An organic compound 'A' with molecular formula CH4O reacts with HI to give an alkyl iodide which than 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' on oxidation followed by reaction with CH3MgBr gives 'D' which on hydrolysis gives compound 'E' which is also an alcohol.
What is the name of the compound 'E'?
Answer & explanation
Correct answer: option 2
The correct answer is option 2. 2-Propanol.
Remember
Aldehyde + Grignard → secondary alcohol
Ketone + Grignard → tertiary alcohol
Step 1: Identify A (CH₄O)
CH₄O is methanol (CH₃OH)
CH₃OH + HI → CH₃I (methyl iodide)
Step 2: Formation of B
CH₃I + KCN → CH₃CN (nitrile, B)
Step 3: Reduction of B
CH₃CN → CH₃CH₂NH₂ (primary amine)
Step 4: Reaction with HNO₂
Primary amine + HNO₂ → alcohol
CH₃CH₂NH₂ → CH₃CH₂OH (ethanol)
Step 5: Oxidation to C
Ethanol → Ethanal (CH₃CHO)
Step 6: Reaction with CH₃MgBr
CH₃CHO + CH₃MgBr → secondary alcohol after hydrolysis
Gives: CH₃–CHOH–CH₃
Final compound E:
CH₃–CHOH–CH₃ that is 2-Propanol