Aldehydes & ketones & carboxylic acids practice · Mixed difficulty

Option B ($\mathrm{Pd}/\mathrm{BaS}{\mathrm{O}_4}$) is correct because Palladium supported on Barium sulfate ($\mathrm{Pd}/\mathrm{BaS}{\mathrm{O}_4}$) is the Lindlar's catalyst used in Rosenmund reduction to convert acyl chlorides to aldehydes.

Option C (Racemate) is correct because the product has a chiral carbon but is formed in equal amounts of both enantiomers due to the achiral reactants.

Option B (Ant) is correct because formic acid (HCOOH) was first isolated from ants, hence the name.

Option C (Benzal chloride) is correct because benzaldehyde reacts with PCl5 to replace the carbonyl oxygen with two chlorine atoms, forming benzal chloride ($\mathrm{C_6H_5CHCl_2}$).

Option B (It is reducing agent) is correct because formic acid can be oxidized to carbon dioxide, making it a reducing agent.

Option B ($\mathrm{Zn}-\mathrm{Hg} / \mathrm{HCl}$ and hydrazine $+\mathrm{KOH} /$ ethylene glycol/ $\Delta$ respectively) is correct because Clemmensen reduction uses $\mathrm{Zn}-\mathrm{Hg} / \mathrm{HCl}$ and Wolff-Kishner reduction uses hydrazine $+\mathrm{KOH} /$ ethylene glycol/ $\Delta$.

Option A (Both A and R are correct and R is correct explanation of A) is correct because Fehling's solution can distinguish between benzaldehyde and acetaldehyde, and the C-H bond strength in the CHO group affects their reactivity.

Option D ($\mathrm{CH}_3 \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_2 \mathrm{CHO}$) is correct because ethyne reacts with HgSO4/H2SO4 to form acetaldehyde, which undergoes aldol condensation in the presence of NaOH.

Option C ($\mathrm{C}{\mathrm{H}_3}\mathrm{C}{\mathrm{H}_2}\mathrm{OH}$) is correct because the reaction sequence involves conversion to amide, then amine, and finally alcohol via diazotization.

Option A (${C_6}{H_5} - C{H_2} - OH\,\,\mathrm{and}\,\,H - COONa$) is correct because the reaction is a Cannizzaro reaction where benzaldehyde is reduced to benzyl alcohol and formaldehyde is oxidized to sodium formate.

Option D ($C{H_3}COOH$) is correct because the reaction sequence involves conversion to amide, then nitrile, and finally carboxylic acid.

Option C ($\mathrm{C}{\mathrm{H}_3}\mathrm{C}{\mathrm{H}_2}\mathrm{CH}\left( {\mathrm{Cl}} \right)\mathrm{C}{\mathrm{O}_2}\mathrm{H}$) is the strongest acid because the presence of the electron-withdrawing chlorine atom increases acidity through inductive effect.

Option A (Esterification) is correct because the reaction of an acid with an alcohol to form an ester is known as esterification.

Option C (Benzaldehyde) is correct because dry distillation of calcium benzoate with calcium formate yields benzaldehyde.

Option D ($CH_3COOH$) is correct because alkyl groups increase electron density and reduce acidity, thus increasing the $pK_a$ value.

Option D (NaHCO3) is correct because formic acid, being an acid, will react with NaHCO3 to produce CO2 gas, while formaldehyde will not react.

Option B (Purification of acetaldehyde by $NaHS{O_3}$) is correct because the reaction of acetaldehyde with $NaHS{O_3}$ involves the nucleophilic addition of $HS{O_3}^-$ to the carbonyl carbon.

Option D (Formaldehyde) is correct because it does not contain a methyl ketone or a secondary alcohol that can be oxidized to a methyl ketone.

Option B (3 > 1 > 2) is correct because carboxylic acids have higher boiling points than alcohols, which have higher boiling points than aldehydes due to stronger intermolecular forces.

Option B (Propanol) is correct because complete reduction of acrolein (an unsaturated aldehyde) results in saturation of the double bond and reduction of the aldehyde to an alcohol, forming propanol.

Option D ($\mathrm{C}{\mathrm{H}_2}\left( \mathrm{I} \right)\mathrm{COOH}$) is the weakest acid because iodine is the least electronegative halogen, resulting in the weakest electron-withdrawing inductive effect.

Option A (Ethyl acetate) is correct because acetic anhydride reacts with diethyl ether in the presence of $AlCl_3$ via Friedel-Crafts acylation to form ethyl acetate.

Option C (Aqueous $\mathrm{FeC}{\mathrm{l}_3}$) is correct because picric acid, being a phenol, reacts with $FeCl_3$ to give a colored complex, while benzoic acid does not.

Option B (Cinnamic acid) is correct because it contains an unsaturated double bond that can be oxidized by potassium permanganate, leading to decolorization.

Option D ($\mathrm{C}{\mathrm{H}_3}\mathrm{COC}{\mathrm{H}_3}$) is correct because reduction of acetone with $\mathrm{LiAl}{\mathrm{H}_4}$ gives isopropanol, which gives a positive iodoform test.

Option C (Iodoform reaction) is correct because 2-pentanone is a methyl ketone and gives a positive iodoform test, while 3-pentanone does not.

Option D (Iodoform test) is correct because 2-pentanone is a methyl ketone and will give a positive iodoform test, while 3-pentanone will not.

Option B ($\mathrm{C}{\mathrm{H}_3}\mathrm{COCl}$) is correct because butanol reacting with acetyl chloride forms an ester, which often have fruity odors like banana.

Option D (NaOI) is correct because NaOI (sodium hypoiodite) is used in the haloform reaction, which oxidizes methyl ketones to carboxylic acids.

Option C (urotropine) is correct because the reaction of formaldehyde with ammonia produces hexamethylenetetramine, which is also known as urotropine.