Class 11 Chemistry – Unit 19: Carbonyl Compounds (FBISE)
This section provides complete, exam-oriented notes for Class 11 Chemistry Unit 19 – Carbonyl Compounds strictly according to the Federal Board (FBISE) syllabus. The unit focuses on structure, properties, reactions, and applications of aldehydes and ketones.
Major topics include aldehydes and ketones, structure of carbonyl group, physical properties, nucleophilic addition reactions, oxidation and reduction reactions, important tests, and uses. Reaction mechanisms and exam-oriented comparisons are included for effective learning.
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Notes: Carbonyl Compounds (Aldehydes and Ketones)
1. Definition and Structure
Carbonyl compounds contain the carbonyl functional group ($>C=O$). They are classified into two main types based on what is attached to the carbonyl carbon:
- Aldehydes: The carbonyl carbon is attached to at least one hydrogen atom. General structure: $R-CHO$.
- Ketones: The carbonyl carbon is bonded to two carbon atoms (alkyl or aryl groups). General structure: $R-CO-R'$.
- General Formula: Both share the homologous series formula $C_nH_{2n}O$.
2. Preparation of Aldehydes and Ketones
They are primarily prepared through the oxidation of alcohols using agents like acidified potassium dichromate ($K_2Cr_2O_7 / H_2SO_4$):
- Primary Alcohols: Oxidize to form Aldehydes.
Example: $CH_3CH_2OH + [O] \xrightarrow[50^\circ C]{K_2Cr_2O_7 + H_2SO_4} CH_3CHO + H_2O$
Note: Aldehydes can further oxidize to carboxylic acids ($CH_3COOH$) if not distilled out immediately. - Secondary Alcohols: Oxidize to form Ketones.
Example: $CH_3-CH(OH)-CH_3 + [O] \xrightarrow{K_2Cr_2O_7 + H_2SO_4} CH_3-CO-CH_3 + H_2O$
3. Nucleophilic Addition Reactions
The carbonyl group is polar ($C^{\delta+} = O^{\delta-}$), making the carbon atom electrophilic and susceptible to nucleophilic attack.
Base-Catalysed Mechanism:
- A base ($OH^-$) reacts with the reagent ($H-Nu$) to generate a strong nucleophile ($Nu^-$).
- The nucleophile attacks the electrophilic carbonyl carbon, breaking the $C=O$ $pi$ bond and forming an alkoxide ion.
- The alkoxide ion reacts with water/acid to gain a proton, forming the final addition product.
Addition of Hydrogen Cyanide (HCN):
HCN reacts with aldehydes and ketones to form cyanohydrins. This is a vital synthetic step because it increases the carbon chain length by one atom.
- Formaldehyde $\rightarrow$ Formaldehyde cyanohydrin
- Acetaldehyde $\rightarrow$ Acetaldehyde cyanohydrin
- Acetone $\rightarrow$ Acetone cyanohydrin
Questions and Answers
Q1: What is the characteristic color change when an alcohol is oxidized using acidified potassium dichromate?
A: The solution changes color from orange to green. This indicates the reduction of $Cr^{6+}$ to $Cr^{3+}$.
Q2: Why is it necessary to keep the reaction temperature near the boiling point of the aldehyde during the oxidation of a primary alcohol?
A: Aldehydes have lower boiling points than the alcohols from which they are formed. By keeping the temperature slightly above $50^\circ C$, the aldehyde can be distilled out as soon as it forms, preventing it from being further oxidized into a carboxylic acid.
Q3: Explain why the carbonyl carbon is subject to nucleophilic attack.
A: Due to the high electronegativity of oxygen compared to carbon, the $C=O$ bond is highly polarized. The carbon atom acquires a partial positive charge ($C^{\delta+}$), making it an electrophilic center that attracts nucleophiles ($Nu^-$).
Q4: Write the chemical equation for the formation of HCN used in the synthesis of cyanohydrins.
A: HCN is usually generated in situ by the reaction of sodium cyanide and HCl:
$NaCN + HCl \rightarrow HCN + NaCl$
Q5: What is the resulting product when Acetone reacts with HCN?
A: The product is Acetone cyanohydrin. The structure is $CH_3-C(OH)(CN)-CH_3$.
Carbonyl Compounds: Summary Notes
1. General Mechanism of Nucleophilic Addition
The carbonyl group $(C=O)$ is polar due to the electronegativity difference between carbon $(\delta+)$ and oxygen $(\delta-)$. In acid-catalyzed reactions, the oxygen is protonated first, which increases the electrophilic character of the carbonyl carbon, making it more susceptible to attack by even weak nucleophiles.
2. Reaction with 2,4-Dinitrophenylhydrazine (2,4-DNPH)
- Purpose: A condensation reaction used as an identifying test for the presence of the carbonyl group in both aldehydes and ketones.
- Observation: Formation of a yellow or orange precipitate (2,4-dinitrophenylhydrazone).
- Conditions: Requires an acid catalyst.
3. Reduction of Aldehydes and Ketones
Carbonyl compounds can be reduced to alcohols using hydride-providing reagents:
- Reagents:
- $LiAlH_4$ (Lithium Aluminum Hydride): A strong reducing agent.
- $NaBH_4$ (Sodium Borohydride): A weaker, more selective reducing agent.
- Products:
- Aldehydes (e.g., Methanal, Ethanal) are reduced to Primary Alcohols.
- Ketones (e.g., Propanone) are reduced to Secondary Alcohols.
4. Haloform Reaction (Iodoform Test)
- Reagents: Iodine $(I_2)$ and Sodium Hydroxide $(NaOH)$.
- Specific Requirement: The compound must contain a methyl carbonyl group $(CH_3-CO-)$.
- In aldehydes, only Ethanal gives a positive test.
- In ketones, only methyl ketones (e.g., Propanone) give a positive test.
- Observation: Appearance of a pale yellow precipitate of tri-iodomethane $(CHI_3)$.
- Utility: Used to distinguish methyl ketones from other ketones, or ethanal from other aldehydes/alcohols.
Relevant Questions & Answers
Q1: What is the visual result of a positive 2,4-DNPH test?
A: A positive test is indicated by the formation of a yellow or orange precipitate.
Q2: Why is $LiAlH_4$ used in reduction reactions of carbonyls?
A: $LiAlH_4$ acts as a source of hydride ions $(H^-)$, which act as nucleophiles to attack the electrophilic carbonyl carbon. It is a strong reducing agent capable of converting aldehydes and ketones into primary and secondary alcohols, respectively.
Q3: Which aldehyde is the only one to give a positive Iodoform test?
A: Ethanal (acetaldehyde) is the only aldehyde that gives a positive result because it is the only one containing the required $CH_3-CO-$ group.
Q4: Write the chemical equation for the reduction of Propanone.
A: $CH_3COCH_3 + 2[H] \xrightarrow{NaBH_4} CH_3CH(OH)CH_3$ (2-Propanol).
Q5: How does an acid catalyst assist in nucleophilic addition?
A: The acid protonates the carbonyl oxygen, increasing the positive charge (electrophilic character) on the carbonyl carbon. This makes it easier for weaker nucleophiles to attack the carbon atom.
Organic Chemistry Notes: Carbonyl Compounds & Carboxylic Acids
1. Nucleophilic Addition Reactions
Carbonyl compounds (aldehydes and ketones) undergo nucleophilic addition due to the polarity of the $C=O$ bond. A common example is the reaction with hydrogen cyanide (HCN) to form cyanohydrins.
- Reagents: $NaCN$ or $KCN$ with $HCl$.
- Mechanism:
- Step 1: The nucleophilic cyanide ion ($CN^-$) attacks the electrophilic carbonyl carbon, breaking the $\pi$ bond and creating an intermediate alkoxide ion.
- Step 2: Protonation of the alkoxide oxygen by an acid ($H^+$) results in the final cyanohydrin product.
2. Reaction with Primary Nitrogen Nucleophiles
Aldehydes and ketones react with primary amines ($R''-NH_2$) to form imines and water.
$$\text{General Reaction: } R-CO-R' + R''-NH_2 \rightarrow R-C(=NR'')-R' + H_2O$$
3. Oxidation Reactions
A. Oxidation of Aldehydes
Aldehydes are easily oxidized to carboxylic acids using strong agents like $K_2Cr_2O_7 / H_2SO_4$ or $KMnO_4 / H_2SO_4$.
B. Oxidation of Ketones
Ketones are resistant to oxidation and require vigorous conditions (strong oxidizing agents and heat). The carbon chain breaks, and a mixture of carboxylic acids is formed.
- Symmetrical Ketones: One carbon adjacent to the carbonyl group is oxidized.
- Unsymmetrical Ketones (Popoff’s Rule): The carbonyl group remains with the smaller alkyl group during cleavage.
4. Distinguishing Aldehydes and Ketones
| Test | Observation with Aldehyde | Observation with Ketone |
|---|---|---|
| Fehling’s Solution | Red precipitate of $Cu_2O$ | No reaction |
| Tollen’s Reagent | Silver mirror ($Ag$ deposit) | No reaction |
5. Carboxylic Acids
Organic compounds containing the carboxyl group ($-COOH$) are called carboxylic acids. Their general formula is $R-COOH$.
Preparation Methods:
- Hydrolysis of Nitriles: Boiling an alkyl nitrile ($R-CN$) with mineral acid or alkali yields a carboxylic acid with one more carbon than the original alkyl halide.
- Oxidation of Primary Alcohols: Primary alcohols are oxidized to aldehydes and then further to carboxylic acids using $KMnO_4$ or $K_2Cr_2O_7$.
- Hydrolysis of Esters: Esters can be hydrolyzed by dilute acid or alkali. Basic hydrolysis is known as saponification.
Relevant Questions & Answers
Q1: Describe the chemical test to distinguish between Propanal and Propanone.
A: Use Tollen's Reagent (ammoniacal silver nitrate). Propanal (an aldehyde) will react to form a shiny silver mirror on the inner walls of the test tube. Propanone (a ketone) will show no reaction because it lacks the reducing hydrogen atom attached to the carbonyl carbon.
Q2: What is the specific product formed when Butanone is oxidized by $K_2Cr_2O_7 / H_2SO_4$?
A: Oxidation of Butanone ($CH_3-CO-CH_2-CH_3$) follows the rule where the carbonyl group stays with the smaller alkyl group. This results in the formation of two molecules of Ethanoic acid ($CH_3COOH$).
Q3: Explain why the hydrolysis of nitriles is a useful synthetic tool.
A: It is useful because the resulting carboxylic acid contains one more carbon atom than the original alkyl halide from which the nitrile was derived. This allows for the extension of the carbon chain.
Q4: Concept Assessment: How can you convert $CH_3Cl$ into $CH_3COOH$?
A: The conversion involves two main steps:
- Step 1 (Nucleophilic Substitution): React $CH_3Cl$ with aqueous $KCN$ to form Ethanenitrile ($CH_3CN$).$$CH_3Cl + KCN \rightarrow CH_3CN + KCl$$
- Step 2 (Acid Hydrolysis): Boil the $CH_3CN$ with dilute $HCl$ to produce Ethanoic acid.$$CH_3CN + 2H_2O + HCl \rightarrow CH_3COOH + NH_4Cl$$
19.7 & 19.8 Reactivity and Reactions of Carboxylic Acids
The carboxyl group ($-COOH$) combines the chemistry of both the carbonyl group ($>C=O$) and the hydroxyl group ($-OH$). Its reactivity is primarily categorized by which part of the molecule is involved in the reaction.
Types of Reactions
- Involving H-atom of the -OH group: Results in salt formation.
- Involving the -OH group: The hydroxyl group is replaced by another group (e.g., Esterification).
- Involving the Carboxyl group as a whole: Reduction to alcohols.
Key Reaction Examples
- Acidic Properties: Carboxylic acids are weaker than mineral acids. They dissociate in water:
$$R-COOH + H_2O \rightleftharpoons R-COO^- + H_3O^+$$ - Salt Formation:
- With Bases: $$CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O$$
- With Carbonates: $$2CH_3COOH + Na_2CO_3 \rightarrow 2CH_3COONa + H_2O + CO_2$$
- Esterification (Acid-Catalyzed): Carboxylic acids react with alcohols in the presence of conc. $H_2SO_4$ to form esters and water.
$$CH_3COOH + C_2H_5OH \xrightarrow{H_2SO_4} CH_3COOC_2H_5 + H_2O$$ - Reduction: Carboxylic acids are reduced to primary alcohols using Lithium Aluminum Hydride ($LiAlH_4$).
$$CH_3COOH + 4[H] \xrightarrow{LiAlH_4} CH_3CH_2OH + H_2O$$
19.9 Hydrolysis of Esters
Esters can be broken down into their constituent acid and alcohol using dilute acid or alkali.
- Acidic Hydrolysis: Reversible reaction producing the free acid and alcohol.
- Basic Hydrolysis (Saponification): Irreversible reaction producing the alkali salt of the carboxylic acid (soap) and alcohol.
Concept Assessment Exercise 19.2: Solutions
| Conversion Task | Reagents & Reaction |
|---|---|
| 1. Ethanoic acid into ethanol | Reduction using $LiAlH_4$. $$CH_3COOH + 4[H] \xrightarrow{LiAlH_4} CH_3CH_2OH + H_2O$$ |
| 2. Ethanoic acid into ethyl acetate | Esterification with ethanol and conc. $H_2SO_4$. $$CH_3COOH + C_2H_5OH \xrightarrow{H^+} CH_3COOC_2H_5 + H_2O$$ |
| 3. Ethanol into ethanoic acid | Oxidation using $K_2Cr_2O_7 / H_2SO_4$. $$CH_3CH_2OH + 2[O] \rightarrow CH_3COOH + H_2O$$ |
Review Questions & Answers
Q1: Why is the basic hydrolysis of esters called saponification?A: It is called saponification because it leads to the formation of the alkali salt of a carboxylic acid, which is the chemical definition of soap.
Q2: Why is $LiAlH_4$ used specifically for reducing carboxylic acids?
A: Carboxylic acids have low reactivity toward reduction; they are less reactive than aldehydes or ketones. Therefore, a powerful reducing agent like $LiAlH_4$ is required to convert them into primary alcohols.
Q3: What is the role of $H_2SO_4$ in the preparation of esters?
A: It acts as a catalyst and a dehydrating agent. It protonates the carbonyl oxygen to make the carbon more electrophilic and helps remove water to shift the equilibrium toward the ester product.