Aldehydes & ketones & carboxylic acids practice · Mixed difficulty

Option A (Acetone) is correct because acetone reacts with HCl to form the given product through an addition reaction.

Option B (Higher in $p -$nitrophenol) is correct because $p$-nitrophenol is a stronger acid than salicylaldehyde due to the electron-withdrawing nitro group, making it more soluble in a base.

Option B (CH3CHO) is correct because chlorination of acetaldehyde yields chloral, which can be condensed with chlorobenzene to form DDT.

Option C (2 water molecules) is correct because hydrated oxalic acid has the formula $H_2C_2O_4 \cdot 2H_2O$, indicating two water molecules.

Option A (Formation of intermolecular H-bonding) is correct because carboxylic acids form stronger intermolecular hydrogen bonds than aldehydes, ketones, and alcohols, leading to higher boiling points.

Option C (Picric acid) is correct because picric acid is a phenol with nitro groups, not a carboxylic acid.

Option B (Ester) is correct because the reaction of a silver salt of a fatty acid with an alkyl halide results in the formation of an ester, known as the Hunsdiecker reaction.

Option C ($\mathrm{HCB}{\mathrm{r}_2}\mathrm{CB}{\mathrm{r}_2}\mathrm{COOH}$) is correct because decarboxylation removes the COOH group, leaving 1,1,2,2-tetrabromoethane.

Option B ($\mathrm{N}{\mathrm{H}_3}$) is correct because amides are formed by the reaction of acid chlorides with ammonia.

Option B (Bouvealt - Blanc reduction) is correct because the reduction of an ester with ${Na} / {C}_2 {H}_5 {OH}$ is specifically known as the Bouvealt-Blanc reduction.

Option C ($\mathrm{p - }$ rosaniline hydrochloride) is correct because magenta is a dye that is chemically $\mathrm{p - }$ rosaniline hydrochloride.

Option C ($\mathrm{C}{\mathrm{H}_3}\mathrm{COCl}$) is correct because acetic acid reacts with $\mathrm{PC}{\mathrm{l}_5}$ to form acetyl chloride.

Option B (Alcoholysis) is correct because alcoholysis is the cleavage of a chemical bond by the addition of an alcohol.

Option B (Formamide) is correct because formamide is a liquid at room temperature, while the others are solids or gases.

Option B ($\mathrm{C}{\mathrm{H}_3}\mathrm{CH} = \mathrm{N} - \mathrm{OH}$) is correct because acetaldoxime is the oxime derivative of acetaldehyde, formed by the reaction of acetaldehyde with hydroxylamine.

Option D (Biurate test) is correct because the biurate test is specifically used to detect the presence of urea.

Option A (Baeyer-Villiger oxidation) is correct because the reaction of acetophenone with perbenzoic acid to give phenyl acetate is a Baeyer-Villiger oxidation.

Option D (Both react with $\mathrm{NaOH}$ to form polymers) is correct because benzaldehyde does not undergo polymerization with NaOH, it undergoes Cannizzaro reaction.

Option A ($\mathrm{C}{\mathrm{H}_3} - \mathrm{CO} - \mathrm{C}{\mathrm{H}_3}$) is correct because it is a ketone that does not give a silver mirror test but forms an oxime.

Option C (Carboxylic acid) is correct because the reaction sequence converts an alkyl halide to a carboxylic acid.

Option D (Acetaldehyde) is correct because Cannizzaro's reaction requires aldehydes without alpha-hydrogens, and acetaldehyde has alpha-hydrogens.

Option C (Esters of unsaturated fatty acids are reduced to those of saturated acids) is correct because hydrogenation involves the addition of hydrogen to double bonds, converting unsaturated fatty acids to saturated ones.

Option B (Secondary alcohols) is correct because the oxidation of secondary alcohols yields ketones.

Option B ($\mathrm{PCl}_3, \mathrm{KCN} \& \mathrm{H}_3 \mathrm{O}^{+}$) is correct because it converts the alcohol to a halide, then to a nitrile, and finally hydrolyzes the nitrile to a carboxylic acid.

Option D ($p -$nitrobenzoic acid) is the strongest acid because the nitro group is a strong electron-withdrawing group that stabilizes the conjugate base most effectively when in the para position.

Option B (Calcium acetate + calcium propanoate) is correct because combining these two salts and heating will produce 2-butanone.

Option A (I>II>III>IV) is correct because acid chlorides are most reactive towards hydrolysis, followed by anhydrides, esters, and then amides due to the leaving group ability.

Option A ($LiAlH_4$) is the most common and strongest reducing agent among the options, but the question is flawed as all options can reduce carboxylic acids to primary alcohols under specific conditions.

Option B (40% aq. sol. Of HCHO) is correct because formalin is a 40% aqueous solution of formaldehyde.

Option B (MeCHO) is correct because aldehydes are generally more reactive towards nucleophilic attack than esters or acid chlorides due to less steric hindrance and no resonance stabilization.