The reaction in which a nucleophile replaces an existing group (the leaving group) in a molecule is called a nucleophilic substitution reaction. Haloalkanes are the substrates in these reactions. In this type of reaction, a nucleophile reacts with a haloalkane (the substrate) having a partial positive charge on the carbon atom bonded to the halogen. A substitution reaction takes place and the halogen atom, called the leaving group, departs as a halide ion. Since the substitution reaction is initiated by a nucleophile, it is called a nucleophilic substitution reaction.
It is one of the most useful classes of organic reactions of alkyl halides in which the halogen is bonded to an sp3 hybridised carbon.
What product is formed when ethyl bromide reacts with sodium ethoxide?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → Diethyl ether
As the passage describes, in nucleophilic substitution a nucleophile attacks the carbon bonded to the halogen and the halogen leaves as a halide ion. Sodium ethoxide supplies the ethoxide ion, \( \ce{C2H5O^-} \), as the nucleophile. It replaces the bromine of ethyl bromide, so an ethyl group and an ethoxy group are joined through oxygen:
\( \displaystyle \ce{CH3CH2-Br + CH3CH2-O^- Na^+ -> CH3CH2-O-CH2CH3 + NaBr} \)
The product is ethoxyethane, diethyl ether. A haloalkane reacting with a sodium alkoxide in this way is the Williamson synthesis of ethers.
Why the other options are incorrect
Option (1) Ethanol — Incorrect. Ethanol forms when the nucleophile is \( \ce{OH^-} \) (aqueous NaOH or KOH). Here the nucleophile is \( \ce{C2H5O^-} \), so the oxygen ends up between two ethyl groups, not bonded to H.
Option (2) Butane — Incorrect. Butane needs a new C–C bond between two ethyl groups, as in the Wurtz reaction with sodium metal in dry ether. The alkoxide forms a C–O bond.
Option (4) Methyl propyl ether — Incorrect. It has the same formula, \( \ce{C4H10O} \), but needs a methyl and a propyl group. Both groups here come from ethyl compounds.
NCERT (2026-27 edition):
2. Williamson synthesis – “It is an important laboratory method for the preparation of symmetrical and unsymmetrical ethers. In this method, an alkyl halide is allowed to react with sodium alkoxide.”
\( \displaystyle \ce{R-X + R'-O^- Na^+ -> R-O-R' + NaX} \)