Carbonyl Compounds and Carboxylic Acids — Practice Quiz

Three Cl atoms have strong –I effect, withdrawing electrons from the carboxylate ion, stabilising the negative charge and making proton donation easier

Step 1: Acidity of a carboxylic acid depends on the stability of the carboxylate ion (RCOO⁻) formed after proton donation. More stable the carboxylate ion, stronger the acid. Step 2: Cl is highly electronegative and exerts a strong –I (electron-withdrawing inductive) effect. In CCl₃COOH, three Cl atoms pull electron density away from the –COO⁻ group through the C–C bond. Step 3: This disperses (spreads out) the negative charge on the carboxylate ion, stabilising it greatly. A more stabilised carboxylate ion means the equilibrium shifts towards ionisation, increasing Ka (decreasing pKa). Step 4

OH⁻ removes an α-hydrogen from acetaldehyde to form a carbanion (enolate ion)

Step 1: Aldol condensation requires an α-hydrogen (H on the carbon adjacent to C=O). Acetaldehyde has α-hydrogen on its CH₃ group. Step 2: In the FIRST step, the base (OH⁻) abstracts the α-hydrogen as a proton: HO⁻ + H–CH₂–CHO → –CH₂–CHO + H₂O. This forms a carbanion (enolate ion). Step 3: In the SECOND step, this nucleophilic carbanion attacks the carbonyl carbon of another acetaldehyde molecule (electrophile), forming an alkoxide intermediate. Step 4: In the THIRD step, the alkoxide ion is protonated by water to give 3-hydroxybutanal (acetaldol): CH₃–CH(OH)–CH₂–CHO. Step 5: Option B describe

SnCl₂/HCl; acetaldehyde (CH₃CHO) is formed via imine intermediate

Step 1: Stephen's reaction is used to prepare aldehydes from alkyl nitriles (cyanides). The reagent is SnCl₂/HCl (stannous chloride in hydrochloric acid). Step 2: The nitrile is reduced to an imine (aldimine): CH₃–C≡N + SnCl₂/HCl → CH₃–CH=NH (imine/aldimine). Step 3: The imine intermediate on hydrolysis (with H₂O in acidic medium) gives the aldehyde: CH₃–CH=NH + H₂O → CH₃CHO + NH₃. Step 4: The net reaction converts acetonitrile to acetaldehyde (one carbon more than a methyl group, but same carbon count as nitrile). Step 5: LiAlH₄ would fully reduce nitrile to primary amine (not aldehyde). H₂/P

A β-keto ester (ethyl acetoacetate: CH₃–CO–CH₂–COOEt)

Step 1: Claisen condensation is a self-condensation of esters containing at least one α-hydrogen, in the presence of a strong base (sodium ethoxide). Step 2: C₂H₅ONa removes an α-hydrogen from one ethyl acetate molecule to form an enolate ion: –CH₂–COOEt. Step 3: This enolate acts as a nucleophile and attacks the carbonyl carbon of another ethyl acetate molecule, displacing the –OEt group (this is a nucleophilic acyl substitution, not nucleophilic addition). Step 4: The product formed is ethyl acetoacetate (ethyl 3-oxobutanoate): CH₃–CO–CH₂–COOC₂H₅. This is a β-keto ester (keto group at β-posi