Aldehyde ketones & carboxylic Acid MCQs for NEET — Chemistry Questions with Answers

The slowest step in the Cannizzaro reaction is the hydride shift. This involves the migration of a hydride ion from the alkoxide anion to the carbonyl carbon, forming an aldehyde and an alcohol.

${\mathrm{CH}}_2=\mathrm{CH}-\mathrm{CH}=\mathrm{O}\to \stackrel{\oplus }{\mathrm{C}}{\mathrm{H}}_2-\mathrm{CH}=\mathrm{CH}-{\mathrm{O}}^{-}\stackrel{\mathrm{HCN}}{\to }\underset{\mathrm{CN}}{\underset{|}{{\mathrm{CH}}_2}}-\mathrm{CH}=\mathrm{CH}-\mathrm{OH}\stackrel{\mathrm{Tautomerisation}}{\to }\underset{\mathrm{CN}}{\underset{|}{{\mathrm{CH}}_2}}-{\mathrm{CH}}_2-\mathrm{CHO}$

Order of reactivity of acid deravative towards nucleophilic substitution is :

CH₃COCl >(CH₃CO)₂O > CH₃COOC₂H_{5 >}CH₃CONH₂

PhCOOH dissolves in aq. NaHCO₃ solution.

CI has more — I and less + M effect.

Maleic acid is a cis-isomer of butenedioic acid, which remains unaffected upon heating. The other options, malonic acid, fumaric acid (trans-isomer), and succinic acid, undergo decarboxylation upon heating, releasing carbon dioxide.

(b) ${\mathrm{PhCOCH}}_3\underset{\mathrm{HCl}}{\overset{{\mathrm{NaNO}}_2}{\to }}\left[{\mathrm{PhCOCH}}_2\mathrm{NO}\right]\to \mathrm{PhCOCH}=\mathrm{NOH}\underset{\mathrm{dehydrating} \mathrm{agent}}{\overset{{\mathrm{Ac}}_2\mathrm{O}}{\to }}\mathrm{PhCOCN}\stackrel{{\mathrm{H}}^{+}/{\mathrm{H}}_2\mathrm{O}}{\to }{\mathrm{PhCOCO}}_2\mathrm{H}$

$$\begin{gathered} {\mathrm{CH}}_3-\mathrm{C}\equiv \mathrm{CH}\underset{{\mathrm{H}}_2\mathrm{O}}{\overset{{\mathrm{Hg}}^{2+}/{\mathrm{H}}^{+}}{\to }}{\mathrm{CH}}_3-\stackrel{\stackrel{\mathrm{OH}}{|}}{\mathrm{C}}={\mathrm{CH}}_2\rightleftharpoons {\mathrm{CH}}_3-\stackrel{\stackrel{\mathrm{O}}{\left|\right|}}{\mathrm{C}}-{\mathrm{CH}}_3 \\ \mathrm{A} \mathrm{ketone} (\mathrm{gives} +\mathrm{ve} \mathrm{iodoform} \mathrm{test} \mathrm{with} \mathrm{NaoI} ) \end{gathered}$$

$$\begin{gathered} {\mathrm{CH}}_3-\mathrm{C}\equiv \mathrm{CH}\underset{\left(2\right){\mathrm{H}}_2{\mathrm{O}}_2/{\mathrm{OH}}^{-}}{\overset{\left(1\right){\mathrm{BH}}_3\mathrm{THF}}{\to }}{\mathrm{CH}}_3-\mathrm{CH}={\mathrm{CH}}_2-\mathrm{OH}\rightleftharpoons {\mathrm{CH}}_3-{\mathrm{CH}}_{2_{}}=\mathrm{O} \\ (\mathrm{an} \mathrm{aldehyde}) \\ (-\mathrm{ve} \mathrm{Iodoform} \mathrm{test} \mathrm{beacuse} -\underset{\stackrel{\left|\right|}{\mathrm{O}}}{\mathrm{C}}-{\mathrm{CH}}_3 \mathrm{group} \mathrm{is} \mathrm{absent}) \end{gathered}$$

(b) Iodoform test.

(b.1) Oxidative ozonolysis by not KMnO₄

(b.2) Iodoform test. Only methyl ketones. No reaction with double bond.

(b.3) H₂/Pt can reduce only double bond.

(b.4) LAH oxidizes ketone to $2°$ alcohol not to carboxylic acid.