Which of the following alkyl halides, on conversion into the Grignard reagent followed by treatment with water, gives n-pentane?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → All of these
A haloalkane reacts with magnesium in dry ether to give the Grignard reagent, RMgBr. Water then gives up a proton to R, so the bromine is replaced by hydrogen and the hydrocarbon RH is formed:
\( \displaystyle \ce{R-Br + Mg} \xrightarrow{\text{dry ether}} \ce{RMgBr} \xrightarrow{\ce{H2O}} \ce{R-H + Mg(OH)Br} \)
All three bromides have the same five-carbon straight chain; they differ only in the carbon that carries Br:
1-Bromopentane — \( \ce{CH3CH2CH2CH2CH2Br} \) → \( \ce{CH3CH2CH2CH2CH2MgBr} \) → \( \ce{CH3CH2CH2CH2CH3} \), n-pentane.
2-Bromopentane — \( \ce{CH3CH(Br)CH2CH2CH3} \) → \( \ce{CH3CH(MgBr)CH2CH2CH3} \) → \( \ce{CH3CH2CH2CH2CH3} \), n-pentane.
3-Bromopentane — \( \ce{CH3CH2CH(Br)CH2CH3} \) → \( \ce{CH3CH2CH(MgBr)CH2CH3} \) → \( \ce{CH3CH2CH2CH2CH3} \), n-pentane.
All three give n-pentane.
NCERT (2026-27 edition):
6.7.1 Reactions of Haloalkanes: 3. Reaction with metals – “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.”