Aldehyde ketones & carboxylic Acid MCQs for NEET — Chemistry Questions with Answers

To solve this, we need to determine the original compound based on the molecular weights given.

1. Molecular weight of the original compound: 44.
2. Molecular weight after complete oxidation: 60.

The molecular formula corresponding to a molecular weight of 44 is likely $C_2H_4O$. Complete oxidation of $C_2H_4O$ yields $CO_2$ and $H_2O$.

The original compound is an aldehyde (ethanal) because on oxidation, an aldehyde converts into an acid (ethanoic acid) with a molecular weight of 60. Therefore, the correct answer is 'an aldehyde'.

The silver mirror test, also known as Tollen's test, is used to distinguish between aldehydes and acids. Aldehydes reduce Tollen's reagent (ammoniacal silver nitrate) to metallic silver, forming a shiny silver mirror on the surface of the test tube, while acids do not.

The Wolff-Kishner reduction is used to reduce aldehydes and ketones to alkanes. It specifically reduces the -CHO group (aldehyde group) to a -CH2 group, effectively removing the oxygen.

Cyanohydrins are formed by the reaction of aldehydes or ketones with hydrogen cyanide (HCN). The cyanohydrin of acetaldehyde (CH3CHO) can be hydrolyzed to form lactic acid (2-hydroxypropanoic acid).

Carboxylic acids are more acidic than phenols and alcohols because their conjugate base (carboxylate ion) is resonance-stabilized. This means that the negative charge can be delocalized over two oxygen atoms, making the conjugate base more stable. Phenols and alcohols do not have this level of resonance stabilization for their conjugate bases.

When $CH_2=CH-COOH$ (acrylic acid) is reduced with $LiAlH_4$, the double bond and the carboxylic acid group are both reduced. This reaction converts the carboxylic acid group ($-COOH$) to an alcohol ($-CH_2OH$) and the double bond ($C=C$) to a single bond ($C-C$), resulting in $CH_3-CH_2-CH_2OH$ (propyl alcohol).

Formic acid (HCOOH) has the lowest pKa value among the given options. This is because the formate ion (conjugate base of formic acid) is more stabilized by the absence of electron-donating groups, making formic acid more acidic compared to other carboxylic acids listed.

The Cannizzaro reaction is a disproportionation reaction in which a non-enolizable aldehyde undergoes self-oxidation and self-reduction in the presence of a strong base to produce a primary alcohol and a carboxylic acid. This is a classic example of disproportionation, where the same substance is simultaneously oxidized and reduced.