Compound A (C₆H₁₂O₂) on reduction with LiAlH₄

Compound A (C₆H₁₂O₂) on reduction with LiAlH₄

Que. A compound A (C₆H₁₂O₂) on reduction with LiAlH₄ yielded two compounds B and C. The compound B on oxidation gave D which on treatment with aq. alkali and subsequent heating furnished E. The later catalytic hydrogenation gave C. Compound D was oxidized further to give F which was found to be a monobasic acid (mol. wt. = 60) find the structures of A, B, C, D, E, and F.
Ans. Upon reduction with LiAlH₄, compound A (C₆H₁₂O₂) gives two compounds, hence it must be an ester. Compound F is monobasic with mol. wt. 60. It is CH₃COOH, hence compound D must be CH₃CHO. As compound D is obtained by oxidation of compound B, hence compound B shall be alcohol i.e., CH₃CH₂OH. As compound A on reduction with LiAlH₄ gives CH₃CH₂OH, hence it must be CH₃COOC₄H₉ (n-butyl acetate).
Step 1: Upon reduction with LiAlH₄, compound A (C₆H₁₂O₂) gives two compounds, hence it must be an ester and the products i.e. B and C are Ethanol and Butanol respectively

C₃H₄ COOC₂H₅ → CH₃CH₂OH + CH₃CH₂CH₂CH₂OH
        A                         B                     C

Step 2: B (Ethanol) upon oxidation produce Ethanal (CH₃CHO) which on further oxidation gives Ethanoic acid (CH₃COOH)
CH₃CH₂OH → CH₃CHO → CH₃COOH
        C                         D

Step 3: Ethanal upon treatment with aqueous alkali (aq NaOH) produces an Aldol. Further heating of this Aldol gives E (Crotonaldehyde) this on treatment with H₂/Ni give butanol
CH₃CHO → CH₃ – CH(OH) – CH₂CHO → CH₃CH = CHCHO → CH₃CH₂CH₂CH₂OH
        D                          Aldol                                          E

# Aldol represents the coexistence of both alcoholic and aldehydic groups in a compound

# Crotonaldehyde represents a molecule that contains both alkene and an aldehydic group in a compound CH₃CH = CHCHO (But-2-enal).