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Chemistry · 2023 · 5 marks
Aldehydes, Ketones and Carboxylic AcidsChemical Reactions of Aldehydes, Ketones and Carboxylic Acids5 marksAnalyselong answer
CBSE 2023 · Region 5 · Set 1 · Q34
(a) (i) Write the reaction involved in Cannizaro's reaction. 1 + 1 + 3 = 5 (ii) Why are the boiling point of aldehydes and ketones lower than that of…
(i)
Write the reaction involved in Cannizaro's reaction. $\displaystyle \mathbf{1}+\mathbf{1}+\mathbf{3}=\mathbf{5}$
(ii)
Why are the boiling point of aldehydes and ketones lower than that of corresponding carboxylic acids ?
(iii)
An organic compound 'A' with molecular formula $\displaystyle \mathrm{C}_{5} \mathrm{H}_{8} \mathrm{O}_{2}$ is reduced to n-pentane with hydrazine followed by heating with NaOH and Glycol. 'A' forms a dioxime with hydroxylamine and gives a positive Iodoform and Tollen's test. Identify 'A' and give its reaction for Iodoform and Tollen's test.
(i)
Give a chemical test to distinguish between ethanal acid and ethanoic acid. $\displaystyle \mathbf{1}+\mathbf{1}+\mathbf{3}=\mathbf{5}$
(ii)
Why is the $\displaystyle \alpha$-hydrogens of aldehydes and ketones are acidic in nature ?
(iii)
An organic compound 'A' with molecular formula $\displaystyle \mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}_{2}$ undergoes acid hydrolysis to form two compounds 'B' and 'C'. Oxidation of 'C' with acidified potassium permanganate also produces 'B'. Sodium salt of 'B' on heating with soda lime gives methane.
(1)
Identify 'A', 'B' and 'C'.
(2)
Out of 'B' and 'C', which will have higher boiling point ? Give reason.
Marking-scheme solution
(i)
(ii)
Carboxylic acids have strong hydrogen bonding whereas aldehydes and ketones have weak dipole-dipole interactions.
(iii)
A = CH₃ – CO – CH₂ – CH₂ – CHO / $\displaystyle 4$-oxopentanal
CH₃ – CO – CH₂CH₂CHO --(NaOH + I₂, heat)--> CHI₃ (Yellow ppt)
CH₃ – CO – CH₂CH₂CHO --([Ag(NH₃)₂]⁺, OH⁻, warm)--> CH₃ – CO – CH₂ – CH₂ – COO⁻ + Ag↓
(i)
Add NaHCO₃ solution to both compounds, ethanoic acid will give the brisk effervescence of CO₂ while ethanal does not.
(ii)
due to resonance stabilization of the conjugate base formed / the strong electron-withdrawing effect of the carbonyl group.
(iii)(1)
A = CH₃COOC₂H₅ / Ethyl ethanoate / Ethyl acetate,
B = CH₃COOH / Ethanoic acid / Acetic acid,
C = CH₃CH₂OH / Ethanol / Ethyl alcohol.
(2)
B, due to the more extensive association of carboxylic acid molecules through strong hydrogen bonding.
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