NEET Practice Questions (MCQs) with Answers & Solutions

The reactivity of carbonyl compounds in nucleophilic addition reactions decreases with the increase in steric hindrance and electron-donating alkyl groups. Hence, the order of reactivity is: Butanone (IV) < Propanone (III) < Propanal (II) < Ethanal (I).

When ketones react with magnesium amalgam (Mg-Hg) over water, pinacols are formed. This reaction is known as the pinacol coupling reaction. The ketone is reduced to a diol (pinacol) through an intermediate radical mechanism.

In the presence of a dilute base, benzaldehyde (C6H5CHO) and acetaldehyde (CH3CHO) undergo a crossed aldol condensation to form cinnamaldehyde (C6H5-CH=CH-CHO). The reaction involves the formation of a β-hydroxy aldehyde intermediate, which then dehydrates to give the α,β-unsaturated aldehyde product.

The weakest acid among the given options is isobutyric acid ((CH3)2CH COOH). The presence of electron-donating alkyl groups (CH3) reduces the acidity of the carboxylic acid group by increasing electron density around it and thus making it less likely to donate a proton.

The most acidic compound among the given options is o-hydroxybenzoic acid. This is because of the intramolecular hydrogen bonding which stabilizes the conjugate base, making it easier for the compound to lose a proton (H+). In o-hydroxybenzoic acid, the hydroxyl group at the ortho position relative to the carboxyl group helps to stabilize the negative charge on the oxygen atom after deprotonation, increasing its acidity.

Boron forms covalent compounds due to its higher ionisation enthalpy and small size. The high ionisation enthalpy makes it difficult for boron to lose electrons and form ionic bonds, while the small size allows it to form stable covalent bonds by sharing electrons with other atoms.

In diborane (B2H6), the two types of hydrogen-boron-hydrogen (H-B-H) bond angles are approximately 95 degrees and 120 degrees. The 95-degree angle corresponds to the bridge hydrogen atoms, while the 120-degree angle corresponds to the terminal hydrogen atoms.

The Order is due to inert pair effect

$ 2H_3 BO_3 \xrightarrow [] { { \triangle} } B_2O_3 + 3H_2O $