Skip to content
Question of 87

Q.(a) Write the structures of the products of the following reactions:

(i) CH3–COONa→ΔNaOH & CaOCH_3\text{--COONa} \xrightarrow[\Delta]{\text{NaOH \& CaO}} ?
(ii) CH3–CHO+HCN→CH_3\text{--CHO} + HCN \rightarrow ? OR
(b) Carboxylic acids are higher boiling liquids than aldehydes, ketones and even alcohols of comparable molecular masses. Why?
(c) Which acid in the following pair is stronger and why? CH3CO2HCH_3CO_2H or CH2FCO2HCH_2FCO_2H
Meghalaya MboseMBOSE Meghalaya Intermediate Board 2021Subjective· 2mImportance★★★★★
0% · 0/87 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Soda-lime decarboxylation of sodium acetate gives methane; HCN addition to acetaldehyde gives a cyanohydrin. (Alternative: carboxylic-acid dimerisation via H-bonding explains their unusually high boiling points, and inductive electron withdrawal explains relative acid strengths.)

(a)(i) Soda-lime decarboxylation:

Heating the sodium salt of a carboxylic acid with soda lime (NaOH + CaO) removes the carboxyl group as Na2CO3Na_2CO_3, leaving the alkane with one less carbon:

CH3COONa+NaOH→ΔCaOCH4↑+Na2CO3CH_3COONa + NaOH \xrightarrow[\Delta]{CaO} CH_4\uparrow + Na_2CO_3

Product: methane, CH4CH_4.

(a)(ii) Cyanohydrin formation:

Aldehydes react with hydrogen cyanide (nucleophilic addition of CN−CN^- to the electrophilic carbonyl carbon, followed by protonation) to give a cyanohydrin:

CH3CHO+HCN→CH3–CH(OH)–CNCH_3CHO + HCN \rightarrow CH_3\text{--}CH(OH)\text{--}CN

Product: acetaldehyde cyanohydrin (2-hydroxypropanenitrile), CH3CH(OH)CNCH_3CH(OH)CN.

Alternative (Or):

(b) Why carboxylic acids boil higher than aldehydes/ketones/alcohols of similar mass:

Carboxylic acid molecules associate strongly in the liquid/vapour state, forming a cyclic dimer held together by two intermolecular hydrogen bonds simultaneously (each C=O⋯H–O\text{C=O}\cdots H\text{--O}):

2 R–COOH⇌(R–COOH)2 (cyclic H-bonded dimer)2\,R\text{--COOH} \rightleftharpoons (R\text{--COOH})_2\ \text{(cyclic H-bonded dimer)}

Breaking these two hydrogen bonds to vaporise the acid requires considerably more energy than vaporising an alcohol (which forms only single, not doubled, H-bonds) or an aldehyde/ketone (which cannot hydrogen-bond with itself at all, having no O–H). This extra energy requirement raises the boiling point of carboxylic acids well above that of alcohols, aldehydes or ketones of comparable molecular mass.

(c) CH3CO2HCH_3CO_2H vs CH2FCO2HCH_2FCO_2H:

…

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.