Bromoethane is treated separately with the reagents given in List-I. Match List-I with List-II, which gives the main organic product of each reaction.
| List-I (Reagent and condition) | List-II (Main organic product) |
|---|---|
| (A) Alcoholic KOH, heat | (I) Butane |
| (B) Aqueous KOH | (II) Ethanol |
| (C) Magnesium in dry ether, and then water | (III) Ethene |
| (D) Sodium in dry ether | (IV) Ethane |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → A-III, B-II, C-IV, D-I
The same haloalkane gives different products with different reagents. KOH in alcohol acts as a base and causes elimination; KOH in water gives OH− ions that cause substitution. With metals in dry ether, magnesium forms a Grignard reagent and sodium joins two alkyl groups (Wurtz reaction).
(A) Alcoholic KOH, heat — This matches (III) Ethene. A hydrogen atom from the β-carbon and the bromine atom from the α-carbon are removed (β-elimination), and an alkene is formed:
\( \displaystyle \ce{CH3CH2Br + KOH} \xrightarrow[\Delta]{\text{alcohol}} \ce{CH2=CH2 + KBr + H2O} \)
(B) Aqueous KOH — This matches (II) Ethanol. The hydroxide ion replaces the bromine atom by nucleophilic substitution, and an alcohol is formed:
\( \displaystyle \ce{CH3CH2Br + KOH(aq) -> CH3CH2OH + KBr} \)
(C) Magnesium in dry ether, and then water — This matches (IV) Ethane. Magnesium gives ethylmagnesium bromide. Water is a source of protons, so it turns the Grignard reagent into the hydrocarbon with the same number of carbon atoms:
\( \displaystyle \ce{CH3CH2Br + Mg} \xrightarrow{\text{dry ether}} \ce{CH3CH2MgBr} \)
\( \displaystyle \ce{CH3CH2MgBr + H2O -> CH3CH3 + Mg(OH)Br} \)
(D) Sodium in dry ether — This matches (I) Butane. In the Wurtz reaction two ethyl groups join, so the hydrocarbon has double the number of carbon atoms (four):
\( \displaystyle \ce{2CH3CH2Br + 2Na} \xrightarrow{\text{dry ether}} \ce{CH3CH2CH2CH3 + 2NaBr} \)
NCERT (2026-27 edition):
6.7.1 Reactions of Haloalkanes – “When a haloalkane with β-hydrogen atom is heated with alcoholic solution of potassium hydroxide, there is elimination of hydrogen atom from β-carbon and a halogen atom from the α-carbon atom. As a result, an alkene is formed as a product.”
“The reaction between CH3Cl and hydroxide ion to yield methanol and chloride ion follows a second order kinetics, i.e., the rate depends upon the concentration of both the reactants.”
“…These reagents are obtained by the reaction of haloalkanes with magnesium metal in dry ether.” … “Grignard reagents are highly reactive and react with any source of proton to give hydrocarbons. Even water, alcohols, amines are sufficiently acidic to convert them to corresponding hydrocarbons.”
“Alkyl halides react with sodium in dry ether to give hydrocarbons containing double the number of carbon atoms present in the halide. This reaction is known as Wurtz reaction.”