Chapter 12: Aldehydes, Ketones, & Carboxylic Acids
Highly Detailed Mark-wise Solutions for Board Exam Preparation
1 Mark Questions (Very Short Answer)
Answer: The IUPAC name of acetone ($CH_3COCH_3$) is Propan-2-one.
Answer: The catalyst used is Palladium supported on Barium sulfate ($Pd/BaSO_4$). It is poisoned by the addition of sulfur or quinoline (to prevent further reduction of the aldehyde to an alcohol).
Answer: Tollens' Test (Silver Mirror Test) or Fehling's Test. Acetaldehyde gives a positive test (forms a silver mirror or red precipitate respectively), whereas Acetone does not react.
Answer: The aldehyde must not contain any $\alpha$-hydrogen atoms. Examples: Formaldehyde ($HCHO$) or Benzaldehyde ($C_6H_5CHO$).
2 Mark Questions (Short Answer-I)
Answer:
In the Stephen reaction, alkyl nitriles (cyanides) are reduced to imine hydrochlorides using Stannous chloride ($SnCl_2$) and concentrated hydrochloric acid ($HCl$). The intermediate imine upon acid hydrolysis yields the corresponding aldehyde.
$R-C \equiv N + 2[H] + HCl \xrightarrow{SnCl_2 / HCl} R-CH=NH\cdot HCl$
$R-CH=NH\cdot HCl + H_2O \xrightarrow{\Delta} R-CHO + NH_4Cl$
Answer:
Wolff-Kishner reduction is a chemical reaction used to convert the carbonyl group ($>C=O$) of aldehydes or ketones into a methylene group ($>CH_2$). The carbonyl compound is treated with hydrazine ($NH_2NH_2$) to form a hydrazone, which is then heated with a strong base like KOH in a high-boiling solvent like ethylene glycol to yield an alkane.
Example (Reduction of Propanone to Propane):
$CH_3-CO-CH_3 + NH_2NH_2 \xrightarrow{-H_2O} CH_3-C(=NNH_2)-CH_3$
$CH_3-C(=NNH_2)-CH_3 \xrightarrow{KOH / \text{Ethylene Glycol}, \Delta} CH_3-CH_2-CH_3 + N_2 \uparrow$
Answer:
The Etard reaction is used to specifically prepare benzaldehyde from toluene. Toluene is oxidized using a mild oxidizing agent, Chromyl chloride ($CrO_2Cl_2$), in a non-polar solvent like Carbon disulfide ($CS_2$). It first forms a brown chromium complex, which on subsequent acid hydrolysis yields Benzaldehyde. The use of chromyl chloride prevents further oxidation to benzoic acid.
$C_6H_5-CH_3 + 2CrO_2Cl_2 \xrightarrow{CS_2} \text{Brown Complex} \xrightarrow{H_3O^+} C_6H_5-CHO$
3 Mark Questions (Short Answer-II / Mechanisms)
Answer:
Condition:
The aldehyde or ketone must contain at least one $\alpha$-hydrogen atom.
Explanation & Example:
When two molecules of an aldehyde or ketone containing an $\alpha$-hydrogen are treated with a dilute alkali (like dil. $NaOH$, $Na_2CO_3$, or $Ba(OH)_2$), they undergo a self-addition reaction to form a $\beta$-hydroxy aldehyde (Aldol) or $\beta$-hydroxy ketone (Ketol). Upon warming, the aldol readily undergoes dehydration (loss of a water molecule) to form an $\alpha,\beta$-unsaturated carbonyl compound.
Example using Acetaldehyde ($CH_3CHO$):
1. Addition Step (Formation of Aldol):
$CH_3-CHO + H-CH_2-CHO \xrightarrow{\text{dil. NaOH}} CH_3-CH(OH)-CH_2-CHO$
(3-Hydroxybutanal / Acetaldol)
2. Condensation Step (Dehydration):
$CH_3-CH(OH)-CH_2-CHO \xrightarrow{\Delta, -H_2O} CH_3-CH=CH-CHO$
(But-2-enal / Crotonaldehyde)
Answer:
Cannizzaro Reaction:
Aldehydes that do not have an $\alpha$-hydrogen atom undergo a self-oxidation and reduction reaction (disproportionation) when heated with a strong, concentrated alkali (like 50% NaOH or KOH). In this reaction, one molecule of the aldehyde is reduced to the corresponding primary alcohol, while the second molecule is oxidized to the salt of the corresponding carboxylic acid.
Equation (Formaldehyde):
$2 HCHO + \text{conc. } NaOH \xrightarrow{\Delta} CH_3OH + HCOONa$
(Formaldehyde) (Methanol) (Sod. formate)
Reasoning:
Formaldehyde ($H-CHO$) contains only one carbon atom and lacks an $\alpha$-carbon entirely, hence it has no $\alpha$-hydrogen atoms. Therefore, it undergoes the Cannizzaro reaction. Acetaldehyde ($CH_3-CHO$) has an $\alpha$-carbon attached to three $\alpha$-hydrogen atoms. Because these $\alpha$-hydrogens are acidic, acetaldehyde reacts with base to form an enolate ion, leading to Aldol condensation rather than the Cannizzaro reaction.
4 Mark Questions (Long Answer / Combined)
Note: In the HSC Board Exam, 4-mark questions in Chemistry are frequently split into sub-questions (e.g., a 2-mark name reaction combined with a 2-mark reasoning question).
(b) Explain why Chloroacetic acid is a stronger acid than Acetic acid. [2 Marks] March 2017, Oct 2020, March 2023
Answer (a): HVZ Reaction
Carboxylic acids containing an $\alpha$-hydrogen atom react with chlorine or bromine in the presence of a small amount of Red Phosphorus to form $\alpha$-halocarboxylic acids. This reaction is known as the Hell-Volhard-Zelinsky (HVZ) reaction.
$CH_3-COOH + Cl_2 \xrightarrow{\text{Red P / } H_2O} Cl-CH_2-COOH + HCl$
(Acetic acid) ($\alpha$-Chloroacetic acid)
Answer (b): Reasoning on Acidic Strength
The acidic strength of a carboxylic acid depends on the stability of the carboxylate ion formed after losing a proton ($H^+$).
- Chloroacetic acid ($Cl-CH_2-COOH$): Chlorine is a highly electronegative atom that exerts a strong -I effect (electron-withdrawing inductive effect). It pulls electron density away from the $O-H$ bond, facilitating the release of the $H^+$ ion. Additionally, it stabilizes the resulting chloroacetate ion by dispersing the negative charge. Hence, it is a stronger acid.
- Acetic acid ($CH_3-COOH$): The methyl group ($-CH_3$) exerts a +I effect (electron-donating effect). It pushes electron density toward the carboxyl group, making the release of $H^+$ harder and destabilizing the resulting acetate ion by intensifying the negative charge. Hence, it is a weaker acid.
(b) How is Acetic acid prepared from Grignard reagent? Write the equation. [2 Marks] Oct 2013, March 2018
Answer (a): Preparation of Benzoic Acid from Toluene
Toluene is converted into benzoic acid by vigorous oxidation using a strong oxidizing agent like alkaline potassium permanganate ($KMnO_4/KOH$) followed by acid hydrolysis. The entire methyl side chain is oxidized to a carboxyl group.
$C_6H_5-CH_3 \xrightarrow{KMnO_4 / KOH, \Delta} C_6H_5-COO^-K^+ \xrightarrow{H_3O^+} C_6H_5-COOH$
(Toluene) (Potassium benzoate) (Benzoic acid)
Answer (b): Preparation from Grignard Reagent
Acetic acid can be prepared by treating Methyl magnesium bromide (a Grignard reagent) with solid carbon dioxide (Dry Ice) in the presence of dry ether. A magnesium salt complex is formed, which upon acid hydrolysis yields acetic acid.
$CH_3-Mg-Br + O=C=O \xrightarrow{\text{Dry Ether}} CH_3-COOMgBr$
$CH_3-COOMgBr + H_2O \xrightarrow{H^+} CH_3-COOH \text{ (Acetic acid)} + Mg(OH)Br$
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