Aldehydes & ketones & carboxylic acids practice · Mixed difficulty

Option D (I is false and II is true) is correct because sulphonic acids are more acidic than carboxylic acids, and carboxylic acids can be obtained by the oxidation of aldehydes with Tollens reagent.

Option C (II > I > III) is correct because the enol content is directly related to the acidity of the alpha-hydrogens and the stability of the resulting enol form.

Option B (Benzaldehyde) is correct because benzaldehyde, having no alpha-hydrogens, undergoes Cannizzaro reaction with 50% NaOH to yield benzyl alcohol and sodium benzoate.

Option A is correct because Size of ${BK}^{+3}$ is less than of ${NP}^{+3}$ due to actinoid contraction. As we know that in a period from left to right ionic radius decreases and in actinide series it is due to actinoid contraction.

Option B (Acetophenone) is correct because Acetophenone is known as hypnone due to its hypnotic properties.

Option C is correct because benzaldehyde lacks an alpha-hydrogen, which is required for aldol condensation to occur.

Option A ($\mathrm{Zn}-\mathrm{Hg}+$ Conc. $\mathrm{HCl}$) is correct because Clemmensen reduction uses zinc amalgam and concentrated hydrochloric acid.

Option A is correct. Solubility decreases with increasing molecular weight due to the increasing hydrophobic character of alkyl or aryl groups. Methanal (A) is the most soluble, followed by propanaldehyde (B), benzaldehyde (C), and acetophenone (D).

Option C (${\mathrm{C}_6}{\mathrm{H}_5}\mathrm{CHO}$) is correct because benzaldehyde has a characteristic bitter almond odor and is used in the manufacture of perfumes.

The correct answer is A (Acetic acid) because hydrolysis of acetamide in the presence of a base produces acetic acid and ammonia.

Option C ($N{O_2}C{H_2}COOH > NCC{H_2}COOH > FC{H_2}COOH > ClC{H_2}COOH$) is correct because the electron-withdrawing effect (-I effect) of the substituents decreases in the order $NO_2 > CN > F > Cl$.

Option A ($-\mathrm{N}{\mathrm{H}_2}$) is correct because the amino group is the basic group in glycine.

Option B is correct because the reactivity of carbonyl compounds towards nucleophilic reagents decreases as the steric hindrance and electron density around the carbonyl carbon increase.

Option D ($\mathrm{N}{\mathrm{H}_2}--\mathrm{N}{\mathrm{H}_2}/{\mathrm{C}_2}{\mathrm{H}_5}\mathrm{ONa}$) is correct because it represents the Wolff-Kishner reduction, which reduces aldehydes and ketones to hydrocarbons.

Option A (I>II>III>IV) is correct because acid chlorides are most reactive towards hydrolysis, followed by anhydrides, esters, and amides.

Option D (Oxidation of secondary alcohol) is correct because ketones are produced by the oxidation of secondary alcohols.

Option A ($\mathrm{N}$) is correct because Nessler's reagent has the formula $K_2HgI_4$, and therefore does not contain nitrogen.

Option B (Both Assertion and Reason are true but Reason is not the correct explanation of Assertion) is correct because while both statements are true, the better leaving group is not the ONLY reason for the higher reactivity of acyl halides.

Option D (Etard's reaction: ${CrO}_3 /\left({CH}_3 {CO}\right)_2 {O}$) is the incorrect match because Etard's reaction uses ${CrO}_2 {Cl}_2 / {CS}_2$ as the reagent.

Option A (Acetamide, carbon dioxide and ammonia) is correct because acetic acid reacts with urea upon heating to produce acetamide, carbon dioxide, and ammonia.

Option B ($\mathrm{Cu}_2 \mathrm{O}$) is correct because Fehling's reagent oxidizes aldehydes to carboxylate ions, and $\mathrm{Cu}^{2+}$ is reduced to reddish-brown $\mathrm{Cu}_2 \mathrm{O}$.

Option A (Statement-I is true and statement-II is false) is correct because ketones cannot be prepared by Rosenmund reduction, and aldehydes are more reactive than ketones towards nucleophilic addition.

Option A (an aqueous copper sulphate and alkaline sodium potassium tartarate (Rochelle salt) respectively) is correct because Fehling's solution A is aqueous copper sulfate, and Fehling's solution B is alkaline sodium potassium tartrate (Rochelle salt).

Option A (Benzaldehyde) is correct because it undergoes Cannizzaro reaction to form benzyl alcohol and benzoic acid in the presence of conc. NaOH.

Option A is correct because acetic acid forms dimers in benzene through hydrogen bonding, effectively doubling its molecular weight.

Option B (Both assertion and reason are correct statements, but reason is not the correct explanation of the assertion) is correct. Carboxylic acids have higher boiling points than alcohols due to more extensive hydrogen bonding, and they can form dimers, but dimerization isn't the primary reason for the boiling point difference.

Option D is correct because Cannizzaro's reaction requires aldehydes without alpha-hydrogens, and while option C has alpha-hydrogens, it is an exception and undergoes Cannizzaro's reaction.

Option A (70) is correct because the equivalent weight of the acid is calculated using the equivalent weights of the silver salt and silver.

Option B ($\mathrm{AgCl}$) is correct because the Carius method involves reacting halogen compounds with nitric acid and silver nitrate to form silver halide precipitates.

Option D (${\left( {\mathrm{C}{\mathrm{H}_3}} \right)_2}\mathrm{CHOH}$) is correct because the reduction of acetone (a ketone) results in the formation of isopropyl alcohol (2-propanol), which has the structure ${\left( {\mathrm{C}{\mathrm{H}_3}} \right)_2}\mathrm{CHOH}$.