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'?

2-Propanol

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