Explain the following with an example. (i) Kolbe's reaction. (ii) Reimer-Tiemann reaction. (iii) Williamson et

Exercises · 7.18

Q. Explain the following with an example.

(i) Kolbe's reaction.

(ii) Reimer-Tiemann reaction.

(iii) Williamson ether synthesis.

(iv) Unsymmetrical 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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All four reactions are key name reactions in organic chemistry. Kolbe's reaction and Reimer-Tiemann reaction are methods to introduce a functional group onto a phenol ring. Williamson ether synthesis is the most reliable way to make ethers, and unsymmetrical ethers are simply ethers with two different alkyl/aryl groups attached to oxygen.

Let's go through each one with a clear explanation and an example.

(i) Kolbe's Reaction

Concept: This is a method to introduce a carboxyl group ( $-COOH$ ) directly onto the benzene ring of phenol. Phenol is activated towards electrophilic substitution, and under specific conditions, carbon dioxide acts as the electrophile.

Why it works: Phenoxide ion (formed by treating phenol with NaOH) is even more activated than phenol itself. The negative charge on oxygen makes the ortho positions extremely electron-rich. When $CO_2$ is forced in under high pressure and heat, it attacks at the ortho position. The product, after acidification, is salicylic acid (ortho-hydroxybenzoic acid).

Example:

Phenol ( $C_6H_5OH$ ) is treated with sodium hydroxide to form sodium phenoxide ( $C_6H_5ONa$ ).
This is then heated with carbon dioxide ( $CO_2$ ) under pressure (about 4-7 atm) at 125-150°C.
The intermediate is a sodium salt of salicylic acid.
Acidification with dilute $HCl$ gives salicylic acid.

C_6H_5ONa + CO_2 \xrightarrow{125-150^\circ C, \text{pressure}} \text{Intermediate} \xrightarrow{H^+} \text{o-Hydroxybenzoic acid (Salicylic acid)}

Watch out

A common mistake is to think the product is meta-substituted. Remember, the phenoxide ion directs ortho/para, but steric hindrance from the incoming $CO_2$ group usually forces the product to be ortho (salicylic acid). A small amount of para product is also formed, but the main product is ortho.

(ii) Reimer-Tiemann Reaction

Concept: This is a method to introduce an aldehyde group ( $-CHO$ ) onto the benzene ring of phenol. It uses chloroform ( $CHCl_3$ ) in the presence of a strong base.

Why it works: The base (usually NaOH) generates a highly reactive intermediate called dichlorocarbene ( $:CCl_2$ ) from chloroform. This carbene is an electrophile that attacks the electron-rich ortho position of the phenoxide ion. The resulting intermediate then undergoes hydrolysis to give the aldehyde.

Example:

Phenol is heated with chloroform ( $CHCl_3$ ) and aqueous sodium hydroxide ( $NaOH$ ) at about 60-70°C.
The product, after acidification, is salicylaldehyde (ortho-hydroxybenzaldehyde).

C_6H_5OH + CHCl_3 + 3NaOH \xrightarrow{60-70^\circ C} \text{Salicylaldehyde} + 3NaCl + 2H_2O

Tip

The key difference from Kolbe's reaction is the reagent: Kolbe uses $CO_2$ to give a carboxylic acid, while Reimer-Tiemann uses $CHCl_3$ to give an aldehyde. Both reactions are ortho-selective for the same reason — the phenoxide ion directs the electrophile to the ortho position.

(iii) Williamson Ether Synthesis

Concept: This is the most general and reliable method for preparing both symmetrical and unsymmetrical ethers. It involves an $S_N2$ reaction between an alkoxide ion (or phenoxide ion) and a primary alkyl halide.

Why it works: The alkoxide ion ( $RO^-$ ) is a strong nucleophile. It attacks the electrophilic carbon of the alkyl halide ( $R'X$ ) in a backside attack, displacing the halide ion ( $X^-$ ). Because it's an $S_N2$ reaction, the alkyl halide must be primary (or methyl) for best yields. Secondary halides give some elimination, and tertiary halides give almost exclusively elimination (alkene).

Example:

To make ethoxyethane (diethyl ether):

Sodium ethoxide ( $C_2H_5ONa$ ) is prepared by reacting ethanol with sodium metal.
This is then reacted with ethyl bromide ( $C_2H_5Br$ ).
The product is diethyl ether.

C_2H_5ONa + C_2H_5Br \xrightarrow{S_N2} C_2H_5-O-C_2H_5 + NaBr

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Here is a clear, concept-first breakdown of the requested reactions, with a special focus on the Williamson Ether Synthesis method.

(iii) Williamson Ether Synthesis

Method Name: Nucleophilic Substitution (SN2) of an Alkoxide on an Alkyl Halide.

Concept: This is the most reliable laboratory method for making both symmetrical and unsymmetrical ethers. The key idea is that an alkoxide ion (a strong nucleophile and strong base) attacks a primary alkyl halide (the electrophile) to form an ether.

Steps:

Generate the Alkoxide: React an alcohol ( $ROH$ ) with a strong base like sodium metal ( $Na$ ) or sodium hydride ( $NaH$ ). This deprotonates the alcohol to form the alkoxide ion ( $RO^-$ ).

ROH + Na \rightarrow RO^-Na^+ + \frac{1}{2}H_2

Perform the SN2 Attack: The alkoxide ion ( $RO^-$ ) attacks the carbon atom bonded to the leaving group (usually $Cl$ , $Br$ , or $I$ ) on the alkyl halide ( $R'X$ ). This is a back-side attack, typical of SN2.

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

Critical Rule for Exam Success: To avoid elimination (forming an alkene instead of an ether), the alkyl halide must be primary ( $1^\circ$ ). If you use a secondary or tertiary halide, the strong alkoxide base will cause an E2 elimination reaction instead.

Example: Synthesis of Ethoxyethane (Diethyl ether) from ethanol and bromoethane.

Step 1: $CH_3CH_2OH + Na \rightarrow CH_3CH_2O^-Na^+ + \frac{1}{2}H_2$ (Sodium ethoxide)
Step 2: $CH_3CH_2O^-Na^+ + CH_3CH_2Br \rightarrow CH_3CH_2-O-CH_2CH_3 + NaBr$

(i) Kolbe’s Reaction

Method Name: Kolbe-Schmitt Reaction (Electrophilic Aromatic Substitution).

Concept: This reaction is used to introduce a carboxyl group ( $-COOH$ ) directly onto the benzene ring of phenol. The key is that the phenoxide ion (formed by treating phenol with a strong base) is so strongly activating that it can attack the weak electrophile, carbon dioxide ( $CO_2$ ).

Steps:

Activation: Phenol ( $C_6H_5OH$ ) is treated with sodium hydroxide ( $NaOH$ ) to form the more reactive sodium phenoxide ( $C_6H_5O^-Na^+$ ).
Carboxylation: The phenoxide ion is heated with carbon dioxide ( $CO_2$ ) under pressure (about 125°C, 4-7 atm). The $CO_2$ acts as an electrophile and attacks the ortho position of the ring.
Acidification: The resulting sodium salicylate is treated with a dilute acid (like $HCl$ ) to give the final product, salicylic acid (2-hydroxybenzoic acid).

Example:

C_6H_5OH \xrightarrow{NaOH} C_6H_5O^-Na^+ \xrightarrow{CO_2, 125^\circ C, \text{pressure}} \text{o-hydroxybenzoate} \xrightarrow{H^+} \text{Salicylic acid}

(ii) Reimer-Tiemann Reaction

Method Name: Carbene Insertion (Electrophilic Aromatic Substitution via Dichlorocarbene).

Concept: This reaction is used to introduce an aldehyde group ( $-CHO$ ) onto the benzene ring of phenol, specifically at the ortho position. The key intermediate is a highly reactive species called dichlorocarbene ( $:CCl_2$ ) .

Steps: …

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Here is a breakdown of the common mistakes students make regarding Williamson Ether Synthesis and how to avoid them, tailored for Indian exam preparation (JEE, NEET, CBSE).

The Core Concept (The "Why")

Williamson Ether Synthesis is the most reliable lab method to make ethers. It involves an SN2 reaction between an alkoxide ion (strong nucleophile) and a primary alkyl halide (good electrophile).

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

The key to success is choosing the correct combination of alkyl halide and alkoxide to avoid elimination (E2) side reactions.

Common Mistake #1: Using a Tertiary Alkyl Halide

The Mistake: Students try to make an ether like tert-butyl methyl ether by reacting sodium methoxide ( $\text{CH}_3\text{ONa}$ ) with tert-butyl bromide ( $(\text{CH}_3)_3\text{CBr}$ ).
Why it's wrong: The alkyl halide is tertiary. In an SN2 reaction, the nucleophile attacks from the back. A tertiary carbon is too crowded (steric hindrance). The alkoxide, being a strong base, will instead pull a proton from the tert-butyl group, causing an Elimination (E2) reaction. You get an alkene, not an ether.
How to Avoid: Never use a tertiary alkyl halide. If you need a tert-butyl ether, you must reverse the roles: use tert-butoxide ( $(\text{CH}_3)_3\text{CO}^- \text{K}^+$ $(CH_{3})_{3} CO^{-} K^{+}$ ) with a primary alkyl halide (like methyl iodide).
- Correct: $(\text{CH}_3)_3\text{CO}^- \text{K}^+ + \text{CH}_3\text{I} \rightarrow (\text{CH}_3)_3\text{C-O-CH}_3$
- Incorrect: $\text{CH}_3\text{O}^- \text{Na}^+ + (\text{CH}_3)_3\text{C-Br} \rightarrow \text{Alkene}$

Common Mistake #2: Forgetting the Alkoxide Preparation Step

The Mistake: Students write the reaction as if the alcohol and alkyl halide react directly (e.g., $\text{ROH} + \text{R'X} \rightarrow \text{ROR'}$ ).
Why it's wrong: Alcohols ( $\text{ROH}$ ) are weak nucleophiles. They cannot effectively attack an alkyl halide in an SN2 reaction. The reaction will not proceed.
How to Avoid: Always show the pre-formation of the alkoxide. The alcohol is first treated with a strong base (like Na, NaH, or KOH) to generate the alkoxide ion.
- Correct Sequence:
  1. $\text{ROH} + \text{Na} \rightarrow \text{RONa} + \frac{1}{2}\text{H}_2$
  2. $\text{RONa} + \text{R'X} \rightarrow \text{ROR'} + \text{NaX}$
- Exam Tip: In a reaction mechanism question, you must show the deprotonation step.

Common Mistake #3: Confusing the Reagent for Making Phenolic Ethers

The Mistake: Students try to make anisole ( $\text{C}_6\text{H}_5\text{OCH}_3$ ) by reacting phenol ( $\text{C}_6\text{H}_5\text{OH}$ ) with $\text{CH}_3\text{Br}$ directly.
Why it's wrong: Phenol is a weak acid, but it's still an alcohol. The $\text{OH}$ group is a poor leaving group and a weak nucleophile. You need to convert it to the phenoxide ion ( $\text{C}_6\text{H}_5\text{O}^-$ ), which is a much stronger nucleophile.
How to Avoid: Treat phenol with a base (like $\text{NaOH}$ $NaOH$ or $\text{KOH}$ $KOH$ ) first to form sodium phenoxide. Then react with the alkyl halide.
- Correct: $\text{C}_6\text{H}_5\text{OH} + \text{NaOH} \rightarrow \text{C}_6\text{H}_5\text{ONa} + \text{H}_2\text{O}$ ; then $\text{C}_6\text{H}_5\text{ONa} + \text{CH}_3\text{Br} \rightarrow \text{C}_6\text{H}_5\text{OCH}_3 + \text{NaBr}$

Common Mistake #4: Not Recognizing the "Best" Route for Unsymmetrical Ethers

The Mistake: For an unsymmetrical ether like $\text{CH}_3\text{CH}_2\text{-O-CH}(\text{CH}_3)_2$ (ethyl isopropyl ether), students choose the wrong combination of alkoxide and alkyl halide.
Why it's wrong: There are two possible routes. One uses a primary alkoxide with a secondary halide (SN2 is possible, but slow). The other uses a secondary alkoxide with a primary halide (SN2 is fast). If you pick the route with a secondary alkyl halide, you risk elimination.
How to Avoid: Always use the smaller (less hindered) alkyl halide and the larger (more substituted) alkoxide. This minimizes steric hindrance at the SN2 transition state. …

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Q.Explain the following with an example.

(i) Kolbe's Reaction

(ii) Reimer-Tiemann Reaction

(iii) Williamson Ether Synthesis

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