MP Board · Class 12 · Chemistry · Aldehydes, Ketones and Carboxylic AcidsAn organic compound 'A' with molecular formula $C2H4O$ on oxidation gives compound 'B' with molecular formula $C2H4O2$. Compound 'A' undergoes Tollen's test and Fehling's test. Compound 'B' reacts with methanol in the presence of concentrated $H2SO4$ to form a sweet-smelling compound 'C'. Identify compounds 'A', 'B', and 'C', write the chemical equations for all the reactions involved, and explain the acidic nature of carboxylic acids compared to alcohols.

Identification of Compounds

1. Compound A: Given molecular formula is $C_2H_4O$. It reduces Tollen's reagent and Fehling's solution, which confirms it is an aldehyde. With 2 carbon atoms, Compound 'A' is Acetaldehyde ($CH_3CHO$).
2. Compound B: On oxidation, $CH_3CHO$ gives a carboxylic acid with 2 carbon atoms, which is Acetic acid ($CH_3COOH$) having molecular formula $C_2H_4O_2$.
3. Compound C: Acetic acid reacts with methanol ($CH_3OH$) in the presence of concentrated $H_2SO_4$ to form an ester, Methyl acetate ($CH_3COOCH_3$), which has a sweet fruity smell.

Chemical Equations

1. Oxidation of A to B: $$CH_3CHO + [O] \xrightarrow{KMnO_4 \text{ or } K_2Cr_2O_7} CH_3COOH$$
2. Tollen's Test for A: $$CH_3CHO + 2[Ag(NH_3)_2]^+ + 3OH^- \rightarrow CH_3COO^- + 2Ag\downarrow + 4NH_3 + 2H_2O$$
3. Esterification (Formation of C): $$CH_3COOH + CH_3OH \xrightarrow{conc. H_2SO_4} CH_3COOCH_3 + H_2O$$

Acidic Nature of Carboxylic Acids vs. Alcohols\nCarboxylic acids are much stronger acids than alcohols because:

• Resonance Stabilization of Carboxylate Ion: When a carboxylic acid loses a proton ($H^+$), it forms a carboxylate ion ($RCOO^-$). This ion is stabilized by resonance, where the negative charge is delocalized over two electronegative oxygen atoms.
• Alkoxide Ion Instability: Alcohols lose a proton to form alkoxide ions ($RO^-$), where the negative charge is localized on a single oxygen atom, making it unstable and less willing to release the proton.