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.
In the reaction sequence given in the passage, what is the compound ‘E’?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Propan-2-ol
In the passage, A (\( \ce{CH4O} \)) is methanol. Following each step:
A to methyl iodide: HI replaces the –OH group by I.
\( \displaystyle \ce{CH3OH + HI -> CH3I + H2O} \)
Methyl iodide to B: the cyanide ion of KCN replaces I, giving acetonitrile (B).
\( \displaystyle \ce{CH3I + KCN -> CH3CN + KI} \)
B to the alcohol: reduction of the nitrile gives ethylamine, and nitrous acid turns this primary amine into ethanol with loss of nitrogen gas.
\( \displaystyle \ce{CH3CN + 2H2} \xrightarrow{\ce{Ni}} \ce{CH3CH2NH2} \)
\( \displaystyle \ce{CH3CH2NH2 + HNO2 -> CH3CH2OH + N2 + H2O} \)
Alcohol to C: mild oxidation of the primary alcohol gives the aldehyde ethanal (C).
\( \displaystyle \ce{CH3CH2OH} \xrightarrow{\ce{PCC}} \ce{CH3CHO} \)
C to D to E: \( \ce{CH3MgBr} \) adds to the C=O group of ethanal to give the adduct D; hydrolysis of D gives propan-2-ol (E).
\( \displaystyle \ce{CH3CHO + CH3MgBr -> (CH3)2CH-OMgBr} \)
\( \displaystyle \ce{(CH3)2CH-OMgBr + H2O -> (CH3)2CHOH + Mg(OH)Br} \)
The new \( \ce{CH3} \) group and the –OH both end up on the old carbonyl carbon, which now carries two \( \ce{CH3} \) groups: a secondary alcohol, as expected for a Grignard reagent with an aldehyde other than methanal.
Why the other options are incorrect
Option (1) Propan-1-ol — Incorrect. A Grignard reagent gives a primary alcohol only with methanal; here the –OH stays on the middle carbon, not on an end carbon.
Option (3) tert-Butyl alcohol — Incorrect. A tertiary alcohol needs a ketone (propanone with \( \ce{CH3MgBr} \)); C is an aldehyde.
Option (4) Butan-2-ol — Incorrect. It has four carbon atoms; ethanal (2 C) plus \( \ce{CH3MgBr} \) (1 C) gives only three.
NCERT (2026-27 edition):
3. From Grignard reagents – “… The first step of the reaction is the nucleophilic addition of Grignard reagent to the carbonyl group to form an adduct. Hydrolysis of the adduct yields an alcohol.”
“… You will notice that the reaction produces a primary alcohol with methanal, a secondary alcohol with other aldehydes and tertiary alcohol with ketones.”