Oxidation, Reduction & Name Reactions
Module 3 | CBSE Class 12 Chemistry | Organic Chemistry
1. Reduction Reactions
Aldehydes and ketones can be reduced to corresponding primary and secondary alcohols respectively by using sodium borohydride (NaBH4) or lithium aluminium hydride (LiAlH4) as well as by catalytic hydrogenation.
1.1 Reduction to Hydrocarbons (Very Important)
The carbonyl group (>C=O) of aldehydes and ketones can be entirely reduced to a methylene group (>CH2) to form alkanes. There are two legendary name reactions for this conversion:
A. Clemmensen Reduction
The carbonyl compound is heated with zinc amalgam (Zn-Hg) and concentrated Hydrochloric acid (HCl).
B. Wolff-Kishner Reduction
The carbonyl compound is first treated with hydrazine (NH2NH2) to form a hydrazone, which is then heated with sodium or potassium hydroxide in a high boiling solvent like ethylene glycol.
2. Oxidation Reactions & Distinguishing Tests
Aldehydes and ketones differ greatly in their oxidation reactions. Aldehydes are easily oxidized to carboxylic acids even by mild oxidizing agents because of the presence of a hydrogen atom attached to the carbonyl carbon. Ketones require vigorous conditions (strong oxidizing agents and elevated temperatures) and their oxidation involves C-C bond cleavage to afford a mixture of carboxylic acids having lesser number of carbon atoms.
2.1 Tollens' Test (Silver Mirror Test)
Tollens' reagent is a freshly prepared ammoniacal silver nitrate solution [Ag(NH3)2]+. On warming an aldehyde with Tollens' reagent, a bright silver mirror is produced on the inner walls of the test tube due to the formation of silver metal.
2.2 Fehling's Test
Fehling's reagent comprises two solutions: Fehling solution A (aqueous copper sulphate) and Fehling solution B (alkaline sodium potassium tartrate, known as Rochelle salt). When an aldehyde is heated with Fehling's reagent, a reddish-brown precipitate is obtained.
2.3 Haloform / Iodoform Reaction
Aldehydes and ketones having at least one methyl group linked to the carbonyl carbon atom (Methyl Ketones, CH3-CO- group) are oxidized by sodium hypohalite (NaOX) to sodium salts of corresponding carboxylic acids having one carbon atom less than the carbonyl compound. The methyl group is converted to haloform.
When Sodium hypoiodite (NaOI, prepared in situ from NaOH and I2) is used, a yellow precipitate of Iodoform (CHI3) is produced.
1. All methyl ketones (e.g., Propanone, Acetophenone, Butan-2-one).
2. Ethanal (Acetaldehyde) - the only aldehyde to give this test.
3. Alcohols containing the CH3-CH(OH)- group (e.g., Ethanol, Propan-2-ol) because NaOI oxidizes them to methyl ketones first.
3. Reactions due to α-Hydrogen
Aldehydes and ketones undergo a number of reactions due to the acidic nature of the α-hydrogen. The acidity of α-hydrogen atoms of carbonyl compounds is due to the strong electron-withdrawing effect of the carbonyl group and resonance stabilization of the conjugate base (enolate ion).
3.1 Aldol Condensation (Highly Tested)
Aldehydes and ketones having at least one α-hydrogen undergo a reaction in the presence of dilute alkali (like dilute NaOH or Ba(OH)2) to form β-hydroxy aldehydes (aldol) or β-hydroxy ketones (ketol), respectively.
Upon heating, these aldols/ketols readily lose water to form α,β-unsaturated carbonyl compounds. This overall process is called Aldol Condensation.
2 CH3CHO →(dil. NaOH) CH3-CH(OH)-CH2-CHO (3-Hydroxybutanal / Aldol)
CH3-CH(OH)-CH2-CHO →(Δ, -H2O) CH3-CH=CH-CHO (But-2-enal)
3.2 Cross Aldol Condensation
When aldol condensation is carried out between two different aldehydes and/or ketones, it is called cross aldol condensation. If both of them contain α-hydrogen atoms, it gives a mixture of four products.
4. Other Important Reactions
4.1 Cannizzaro Reaction
Aldehydes which do not have an α-hydrogen atom (e.g., Formaldehyde HCHO, Benzaldehyde C6H5CHO) undergo self-oxidation and reduction (disproportionation) on heating with concentrated alkali (like 50% NaOH or KOH).
In this reaction, one molecule of the aldehyde is reduced to an alcohol while another is oxidized to the carboxylic acid salt.
2 C6H5CHO + Conc. NaOH → C6H5CH2OH (Benzyl alcohol) + C6H5COONa (Sodium benzoate)
4.2 Electrophilic Substitution
Aromatic aldehydes and ketones undergo electrophilic substitution at the ring. The carbonyl group (>C=O) acts as a deactivating and meta-directing group (due to -R effect).
Example (Nitration): Benzaldehyde + conc. HNO3 + conc. H2SO4 (273-283 K) → m-Nitrobenzaldehyde.
5. NCERT Solved Examples (Step-by-Step)
NCERT Example 12.5: Predict the products formed when cyclohexanecarbaldehyde reacts with following reagents:
(i) PhMgBr and then H3O+
(ii) Tollens’ reagent
(i) The given compound is an aldehyde (C6H11-CHO). Grignard reagent (Phenylmagnesium bromide) adds to the carbonyl carbon. Hydrolysis yields a secondary alcohol.
Product: Cyclohexylphenylcarbinol (or 1-Cyclohexyl-1-phenylmethanol).
(ii) Cyclohexanecarbaldehyde is an aliphatic aldehyde. It will be oxidized by Tollens' reagent to the corresponding carboxylate ion, and silver mirror will be formed.
Product: Cyclohexanecarboxylate ion (C6H11-COO-) + Ag↓.
6. Previous Year Questions (PYQs) & Exhaustive Question Bank
Part A: Distinguishing Tests (3 Marks - Guaranteed Board Question)
Q1. Give simple chemical tests to distinguish between the following pairs of compounds:
(a) Propanal and Propanone
(b) Pentan-2-one and Pentan-3-one
(c) Benzaldehyde and Acetophenone
(a) Tollens' Test: Propanal (an aldehyde) forms a bright silver mirror when heated with Tollens' reagent. Propanone (a ketone) does not respond to this test.
(b) Iodoform Test: Pentan-2-one has a methyl ketone group (CH3-CO-) and gives a yellow precipitate of Iodoform (CHI3) on heating with NaOH and I2. Pentan-3-one does not have a terminal methyl group attached to the carbonyl and does not give this test.
(c) Tollens' Test: Benzaldehyde (an aldehyde) forms a silver mirror with Tollens' reagent. Acetophenone (a ketone) does not.
(Alternatively, Iodoform test can be used: Acetophenone gives a yellow ppt, Benzaldehyde does not).
Part B: Conceptual (1-2 Marks)
Q2. Why is Benzaldehyde less reactive than Propanal towards nucleophilic addition reactions?
Q3. Why does formaldehye undergo Cannizzaro reaction while acetaldehyde undergoes Aldol condensation?
Part C: Assertion-Reason Type (1 Mark)
Q4. Assertion (A): Benzaldehyde does not give Fehling's test.
Reason (R): Benzaldehyde does not contain an alpha-hydrogen atom.
Benzaldehyde does not give Fehling's test because Fehling's solution is a relatively weaker oxidizing agent compared to Tollens' reagent and cannot oxidize aromatic aldehydes. The absence of an alpha-hydrogen is the reason it undergoes Cannizzaro reaction, not the reason it fails Fehling's test.
Part D: Synthesis & Reactions (3 Marks)
Q5. Write the structures of the main products when Propanone (CH3COCH3) reacts with the following reagents:
(i) Zn-Hg / conc. HCl
(ii) Hydrazine followed by heating with KOH in ethylene glycol
(iii) Dilute Ba(OH)2
(i) This is Clemmensen reduction. The >C=O group is reduced to >CH2.
Product: CH3-CH2-CH3 (Propane).
(ii) This is Wolff-Kishner reduction. It also reduces >C=O to >CH2.
Product: CH3-CH2-CH3 (Propane).
(iii) This is Aldol condensation. Two molecules of propanone condense.
Product: CH3-C(OH)(CH3)-CH2-CO-CH3 (4-Hydroxy-4-methylpentan-2-one / Ketol).
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