Prepare for PTB exams with solved exercises on Aldehydes and Ketones. Focus on key topics like structure, properties, reactions, and applications to deepen your understanding and improve exam performance.
Q. 4: Give one laboratory and one industrial method for the preparation of formaldehyde.
Laboratory method:
Formaldehyde can be prepared in the lab by the oxidation of methanol. This is done by passing methanol vapor mixed with air over a heated copper or silver catalyst at around 400°C. The reaction is as follows:
[ \text{CH}_3\text{OH} + \frac{1}{2}\text{O}_2 \xrightarrow{Cu/Ag} \text{HCHO} + \text{H}_2\text{O} ]
Industrial method:
On an industrial scale, formaldehyde is produced by the catalytic oxidation of methanol. Methanol vapor is passed over a catalyst of iron oxide-molybdenum oxide (Fe-Mo) or silver at 300-400°C:
[ \text{CH}_3\text{OH} + \frac{1}{2}\text{O}_2 \rightarrow \text{HCHO} + \text{H}_2\text{O} ]
This method is highly efficient and commonly used in formaldehyde production.
Q. 5: How does formaldehyde react with the following reagents?
(i) CH₃MgI (Grignard reagent):
Formaldehyde reacts with methyl magnesium iodide to give a primary alcohol upon hydrolysis.
[ \text{HCHO} + \text{CH}_3\text{MgI} \xrightarrow{\text{H}_2\text{O}} \text{CH}_3\text{CH}_2\text{OH} ]
(ii) HCN:
Formaldehyde reacts with hydrogen cyanide (HCN) to form cyanohydrin.
[ \text{HCHO} + \text{HCN} \rightarrow \text{HOCH}_2\text{CN} ]
(iii) NaHSO₃ (Sodium bisulfite):
Formaldehyde reacts with sodium bisulfite to form a bisulfite addition compound.
[ \text{HCHO} + \text{NaHSO}_3 \rightarrow \text{HOCH}_2\text{SO}_3\text{Na} ]
(iv) Conc. NaOH (Cannizzaro Reaction):
In the presence of concentrated NaOH, formaldehyde undergoes the Cannizzaro reaction, where one molecule is reduced to methanol and another is oxidized to formate.
[ 2 \text{HCHO} + \text{NaOH} \rightarrow \text{HCOONa} + \text{CH}_3\text{OH} ]
(v) NaBH₄/H₂O (Sodium borohydride):
Formaldehyde is reduced to methanol when treated with sodium borohydride.
[ \text{HCHO} + \text{NaBH}_4 \xrightarrow{\text{H}_2\text{O}} \text{CH}_3\text{OH} ]
(vi) Tollens’ reagent:
Formaldehyde is oxidized by Tollens’ reagent (ammoniacal silver nitrate solution) to form formic acid.
[ \text{HCHO} + 2 \text{Ag(NH}_3\text{)}_2\text{OH} \rightarrow \text{HCOOH} + 2 \text{Ag} + \text{H}_2\text{O} + 2 \text{NH}_3 ]
(vii) Fehling’s reagent:
Formaldehyde reduces Fehling’s reagent, precipitating copper(I) oxide (Cu₂O).
[ \text{HCHO} + 2 \text{Cu}^{2+} + 5 \text{OH}^- \rightarrow \text{HCOO}^- + \text{Cu}_2\text{O} + 3 \text{H}_2\text{O} ]
Q. 6: Give one laboratory and one industrial method for the preparation of acetaldehyde.
Laboratory method:
Acetaldehyde can be prepared in the lab by the oxidation of ethanol using potassium dichromate and dilute sulfuric acid.
[ \text{C}_2\text{H}_5\text{OH} + \text{K}_2\text{Cr}_2\text{O}_7 + \text{H}_2\text{SO}_4 \rightarrow \text{CH}_3\text{CHO} + \text{Cr}_2\text{(SO}_4\text{)}_3 + \text{K}_2\text{SO}_4 + \text{H}_2\text{O} ]
Industrial method:
Acetaldehyde is produced industrially by the Wacker process, which involves the oxidation of ethylene using a palladium(II) chloride catalyst.
[ \text{C}_2\text{H}_4 + \frac{1}{2}\text{O}_2 \xrightarrow{PdCl_2, CuCl_2} \text{CH}_3\text{CHO} ]
Q. 7: How does acetaldehyde react with the following reagents?
(i) C₂H₅MgI (Ethyl magnesium iodide):
Acetaldehyde reacts with ethyl magnesium iodide to give a secondary alcohol after hydrolysis.
[ \text{CH}_3\text{CHO} + \text{C}_2\text{H}_5\text{MgI} \xrightarrow{\text{H}_2\text{O}} \text{CH}_3\text{CH}_2\text{CH}_2\text{OH} ]
(ii) HCN:
Acetaldehyde reacts with HCN to form α-hydroxypropionitrile (cyanohydrin).
[ \text{CH}_3\text{CHO} + \text{HCN} \rightarrow \text{CH}_3\text{CH(OH)CN} ]
(iii) NaHSO₃:
Acetaldehyde reacts with sodium bisulfite to form a bisulfite addition compound.
[ \text{CH}_3\text{CHO} + \text{NaHSO}_3 \rightarrow \text{CH}_3\text{CH(OH)SO}_3\text{Na} ]
(iv) Dilute NaOH (Aldol condensation):
In dilute NaOH, acetaldehyde undergoes aldol condensation, forming 3-hydroxybutanal.
[ 2 \text{CH}_3\text{CHO} \xrightarrow{\text{NaOH}} \text{CH}_3\text{CH(OH)CH}_2\text{CHO} ]
(v) LiAlH₄:
Acetaldehyde is reduced to ethanol by lithium aluminum hydride.
[ \text{CH}_3\text{CHO} + \text{LiAlH}_4 \rightarrow \text{CH}_3\text{CH}_2\text{OH} ]
(vi) NaBH₄/H₂O:
Acetaldehyde is reduced to ethanol by sodium borohydride.
[ \text{CH}_3\text{CHO} + \text{NaBH}_4 \rightarrow \text{CH}_3\text{CH}_2\text{OH} ]
(vii) NH₂OH (Hydroxylamine):
Acetaldehyde reacts with hydroxylamine to form acetaldehyde oxime.
[ \text{CH}_3\text{CHO} + \text{NH}_2\text{OH} \rightarrow \text{CH}_3\text{CH(OH)NOH} ]
(viii) K₂Cr₂O₇/H₂SO₄:
Acetaldehyde is oxidized by potassium dichromate to form acetic acid.
[ \text{CH}_3\text{CHO} + \text{K}_2\text{Cr}_2\text{O}_7 + \text{H}_2\text{SO}_4 \rightarrow \text{CH}_3\text{COOH} + \text{Cr}_2\text{(SO}_4\text{)}_3 ]
Q. 8: Describe briefly the mechanism of nucleophilic addition to a carbonyl compound.
Nucleophilic addition to a carbonyl compound typically involves the following steps:
- Attack of the nucleophile: The nucleophile attacks the electrophilic carbon of the carbonyl group (C=O), breaking the π-bond and pushing electrons onto oxygen. [ \text{Nu}^- + \text{R}_2\text{C=O} \rightarrow \text{R}_2\text{C(OH)Nu} ]
- Formation of a tetrahedral intermediate: The nucleophile forms a bond with the carbon, and the oxygen temporarily carries a negative charge as part of the intermediate.
- Protonation: In the final step, a proton is typically transferred to the oxygen to stabilize the compound.
The reaction proceeds faster if the carbonyl group is more electrophilic
(i.e., aldehydes are more reactive than ketones).
Q. 9: Explain with mechanism the addition of ethylmagnesium bromide to acetaldehyde. What is the importance of this reaction?
Mechanism:
Ethylmagnesium bromide (Grignard reagent) reacts with acetaldehyde as follows:
- The ethyl group (acting as a nucleophile) from ethylmagnesium bromide attacks the carbonyl carbon in acetaldehyde. [ \text{CH}_3\text{CHO} + \text{C}_2\text{H}_5\text{MgBr} \rightarrow \text{CH}_3\text{CH(C}_2\text{H}_5)\text{OMgBr} ]
- The resulting alkoxide ion is then hydrolyzed with water to give butanol. [ \text{CH}_3\text{CH(C}_2\text{H}_5)\text{OMgBr} + \text{H}_2\text{O} \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{OH} ]
Importance:
This reaction is a key step in organic synthesis for forming new carbon-carbon bonds, allowing the formation of alcohols from aldehydes.
Q. 10: Explain with mechanism the addition of sodium bisulphite to acetone. What is the utility of this reaction?
Mechanism:
Sodium bisulfite reacts with acetone by nucleophilic attack on the carbonyl carbon.
- The sulfur atom in bisulfite (HSO₃⁻) acts as the nucleophile and attacks the carbonyl carbon. [ \text{CH}_3\text{COCH}_3 + \text{HSO}_3^- \rightarrow \text{CH}_3\text{C(OH)(SO}_3\text{H)}\text{CH}_3} ]
- A bisulfite addition product (also called a hydrosulfite compound) is formed.
Utility:
The formation of bisulfite addition compounds is used for the purification of aldehydes and ketones, as they can easily be crystallized and later regenerated by acid treatment.
Q. 11: Describe with mechanism aldol condensation. Why formaldehyde does not give this reaction?
Aldol condensation mechanism:
- Formation of enolate ion: Under basic conditions, an aldehyde like acetaldehyde forms an enolate ion. [ \text{CH}_3\text{CHO} \xrightarrow{\text{OH}^-} \text{CH}_2\text{CHO}^- ]
- Nucleophilic attack: The enolate ion attacks the carbonyl carbon of another acetaldehyde molecule, forming 3-hydroxybutanal (aldol). [ \text{CH}_3\text{CH=O} + \text{CH}_2\text{CHO}^- \rightarrow \text{CH}_3\text{CH(OH)CH}_2\text{CHO} ]
- Dehydration: Under higher temperatures, the aldol product can lose water to form an α,β-unsaturated aldehyde. [ \text{CH}_3\text{CH(OH)CH}_2\text{CHO} \rightarrow \text{CH}_3\text{CH=CHCHO} + \text{H}_2\text{O} ]
Why formaldehyde does not undergo aldol condensation:
Formaldehyde lacks an α-hydrogen, which is necessary to form the enolate ion in the first step of the aldol condensation. Hence, it cannot participate in the reaction.
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Q. 12: What types of aldehydes give Cannizzaro’s reaction? Give its mechanism.
Aldehydes without α-hydrogen atoms undergo Cannizzaro’s reaction. Examples include formaldehyde and benzaldehyde.
Mechanism:
- Nucleophilic attack: In the presence of a strong base (like NaOH), a hydroxide ion attacks the carbonyl carbon of one aldehyde molecule, forming an alkoxide ion. [ \text{RCHO} + \text{OH}^- \rightarrow \text{RCH(OH)}^- ]
- Hydride transfer: The alkoxide ion then transfers a hydride ion to another aldehyde molecule, reducing it to an alcohol and oxidizing the original molecule to a carboxylate ion. [ \text{RCH(OH)}^- + \text{RCHO} \rightarrow \text{RCH}_2\text{OH} + \text{RCOO}^- ]
- Final products: The products are a primary alcohol and a carboxylate ion. [ 2 \text{RCHO} \rightarrow \text{RCH}_2\text{OH} + \text{RCOO}^- ]
Q. 13: Explain the mechanism of the reaction of phenylhydrazine with acetone.
Mechanism:
- Formation of hydrazone: Phenylhydrazine (C₆H₅NHNH₂) reacts with acetone, a carbonyl compound, in a condensation reaction to form a hydrazone. The lone pair on nitrogen of phenylhydrazine attacks the carbonyl carbon of acetone. [ \text{CH}_3\text{COCH}_3 + \text{C}_6\text{H}_5\text{NHNH}_2 \rightarrow \text{CH}_3\text{C=NNHC}_6\text{H}_5 + \text{H}_2\text{O} ]
- Removal of water: Water is eliminated, and a double bond is formed between the carbon and nitrogen.
The product is acetone phenylhydrazone.
Q. 14: Using ethyne as a starting material, how would you get acetaldehyde, acetone, and ethyl alcohol?
Acetaldehyde:
- Ethyne (acetylene) undergoes hydration (Kucherov’s reaction) in the presence of mercury(II) sulfate and sulfuric acid, forming acetaldehyde. [ \text{CH≡CH} + \text{H}_2\text{O} \xrightarrow{\text{Hg}^{2+}, \text{H}_2\text{SO}_4} \text{CH}_3\text{CHO} ]
Acetone:
- Convert acetaldehyde to acetone by the oxidation of isopropanol, which can be formed from ethylene through Markovnikov hydration. [ \text{CH}_2\text{CH}_2 + \text{H}_2\text{O} \rightarrow \text{CH}_3\text{CHOHCH}_3 ]
Ethyl alcohol:
- Hydrate ethyne using sulfuric acid to form acetaldehyde, then reduce acetaldehyde using a reducing agent such as sodium borohydride (NaBH₄) to form ethyl alcohol. [ \text{CH}_3\text{CHO} + \text{NaBH}_4 \rightarrow \text{CH}_3\text{CH}_2\text{OH} ]
Q. 15: Give the mechanism of addition of HCN to acetone.
Mechanism:
- Nucleophilic attack: The cyanide ion (CN⁻) attacks the electrophilic carbonyl carbon in acetone. [ \text{CH}_3\text{COCH}_3 + \text{CN}^- \rightarrow \text{CH}_3\text{C(CN)OHCH}_3^- ]
- Protonation: The negatively charged oxygen in the intermediate is protonated by water, forming acetone cyanohydrin. [ \text{CH}_3\text{C(CN)OHCH}_3^- + \text{H}_2\text{O} \rightarrow \text{CH}_3\text{C(CN)OHCH}_3 ]
Q. 16: How would you bring about the following conversions?
(i) Acetone into t-butyl alcohol:
- React acetone with methylmagnesium bromide (Grignard reagent) followed by hydrolysis to form t-butyl alcohol. [ \text{CH}_3\text{COCH}_3 + \text{CH}_3\text{MgBr} \rightarrow \text{(CH}_3)_3\text{COH} ]
(ii) Propanal into 1-propanol:
- Reduce propanal with sodium borohydride or lithium aluminum hydride to form 1-propanol. [ \text{CH}_3\text{CH}_2\text{CHO} \xrightarrow{\text{NaBH}_4} \text{CH}_3\text{CH}_2\text{CH}_2\text{OH} ]
(iii) Propanone into 2-propanol:
- Reduce propanone (acetone) using sodium borohydride or lithium aluminum hydride to form 2-propanol. [ \text{CH}_3\text{COCH}_3 \xrightarrow{\text{NaBH}_4} \text{CH}_3\text{CHOHCH}_3 ]
(iv) Ethanol into propanone:
- Oxidize ethanol to acetaldehyde, followed by a Grignard reaction with methylmagnesium bromide to form isopropyl alcohol. Oxidation of isopropyl alcohol gives propanone. [ \text{CH}_3\text{CH}_2\text{OH} \rightarrow \text{CH}_3\text{CHO} \xrightarrow{\text{CH}_3\text{MgBr}} \text{CH}_3\text{CHOHCH}_3 ]
(v) Ethyne into ethanol:
- Hydration of ethyne in the presence of sulfuric acid and mercury sulfate produces acetaldehyde, which can be reduced to ethanol. [ \text{CH≡CH} \xrightarrow{\text{Hg}^{2+}} \text{CH}_3\text{CHO} \xrightarrow{\text{NaBH}_4} \text{CH}_3\text{CH}_2\text{OH} ]
(vi) Ethanol into ethene:
- Dehydrate ethanol by heating it with concentrated sulfuric acid. [ \text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\text{H}_2\text{SO}_4} \text{CH}_2=CH}_2 ]
(vii) Ethanol into ethanoic acid:
- Oxidize ethanol using potassium permanganate or potassium dichromate to form ethanoic acid. [ \text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\text{KMnO}_4/\text{H}_2\text{SO}_4} \text{CH}_3\text{COOH} ]
(viii) Methanol into ethanol:
- Convert methanol to formaldehyde, followed by the reaction with a Grignard reagent (methylmagnesium bromide), and then hydrolysis. [ \text{CH}_3\text{OH} \rightarrow \text{CH}_3\text{CHO} \xrightarrow{\text{CH}_3\text{MgBr}} \text{CH}_3\text{CH}_2\text{OH} ]
Q. 17: How will you distinguish between:
(i) Methanal and Ethanal:
Methanal gives Cannizzaro’s reaction, whereas ethanal gives aldol condensation under basic conditions.
(ii) Ethanol and Propanone:
Ethanol reacts with iodoform reagent (NaOH/I₂) to give a yellow precipitate, whereas propanone does not.
(iii) Ethanol and Propanal:
Ethanol does not react with Tollens’ reagent, whereas propanal gives a silver mirror.
(iv) Acetone and Ethyl Alcohol:
Acetone gives a positive iodoform test, whereas ethyl alcohol does not.
(v) Butanone and 3-Pentanone:
Butanone gives a positive iodoform test, whereas 3-pentanone does not.
(vi) Acetaldehyde and Benzaldehyde:
Acetaldehyde gives a positive Fehling’s test, but benzaldehyde does not.
Q. 18: Discuss the oxidation of:
(a) Aldehydes:
- With K₂Cr₂O₇/H₂SO₄: Aldehydes are oxidized to carboxylic acids
.
- With Tollens’ reagent: Aldehydes give a silver mirror.
- With Fehling’s solution: Aldehydes reduce Fehling’s solution to a red precipitate of Cu₂O.
(b) Ketones:
- Ketones resist oxidation under mild conditions but can be cleaved into carboxylic acids by strong oxidizing agents.
Q. 19: Discuss reduction of:
(a) Aldehydes:
Aldehydes are reduced to primary alcohols by NaBH₄ or LiAlH₄.
(b) Ketones:
Ketones are reduced to secondary alcohols by NaBH₄ or LiAlH₄.
Q. 20: Give three uses for each of formaldehyde and acetaldehyde.
Formaldehyde:
- Used in the production of polymers such as bakelite.
- Used as a preservative in laboratories.
- Used in the manufacture of disinfectants.
Acetaldehyde:
- Used as an intermediate in the production of acetic acid.
- Used in the manufacture of perfumes.
- Used in the synthesis of drugs.
