(a)(i) Write the reaction involved in Cannizzaro's reaction.
(ii) Why are the boiling points of aldehydes and ketones lower than those of the corresponding carboxylic acids?
(iii) An organic compound 'A' with molecular formula C₅H₈O₂ is reduced to n-pentane with hydrazine followed by heating with NaOH and glycol. 'A' forms a dioxime with hydroxylamine and gives positive iodoform and Tollens' tests. Identify 'A' and give its reactions for the iodoform and Tollens' tests.
(ii) Why are the boiling points of aldehydes and ketones lower than those of the corresponding carboxylic acids?
(iii) An organic compound 'A' with molecular formula C₅H₈O₂ is reduced to n-pentane with hydrazine followed by heating with NaOH and glycol. 'A' forms a dioxime with hydroxylamine and gives positive iodoform and Tollens' tests. Identify 'A' and give its reactions for the iodoform and Tollens' tests.
Solution
(i)
(ii) Carboxylic acids exist as hydrogen-bonded dimers (two O–H···O bridges) — far more energy is needed to separate them than the dipole-bound aldehydes/ketones.
(iii) Wolff–Kishner to \textbf{n-pentane} ⇒ a straight C skeleton with two C=O (dioxime!).
Tollens-positive ⇒ a –CHO; iodoform-positive ⇒ a CHCO–.
Iodoform: CHCO– end + 3I/4NaOH → CHI↓ + NaOOC–CHCHCHO.
Tollens: –CHO end + 2[Ag(NH)]/OH → silver mirror + the carboxylate.
(ii) Carboxylic acids exist as hydrogen-bonded dimers (two O–H···O bridges) — far more energy is needed to separate them than the dipole-bound aldehydes/ketones.
(iii) Wolff–Kishner to \textbf{n-pentane} ⇒ a straight C skeleton with two C=O (dioxime!).
Tollens-positive ⇒ a –CHO; iodoform-positive ⇒ a CHCO–.
Iodoform: CHCO– end + 3I/4NaOH → CHI↓ + NaOOC–CHCHCHO.
Tollens: –CHO end + 2[Ag(NH)]/OH → silver mirror + the carboxylate.
CBSE Class 12 Board · 2023 · Subjective