The Cannizzaro reaction is a type of redox reaction where aldehydes without alpha-hydrogen atoms undergo disproportionation in the presence of a strong base. In this reaction, one molecule of aldehyde is reduced to alcohol, while another is oxidized to the corresponding carboxylate anion.
Let's analyze each compound:
- (A) 2-Chlorobutanal: This compound has a chlorine atom attached to the second carbon, which makes it unable to undergo the Cannizzaro reaction because it lacks the alpha-hydrogen required for the reaction to proceed. Therefore, 2-Chlorobutanal does not undergo Cannizzaro's reaction.
- (B) 2,2-Dimethylpropanal: This compound has no alpha-hydrogens as the two methyl groups at the alpha position prevent the formation of the enolate ion required for the Cannizzaro reaction. Hence, this compound would not undergo Cannizzaro's reaction either.
- (C) Benzaldehyde: Benzaldehyde contains an aldehyde group attached to a phenyl group, and it has an alpha-hydrogen. Therefore, it can undergo the Cannizzaro reaction, so it does not belong to the pair that cannot undergo the reaction.
- (D) 2-Phenyl ethanol: This compound is an alcohol, not an aldehyde, and thus cannot undergo the Cannizzaro reaction, which requires an aldehyde group. Thus, the two compounds that do not undergo Cannizzaro's reaction are (A) 2-Chlorobutanal and (D) 2-Phenyl ethanol.
, \((\text{IV}) \)If potential (in volt) in a region is expressed as $ V(x, y, z) = 6xy - y + 2yz $, the electric field (in} $ \text{N/C} $ at point (1, 0, 1) is:
Three point charges are located on a circular arc at A, B and C as shown in the figure below. The total electric field at the centre of the arc (C) is 
Five charges, 'q' each are placed at the corners of a regular pentagon of side 'a' as shown in figure. First, charge from 'A' is removed with other charges intact, then charge at 'A' is replaced with an equal opposite charge. The ratio of magnitudes of electric fields at O, without charge at A and that with equal and opposite charge at A is 