What will be the major product when m-cresol is reacted with propargyl bromide (HC≡C—CH2Br) in prese

Q: Why doesn't the base attack the propargyl bromide directly?

Potassium carbonate is a relatively weak, bulky base. Acid-base reactions (like abstracting a proton from phenol) are extremely fast and have a much lower activation energy compared to a substitution or elimination reaction on an alkyl halide.

Q: Why does the reaction happen at the oxygen and not on the benzene ring?

This is a great question! While the phenoxide ion has resonance structures with negative charge on the ortho and para carbons, O-alkylation is heavily favored in polar aprotic solvents like acetone. The solvent doesn't hydrogen-bond to the oxygen, leaving it highly reactive and accessible for the $\text{S}_\text{N}2$ attack.

Q: What is the specific role of acetone in this reaction?

Acetone acts as a polar aprotic solvent. It dissolves the ionic intermediates but does not solvate the nucleophile (the phenoxide oxygen) strongly. This lack of a 'solvent cage' around the oxygen significantly increases its nucleophilicity, driving the $\text{S}_\text{N}2$ reaction forward.

Q: Could this reaction proceed via an $\text{S}_\text{N}1$ mechanism?

No, it is highly unlikely. Propargyl bromide is a primary alkyl halide, which strongly resists forming a primary carbocation (even though it would be resonance stabilized, $\text{S}_\text{N}2$ is much faster). Furthermore, the presence of a strong nucleophile like phenoxide strongly dictates an $\text{S}_\text{N}2$ pathway.

Q: What would happen if we used water as the solvent instead of acetone?

Water is a polar protic solvent. It would form strong hydrogen bonds with the negatively charged oxygen of the phenoxide ion, effectively 'caging' it. This would drastically reduce the oxygen's nucleophilicity, slowing down the reaction and potentially increasing the amount of C-alkylated side products.

Question

What will be the major product when $m$-cresol is reacted with propargyl bromide ($	ext{HC}\equiv	ext{C}—	ext{CH}_2	ext{Br}$) in presence of $	ext{K}_2	ext{CO}_3$ in acetone?

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine you are in a laboratory, and you are handed a flask containing $m$-cresol, a classic phenol derivative. To this, you add propargyl bromide ($	ext{HC}\equiv	ext{C}—	ext{CH}_2	ext{Br}$), potassium carbonate ($	ext{K}_2	ext{CO}_3$), and acetone as the solvent. What exactly is going to happen here?

This setup is the textbook recipe for a very famous organic reaction: the **Williamson Ether Synthesis**. To predict the major product, we need to break down the reaction into two logical, atomic steps.

### Step 1: The Acid-Base Reaction
Phenols are uniquely acidic compared to regular alcohols because their conjugate base is stabilized by resonance. Potassium carbonate ($	ext{K}_2	ext{CO}_3$) acts as a mild base in this environment.

The very first thing that occurs is an acid-base reaction. The carbonate ion abstracts the acidic phenolic proton from the hydroxyl group of $m$-cresol.

This rapid proton transfer generates the **phenoxide ion**. Notice the negative charge on the oxygen atom. Because this negative charge is delocalized over the aromatic benzene ring, the ion is highly stable, yet it remains a potent nucleophile ready for the next step.

### Step 2: The Nucleophilic Attack ($	ext{S}_	ext{N}2$)
Now, our nucleophilic phenoxide ion sets its sights on the second reagent: propargyl bromide. The carbon atom attached to the bromine is highly electrophilic due to the electron-withdrawing nature of the halogen.

Here is where the solvent plays a critical role. Acetone is a **polar aprotic solvent**. It dissolves the reactants perfectly but does not heavily solvate (or cage) the nucleophilic oxygen with hydrogen bonds. This leaves the oxygen "naked" and highly reactive.

The phenoxide oxygen attacks the $	ext{CH}_2$ carbon of propargyl bromide from the backside, initiating an **$	ext{S}_	ext{N}2$ (Substitution Nucleophilic Bimolecular)** mechanism. As the new carbon-oxygen bond forms, the carbon-bromine bond simultaneously breaks, expelling the bromide ion ($	ext{Br}^-$) as a leaving group.

### The Final Product and the C- vs O-Alkylation Debate
The result of this elegant dance of electrons is the formation of a new ether linkage ($	ext{C-O-C}$). The propargyl group is now directly attached to the oxygen atom, yielding **propargyl $m$-tolyl ether**.

You might wonder: since the phenoxide ion has resonance structures that place negative charge on the ortho and para carbon atoms of the ring, why didn't the alkyl group attach there? This is the classic **C-alkylation vs. O-alkylation** dilemma.

The answer again lies in the solvent. Polar aprotic solvents like acetone strongly favor **O-alkylation** because they do not hinder the highly electronegative oxygen atom. If a polar protic solvent like water or alcohol were used, the oxygen would be heavily hydrogen-bonded, potentially forcing the reaction to occur at the less hindered carbon atoms of the ring.

By carefully observing the reagents and the solvent, we can confidently conclude that the major product is the ether shown in option (a).

Animated Solution for Chemistry - Alcohols, Phenols, Ethers: What will be the major product when $m$ -cresol is reacted with propargyl bromide ( $HC \equiv C — CH_{2} Br$ ) in presence of $K_{2} CO_{3}$ in acetone?

Select Answer:

Visualized Solution

\text{Analyzing the Reactants}

\text{Acid-Base Reaction}

\text{Formation of Phenoxide Ion}

\text{Nucleophilic Attack (S}_\text{N}\text{2)}

\text{Formation of Ether}

\text{Final Product}

\text{The Way Forward}

The Sigma Insight: Ethers

Analyzing the Setup Imagine you are in a laboratory, and you are handed a flask containing $m$ -cresol, a classic phenol derivative

Step 1

Step 2

The Final Product and the C- vs O-Alkylation Debate The result of this elegant dance of electrons is the formation of a new ether linkage ( $C-O-C$ )