Chapter 12: Aldehydes, Ketones, & Carboxylic Acids Mock Test
Time: 1 Hour | Maximum Marks: 25
- All questions are compulsory.
- Section A contains Q1 (Multiple Choice) and Q2 (Very Short Answer).
- Section B contains Short Answer Type I questions (2 marks each). Attempt any 4.
- Section C contains Short Answer Type II questions (3 marks each). Attempt any 2.
- Section D contains Long Answer questions (4 marks each). Attempt any 1.
SECTION A
Q1. Select and write the most appropriate answer from the given alternatives: [4 Marks]
-
Which of the following compounds does NOT undergo the Cannizzaro reaction?
(A) Formaldehyde(B) Acetaldehyde(C) Benzaldehyde(D) Trichloroacetaldehyde
-
The reagent used in the Clemmensen reduction of carbonyl compounds is:
(A) $NH_2NH_2 / KOH$(B) $Zn-Hg / \text{conc. } HCl$(C) $H_2 / Pd-BaSO_4$(D) $LiAlH_4$
-
Which of the following carboxylic acids is the strongest?
(A) $CH_3COOH$(B) $CH_2ClCOOH$(C) $CHCl_2COOH$(D) $CCl_3COOH$
-
Tollens' reagent is basically an ammoniacal solution of:
(A) Silver nitrate(B) Copper sulfate(C) Copper chloride(D) Potassium dichromate
Q2. Answer the following questions in one sentence: [3 Marks]
- Write the IUPAC name of acetone.
- State the condition necessary for an aldehyde to undergo Aldol condensation.
- What is the product formed when methyl magnesium bromide (Grignard reagent) is treated with dry ice followed by acid hydrolysis?
SECTION B
Attempt any FOUR of the following: [8 Marks]
- Explain the Cannizzaro reaction with a suitable example.
- How is benzaldehyde prepared from toluene using Etard reaction?
- Explain why aldehydes are more reactive than ketones towards nucleophilic addition reactions.
- Write a short note on the Stephen reaction.
- Why is chloroacetic acid stronger than acetic acid?
SECTION C
Attempt any TWO of the following: [6 Marks]
- Explain the Aldol condensation of acetaldehyde with a chemical equation.
- Describe the Hell-Volhard-Zelinsky (HVZ) reaction with a suitable example.
- Write the reactions involved when:
- Acetyl chloride is treated with $H_2 / Pd-BaSO_4$.
- Acetaldehyde is treated with Tollens' reagent.
- Acetone is treated with Zinc amalgam and conc. $HCl$.
SECTION D
Attempt any ONE of the following: [4 Marks]
- (a) Explain Wolff-Kishner reduction with a chemical equation. [2 Marks]
(b) Distinguish between aldehydes and ketones using Fehling's solution. [2 Marks] - (a) Write a short note on Gatterman-Koch formylation. [2 Marks]
(b) What is the action of $PCl_5$ on acetic acid? Write the chemical equation. [2 Marks]
Solutions & Marking Scheme
SECTION A [7 Marks]
Q1. Multiple Choice Answers:
1. (B) Acetaldehyde [1 Mark. It contains an $\alpha$-hydrogen, so it undergoes Aldol condensation instead]
2. (B) $Zn-Hg / \text{conc. } HCl$ [1 Mark for correct option]
3. (D) $CCl_3COOH$ [1 Mark. Max number of electron-withdrawing Cl groups increases acidity via -I effect]
4. (A) Silver nitrate [1 Mark for correct option]
Q2. Very Short Answers:
1. IUPAC name of acetone:
Propan-2-one. [1 Mark]
2. Condition for Aldol condensation:
The aldehyde must contain at least one $\alpha$-hydrogen atom. [1 Mark]
3. Product of Grignard + Dry Ice:
Acetic acid (Ethanoic acid). [1 Mark]
SECTION B [8 Marks]
Q3. Cannizzaro Reaction:
Aldehydes that lack an $\alpha$-hydrogen atom undergo self-oxidation and reduction (disproportionation) when heated with concentrated alkali (like 50% NaOH). One molecule is oxidized to a carboxylic acid salt, while the other is reduced to a primary alcohol. [1 Mark]
$2HCHO + \text{conc. } NaOH \xrightarrow{\Delta} CH_3OH \text{ (Methanol)} + HCOONa \text{ (Sod. formate)}$
[1 Mark for correct equation]Q4. Etard Reaction:
Toluene is oxidized to benzaldehyde using Chromyl chloride ($CrO_2Cl_2$) in a non-polar solvent like $CS_2$. It forms a brown chromium complex, which on acid hydrolysis yields benzaldehyde. [1 Mark]
$C_6H_5-CH_3 + 2CrO_2Cl_2 \xrightarrow{CS_2} \text{Brown Complex} \xrightarrow{H_3O^+} C_6H_5CHO \text{ (Benzaldehyde)}$
[1 Mark for reaction]Q5. Reactivity of Aldehydes vs Ketones:
Aldehydes are more reactive than ketones towards nucleophilic addition due to two reasons:
- Steric Hindrance: Ketones have two bulky alkyl groups that hinder the approach of the nucleophile to the carbonyl carbon, whereas aldehydes have only one alkyl group and a small hydrogen atom. [1 Mark]
- Electronic Factor (+I Effect): The two alkyl groups in ketones are electron-donating. They reduce the positive charge ($\delta^+$) on the carbonyl carbon more effectively than the single alkyl group in aldehydes, thereby decreasing its electrophilicity. [1 Mark]
Q6. 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. [1 Mark]
$R-C \equiv N + 2[H] \xrightarrow{SnCl_2 / HCl} R-CH=NH\cdot HCl \xrightarrow{H_2O, \Delta} R-CHO + NH_4Cl$
[1 Mark for reaction]Q7. Chloroacetic acid vs Acetic acid:
Chloroacetic acid ($Cl-CH_2-COOH$) contains a highly electronegative Chlorine atom. Chlorine exerts a strong electron-withdrawing inductive effect (-I effect). This pulls the electron density away from the O-H bond, facilitating the easy release of $H^+$. It also stabilizes the resulting chloroacetate ion by dispersing the negative charge. [1.5 Marks]
Acetic acid ($CH_3-COOH$) contains an electron-donating methyl group (+I effect), which destabilizes the acetate ion. Hence, chloroacetic acid is stronger. [0.5 Mark]
SECTION C [6 Marks]
Q8. Aldol Condensation of Acetaldehyde:
When two molecules of acetaldehyde (which contains $\alpha$-hydrogens) are treated with dilute alkali (e.g., dil. NaOH), they condense to form a $\beta$-hydroxy aldehyde called Acetaldol. Upon heating, acetaldol loses a water molecule to form an $\alpha,\beta$-unsaturated aldehyde (Crotonaldehyde). [1 Mark]
Reaction:
Step 1: $2CH_3CHO \xrightarrow{\text{dil. NaOH}} CH_3-CH(OH)-CH_2-CHO \text{ (Acetaldol)}$
Step 2: $CH_3-CH(OH)-CH_2-CHO \xrightarrow{\Delta, -H_2O} CH_3-CH=CH-CHO \text{ (Crotonaldehyde/But-2-enal)}$
Q9. Hell-Volhard-Zelinsky (HVZ) Reaction:
Carboxylic acids having an $\alpha$-hydrogen atom react with chlorine or bromine in the presence of a small amount of Red Phosphorus to give $\alpha$-halocarboxylic acids. This is known as the HVZ reaction. [1.5 Marks]
Example:
$CH_3-COOH + Cl_2 \xrightarrow{\text{Red P / } H_2O} Cl-CH_2-COOH + HCl$
(Acetic acid) ($\alpha$-Chloroacetic acid)
Q10. Chemical Reactions:
- (i) Rosenmund Reduction:
$CH_3COCl + H_2 \xrightarrow{Pd/BaSO_4} CH_3CHO \text{ (Acetaldehyde)} + HCl$ [1 Mark] - (ii) Tollens' Test:
$CH_3CHO + 2[Ag(NH_3)_2]^+ + 3OH^- \rightarrow CH_3COO^- + 2Ag \downarrow \text{ (Silver mirror)} + 4NH_3 + 2H_2O$ [1 Mark] - (iii) Clemmensen Reduction:
$CH_3COCH_3 + 4[H] \xrightarrow{Zn-Hg / \text{conc. } HCl} CH_3CH_2CH_3 \text{ (Propane)} + H_2O$ [1 Mark]
SECTION D [4 Marks]
Q11. (a) Wolff-Kishner Reduction [2 Marks] (b) Fehling's Test [2 Marks]
(a) Wolff-Kishner Reduction:
The carbonyl group ($>C=O$) of aldehydes or ketones is reduced to a methylene group ($-CH_2-$) by heating with Hydrazine ($NH_2NH_2$) and KOH in a high-boiling solvent like ethylene glycol. [1 Mark]
$CH_3COCH_3 + NH_2NH_2 \xrightarrow{-H_2O} CH_3C(=NNH_2)CH_3 \xrightarrow{KOH / \text{Glycol}, \Delta} CH_3CH_2CH_3 \text{ (Propane)} + N_2$
[1 Mark for reaction](b) Fehling's Test:
Fehling's solution is a mixture of aqueous $CuSO_4$ and Rochelle salt (in NaOH). When an aliphatic aldehyde is heated with Fehling's solution, it reduces the $Cu^{2+}$ ions to form a red precipitate of cuprous oxide ($Cu_2O$), while the aldehyde is oxidized to a carboxylate ion. [1 Mark]
Ketones do not respond to this test and do not form any red precipitate. This serves as a distinguishing test. [1 Mark]
Q12. (a) Gatterman-Koch Formylation [2 Marks] (b) Action of $PCl_5$ [2 Marks]
(a) Gatterman-Koch Formylation:
Benzene or its derivatives are treated with a mixture of carbon monoxide (CO) and hydrogen chloride (HCl) gas under high pressure in the presence of an anhydrous $AlCl_3$ and cuprous chloride ($CuCl$) catalyst to yield Benzaldehyde. [1 Mark]
$C_6H_6 + CO + HCl \xrightarrow{\text{anh. } AlCl_3 / CuCl} C_6H_5CHO + HCl$ [1 Mark]
(b) Action of $PCl_5$ on Acetic Acid:
Acetic acid reacts with Phosphorus pentachloride ($PCl_5$) to form Acetyl chloride (Ethanoyl chloride). The $-OH$ group of the carboxylic acid is replaced by a $-Cl$ atom. [1 Mark]
$CH_3COOH + PCl_5 \rightarrow CH_3COCl \text{ (Acetyl chloride)} + POCl_3 + HCl$ [1 Mark]
📚 Also Read
Lecture Notes🏆 Complete Maharashtra HSC Class 12 Chemistry Preparation
Prepare for the Maharashtra HSC Class 12 Chemistry Board Exam with chapter-wise revision notes, important questions, PYQs, formula sheets, mock tests, quick revision resources and exam-oriented study material. Everything you need to score high in one comprehensive learning hub.
🚀 Explore the Complete Maharashtra HSC Chemistry Hub
No comments:
Post a Comment