Q.(a) Write the mechanism of the following reaction: $2\,CH_3CH_2OH \xrightarrow[413\,K]{H^+} CH_3CH_2OCH_2CH_3 + H_2O$

Showing the 12 most recent of 13 on this concept.

1. CBSE 2026Set V11 mark
   Q.In Williamson ether synthesis, an alkyl halide is allowed to react with ________.

   ›Reveal solutionSolution

   Williamson synthesis couples an alkyl halide with a sodium alkoxide (or phenoxide) to give an ether.

   In Williamson ether synthesis, an alkyl halide is treated with a sodium alkoxide (or sodium phenoxide). The alkoxide ion acts as a nucleophile and displaces the halide by an $S_N2$ reaction, forming the ether: …

   🔒 Start your 14-day free trial to see the full solution →
2. CBSE 2026Set ANNUAL1 markMCQ
   Q.CH3CH2O(-) Na(+) + CH3Cl --> P. Product 'P' is

   (a) CH3CH2CH3

   (b) CH3CH2OCH3

   (c) CH3CH=CH2

   (d) CH3CH2OH

   ›Reveal solutionSolution

   The Williamson ether synthesis reacts an alkoxide ion with a primary alkyl halide via SN2 substitution to form an ether.

   CH3CH2O-Na+ (sodium ethoxide) is a strong nucleophile. It attacks the electrophilic carbon of CH3Cl (methyl chloride, a primary halide well-suited to SN2) from the backside, displacing the chloride leaving group:

   CH3CH2O- + CH3-Cl -> CH3CH2-O-CH3 + Cl-

   …

   🔒 Start your 14-day free trial to see the full solution →
3. CBSE 2025Set 56/6/11 markMCQ
   Q.Williamson synthesis of preparing unsymmetrical ether is : (A) $S_N1$ reaction (B) $S_N2$ reaction (C) Electrophilic addition reaction (D) Elimination reaction

   ›Reveal solutionSolution

   Williamson ether synthesis is a classic $S_N2$ reaction where an alkoxide ion attacks a primary alkyl halide, making it the correct answer for preparing unsymmetrical ethers.

   Williamson synthesis is one of the most reliable methods for making ethers, especially unsymmetrical ones. The key insight is that you need to control which carbon gets the oxygen — and that control comes from the reaction mechanism itself.

   The reaction works by treating an alcohol with a strong base (like NaH or Na metal) to generate an alkoxide ion. This alkoxide is a strong nucleophile. You then add an alkyl halide (or tosylate), and the alkoxide attacks the electrophilic carbon of the alkyl halide, displacing the halide ion.

   Why does this have to be $S_N2$? Because the alkoxide is a strong base and a good nucleophile, and the alkyl halide is typically primary (to avoid elimination side reactions). The reaction proceeds through a single step — backside attack with inversion of configuration — which is the hallmark of $S_N2$.

   Let's walk through the reasoning step by step.

   1. Identify the reactants and products. In Williamson synthesis, you have:

   $R-O^- \text{ (alkoxide)} + R'-X \text{ (alkyl halide)} \rightarrow R-O-R' + X^-$

   The oxygen from the alkoxide ends up bonded to the carbon that originally held the halogen. This is a substitution — the halide is replaced by the alkoxide.

   2. Consider the mechanism.

      The alkoxide ion is negatively charged and electron-rich. It attacks the carbon bearing the halogen, which is partially positive due to the electronegativity of the halogen. This attack happens from the opposite side of the leaving group, in a single concerted step. There is no carbocation intermediate.

   3. Rule out $S_N1$.

      $S_N1$ reactions proceed through a carbocation intermediate. In Williamson synthesis, if you use a secondary or tertiary alkyl halide, elimination (forming an alkene) becomes a major side reaction. The method works best with primary alkyl halides, where $S_N2$ is favoured and $S_N1$ is essentially impossible (primary carbocations are too unstable).

   4. Rule out electrophilic addition.

      Electrophilic addition involves adding something across a double bond. There is no double bond in the reactants here — just an alkoxide and an alkyl halide. This mechanism is irrelevant.

   5. Rule out elimination. …

   🔒 Start your 14-day free trial to see the full solution →
4. CBSE 2025Set ANNUAL1 markMCQ
   Q.The compound formed by Williamson synthesis is:

   (a) Ether

   (b) Alcohol

   (c) Alkylhalide

   (d) Sodium alkoxide

   ›Reveal solutionSolution

   Williamson synthesis reacts a sodium alkoxide with a primary alkyl halide (SN2) to form an ether.

   R-ONa + R'-X → R-O-R' + NaX

   A sodium (or potassium) alkoxide, generated by treating an alcohol with sodium metal, acts as a strong nucleophile and displaces the halide ion from a primary alkyl halide in an SN2 reaction, forming an ether linkage (R-O-R'). This method can make both symmetrical ethers (same R and R') and uns …

   🔒 Start your 14-day free trial to see the full solution →
5. CBSE 2025Set ANNUAL1 markMCQ
   Q.Which reaction is used for the preparation of symmetric and asymmetric ethers?

   (a) Finkelstein reaction

   (b) Reimer-Tiemann reaction

   (c) Williamson synthesis

   (d) Etard reaction

   ›Reveal solutionSolution

   The Williamson ether synthesis (SN2 reaction of an alkoxide with an alkyl halide) is the general method for preparing both symmetrical and unsymmetrical ethers.

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

   …

   🔒 Start your 14-day free trial to see the full solution →
6. CBSE 2024Set 56/1/11 markMCQ
   Q.The reaction of an alkyl halide with sodium alkoxide forming ether is known as: (A) Wurtz reaction (B) Reimer-Tiemann reaction (C) Williamson synthesis (D) Kolbe reaction

   ›Reveal solutionSolution

   The reaction of an alkyl halide with a sodium alkoxide to form an ether is a classic example of nucleophilic substitution, specifically known as the Williamson ether synthesis. The correct option is (C).

   The question describes a fundamental reaction in organic chemistry for synthesizing ethers. To understand why a specific name is associated with it, we need to break down the roles of the reactants and the type of transformation occurring.

   Concept and Intuition: Nucleophilic Substitution for Ether Formation

   Ethers have the general structure $R-O-R'$, where $R$ and $R'$ are alkyl or aryl groups. To form an ether, we need to create a new carbon-oxygen bond. One very effective way to do this is through a nucleophilic substitution reaction.

   Imagine an alkyl halide, $R-X$, where $X$ is a good leaving group (like $\text{Cl}$, $\text{Br}$, $\text{I}$). The carbon atom bonded to the halogen is partially positive ($\delta^+$) because the halogen is electronegative. This makes the carbon an electrophilic center, meaning it's susceptible to attack by electron-rich species.

   Now consider a sodium alkoxide, $R'-O^- Na^+$. The alkoxide ion ($R'-O^-$) is a strong nucleophile because the oxygen atom carries a full negative charge and has lone pairs of electrons available to donate. It's also a strong base.

   When these two species meet, the nucleophilic alkoxide oxygen attacks the electrophilic carbon of the alkyl halide, displacing the halide ion. This is a classic $S_N2$ (bimolecular nucleophilic substitution) pathway, leading directly to the formation of an ether.

   Step-by-Step Explanation

   1. Identify the Reactants and Product:

      • Alkyl halide: An organic compound with a halogen atom ($X$) bonded to an alkyl group ($R$). General formula: $R-X$.
      • Sodium alkoxide: A salt formed from an alcohol and sodium, containing an alkoxide ion ($R'-O^-$) and a sodium cation ($Na^+$). General formula: $R'-O^- Na^+$.
      • Ether: An organic compound with an oxygen atom bonded to two alkyl or aryl groups. General formula: $R-O-R'$.

   2. Role of Reactants:

      • The alkoxide ion ($R'-O^-$) acts as a strong nucleophile, seeking an electron-deficient center.
      • The alkyl halide ($R-X$) acts as an electrophile, with the carbon atom bonded to the halogen being the site of nucleophilic attack. The halogen ($X$) serves as a good leaving group.

   3. Mechanism of Reaction:

      The reaction proceeds via an $S_N2$ mechanism. The alkoxide nucleophile attacks the carbon atom bearing the halogen from the backside, simultaneously displacing the halide ion. This is a concerted, one-step process.

   $R-X + R'-O^- Na^+ \longrightarrow R-O-R' + NaX$

   For example, if we react bromoethane with sodium methoxide:
   

   $\text{CH}_3\text{CH}_2-\text{Br} + \text{CH}_3-\text{O}^- \text{Na}^+ \longrightarrow \text{CH}_3\text{CH}_2-\text{O}-\text{CH}_3 + \text{NaBr}$

   (Bromoethane) + (Sodium methoxide) $\longrightarrow$ (Ethyl methyl ether) + (Sodium bromide) …
   

   🔒 Start your 14-day free trial to see the full solution →
7. CBSE 2024Set D1 markMCQ
   Q.When ethyl bromide is treated with dry silver oxide, then we get

   (a) Diethyl ether

   (b) Ethanal

   (c) Ethane

   (d) Ethene

   ›Reveal solutionSolution

   2 C2H5Br + Ag2O -> C2H5-O-C2H5 + 2 AgBr (diethyl ether).

   Dry silver oxide acts as a source of oxide/hydroxide and promotes ether formation from alkyl halides:

   2 C2H5Br + Ag2O (dry) -> C2H5-O-C2H5 + 2 AgBr

   …

   🔒 Start your 14-day free trial to see the full solution →
8. CBSE 2024Set D1 markMCQ
   Q.Alkyl halides form ethers by reacting with which of the following?

   (a) Dry Ag2O

   (b) Moist Ag2O

   (c) Dry ZnO

   (d) Moist ZnO

   ›Reveal solutionSolution

   Alkyl halides + dry Ag2O -> ethers.

   When an alkyl halide is warmed with dry silver oxide, an ether is formed:

   2 R-X + Ag2O (dry) -> R-O-R + 2 AgX

   The silver ion helps remove the halide, and the oxide oxygen bridges two alkyl groups to give the ether.

   …

   🔒 Start your 14-day free trial to see the full solution →
9. CBSE 2023Set ANNUAL1 mark
   Q.In Williamson synthesis the ________ reacts with sodium alkoxide and give dialkyl ether.

   ›Reveal solutionSolution

   Williamson synthesis prepares an ether by reacting a sodium alkoxide with an alkyl halide.

   In Williamson's ether synthesis, an alkyl halide (R-X) undergoes nucleophilic substitution (SN2) with a sodium alkoxide (R'-ONa): R-X + R'-ONa -> R-O-R' + NaX. Pri …

   🔒 Start your 14-day free trial to see the full solution →
10. CBSE 2023Set ANNUAL1 mark
    Q.Write chemical equation to prepare salicylic acid from phenol.

    ›Reveal solutionSolution

    Salicylic acid is made from phenol via Kolbe's reaction: phenol is first converted to sodium phenoxide, which reacts with CO2 and is then acidified.

    Step 1: Phenol is treated with NaOH to form sodium phenoxide (more nucleophilic than phenol itself):

    C6H5OH + NaOH -> C6H5ONa + H2O

    Step 2 (Kolbe's reaction): Sodium phenoxide is heated with CO2 at about 400 K under 4-7 atm pressure. The electrophilic carbon of CO2 is attacked at the ortho position (electrophilic substitution), and after tautomerisation, sodium salicylate is obtained:

    C6H5ONa + CO2 --(400 K, 4-7 atm)--> sodium salicylate (2-hydroxybenzoate, Na salt)

    …

    🔒 Start your 14-day free trial to see the full solution →
11. CBSE 2022Set ANNUAL1 markMCQ
    Q.Molecular formula of ethers is:

    (a) $C_nH_{2n+1}O$

    (b) $C_nH_{2n}O$

    (c) $C_nH_{2n+2}O$

    (d) $C_nH_{2n-2}O$

    ›Reveal solutionSolution

    An ether $R-O-R'$ is a saturated, acyclic, oxygen-containing compound; its general molecular formula is $C_nH_{2n+2}O$. Option (C).

    An ether has the structure $R-O-R'$ with only single (saturated) bonds and no ring.

    Check with a simple member: dimethyl ether $CH_3-O-CH_3$ has $n = 2$: formula $C_2H_6O$. Substituting into $C_nH_{2n+2}O$ gives $C_2H_{6}O$ ✓.

    …

    🔒 Start your 14-day free trial to see the full solution →
12. CBSE 2020Set ANNUAL1 markMCQ
    Q.C2H5Br + C2H5ONa -> C2H5OC2H5 + NaBr. The name of the above reaction is

    (a) Riemer-Tiemann reaction

    (b) Aldol condensation

    (c) Williamson synthesis

    (d) Kolbe's reaction

    ›Reveal solutionSolution

    C2H5Br + C2H5ONa -> C2H5-O-C2H5 + NaBr is the Williamson ether synthesis.

    The Williamson synthesis is the standard method for preparing symmetrical and unsymmetrical ethers: an alkyl halide is treated with sodium alkoxide (formed by reacting an alcohol with sodium metal). The alkoxide ion (a strong nucleophile) attacks the electrophilic carbon of the haloalkane in an SN2 mechanism, displacing the halide ion (leaving group) and forming the C-O-C ether linkage:

    C2H5ONa + C2H5Br -> C2H5-O-C2H5 (diethyl ether) + NaBr

    …

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