Why is the reactivity of all the three classes of alcohols with conc. HCl and ZnCl₂ (Lucas reagent) different?

NCERT Exemplar · Q40

ZnCl_2

Andhra Pradesh BieapShort· 2mImportance★★★★★

68% · 92/135 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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.

--- …

Start your 14-day free trial to unlock the full solution →

The Lucas test distinguishes alcohols by their carbocation stability: tertiary alcohols react instantly (cloudy), secondary in 5–10 minutes, and primary do not react at room temperature — all because the rate-determining step is the formation of a carbocation, whose stability follows the order $3^\circ > 2^\circ > 1^\circ$ .

The Lucas test is a classic example of how reaction mechanism dictates observable differences in reactivity. The reagent — concentrated HCl with anhydrous $\text{ZnCl}_2$ — converts alcohols to alkyl chlorides. But why does a tertiary alcohol turn cloudy in seconds while a primary alcohol sits clear for hours?

The answer lies in the carbocation intermediate.

1. What the Lucas reagent actually does

$\text{ZnCl}_2$ is a Lewis acid. It coordinates to the oxygen of the alcohol, making the $\text{C–O}$ bond more polar and easier to break. The overall reaction is:

\text{R–OH} + \text{HCl} \xrightarrow{\text{ZnCl}_2} \text{R–Cl} + \text{H}_2\text{O}

But the mechanism is $S_N1$ for tertiary and secondary alcohols, and $S_N2$ for primary alcohols — and that difference is everything.

\text{Reactivity order: } 3^\circ > 2^\circ > 1^\circ

2. The rate-determining step: carbocation formation

For an $S_N1$ reaction, the slow step is:

\text{R–OH} + \text{ZnCl}_2 \rightarrow \text{R}^+ + \text{ZnCl}_2(\text{OH})^-

The alcohol first gets protonated (or coordinated to $\text{Zn}^{2+}$ ), then loses water to form a carbocation. The stability of this carbocation determines how fast the reaction proceeds.

Tertiary carbocation: three alkyl groups donate electron density via hyperconjugation and inductive effect. It is highly stable.
Secondary carbocation: two alkyl groups — moderately stable.
Primary carbocation: only one alkyl group — so unstable it barely forms at room temperature.

Watch out

A common mistake is to think primary alcohols never react with Lucas reagent. They do — but only at higher temperatures or over very long times. At room temperature, the reaction is too slow to observe.

3. What you actually see in the lab

The test is simple: add Lucas reagent to the alcohol and shake. The alkyl chloride product is insoluble in the aqueous medium, so it appears as a cloudy emulsion or a separate oily layer.

Alcohol class	Time to cloudiness	Reason
Tertiary	Immediate (seconds)	Stable carbocation forms instantly
Secondary	5–10 minutes	Moderate carbocation stability
Primary	No cloudiness (hours)	Carbocation too unstable; $S_N2$ is slow

Tip

If you see immediate cloudiness, the alcohol is definitely tertiary. If it takes a few minutes, it's secondary. If it stays clear for 30 minutes, it's primary — or methanol/ethanol, which are even slower.

4. Why primary alcohols fail at room temperature

Primary alcohols can react via $S_N2$ , where the nucleophile ( $\text{Cl}^-$ ) attacks the carbon directly. But $S_N2$ requires a backside attack on an unhindered carbon. In Lucas reagent, the $\text{Cl}^-$ concentration is high, but the $\text{ZnCl}_2$ first coordinates to oxygen — making the carbon more electrophilic. Even so, the $S_N2$ pathway is much slower than the $S_N1$ pathway for tertiary alcohols.

For a primary alcohol to react via $S_N1$ , it would need to form a primary carbocation — which is so unstable that it essentially never happens in solution. So the reaction proceeds via $S_N2$ , which is slow at room temperature. …

Start your 14-day free trial to unlock the full solution →

Here are the common mistakes students make when explaining the reactivity of alcohols with Lucas reagent ( $\text{conc. HCl} + \text{ZnCl}_2$ ), along with how to avoid each.

Mistake 1: Forgetting the Mechanism (Why $\text{ZnCl}_2$ is needed)

The mistake: Students say “tertiary alcohols react fastest because they are more stable” without mentioning how the reaction happens. They often think $\text{ZnCl}_2$ is just a catalyst that speeds everything equally.
Why it’s wrong: The reactivity difference is due to the carbocation intermediate formed in the $S_N1$ mechanism. $\text{ZnCl}_2$ helps convert the $\text{OH}$ group into a better leaving group ( $\text{OH}_2^+$ or $\text{ZnCl}_2\text{OH}^-$ ). Without it, primary alcohols barely react.
How to avoid: Always write the stepwise mechanism:
1. $\text{ZnCl}_2$ coordinates to $\text{OH}$ → makes $\text{OH}$ a better leaving group.
2. $\text{C–O}$ bond breaks → carbocation forms.
3. $\text{Cl}^-$ attacks the carbocation.
Key point: The rate depends on carbocation stability: Tertiary > Secondary > Primary (tertiary carbocation is most stable, so it forms fastest).

Mistake 2: Confusing Reactivity Order with Nucleophilicity

The mistake: Students think “primary alcohols react fastest because $\text{OH}$ is a good leaving group” or “primary alcohols are more reactive in $S_N2$ ”.
Why it’s wrong: Lucas test is $S_N1$ (not $S_N2$ ). In $S_N1$ , the rate depends on carbocation stability, not on steric hindrance. Primary carbocations are highly unstable, so primary alcohols react very slowly (or not at all) under Lucas conditions.
How to avoid: Memorise the correct order for Lucas test: Tertiary > Secondary > Primary (Tertiary: immediate cloudiness; Secondary: 5–10 min; Primary: no reaction at room temp).

Mistake 3: Ignoring the Role of $\text{ZnCl}_2$ for Primary Alcohols

The mistake: Students say “primary alcohols do not react at all with Lucas reagent”.
Why it’s wrong: Primary alcohols can react if heated or if $\text{ZnCl}_2$ concentration is high, but at room temperature they react very slowly (hours). The $\text{ZnCl}_2$ is crucial to make the $\text{OH}$ a better leaving group, but even then, the primary carbocation is too unstable to form easily.
How to avoid: State clearly: “Primary alcohols do not give a visible reaction at room temperature because the primary carbocation is too unstable. $\text{ZnCl}_2$ helps, but not enough to overcome the high activation energy.”

Mistake 4: Mixing up Lucas Test with Oxidation or Dehydration

The mistake: Students compare Lucas test reactivity with oxidation (e.g., $\text{K}_2\text{Cr}_2\text{O}_7$ ) or dehydration (e.g., $\text{H}_2\text{SO}_4$ , heat).
Why it’s wrong: Lucas test is a substitution reaction ( $\text{OH} \to \text{Cl}$ ), not oxidation or elimination. The reactivity order for oxidation is Primary > Secondary > Tertiary (opposite of Lucas). For dehydration, the order is similar to Lucas (tertiary fastest), but the mechanism is different.
How to avoid: Keep a separate table in your notes:
- Lucas test (substitution): Tertiary > Secondary > Primary
- Oxidation: Primary > Secondary > Tertiary (tertiary resists oxidation)
- Dehydration: Tertiary > Secondary > Primary

Mistake 5: Forgetting Solubility and Cloudiness Observation …

Start your 14-day free trial to unlock the full solution →

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

AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
$(CH_3)_3CONa + CH_3Br \rightarrow (CH_3)_3COCH_3$
›Reveal solutionSolution
Aryl halides do not undergo ordinary nucleophilic substitution the way alkyl halides

do, so bromobenzene + sodium methoxide will not simply hand you anisole. Answer: (D).

Concept and Intuition

(A) $(CH_3)_3CONa + CH_3Br \rightarrow (CH_3)_3COCH_3$ : this is Williamson ether synthesis. The rule of thumb is that the alkyl halide should be unhindered (methyl or primary) for a clean $S_N2$ ; the bulk of the alkoxide doesn't matter much since it's the nucleophile, not the electrophile. Methyl bromide is a perfect $S_N2$ substrate, so this reaction proceeds cleanly to give methyl tert-butyl ether. Correct.

(B) $(C_2H_5)_2O \xrightarrow[\Delta]{excess\ HI} 2C_2H_5I + H_2O$ : with excess hot HI, a symmetrical ether is cleaved completely, both alkyl-oxygen bonds broken, to give two equivalents of alkyl iodide. Correct textbook fact.

(C) $(CH_3)_3COC_2H_5 \xrightarrow[\Delta]{HI} (CH_3)_3CI + C_2H_5OH$ : this is a mixed ether with one tertiary and one primary alkyl group. Cleavage proceeds via $S_N1$ at the carbon that gives the more stable carbocation (tertiary), so the tert-butyl group leaves as the iodide and the ethyl-oxygen fragment is released as ethanol. This is the standard textbook outcome. Correct.

(D) Aryl halides like bromobenzene have their C–X bond strengthened by resonance with the ring (partial double-bond character) and the carbon is $sp^2$ , blocking backside $S_N2$ attack. Nucleophilic substitution on an unactivated aryl halide by a simple alkoxide under ordinary conditions simply does not happen — it requires either very forcing conditions (high temperature and pressure, as in phenol manufacture from chlorobenzene) or a strong base capable of a benzyne (elimination-addition) pathway (e.g. $NaNH_2$ ). Plain $CH_3ONa$ at ordinary conditions will not convert bromobenzene to anisole. This statement is therefore NOT correct.

Step-by-Step Solution …
Start your 14-day free trial to see the full solution →
AP EAPCET 2026Set ap-2026-05-20-AN1 markMCQ
$Y$
›Reveal solutionSolution
Tertiary C-H oxidation gives tert-butanol, then Williamson ether synthesis gives MTBE; the only remaining subtlety is citing the substituent prefixes in the correct alphabetical order.

Concept and Intuition

Tertiary C-H bonds are the weakest and most easily oxidised C-H bonds in an alkane (the resulting radical/cation is most stabilised), so a strong oxidant like $KMnO_4$ selectively converts isobutane's one tertiary hydrogen into a tertiary alcohol rather than attacking the primary methyl hydrogens. Converting that alcohol to its sodium alkoxide and reacting with a primary alkyl halide (Williamson ether synthesis, $S_N2$ at the primary carbon of $CH_3Br$ ) builds the ether cleanly, since $S_N2$ works well on primary halides.

Step-by-Step Solution

Isobutane $(CH_3)_3CH$ + $KMnO_4$ → oxidation at the sole tertiary C-H → X = tert-butyl alcohol, $(CH_3)_3C-OH$ (2-methylpropan-2-ol).

X + Na → sodium tert-butoxide, $(CH_3)_3C-O^-Na^+$ (Na displaces the O-H proton).

Sodium tert-butoxide + $CH_3Br$ → Williamson ether synthesis (the alkoxide's oxygen performs $S_N2$ on the primary carbon of methyl bromide) → Y = $(CH_3)_3C-O-CH_3$ , methyl tert-butyl ether. …
Start your 14-day free trial to see the full solution →
AP EAPCET 2025Set ap-2025-05-20-AN1 markMCQ
$\mathrm{C_6H_5NH_2}$
›Reveal solutionSolution
Aniline → diazonium salt → phenol (X) → anisole (via Williamson ether synthesis) → 4-bromoanisole (Y, major product of electrophilic bromination directed para by the methoxy group).

Concept and Intuition

This is a multi-step synthesis chaining three classic reactions: diazotisation/hydrolysis (amine → phenol via the diazonium salt), Williamson ether synthesis (phenoxide + alkyl halide → aryl alkyl ether), and electrophilic aromatic bromination directed by a strongly activating, ortho/para-directing methoxy group. Since anisole's methoxy substituent is already present on the ring, the incoming $\mathrm{Br^+}$ (from $\mathrm{Br_2}$ in acetic acid) attacks preferentially para (and some ortho), and para is reported as the major product due to less steric crowding.

Step-by-Step Solution

$\mathrm{C_6H_5NH_2} \xrightarrow{\mathrm{NaNO_2/HCl},\ 273K} \mathrm{C_6H_5N_2^+Cl^-}$ (benzenediazonium chloride) — standard diazotisation of a primary aromatic amine at 0–5°C.

$\mathrm{C_6H_5N_2^+Cl^-} \xrightarrow{\mathrm{H_2O},\ \Delta}$ phenol ( $\mathrm{C_6H_5OH}$ ) + $\mathrm{N_2}$ + HCl — hydrolysis of the diazonium salt on warming. So X = phenol.

Phenol + NaOH → sodium phenoxide ( $\mathrm{C_6H_5O^-Na^+}$ ); phenoxide + $\mathrm{CH_3Br}$ → anisole ( $\mathrm{C_6H_5OCH_3}$ ) via $\mathrm{S_N2}$ (Williamson ether synthesis). …
Start your 14-day free trial to see the full solution →
AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
$\mathrm{Y \xleftarrow{NaOEt} CH_3CH_2CH_2CH_2Br \xrightarrow[(ii)\,H_2O]{(i)\,Mg/dry\ ether} X}$
›Reveal solutionSolution
Grignard + water gives the alkane (not the alcohol you'd get from a carbonyl); sodium ethoxide on a primary halide gives clean $S_N2$ substitution (an ether), not elimination.

Concept and Intuition

A Grignard reagent $R\text{-}MgX$ behaves like a carbanion $R^-$ . Any proton source acidic enough (even water, $pK_a\approx15.7$ , is far more acidic than the alkane conjugate acid) instantly protonates it: $R\text{-}MgX + H_2O \to R\text{-}H + Mg(OH)X$ . This is the classic way to convert a halide into the corresponding hydrocarbon — it is not how you make an alcohol (that needs the Grignard to attack a carbonyl carbon first).

Separately, when a nucleophile/base attacks an alkyl halide, the outcome ( $S_N2$ vs $E2$ ) depends on both the base's bulk and the substrate's branching. Ethoxide ( $\mathrm{NaOEt}$ ) is a comparatively small, strong base. Against a primary substrate with an easily accessible backside carbon (n-butyl bromide), steric hindrance to backside attack is minimal, so $S_N2$ substitution is the major pathway, not elimination.

Step-by-Step Solution

$\mathrm{CH_3CH_2CH_2CH_2Br} + Mg \xrightarrow{dry\ ether} \mathrm{CH_3CH_2CH_2CH_2MgBr}$ (Grignard reagent formed).

$\mathrm{CH_3CH_2CH_2CH_2MgBr} + H_2O \to \mathrm{CH_3CH_2CH_2CH_3} + Mg(OH)Br$ — protonolysis gives butane, so $X=\mathrm{CH_3CH_2CH_2CH_3}$ . …
Start your 14-day free trial to see the full solution →
AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
$\mathrm{(CH_3)_3CONa + CH_3CH_2Br \rightarrow X}$
›Reveal solutionSolution
$S_N2$ vs $E2$ is decided by whether backside attack at the carbon bearing the leaving group is sterically possible — here that means looking at the alkyl halide's own branching, not just the base's bulk.

Concept and Intuition

The classic trap in these paired reactions is to only look at how bulky the base/nucleophile is. What actually controls the outcome is whether the nucleophile can reach the back lobe of the C–LG bond:

If the carbon bearing the leaving group is primary and unhindered (ethyl bromide), $S_N2$ remains fast and dominant even with a bulky base like $t$ -butoxide, because the bulk of the base doesn't block approach to a completely open primary carbon nearly as much as branching at the substrate would.

If the carbon bearing the leaving group is tertiary (tert-butyl bromide), backside attack is essentially impossible regardless of which base is used — the only viable pathway is $E2$ (proton abstraction from a $\beta$ -carbon), so even a small, strong, ionic base like ethoxide gives elimination as the major product.

Step-by-Step Solution

Reaction 1: $(CH_3)_3CONa$ (base/nucleophile) + $\mathrm{CH_3CH_2Br}$ (primary substrate). The electrophilic carbon (in $CH_3CH_2Br$ ) is unhindered, so $t$ -butoxide's oxygen performs $S_N2$ substitution: $X = (CH_3)_3C\text{-}O\text{-}CH_2CH_3$ . …
Start your 14-day free trial to see the full solution →
AP EAPCET 2023Set ap-2023-05-23-FN1 markMCQ
$\mathrm{CH_3CH_2C(CH_3)_2OCH_2CH_2CH_3}$
›Reveal solutionSolution
Williamson ether synthesis needs the bulky/aryl partner as the alkoxide and the unhindered partner as the primary halide, to avoid elimination or a non-reactive aryl-halide substitution.

Concept and Intuition

Williamson synthesis is an $\mathrm{S_N2}$ reaction between an alkoxide and an alkyl halide. Two structural traps show up here: (1) tertiary alkyl halides are poor $\mathrm{S_N2}$ substrates — with a strong nucleophile/base like an alkoxide they instead undergo E2 elimination; and (2) aryl halides (like bromobenzene) cannot undergo $\mathrm{S_N2}$ at all because the aryl carbon is sp2, in-plane, and shielded, and the C–X bond is strengthened by ring resonance.

Step-by-Step Solution

For X, $\mathrm{CH_3CH_2C(CH_3)_2{-}O{-}CH_2CH_2CH_3}$ : the tertiary-pentyl part must come in as the alkoxide (sodium tert-alkoxide), while the primary propyl part comes in as the halide (propyl bromide) — the primary halide undergoes clean $\mathrm{S_N2}$ with the bulky alkoxide, avoiding elimination.

For Y, phenyl propyl ether ( $\mathrm{C_6H_5{-}O{-}CH_2CH_2CH_3}$ ): since aryl halides (bromobenzene) cannot undergo nucleophilic substitution, the correct route is phenoxide (from phenol) attacking the primary alkyl halide (propyl bromide). …
Start your 14-day free trial to see the full solution →
AP EAPCET 2022Set ap-2022-07-11-AN1 markMCQ
$-C(CH_3)_2Br$
›Reveal solutionSolution
Bulky sodium tert-butoxide is a classic E2-promoting, poor-SN2 base; on the primary bromide side chain it eliminates HBr to give a terminal alkene (vinyl group), while the more hindered tertiary bromide is unaffected — matching option (C).

Concept and Intuition

The identity of a base/nucleophile controls whether an alkyl halide undergoes substitution or elimination. Sodium tert-butoxide, $(CH_3)_3CO^-Na^+$ , is strongly basic (as a good alkoxide) but its oxygen is buried under three bulky methyl groups, making it a very poor nucleophile for SN2 (it cannot easily approach a carbon from the back side). This steric bulk is exactly why tert-butoxide is the textbook reagent of choice to force elimination (E2) rather than substitution even on primary alkyl halides, which would normally favour SN2 with a small nucleophile like hydroxide or ethoxide. For a primary bromide such as $-CH_2CH_2Br$ , tert-butoxide therefore removes a $\beta$ -hydrogen (E2) to give the terminal alkene $-CH=CH_2$ , releasing Br $^-$ .

Meanwhile, this same molecule also carries a tertiary, benzylic bromide, $-C(CH_3)_2Br$ , directly on the ring. In problems of this kind, the answer choices make clear that the reaction is selective for the primary position over the tertiary one under the given conditions/timescale — the tertiary center is left as the unreacted starting bromide in the major product.

Step-by-Step Solution

Identify the base: Me $_3$ CONa = sodium tert-butoxide — strong base, very weak/hindered nucleophile.

Recognize its standard textbook behaviour: promotes E2 elimination, especially valuable for driving primary alkyl halides toward alkenes instead of the substitution product they would normally favour with a small nucleophile.

Apply E2 to the primary bromide arm: $-CH_2-CH_2-Br \xrightarrow{-HBr} -CH=CH_2$ (a terminal vinyl group, the only possible elimination product from a 2-carbon primary chain). …
Start your 14-day free trial to see the full solution →
AP EAPCET 2021Set eng-2021-08-25-AN1 markMCQ
$\underset{CH_3}{\overset{CH_3}{CH_3-C-ONa}} + CH_3Cl \longrightarrow\ ?$
›Reveal solutionSolution
A branched sodium alkoxide plus methyl chloride is a textbook Williamson ether synthesis: the alkoxide's own carbon skeleton is retained, and a new methyl group is added onto its oxygen.

Concept and Intuition

Williamson ether synthesis works best when the alkyl halide is primary/unhindered (favours clean $S_N2$ ) — methyl halides are ideal because they have no $\beta$ -hydrogens at all, so competing elimination is impossible regardless of how bulky the incoming alkoxide nucleophile is. The alkoxide's own carbon skeleton is untouched by the reaction (only its oxygen attacks); the methyl group from the alkyl halide simply attaches to that oxygen.

Step-by-Step Solution

Identify the nucleophile: the sodium alkoxide of the branched alcohol, $R_3C$ – $O^-\,Na^+$ (with $R_3C$ its full carbon skeleton, retained unchanged throughout the reaction).

Identify the electrophile: $CH_3Cl$ — small, unhindered, no possibility of elimination (no $\beta$ -H).

$S_N2$ attack: the alkoxide oxygen's lone pair displaces $Cl^-$ from the methyl carbon. …
Start your 14-day free trial to see the full solution →

←Previous questionQ39 — Explain why is O=C=O nonpolar while R-O-R is polar.→Next questionQ41 — Write steps to carry out the conversion of phenol to aspirin.

Q.Why is the reactivity of all the three classes of alcohols with conc. HCl and ZnCl2ZnCl_2ZnCl2​ (Lucas reagent) different?

1. What the Lucas reagent actually does

2. The rate-determining step: carbocation formation

3. What you actually see in the lab

4. Why primary alcohols fail at room temperature

🎓Unlock everything free for 14 days

Q.Why is the reactivity of all the three classes of alcohols with conc. HCl and $ZnCl_2$ (Lucas reagent) different?

Unlock everything free for 14 days