What is the correct order of reactivity of alcohols in the following reaction? R-OH + HCl xrightarrowanhyd. Zn

NCERT Exemplar · Q3

R\text{-}OH + HCl \xrightarrow{\text{anhyd. } ZnCl_2} R\text{-}Cl + H_2O

(i)

1^\circ > 2^\circ > 3^\circ

(ii)

1^\circ < 2^\circ > 3^\circ

(iii)

3^\circ > 2^\circ > 1^\circ

(iv)

3^\circ > 1^\circ > 2^\circ

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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

Students often try to make an ether by reacting two alcohols together. That doesn't work directly — you need one alcohol to become the nucleophile (alkoxide) and the other to become the electrophile (alkyl halide). The Williamson synthesis is asymmetric by design.

The Big Picture

Williamson ether synthesis is the go-to method for making unsymmetrical ethers ( $R-O-R'$ where $R \neq R'$ ). It's reliable, high-yielding, and conceptually clean — as long as you respect the SN2 mechanism and avoid tertiary halides.

The one-line takeaway: An alkoxide attacks an alkyl halide in an SN2 reaction to form an ether — but only if the halide is primary or methyl.

Williamson ether synthesis is the standard method for making ethers, taught in the NCERT/CBSE Class 12 Chemistry chapter on Alcohols, Phenols and Ethers, and ‘Williamson synthesis mechanism’ or ‘Williamson ether synthesis limitations’ are frequently searched important-question topics for board exams, JEE Main and NEET. Knowing why tertiary halides fail in this SN2-based reaction is a common distinguishing question in competitive organic chemistry exams.

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.

4. Why the Alkoxide Must Be the Nucleophile (Not the Halide)

The "Wrong Way" Problem

If you try to use an alcohol as the nucleophile and an alkoxide as the leaving group, it won't work. Why?

The alkoxide is a stronger base than the halide
The halide is a better leaving group than the alkoxide

So the reaction is irreversible in the direction shown:

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

The reverse reaction (where $\text{X}^-$ attacks the ether) would require $\text{X}^-$ to be a nucleophile and $\text{R-O}^-$ to be a leaving group — but $\text{R-O}^-$ is a terrible leaving group (strong base).

Key takeaway: The reaction is driven by the difference in leaving group ability — halides leave easily, alkoxides do not.

Summary: The Three Pillars of Williamson Ether Synthesis

Strong nucleophile (alkoxide, not alcohol) — full negative charge on oxygen
Unhindered electrophile (primary or methyl halide) — $\text{S}_\text{N}2$ requires backside access
Good leaving group (iodide, bromide, or chloride) — halide must depart easily

If any of these conditions is violated, the reaction fails or gives elimination products.

Quick Exam Tip

When asked "Why does Williamson synthesis fail with tertiary halides?" — never say "because it's bulky." Say:

"Tertiary halides undergo E2 elimination instead of $\text{S}_\text{N}2$ because the alkoxide acts as a strong base and the steric hindrance prevents backside attack."

This shows you understand the competition between substitution and elimination — a common exam trap.

The reaction is the Lucas test, where alcohols react with HCl in the presence of anhydrous ZnCl2. The reactivity order follows carbocation stability: tertiary > secondary > primary. The correct option is (iii).

This question is about the Lucas test, a classic qualitative test used to distinguish between primary, secondary, and tertiary alcohols. The reagent is a mixture of concentrated HCl and anhydrous ZnCl2 (called Lucas reagent). The reaction converts an alcohol into an alkyl chloride, and the rate depends entirely on how easily the alcohol can form a carbocation intermediate.

Why does ZnCl2 help? ZnCl2 is a Lewis acid. It coordinates to the oxygen of the alcohol, making the O–H bond more polar and the hydroxyl group a better leaving group (as water). This turns a poor leaving group (OH-) into a good one (H2O). Once water leaves, a carbocation forms — and the stability of that carbocation dictates the reaction speed.

Mechanism matters. The reaction proceeds via an SN1 mechanism for tertiary and secondary alcohols (and for primary alcohols under these conditions, it's SN2, but very slow). In SN1, the rate-determining step is the formation of the carbocation. So the more stable the carbocation, the faster the reaction.
Carbocation stability order. Tertiary carbocations are the most stable (three alkyl groups donate electron density via hyperconjugation and inductive effect). Secondary are less stable, and primary are the least stable (and primary carbocations are so unstable they rarely form at all under normal conditions). So the expected reactivity is:

$3^\circ > 2^\circ > 1^\circ$
What about option (ii)? Option (ii) says $1^\circ < 2^\circ > 3^\circ$ , which would mean secondary is faster than tertiary — that contradicts carbocation stability. So (ii) is wrong.
What about option (iv)? Option (iv) says $3^\circ > 1^\circ > 2^\circ$ , which would put primary above secondary — also impossible given the stability trend.
Why not (i)? Option (i) says $1^\circ > 2^\circ > 3^\circ$ , which is the opposite of the correct order. That would be true only if the reaction were SN2 and the nucleophile attacked the carbon directly — but here, the conditions (anhydrous ZnCl2, HCl) favour SN1 for all but the simplest alcohols.

Watch out

A common mistake is to think that because primary alcohols react fastest in SN2 reactions (like with HBr or HI without ZnCl2), they should also react fastest here. But the Lucas test is specifically designed to exploit carbocation stability — ZnCl2 makes the leaving group so good that even secondary alcohols go through an SN1 pathway. Primary alcohols react so slowly that they are considered unreactive at room temperature in the Lucas test.

Tip

In the Lucas test, the time taken for the solution to turn cloudy (due to the formation of the insoluble alkyl chloride) is the practical measure: tertiary alcohols react instantly, secondary alcohols react in 5–10 minutes, and primary alcohols do not react at room temperature. This directly confirms the order $3^\circ > 2^\circ > 1^\circ$ .

✓Final answer

3^\circ > 2^\circ > 1^\circ

This is a classic Lucas test question, not a Williamson Ether Synthesis. The reaction given is the Lucas test for distinguishing alcohols based on their reactivity with HCl in the presence of anhydrous $ZnCl_2$ .

Let’s break down the common mistakes and how to avoid them.

✗ Mistake 1: Confusing the reaction with Williamson Ether Synthesis

Why it happens:

Students often mix up the Lucas test (alcohol + HCl/ZnCl₂ → alkyl halide) with Williamson ether synthesis (alkoxide + alkyl halide → ether). The names sound similar, but the mechanisms are completely different.

How to avoid:

Williamson Ether Synthesis: $R\text{-}O^- + R'\text{-}X \rightarrow R\text{-}O\text{-}R' + X^-$
Lucas Test: $R\text{-}OH + HCl \xrightarrow{ZnCl_2} R\text{-}Cl + H_2O$
Memorise the reagents: Lucas test uses HCl + anhydrous ZnCl₂. Williamson uses NaH/Na metal + alkyl halide.

✗ Mistake 2: Forgetting the mechanism (SN1 vs SN2)

Why it happens:

Students assume all alcohols react the same way. They don’t check whether the reaction goes via SN1 or SN2.

How to avoid:

Tertiary alcohols ( $3^\circ$ ): React via SN1 — form a stable carbocation → fastest.
Secondary alcohols ( $2^\circ$ ): React via SN1 (with some SN2 character) — moderate.
Primary alcohols ( $1^\circ$ ): React via SN2 — slowest because no stable carbocation forms.
Result: $3^\circ > 2^\circ > 1^\circ$

✗ Mistake 3: Ignoring the role of anhydrous ZnCl₂

Why it happens:

Students think HCl alone is enough. They forget that ZnCl₂ acts as a Lewis acid to help break the C–OH bond.

How to avoid:

ZnCl₂ coordinates with the oxygen of the alcohol, making it a better leaving group (as $OH_2^+$ ).
This is crucial for primary alcohols — without ZnCl₂, they barely react.
For tertiary alcohols, the reaction is fast even without ZnCl₂, but the reagent ensures clarity in the test.

✗ Mistake 4: Misreading the answer options

Why it happens:

Options like (B) $1^\circ < 2^\circ > 3^\circ$ are deliberately confusing. Students pick the first one that looks correct without checking the inequality signs.

How to avoid:

Write the order clearly: $3^\circ > 2^\circ > 1^\circ$
Match exactly with the option: (C) $3^\circ > 2^\circ > 1^\circ$
Double-check that the inequality signs point in the correct direction.

✓ Correct Answer

Option (C): $3^\circ > 2^\circ > 1^\circ$

Reason:

$3^\circ$ alcohols form the most stable carbocation → fastest SN1 reaction.
$2^\circ$ alcohols react slower.
$1^\circ$ alcohols react slowest (SN2, no carbocation stability).

📌 Quick Revision Tip

Alcohol type	Mechanism	Relative rate
$3^\circ$	SN1	Fastest
$2^\circ$	SN1 (some SN2)	Moderate
$1^\circ$	SN2	Slowest

Mnemonic: “Tertiary is Terrific, Secondary is So-so, Primary is Painfully slow” in Lucas test.

←Previous questionQ2 — How many alcohols with molecular formula C₄H₁₀O are chiral in nature? …→Next questionQ4 — CH₃CH₂OH can be converted into CH₃CHO by ______. (i) catalytic hydroge…

Q.What is the correct order of reactivity of alcohols in the following reaction?
$R\text{-}OH + HCl \xrightarrow{\text{anhyd. } ZnCl_2} R\text{-}Cl + H_2O$

Concept understanding — Williamson Ether Synthesis

Williamson Ether Synthesis: From Intuition to Mechanism

The Core Idea

The Mechanism (SN2)

The Practical Rule

How to Choose the Alkoxide

A Common Exam Trap

The Big Picture

Why this formula?

Williamson Ether Synthesis: Why the Key Principles Hold

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

The Problem with Alcohols

The Solution: Deprotonation

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

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

Why Primary Halides Work Best

Why Tertiary Halides Fail

3. Why the Leaving Group Must Be Good

The Role of the Halide

Why Fluoride Fails

4. Why the Alkoxide Must Be the Nucleophile (Not the Halide)

The "Wrong Way" Problem

Summary: The Three Pillars of Williamson Ether Synthesis

Quick Exam Tip

Method: Lucas Test (Reactivity of Alcohols with HCl/ZnCl₂)

Steps to determine the correct order:

Final Answer:

✗ Mistake 1: Confusing the reaction with Williamson Ether Synthesis

✗ Mistake 2: Forgetting the mechanism (SN1 vs SN2)

✗ Mistake 3: Ignoring the role of anhydrous ZnCl₂

✗ Mistake 4: Misreading the answer options

✓ Correct Answer

📌 Quick Revision Tip

Q.What is the correct order of reactivity of alcohols in the following reaction? R-OH+HCl→anhyd. ZnCl2R-Cl+H2OR\text{-}OH + HCl \xrightarrow{\text{anhyd. } ZnCl_2} R\text{-}Cl + H_2OR-OH+HClanhyd. ZnCl2​​R-Cl+H2​O

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Q.What is the correct order of reactivity of alcohols in the following reaction?
$R\text{-}OH + HCl \xrightarrow{\text{anhyd. } ZnCl_2} R\text{-}Cl + H_2O$

Unlock everything free for 14 days