(a) Can sodium ethoxide and t-butyl chloride be used for the preparation of t-butyl ethyl ether ? Give suitabl

Question

Q. (a) Can sodium ethoxide and t-butyl chloride be used for the preparation of t-butyl ethyl ether ? Give suitable explanation. Justify your answer by suggesting the appropriate starting material required for preparation of t-butyl ethyl ether.

(b) Give the IUPAC name of above mentioned 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 key idea is that Williamson ether synthesis fails with tertiary alkyl halides due to elimination dominating over substitution. Sodium ethoxide and t-butyl chloride cannot form t-butyl ethyl ether; instead, the correct approach uses sodium t-butoxide and ethyl chloride. The IUPAC name of the ether is 2-ethoxy-2-methylpropane.

Why Williamson Ether Synthesis Works — and When It Fails

Williamson ether synthesis is the most reliable method for preparing ethers. The reaction is a straightforward $S_N2$ displacement: an alkoxide ion (the nucleophile) attacks an alkyl halide (the electrophile), pushing out the halide ion. The general equation is:

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

The beauty of this reaction lies in its simplicity — but that simplicity comes with a strict condition. The $S_N2$ mechanism demands a backside attack on the carbon bearing the leaving group. This means the alkyl halide must be primary or, at most, methyl. Secondary halides work but give lower yields due to competing elimination. Tertiary halides? They are essentially useless for this reaction.

Watch out

A tertiary alkyl halide like t-butyl chloride ( $(CH_3)_3C-Cl$ ) cannot undergo $S_N2$ substitution. The three bulky methyl groups create severe steric hindrance, blocking any backside attack. Instead, the strong base (alkoxide) will abstract a $\beta$ -hydrogen, triggering E2 elimination to form an alkene.

Step-by-Step Reasoning

1. Identify the target ether and the proposed reactants

The target is t-butyl ethyl ether: $(CH_3)_3C-O-CH_2CH_3$ .

The question asks whether sodium ethoxide ( $CH_3CH_2O^-Na^+$ ) and t-butyl chloride ( $(CH_3)_3C-Cl$ ) can produce this ether.

2. Analyse the proposed reaction

If we mix these two, the alkoxide ion ( $CH_3CH_2O^-$ ) acts as both a strong nucleophile and a strong base. The t-butyl chloride has a tertiary carbon bearing the chlorine. For $S_N2$ to occur, the nucleophile would need to approach from the side opposite the leaving group — but three methyl groups crowd that carbon completely. The activation energy for substitution is prohibitively high.

What actually happens? The ethoxide ion, being a strong base, abstracts a proton from a $\beta$ -carbon of t-butyl chloride:

CH_3CH_2O^- + (CH_3)_3C-Cl \rightarrow CH_3CH_2OH + (CH_3)_2C=CH_2 + Cl^-

The major product is isobutylene (2-methylpropene), not the desired ether. The reaction is dominated by E2 elimination.

Tip

A quick way to spot this pitfall: if the alkyl halide is tertiary, Williamson synthesis will never work with that halide as the electrophile. The alkoxide must always be attached to the less hindered carbon in the final ether.

3. Determine the correct starting materials

To make t-butyl ethyl ether via Williamson synthesis, we must reverse the roles of the two alkyl groups. The alkoxide should come from the more hindered alcohol (t-butanol), and the alkyl halide should be the less hindered one (ethyl halide).

So the correct approach is:

Alkoxide source: Sodium t-butoxide, $(CH_3)_3C-O^-Na^+$ , prepared by reacting t-butanol with sodium metal.
Alkyl halide: Ethyl chloride ( $CH_3CH_2Cl$ ) or ethyl bromide ( $CH_3CH_2Br$ ).

The reaction proceeds cleanly via $S_N2$ :

$(CH_3)_3C-O^-Na^+ + CH_3CH_2Cl \rightarrow (CH_3)_3C-O-CH_2CH_3 + NaCl$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2025Set A1 mark
$Q. Write True or False: C2H5OCH3 is a symmetrical ether.$
›Reveal solutionSolution
A symmetrical ether has two IDENTICAL alkyl/aryl groups on either side of the oxygen; here the two groups (ethyl, methyl) differ, so it is unsymmetrical.

Ethers are classified as:

Simple/symmetrical ether: R–O–R, where both R groups are the same, e.g. C2H5–O–C2H5 (diethyl ether).

Mixed/unsymmetrical ether: R–O–R′, where the two groups differ, e.g. C2H5–O–CH3 (ethyl methyl ether). …
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CBSE 2023Set A1 mark
$Q. Match the following. Column A item: 'R-O-R'. Choose its correct match from Column B:$
(a) Ether
(b) Primary amine
(c) Lactose
(d) C12H22O11
(e) Glucose
(f) Negative ions
(g) C6H5SO2Cl
(h) +7
›Reveal solutionSolution

The general formula R-O-R, where two alkyl/aryl groups are joined by an oxygen atom, represents an ether.

…

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CBSE 2023Set ANNUAL1 markMCQ
$Q. Williamson's method is a very useful method for the preparation of ethers. However it will not work in the preparation of -$
(a) $(CH_3)_2O$
(b) $CH_3OC_2H_5$
(c) $C_6H_5OCH_2CH_3$
(d) $C_6H_5OC_6H_5$
›Reveal solutionSolution
Williamson synthesis needs an alkyl halide for the SN2 step; diphenyl ether would need an aryl halide instead, and aryl halides simply don't undergo this kind of substitution.

The Williamson ether synthesis works by an $S_N2$ reaction: an alkoxide/phenoxide ion (the nucleophile) displaces a halide (leaving group) from an alkyl halide.

(a) $(CH_3)_2O$ : methoxide + methyl halide — both are simple, unhindered primary alkyl systems → works fine.

(b) $CH_3OC_2H_5$ : methoxide/ethoxide + the other's alkyl halide (both primary) → works fine.

(c) $C_6H_5OCH_2CH_3$ (phenetole): sodium phenoxide + ethyl halide (an alkyl halide) → works fine, since the halide being displaced is on the alkyl (ethyl) partner, not the aryl one. …
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CBSE 2020Set 56/1/11 mark
$Q. Write the structures of the products formed when anisole is treated with HI.$
›Reveal solutionSolution
Anisole undergoes ether cleavage with HI to yield phenol and methyl iodide; the mechanism involves nucleophilic attack by iodide on the less hindered carbon of the C–O bond.

Why HI cleaves ethers: the concept behind the reaction

Ethers are generally stable compounds, but hydrogen halides—especially HI—can break the C–O bond through nucleophilic substitution. The reaction works because HI is both a strong acid (protonating the ether oxygen) and a source of iodide, an excellent nucleophile.

In anisole (methoxybenzene, $\ce{C6H5-O-CH3}$ ), we have an aromatic ring attached to one side of the oxygen and a methyl group on the other. The key question is: which C–O bond breaks? The answer lies in understanding that iodide will attack the less hindered, more electrophilic carbon—in this case, the methyl carbon—because $\ce{S_N2}$ attack on the aromatic ring is essentially impossible (the ring carbon is $sp^2$ hybridized and the transition state would be impossibly strained).

Step-by-step mechanism and product formation

Protonation of the ether oxygen HI donates a proton to the lone pair on oxygen, converting anisole into an oxonium ion:

$\ce{C6H5-O-CH3 + HI -> C6H5-\overset{+}{O}H-CH3 + I^-}$
This protonation makes the C–O bonds more polar and the adjacent carbons more electrophilic.

Nucleophilic attack by iodide The iodide ion ( $\ce{I^-}$ ) attacks the methyl carbon in an $\ce{S_N2}$ fashion. The methyl group is unhindered and accessible, whereas the phenyl carbon is part of an aromatic system and cannot undergo backside attack:

$\ce{C6H5-\overset{+}{O}H-CH3 + I^- -> C6H5-OH + CH3I}$
The C–O bond between oxygen and the methyl group breaks, and iodide forms a new bond with carbon.

Product identification The two products are: …
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CBSE 2019Set ANNUAL1 mark
$Q. How will you synthesize the isomeric ether of benzyl alcohol by Williamson synthesis?$
›Reveal solutionSolution

Anisole (methoxybenzene), isomeric with benzyl alcohol, is made by Williamson synthesis from sodium phenoxide and methyl iodide.

Benzyl alcohol ( $C_6H_5CH_2OH$ , $C_7H_8O$ ) has the isomeric ether anisole (methoxybenzene, $C_6H_5-O-CH_3$ , also $C_7H_8O$ ). By the Williamson ether synthesis, an alkoxide/phenoxide displaces a halide from an alkyl halide ( $S_N2$ ); here, sodium phenoxide reacts with methyl …

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CBSE 2018Set ANNUAL1 markMCQ
$Q. Williamson Synthesis is used to prepare :$
(a) Alcohol
(b) Amine
(c) Ketone
(d) Ether
›Reveal solutionSolution

Williamson synthesis is the reaction of a sodium alkoxide with an alkyl halide (SN2) to give an ether.

The Williamson ether synthesis proceeds as:

$R\text{-}O^-Na^+ + R'\text{-}X \longrightarrow R\text{-}O\text{-}R' + NaX$ …

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Q.(a) Can sodium ethoxide and t-butyl chloride be used for the preparation of t-butyl ethyl ether ? Give suitable explanation. Justify your answer by suggesting the appropriate starting material required for preparation of t-butyl ethyl ether.

Why Williamson Ether Synthesis Works — and When It Fails

Step-by-Step Reasoning

1. Identify the target ether and the proposed reactants

2. Analyse the proposed reaction

3. Determine the correct starting materials

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