An organic compound ‘A’ with molecular formula CH4O reacts with HI to give an alkyl iodide, which then reacts with potassium cyanide to produce ‘B’. Compound ‘B’ on reduction followed by reaction with HNO2 gives an alcohol, which on oxidation gives ‘C’. Compound ‘C’ reacts with CH3MgBr to give ‘D’, which on hydrolysis gives compound ‘E’, which is also an alcohol.
Which of the following is a Grignard reagent?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → CH3MgBr
A Grignard reagent is an alkyl magnesium halide, RMgX: an organometallic compound with a carbon–magnesium bond, made by treating a haloalkane with magnesium in dry ether. In \( \ce{CH3MgBr} \) (methylmagnesium bromide), R = \( \ce{CH3} \) and X = Br:
\( \displaystyle \ce{CH3Br + Mg} \xrightarrow{\text{dry ether}} \ce{CH3MgBr} \)
In the passage, \( \ce{CH3MgBr} \) is the reagent that reacts with compound C (ethanal).
Why the other options are incorrect
Option (2) CH3CH2NH2 — Incorrect. Ethylamine is a primary amine with no metal in it. In the passage it is the amine formed when B (\( \ce{CH3CN} \)) is reduced.
Option (3) (CH3)2CH–OMgBr — Incorrect. Here magnesium is bonded to oxygen, not carbon. It is the adduct D formed when \( \ce{CH3MgBr} \) adds to the C=O group of ethanal; its hydrolysis gives the alcohol E.
NCERT (2026-27 edition):
6.7.1 Reactions of Haloalkanes: 3. Reaction with metals – “Most organic chlorides, bromides and iodides react with certain metals to give compounds containing carbon-metal bonds. Such compounds are known as organo-metallic compounds. An important class of organo-metallic compounds discovered by Victor Grignard in 1900 is alkyl magnesium halide, RMgX, referred as Grignard Reagents. These reagents are obtained by the reaction of haloalkanes with magnesium metal in dry ether.”