Aldehydes & ketones & carboxylic acids practice · Mixed difficulty

Option D ($\mathrm{NH}_3$) is correct because it reacts differently with formaldehyde, acetaldehyde, and acetone due to steric and electronic factors.

Option B (HCOOH) is correct because chloroform reacts with aqueous alkali (KOH or NaOH) to produce potassium or sodium formate, which upon acidification yields formic acid (HCOOH).

Option C (Acetic anhydride) is correct because $P_2O_5$ is a dehydrating agent that removes water from acetic acid to form acetic anhydride.

Option C ($\mathrm{C}{\mathrm{H}_3}\mathrm{COC}{\mathrm{H}_3}$ $\to$ mesityl oxide) is correct because condensation involves the joining of two molecules with the loss of a small molecule like water, which occurs in the formation of mesityl oxide from acetone.

Option D (Aldehyde) is correct because aldehydes require 0.5 moles of oxygen per mole to produce a carboxylic acid.

Option D ($C{H_3}COC{H_3} + C{H_3}COOH$) is correct because KMnO4 oxidation cleaves the carbon-carbon bond to form a ketone and a carboxylic acid.

Option A ($\mathrm{BrC}{\mathrm{H}_2}\mathrm{C}{\mathrm{H}_2}\mathrm{C}{\mathrm{H}_2}\mathrm{COOH}$) is correct because ${\mathrm{LiAl}{\mathrm{H}_4}}$ will also reduce the alkyl halide (Br) to an alkane, leading to a different product than the corresponding 1° alcohol.

Option A (${\mathrm{C}_6}{\mathrm{H}_5}\mathrm{CH} = \mathrm{CHCO}{\mathrm{C}_6}{\mathrm{H}_5}$) is correct because this is the product of an aldol condensation between benzaldehyde and acetophenone.

Option C (${\text{C}}{{\text{H}}_{\text{2}}} = {\text{CHCOOH}}$) is correct because the reaction sequence involves alpha-halogenation followed by dehydrohalogenation to form an alpha, beta-unsaturated acid.

Option C (Aldehydic group) is correct because Fehling's solution is used to detect the presence of an aldehydic group; aldehydes reduce the copper(II) ions in Fehling's solution to copper(I) oxide, forming a red precipitate.

Option D (Methyl salicylate) is correct because methyl salicylate is commonly known as \'oil of winter green\' and is used as a flavoring agent.

Option C (3-pentanone) is correct because the reaction is an Oppenauer oxidation, which converts a secondary alcohol to a ketone.

Option C (Hexamethylene tetramine) is correct because formaldehyde reacts with ammonia to form hexamethylenetetramine, also known as urotropine.

Option B (Cannizaro’s reaction) is correct because the Cannizzaro reaction involves the base-induced disproportionation of an aldehyde lacking an alpha-hydrogen, resulting in the formation of a primary alcohol and a carboxylic acid.

Option A (Arndt-Eistert reaction) is correct because the Arndt-Eistert reaction is used to homologate carboxylic acids by inserting a methylene group.

Option B (Maleic acid is stronger than fumaric acid) is correct because maleic acid exhibits intramolecular hydrogen bonding, which stabilizes the monoanion and makes it a stronger acid than fumaric acid.

Option C ($\mathrm{NaOH}/{\mathrm{I}_2}/{\mathrm{H}^ + }$) is correct because this represents the haloform reaction, which converts methyl ketones to carboxylic acids with one less carbon.

Option A (Reduction of carbonyl compound into hydrocarbons) is correct because the Wolff-Kishner reaction reduces carbonyl groups (C=O) to methylene groups (CH2).

Option D (None of these) is correct because Fehling's solution is a complex mixture of copper sulfate, sodium potassium tartrate (Rochelle salt), and sodium hydroxide.

Option A (An aldehyde) is correct because an aldehyde with a molecular weight of 44 (acetaldehyde) would oxidize to a carboxylic acid with a molecular weight of 60 (acetic acid).

Option D (Pyridinium chlorochromate) is correct because PCC is a mild oxidizing agent that converts primary alcohols to aldehydes without further oxidation to carboxylic acids.

Option B (Cannizzaro’s reaction) is correct because Cannizzaro's reaction involves the disproportionation of aldehydes without alpha-hydrogens, leading to the formation of an alcohol and a carboxylic acid.

Option B is the correct representation of the Clemmensen reduction, where a carbonyl group is reduced to a methylene group using Zn-Hg and concentrated HCl.

Option D (Schmidt reaction) is correct because the Schmidt reaction converts a carboxylic acid to an amine with the loss of carbon dioxide.

Option D (Tollen’s reagent) is correct because it can distinguish between aldehydes (benzaldehyde) and ketones (acetone) due to its ability to oxidize aldehydes but not ketones.

Option A (Chloral) is correct because it has the molecular formula ${\mathrm{C}_2}\mathrm{C}{\mathrm{l}_3}\mathrm{OH}$, reduces Fehling's solution, and is obtained by the action of ${\mathrm{C}{\mathrm{l}_2}}$ on ethyl alcohol.

Option B ($\mathrm{C}{\mathrm{H}_2} = CH-COOH$) is correct because the reaction sequence results in dehydrohalogenation followed by the formation of an alpha, beta-unsaturated acid.

Option A ($I > II > III$) is correct because the acidity decreases as the distance between the two carboxylic acid groups increases, due to decreased stabilization of the conjugate base.

Option A ($\mathrm{C}{\mathrm{H}_3}\mathrm{COC}{\mathrm{H}_3}$) is correct because oxidation of acetone results in cleavage of C-C bonds, leading to products with fewer carbon atoms than the original ketone.

Option C ($\mathrm{C}{\mathrm{H}_3}\mathrm{CHC}{\mathrm{l}_2}$) is correct because geminal dihalides react with aqueous KOH to form aldehydes or ketones.