Aldehydes & ketones & carboxylic acids practice · Medium

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 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 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 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 (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 ${\left( {\mathrm{C}{\mathrm{H}_3}} \right)_2}\mathrm{CHOH}$.

Option D (all the above factors) is correct because removing water or increasing the concentration of either reactant will shift the equilibrium towards the product side, increasing the yield of the ester.

Option D (None of the above) is correct because Rosenmund's reduction and Stephen's reduction are both methods used to prepare aldehydes, not ketones.

Option D ($PhCHO$) is correct because Cannizzaro reaction requires the absence of alpha-hydrogens, which is the case for benzaldehyde ($PhCHO$).

Option C (Pyrolysis of calcium acetate) is correct because acetone is prepared by the dry distillation (pyrolysis) of calcium acetate.

Option A ($\mathrm{C}{\mathrm{H}_3}\mathrm{COOH}$) is correct because the reaction of an amide with nitrous acid (generated in situ from NaNO2/HCl) results in the formation of a carboxylic acid.

Option B ($\mathrm{C}{\mathrm{H}_3}\mathrm{N}{\mathrm{H}_2}$) is correct because acetic acid reacts with hydrazoic acid in the presence of $conc.{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4}$ to give methylamine.

Option C (i, ii, iii, and iv) is correct because the oxidation of 2,5-dimethylhexan-3-one leads to the formation of various carboxylic acids due to cleavage at different bonds near the carbonyl group.

Option D (resonance stabilization of their conjugate base) is correct because the carboxylate ion, the conjugate base of carboxylic acids, is resonance-stabilized, making carboxylic acids more acidic than phenols and alcohols.

Option A (Zinc amalgam $+ HCl$) is correct because Clemmensen reduction uses zinc amalgam and hydrochloric acid to reduce carbonyl compounds to alkanes.

Option C is correct because the compound reacts with NaHCO3 to release CO2, indicating the presence of a carboxylic acid group, and reduction with LiAlH4 yields an achiral compound.

Option A (esters) is correct because carboxylic acids are commercially reduced to alcohols by first converting them into esters, which are then reduced via catalytic hydrogenation.

Option D (Statement-1 is False; Statement-2 is True) is correct because electron-withdrawing groups increase the electrophilicity of the carbonyl carbon, making nucleophilic addition more favorable.

Option D (7) is correct because there are 7 structural isomers of carbonyl compounds with the formula $\mathrm{C}_5\mathrm{H}_{10}\mathrm{O}$.

Option C (I is true and II is false) is correct because fluorine is more electronegative than chlorine, making fluoroacetic acid more acidic, and the nitro group has a stronger -I effect than the cyano group.

Option B (Schmidt reaction) is correct because the reaction of a carboxylic acid with hydrazoic acid in the presence of concentrated sulfuric acid to form an amine is the Schmidt reaction.

Option 5 moles is correct because A is the limiting reagent and the stoichiometry of the reaction A + B + C -> D indicates that 1 mole of A produces 1 mole of D.

Option C (Addition-elimination) is correct because the reaction involves both the addition of the amine to the carbonyl group, followed by the elimination of water.

Option C (Haloform reaction) is correct because the haloform reaction involves the conversion of a methyl ketone to a carboxylic acid and a haloform, without forming a new carbon-carbon bond.