CBSE Class 12 Chemistry Chapter 12: Aldehydes, Ketones and Carboxylic Acids NCERT Solutions

Solution: The addition of water to alkynes in the presence of Hg²⁺ and acid catalyst follows Markovnikov's rule. For but-1-yne ($CH\_3\; -\; CH\_2\; -\; CH\; =\; CH\_2$), the water molecule adds such that the hydrogen atom attaches to the carbon with more hydrogen atoms, and the hydroxyl group attaches to the other carbon. This initially forms an unstable enol intermediate ($CH\_3\; -\; CH\_2\; -\; CH(OH)\; =\; CH\_2$). This enol then undergoes tautomerism to form the more stable ketone, butan-2-one ($CH\_3\; -\; CH\_2\; -\; C(=O)\; -\; CH\_3$).

$$CH\_3\; -\; CH\_2\; -\; CH\; =\; CH\_2\; \backslash xrightarrow\{\backslash text\{Hg\}^\{2+\},\; \backslash text\{H\}\_2\backslash text\{O\}\}\; [CH\_3\; -\; CH\_2\; -\; CH(OH)\; =\; CH\_2]\; \backslash xrightarrow\{\backslash text\{Tautomerism\}\}\; CH\_3\; -\; CH\_2\; -\; C(=O)\; -\; CH\_3$$

Answer: (b) Butan-2-one

Solution: The reactivity of carbonyl compounds towards nucleophilic addition reactions depends on two main factors: electronic effects and steric effects.

1. Electronic Factor: The more positive the carbonyl carbon, the greater its reactivity. Electron-donating groups (like alkyl groups) decrease the positive charge on the carbonyl carbon, reducing reactivity. Electron-withdrawing groups increase reactivity.
2. Steric Factor: Less bulky groups around the carbonyl carbon lead to greater reactivity as they offer less hindrance to the incoming nucleophile.

Comparing the given compounds:

• Ethanal ($CH\_3CHO$) has one electron-donating methyl group and a hydrogen atom.
• Propanone ($CH\_3COCH\_3$) has two electron-donating methyl groups, making the carbonyl carbon less positive and more sterically hindered than in ethanal.
• Benzaldehyde ($C\_6H\_5CHO$) has a phenyl group, which can withdraw electrons via resonance but also offers some steric hindrance. However, its reactivity is generally less than ethanal.
• Benzoic acid ($C\_6H\_5COOH$) is a carboxylic acid, and while it has a carbonyl group, its reactivity towards nucleophilic addition is significantly different and generally lower than aldehydes and ketones due to resonance and the presence of the -OH group.

Therefore, ethanal, with the least electron donation and minimal steric hindrance, is the most reactive towards nucleophilic addition. Answer: (a) Ethanal

Solution: The acidic strength of a compound is determined by the stability of its conjugate base formed after the removal of a proton ($H^+$).

1. Ethanol ($CH\_3CH\_2OH$): The ethoxide ion ($CH\_3CH\_2O^-$) is formed. This is a strong base and is destabilized by the electron-donating effect of the ethyl group.
2. Phenol ($C\_6H\_5OH$): The phenoxide ion ($C\_6H\_5O^-$) is formed. This ion is resonance-stabilized by delocalization of the negative charge into the benzene ring, making it less basic and thus phenol more acidic than ethanol.
3. Acetic acid ($CH\_3COOH$): The acetate ion ($CH\_3COO^-$) is formed. It is stabilized by resonance between the two oxygen atoms, making it more stable than the phenoxide ion. Therefore, acetic acid is more acidic than phenol.
4. Chloroacetic acid ($ClCH\_2COOH$): The chloroacetate ion ($ClCH\_2COO^-$) is formed. The electron-withdrawing chlorine atom attached to the alpha-carbon stabilizes the negative charge on the carboxylate group through inductive effect, further increasing the acidity. Thus, chloroacetic acid is more acidic than acetic acid.

The order of increasing acidic strength is: Ethanol < Phenol < Acetic acid < Chloroacetic acid. Answer: (c) ethanol < phenol < acetic acid < chloroacetic acid

Solution: The compound Ph-COO-Ph is diphenyl ester. It can be prepared by the reaction of phenol ($C\_6H\_5OH$) with benzoyl chloride ($C\_6H\_5COCl$). This reaction is typically carried out in the presence of a base, such as pyridine or aqueous NaOH, which acts as a catalyst and neutralizes the HCl formed during the reaction. This is a classic example of esterification, specifically known as the Schotten-Baumann reaction when performed under specific conditions.

$$C\_6H\_5OH\; +\; C\_6H\_5COCl\; \backslash xrightarrow\{\backslash text\{Pyridine\}\}\; C\_6H\_5COOC\_6H\_5\; +\; HCl$$

Option (a) involves a carboxylic acid and a phenol, which typically requires stronger dehydrating agents for esterification. Option (c) uses ZnCl₂, which is a Lewis acid catalyst often used for other reactions. Option (d) involves benzaldehyde, which would lead to different products. Answer: (b) phenol and benzoyl chloride in the presence of pyridine