reaction proceeds through protonation of the alcohol, followed by nucleophilic attack by a second alcohol molecule — a bimolecular substitution (SN2) that requires the carbon centre to be relatively unhindered.

For secondary and tertiary alcohols, the mechanism changes entirely. Let’s see why.

Protonation and carbocation formation

When a secondary or tertiary alcohol is treated with a strong acid (like $\ce{H2SO4}$ or $\ce{H3PO4}$ ) at moderate temperatures, the alcohol is first protonated. The protonated alcohol then loses water to form a carbocation.

For a tertiary alcohol, this happens very easily because the tertiary carbocation is highly stable. For a secondary alcohol, the carbocation is less stable but still forms under acidic conditions.
The carbocation has two competing fates

Once the carbocation exists, it can react in two ways:
- Nucleophilic attack by another alcohol molecule → ether formation (SN1 pathway)
- Loss of a proton from an adjacent carbon → alkene formation (E1 pathway)
The key point: elimination is strongly favoured for secondary and tertiary carbocations.
Why elimination wins

The E1 elimination requires only that a base (often the conjugate base of the acid, like $\ce{HSO4-}$ or even water) abstracts a proton from the carbon next to the carbocation. This is a low-energy, fast process.

In contrast, the SN1 pathway requires a second alcohol molecule to approach the positively charged carbon. For a tertiary carbocation, this approach is sterically hindered — the three alkyl groups block the incoming nucleophile. For a secondary carbocation, the steric hindrance is less severe, but the elimination is still kinetically favoured because the proton is always nearby and easily removed.
Temperature makes it worse

Even if some ether forms at lower temperatures, raising the temperature (which is often needed to drive the reaction) overwhelmingly favours elimination. The alkene is thermodynamically more stable, and the reaction shifts toward it. …

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Concept: Williamson Ether Synthesis

Method Name

Williamson Ether Synthesis — an $\text{S}_\text{N}2$ reaction between an alkoxide ion and a primary alkyl halide (or tosylate).

Steps of the Method

Deprotonate the alcohol React the alcohol ( $\text{ROH}$ ) with a strong base (e.g., $\text{NaH}$ , $\text{Na}$ metal, or $\text{KOH}$ ) to form the alkoxide ion ( $\text{RO}^-$ ).

\text{ROH} + \text{NaH} \rightarrow \text{RO}^- \text{Na}^+ + \text{H}_2

Perform $\text{S}_\text{N}2$ attack The alkoxide ion acts as a nucleophile and attacks the primary alkyl halide ( $\text{R'—X}$ , where X = Cl, Br, I) from the back side, displacing the halide.

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

Isolate the ether The product is a symmetrical or unsymmetrical ether, depending on the choice of $\text{R}$ and $\text{R'}$ .

Key condition: The alkyl halide must be primary (or methyl) to avoid elimination. Secondary and tertiary halides favour $\text{E2}$ elimination over substitution.

Why Acid Dehydration of Secondary/Tertiary Alcohols is Unsuitable

Acid-catalysed dehydration of alcohols (using $\text{H}_2\text{SO}_4$ or $\text{H}_3\text{PO}_4$ ) to form ethers works only for primary alcohols. For secondary or tertiary alcohols, the reaction follows a different path:

Carbocation formation Secondary and tertiary alcohols readily form stable carbocations in acidic medium.

\text{R}_2\text{CHOH} \xrightarrow{\text{H}^+} \text{R}_2\text{CH}^+ + \text{H}_2\text{O}

Elimination dominates Instead of reacting with another alcohol molecule, the carbocation loses a proton to give an alkene (E1 mechanism). $\text{R}_2\text{CH}^+ \rightarrow \text{R}_2\text{C=CH}_2 + \text{H}^+$ …

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Why the Question Matters

The acid dehydration of alcohols is a common method to prepare ethers, but it only works well for primary alcohols. For secondary and tertiary alcohols, the reaction takes a different, unwanted path.

The Core Reason (What You Must Write)

When secondary or tertiary alcohols are heated with concentrated $H_2SO_4$ (or $H_3PO_4$ ), the reaction favours elimination over substitution.

Primary alcohol → $S_N2$ mechanism → ether formation is possible.
Secondary/Tertiary alcohol → E1 elimination dominates → alkene is the major product.

Key exam point: The carbocation intermediate formed from secondary/tertiary alcohols is stable enough to lose a proton (elimination) rather than be attacked by another alcohol molecule (ether formation).

Common Mistakes Students Make

✗ Mistake 1: Saying "the reaction doesn't happen at all"

Why it's wrong: The reaction does happen — but it gives the wrong product (alkene), not the ether.

How to avoid: Always specify: "Ether formation is not suitable because elimination to alkene is favoured."

✗ Mistake 2: Confusing the mechanism (S $_N$ 1 vs S $_N$ 2)

Why it's wrong: Students often say "secondary alcohols undergo S $_N$ 2" — but under acidic conditions, secondary alcohols form carbocations (S $_N$ 1/E1 pathway).

How to avoid: Remember:

Primary → S $_N$ 2 (ether possible)
Secondary/Tertiary → carbocation forms → E1 dominates

✗ Mistake 3: Forgetting to mention the acid catalyst role

Why it's wrong: The acid ( $H_2SO_4$ ) protonates the $OH$ group, making it a better leaving group ( $H_2O$ ). Without mentioning this, the mechanism is incomplete.

How to avoid: Always state: "In presence of conc. $H_2SO_4$ , the alcohol is protonated, water leaves, and a carbocation forms."

✗ Mistake 4: Thinking temperature control can fix it

Why it's wrong: Some students write "use lower temperature to favour ether." But even at moderate temperatures, secondary/tertiary carbocations eliminate easily.

How to avoid: Know that elimination is thermodynamically favoured for these substrates — temperature control cannot suppress it enough.

✗ Mistake 5: Not distinguishing between intermolecular vs intramolecular dehydration …

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Q.Preparation of ethers by acid dehydration of secondary or tertiary alcohols is not a suitable method. Give reason.

Why acid dehydration fails for secondary/tertiary alcohols

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