Q.Ethanol to propane nitrile
Concept understanding — Nucleophilic Substitution Reactions of Haloalkanes
Because the C(delta+)-X(delta-) bond makes the carbon electron-poor, a wide range of nucleophiles displace X- there, each giving a distinct functional group -- this table of named substitutions is tested heavily. Aqueous KOH or moist Ag2O/H2O gives an alcohol (hydrolysis). Alcoholic ammonia gives a primary amine, but with EXCESS haloalkane the amine keeps reacting further (ethylamine -> diethylamine -> triethylamine -> a quaternary ammonium salt) -- ammonolysis. Alcoholic KCN gives an alkyl cyanide/nitrile (attack through carbon), while alcoholic AgCN instead gives the alkyl isocyanide (attack through nitrogen) -- cyanide is an 'ambident nucleophile', with two different atoms able to bond, and which one attacks depends on whether the metal (K, soft/ionic vs Ag, forms a more covalent, nitrogen-first bond) favours the softer carbon end or the harder nitrogen end. The same ambident idea repeats with the nitrite ion: sodium/potassium nitrite gives an alkyl nitrite ester (O-attack, R-O-N=O), while silver nitrite instead gives a nitroalkane (N-attack, R-NO2). Sodium/potassium hydrogen sulphide gives a thiol. Sodium alkoxide gives an ether (Williamson ether synthesis) -- and using two different alkyl partners builds an unsymmetrical/mixed ether. Solvent choice matters for how fast these SN2-type reactions run: a polar APROTIC solvent such as DMF leaves the nucleophile (e.g. CN-) 'naked' and highly reactive (no hydrogen-bonding cage around it), so the same substitution runs distinctly faster in DMF than in a protic solvent like water, ethanol or methanol, which instead solvate and blunt the nucleophile.
Convert ethanol to bromoethane, then substitute with KCN -- the nitrile carbon supplied by CN makes the 2-carbon halide into a 3-carbon nitrile.
Ethanol --HBr (or PBr3)--> bromoethane --KCN, alc.--> propanenitrile.
Step 1. Convert the alcohol to the halide. Ethanol (CH3CH2OH, 2 carbons) is converted to bromoethane using HBr (with, e.g., NaBr/H2SO4 for a clean primary bromide, section 10.3.1) or with PBr3: CH3CH2OH + HBr -> CH3CH2Br + H2O.
Step 2. Substitute with cyanide, adding the third carbon. Bromoethane treated with alcoholic KCN gives the nitrile (Table 10.3, entry 5), and because the nitrile's own carbon (from CN-) becomes part of the product chain, the 2-carbon ethyl group becomes a 3-carbon nitrile: CH3CH2Br + KCN (alc.) -> CH3-CH2-CN (propanenitrile) + KBr.
Ethanol --HBr--> bromoethane --KCN, alc.--> propanenitrile.
Alcohol to halide, then halide to nitrile via ionic KCN, tracking that the nitrile step adds one extra carbon to the chain.
- Forgetting that the product nitrile has ONE MORE carbon than the starting alkyl halide (the CN- ion supplies its own carbon), which is exactly why a 2-carbon ethanol correctly leads to a 3-carbon nitrile, not a 2-carbon one.
- CBSE 2026Set ANNUAL1 markMCQQ.Acetonitrile may be prepared by heating the following reactants:(a) Ethyl chloride with alcoholic KCN(b) Ethyl chloride with alcoholic AgCN(c) Methyl chloride with alcoholic KCN(d) Methyl chloride with alcoholic AgCN
›Reveal solutionSolution
Acetonitrile (CH3CN, 2 carbons) is made by heating methyl chloride with alcoholic KCN.
Acetonitrile is CH3−C≡N, which has only 2 carbon atoms (one from the original methyl group, one from the CN group). It is prepared by heating a haloalkane with alcoholic potassium cyanide (KCN), which provides the nucleophilic CN− ion (via the more covalent, less ionic KCN reacting through carbon to give the nitrile, C-substitution product):
CH3Cl+KCNalc.CH3CN+KCl
Ethyl chloride would give propionitrile (C2H5CN, 3 carbons) instead, and AgCN (being more ionic/covalent through N) tends to favour the isocyanide (isonitrile) product rather than the nitrile. So methyl chloride with alcoholic KCN correctly gives acetonitrile.
✓Final answer(c) Methyl chloride with alcoholic KCN.
- CBSE 2025Set ANNUAL1 markMCQQ.Ethylidene chloride on treatment with aqueous KOH gives :(a) formaldehyde(b) acetaldehyde(c) glyoxal(d) ethylene glycol
›Reveal solutionSolution
CH3-CHCl2 (ethylidene chloride, a gem-dihalide) hydrolyses with aqueous KOH to give acetaldehyde (CH3CHO), because the intermediate gem-diol is unstable and loses water to form a carbonyl group.
Ethylidene chloride is 1,1-dichloroethane, CH3-CHCl2 — both chlorines sit on the SAME carbon (a geminal dihalide), unlike ethylene dichloride (CH2Cl-CH2Cl), which has one chlorine on each carbon.
When a gem-dihalide is treated with aqueous KOH, nucleophilic substitution (OH- replacing Cl-, twice) first gives a geminal diol:
CH3-CHCl2 + 2KOH -> CH3-CH(OH)2 + 2KCl
A carbon bearing two -OH groups on the same carbon (a gem-diol) is unstable and spontaneously eliminates a molecule of water to form a carbonyl (C=O) group:
CH3-CH(OH)2 -> CH3-CHO + H2O
So the overall product is acetaldehyde (CH3CHO), not formaldehyde (which would come from CH2Cl2, methylene chloride), not glyoxal (which needs two carbonyls, from a compound like CHCl2-CHCl2), and not ethylene glycol (which comes from vicinal dihalide CH2Cl-CH2Cl via simple substitution, not elimination of water).
✓Final answerThe correct option is (b) acetaldehyde — hydrolysis of the gem-dihalide CH3CHCl2 gives an unstable gem-diol that loses water to form CH3CHO.
- CBSE 2024Set ANNUAL1 markQ.Complete the reaction: CH3CH2ClAgCNalc.Δ?
›Reveal solutionSolution
AgCN attacks through N (covalent Ag–C bond), giving ethyl isocyanide, CH3CH2NC, not the nitrile.
AgCN is predominantly covalent (Ag–C bond has significant covalent character), so the more electronegative nitrogen atom of the cyanide ion is the one that attacks the alkyl halide's electrophilic carbon (as opposed to KCN, which is ionic and reacts through carbon). This gives the isocyanide (carbylamine) rather than the nitrile:
CH3CH2Cl+AgCNalc.ΔCH3CH2−NC+AgCl
✓Final answerEthyl isocyanide (ethyl carbylamine), CH3CH2−NC.
- CBSE 2022Set ANNUAL1 markQ.Write the product formed when alkyl halide reacts with silver nitrite.
›Reveal solutionSolution
Alkyl halides with silver nitrite give nitroalkanes as the major product.
Silver nitrite (AgNO2) is largely covalent, and in the nitrite ion the nitrogen atom (rather than oxygen) is the stronger nucleophile that attacks the electrophilic carbon of the alkyl halide (SN2), so the major product is the nitroalkane (C-bonded), with the metal halide as byproduct:
R−X+AgNO2→R−NO2 (nitroalkane)+AgX↓
(In contrast, the ionic sodium/potassium nitrite reacts predominantly through oxygen to give an alkyl nitrite, R–O–N=O, as the major product.)
✓Final answerNitroalkane, R–NO2
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