Aldehydes, Ketones and Carboxylic Acids

Due to the formation of intermolecular H-bonding, association occurs in carboxylic acids.

So, they have higher boiling points than aldehydes, ketones and even alcohols of comparable molecular mass.

Since, A gives silver mirror test, it must be an aldehyde and aldehydes are formed by oxidation of 1 o alcohols.

Thus, ‘X’ is a 1 o alcohol, i.e., CH 3 CH 2 OH.

We know, Acidic strength $\propto$ – I effect As oxygen is more electron withdrawing (II) and (III) show greater – I effect than (I).

Thus, (I) is least acidic.

Out of (II) and (III), (II) is more acidic than (III) as distance of O increases from —COOH group and acidic strength decreases.

Keto-enol tautomerism is as follows :

Trans isomer does not react with acetone as removal of H 2 O molecule is difficult.

Carbonyl compounds react with ammonia derivatives in weakly acidic medium as follows :

As the compound X yields phenyl hydrazone and gives negative response to the iodoform test and Tollen’s test so it must contain a C = O group but neither a methyl ketone nor in aldehyde.

Thus, the structure of X will be

Enolic form predominates in compounds containing two carbonyl groups separated by a –CH 2 group.

This is due to following two factors. (i) Presence of conjugation which increases stability. (ii) Formation of intramolecular hydrogen bond between enolic hydroxyl group and second carbonyl group which leads to stablisation of the molecule.

Hence the correct answer is III > II > I

RCHO + NH 2 NH 2 $\to$ RCH $=$ N $-$ NH 2 Such reactions take place in slightly acidic medium and involve nucleophilic addition of the ammonia derivative.

It is Rosenmund’s reaction.