Write the names of reagents and equations for the preparation of the following ethers by Williamson's synthesi

Exercises · 7.24

Q. Write the names of reagents and equations for the preparation of the following ethers by Williamson's synthesis:

(i) 1-Propoxypropane

(ii) Ethoxybenzene

(iii) 2-Methoxy-2-methylpropane

(iv) 1-Methoxyethane

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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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Williamson ether synthesis is an SN2 reaction between an alkoxide ion and a primary alkyl halide (or tosylate). For symmetrical ethers, either halide works; for unsymmetrical ethers, choose the route that avoids elimination. The four ethers are prepared as follows: (i) sodium propoxide + 1-chloropropane,

(ii) sodium ethoxide + bromobenzene (or sodium phenoxide + ethyl bromide),

(iii) sodium 2-methyl-2-propoxide + methyl iodide,

(iv) sodium methoxide + iodoethane (or sodium ethoxide + iodomethane).

Williamson ether synthesis is the most reliable laboratory method for making ethers. The core idea is simple: an alkoxide ion (strong base, good nucleophile) attacks an alkyl halide in an SN2 reaction. The product is an ether, and the byproduct is a salt.

The key constraint is that the alkyl halide must be primary (or methyl) to avoid competing elimination. If you use a secondary or tertiary halide, the alkoxide will deprotonate it instead of substituting it — you'll get an alkene, not an ether.

Let's apply this to each case.

1. 1-Propoxypropane — a symmetrical ether.

This is $\text{CH}_3\text{CH}_2\text{CH}_2\text{—O—CH}_2\text{CH}_2\text{CH}_3$ . Since both alkyl groups are identical, you can make it by reacting sodium propoxide with 1-chloropropane (or 1-bromopropane).

The alkoxide is prepared by dissolving sodium metal in propan-1-ol:

\text{2 CH}_3\text{CH}_2\text{CH}_2\text{OH} + \text{2 Na} \rightarrow \text{2 CH}_3\text{CH}_2\text{CH}_2\text{O}^-\text{Na}^+ + \text{H}_2

Then the SN2 reaction:

\text{CH}_3\text{CH}_2\text{CH}_2\text{O}^-\text{Na}^+ + \text{Cl—CH}_2\text{CH}_2\text{CH}_3 \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{—O—CH}_2\text{CH}_2\text{CH}_3 + \text{NaCl}

Tip

For a symmetrical ether, you could also use the other combination (e.g., sodium propoxide + propyl halide is the same thing). There's no ambiguity here.

2. Ethoxybenzene — an unsymmetrical ether with an aryl group.

This is $\text{C}_6\text{H}_5\text{—O—CH}_2\text{CH}_3$ . The aryl group is attached to oxygen. Here's the critical point: you cannot use bromobenzene as the alkyl halide with sodium ethoxide, because aryl halides do not undergo SN2 reactions (the carbon is sp² hybridised, and the $\pi$ system blocks backside attack). That route fails.

The correct approach is to use sodium phenoxide (the alkoxide from phenol) and ethyl bromide (a primary alkyl halide). Phenol is acidic enough to be deprotonated by NaOH:

\text{C}_6\text{H}_5\text{OH} + \text{NaOH} \rightarrow \text{C}_6\text{H}_5\text{O}^-\text{Na}^+ + \text{H}_2\text{O}

Then:

\text{C}_6\text{H}_5\text{O}^-\text{Na}^+ + \text{Br—CH}_2\text{CH}_3 \rightarrow \text{C}_6\text{H}_5\text{—O—CH}_2\text{CH}_3 + \text{NaBr}

Watch out

A common mistake is to write sodium ethoxide + bromobenzene. That won't work — bromobenzene is inert to SN2. Always put the aryl group on the alkoxide side, not the halide side.

3. 2-Methoxy-2-methylpropane — a tert-butyl methyl ether.

This is $(\text{CH}_3)_3\text{C—O—CH}_3$ . One alkyl group is tertiary, the other is methyl. The tertiary group cannot be on the halide — if you tried to use tert-butyl bromide with sodium methoxide, elimination would dominate (E2), giving isobutylene. So the tertiary group must come from the alkoxide.

Use sodium 2-methyl-2-propoxide (sodium tert-butoxide) and methyl iodide (or methyl bromide). Methyl halides are excellent SN2 substrates with no competing elimination.

Preparation of the alkoxide:

\text{2 (CH}_3)_3\text{COH} + \text{2 Na} \rightarrow \text{2 (CH}_3)_3\text{CO}^-\text{Na}^+ + \text{H}_2

Then:

$(\text{CH}_3)_3\text{CO}^-\text{Na}^+ + \text{I—CH}_3 \rightarrow (\text{CH}_3)_3\text{C—O—CH}_3 + \text{NaI}$ …

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Williamson Ether Synthesis — One Clear Method

Method: Williamson Ether Synthesis (SN2 Pathway)

This is the most reliable laboratory method for preparing both symmetrical and unsymmetrical ethers. The reaction is an SN2 displacement of a halide (or other leaving group) by an alkoxide ion.

Core Principle

\ce{R-O^- + R'-X -> R-O-R' + X^-}

The alkoxide ( $\ce{RO^-}$ ) acts as the nucleophile.
The alkyl halide ( $\ce{R'X}$ ) must be primary (or methyl) to avoid elimination.
For unsymmetrical ethers, choose the less hindered alkyl halide to form the alkoxide from the more hindered alcohol.

Steps for Any Williamson Synthesis

Identify the ether — split it into an alkoxide part and an alkyl halide part.
Choose the alkoxide — derived from the alcohol with the bulkier alkyl group (to avoid elimination).
Choose the alkyl halide — must be primary (or methyl) to favour SN2 over E2.
React the alkoxide (prepared by treating alcohol with Na or NaH) with the alkyl halide.

(i) 1-Propoxypropane (dipropyl ether)

Ether structure: $\ce{CH3CH2CH2-O-CH2CH2CH3}$

Method: Both sides are identical — use propoxide + 1-bromopropane.

Reagents: Sodium propoxide ( $\ce{CH3CH2CH2ONa}$ ) + 1-bromopropane ( $\ce{CH3CH2CH2Br}$ )

Equation:

\ce{CH3CH2CH2ONa + CH3CH2CH2Br -> CH3CH2CH2-O-CH2CH2CH3 + NaBr}

(ii) Ethoxybenzene (phenetole)

Ether structure: $\ce{C6H5-O-CH2CH3}$

Method: Phenol is more acidic than ethanol — use sodium phenoxide + ethyl halide.

Reagents: Sodium phenoxide ( $\ce{C6H5ONa}$ ) + ethyl bromide ( $\ce{CH3CH2Br}$ )

Equation:

\ce{C6H5ONa + CH3CH2Br -> C6H5-O-CH2CH3 + NaBr}

Why this way? If we used sodium ethoxide + bromobenzene, the reaction would fail — aryl halides do not undergo SN2.

(iii) 2-Methoxy-2-methylpropane (methyl tert-butyl ether, MTBE)

Ether structure: $\ce{(CH3)3C-O-CH3}$

Method: The tert-butyl group is tertiary — cannot be the alkyl halide (would give elimination). So:

Alkoxide = sodium tert-butoxide (from tert-butanol)
Alkyl halide = methyl iodide (primary)

Reagents: Sodium tert-butoxide ( $\ce{(CH3)3CONa}$ ) + methyl iodide ( $\ce{CH3I}$ )

Equation: …

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🧠 Core Concept (Why it matters)

Williamson synthesis is an SN2 reaction between an alkoxide ion (strong nucleophile) and a primary alkyl halide (or tosylate). The key principle:

The alkoxide comes from an alcohol + strong base (NaH, Na metal, etc.).
The alkyl halide must be primary (or methyl) to avoid elimination.
Tertiary alkyl halides give elimination (alkene), not ether.

✗ Mistake 1: Using a tertiary alkyl halide

Example: For (iii) 2-Methoxy-2-methylpropane (a tert-butyl methyl ether), students often try:

$(\text{CH}_3)_3\text{C-Br} + \text{CH}_3\text{O}^- \rightarrow (\text{CH}_3)_3\text{C-O-CH}_3$

Why it’s wrong:

The alkoxide is a strong base. With a tertiary halide, E2 elimination dominates — you get isobutylene, not the ether.

✓ How to avoid:

Use the alkoxide of the tertiary alcohol + methyl halide (primary):

(\text{CH}_3)_3\text{C-O}^- \text{Na}^+ + \text{CH}_3\text{I} \rightarrow (\text{CH}_3)_3\text{C-O-CH}_3 + \text{NaI}

Rule: The alkyl halide must be primary (or methyl). The bulky/tertiary group comes from the alkoxide.

✗ Mistake 2: Wrong choice of alkoxide/halide for unsymmetrical ethers

Example: For (ii) Ethoxybenzene (phenetole), students write:

$\text{C}_6\text{H}_5\text{Br} + \text{C}_2\text{H}_5\text{O}^- \rightarrow \text{C}_6\text{H}_5\text{O-C}_2\text{H}_5$

Why it’s wrong:

Aryl halides (like bromobenzene) do not undergo SN2 — the C–X bond has partial double-bond character. No reaction.

✓ How to avoid:

Use sodium phenoxide (alkoxide from phenol) + ethyl halide (primary):

\text{C}_6\text{H}_5\text{O}^- \text{Na}^+ + \text{C}_2\text{H}_5\text{Br} \rightarrow \text{C}_6\text{H}_5\text{O-C}_2\text{H}_5 + \text{NaBr}

Rule: The aryl/aromatic part must come from the phenoxide (alkoxide), not the halide.

✗ Mistake 3: Forgetting to specify reagents (base, solvent, conditions)

Students often write only the organic reactants, missing:

Base used to generate alkoxide (NaH, Na metal, KOH, etc.)
Solvent (often the alcohol itself or dry ether)
Anhydrous conditions (water destroys alkoxide)

✓ How to avoid:

Always write the full reagent set. Example for (i) 1-Propoxypropane:

Reagents: Sodium metal + 1-propanol → sodium propoxide + H₂

Then: sodium propoxide + 1-bromopropane (SN2) → ether

Equation:

2 \text{CH}_3\text{CH}_2\text{CH}_2\text{OH} + 2 \text{Na} \rightarrow 2 \text{CH}_3\text{CH}_2\text{CH}_2\text{O}^- \text{Na}^+ + \text{H}_2

\text{CH}_3\text{CH}_2\text{CH}_2\text{O}^- \text{Na}^+ + \text{CH}_3\text{CH}_2\text{CH}_2\text{Br} \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{-O-CH}_2\text{CH}_2\text{CH}_3 + \text{NaBr}

✗ Mistake 4: Using wrong halide for symmetrical ethers

For (i) 1-Propoxypropane (di-n-propyl ether), students sometimes try:

$2 \text{CH}_3\text{CH}_2\text{CH}_2\text{OH} \xrightarrow{\text{H}_2\text{SO}_4} \text{ether}$

That’s acid-catalysed dehydration — not Williamson. Williamson requires alkoxide + alkyl halide.

✓ How to avoid: …

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Q.Write the names of reagents and equations for the preparation of the following ethers by Williamson's synthesis:

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