Organic Chemistry: Important Reactions

NEET Organic Chemistry: Important Reactions & Concepts with MCQs

This document provides a concise review of essential named reactions, other crucial organic reactions, and fundamental concepts frequently tested in the NEET examination. It also includes a practice set of Multiple Choice Questions (MCQs) to test your understanding.

I. Important Named Reactions

Here’s a compilation of key named reactions in Organic Chemistry, crucial for NEET preparation:

1. Wurtz Reaction

  • Reactants: Two alkyl halides (R−X)
  • Reagents/Conditions: Sodium metal (Na) in dry ether.
  • Products: Symmetrical alkane (R−R).
  • Equation: 2RX+2NaDry Ether​R−R+2NaX
  • Key Point: Cannot be used for unsymmetrical alkanes due to formation of a mixture of products. Not suitable for tertiary alkyl halides due to elimination.

2. Friedel-Crafts Alkylation

  • Reactants: Benzene or activated aromatic compound, alkyl halide.
  • Reagents/Conditions: Anhydrous AlCl3​ (Lewis acid catalyst).
  • Products: Alkylbenzene.
  • Equation: C6​H6​+R−XAnhydrous AlCl3​​C6​H5​−R+HX
  • Key Point: Prone to polyalkylation and rearrangement of alkyl carbocation.

3. Friedel-Crafts Acylation

  • Reactants: Benzene or activated aromatic compound, acid halide (e.g., RCOCl) or acid anhydride ((RCO)2​O).
  • Reagents/Conditions: Anhydrous AlCl3​ (Lewis acid catalyst).
  • Products: Acylbenzene (ketone).
  • Equation: C6​H6​+RCOClAnhydrous AlCl3​​C6​H5​COR+HCl
  • Key Point: Does not suffer from polyacylation. The product ketone can be reduced to an alkylbenzene (ee.g., Clemmensen or Wolff-Kishner).

4. Reimer-Tiemann Reaction

  • Reactants: Phenol.
  • Reagents/Conditions: Chloroform (CHCl3​) and aqueous NaOH or KOH, followed by acidification.
  • Products: Salicylaldehyde (o-hydroxybenzaldehyde).
  • Equation: Phenol+CHCl3​+3NaOH340K​IntermediateH+​Salicylaldehyde
  • Key Point: Electrophile is dichlorocarbene (:CCl2​).

5. Kolbe’s Reaction (Kolbe-Schmidt Reaction)

  • Reactants: Phenol.
  • Reagents/Conditions: Sodium hydroxide (NaOH), then carbon dioxide (CO2​) under pressure, followed by acidification.
  • Products: Salicylic acid (o-hydroxybenzoic acid).
  • Equation: PhenolNaOH​Sodium phenoxideCO2​,400K,4−7atm​IntermediateH+​Salicylic acid
  • Key Point: Electrophilic substitution of phenoxide ion.

6. Williamson Ether Synthesis

  • Reactants: Alkyl halide (primary) and sodium alkoxide (RONa).
  • Reagents/Conditions: Heat.
  • Products: Ether (R−O−R′).
  • Equation: R−X+R′ONa→R−O−R′+NaX
  • Key Point: Best for preparing unsymmetrical ethers, especially if the alkyl halide is primary. Tertiary alkyl halides undergo elimination.

7. Hell-Volhard-Zelinsky (HVZ) Reaction

  • Reactants: Carboxylic acid having α-hydrogen.
  • Reagents/Conditions: X2​ (Cl or Br) in presence of red P, followed by hydrolysis.
  • Products: α-Halo carboxylic acid.
  • Equation: R−CH2​−COOH+X2​Red P​R−CH(X)−COOH+HX
  • Key Point: Only carboxylic acids with at least one α-hydrogen undergo this reaction.

8. Clemmensen Reduction

  • Reactants: Aldehydes or ketones.
  • Reagents/Conditions: Zinc amalgam (Zn(Hg)) and concentrated HCl.
  • Products: Alkanes (carbonyl group reduced to methylene group).
  • Equation: R−CO−R′+4[H]Zn(Hg)/Conc.HCl​R−CH2​−R′+H2​O
  • Key Point: Not suitable for compounds sensitive to acid.

9. Wolff-Kishner Reduction

  • Reactants: Aldehydes or ketones.
  • Reagents/Conditions: Hydrazine (NH2​NH2​), followed by KOH or NaOH and ethylene glycol (solvent), heat.
  • Products: Alkanes (carbonyl group reduced to methylene group).
  • Equation: R−CO−R′+NH2​NH2​KOH/Ethylene Glycol, Heat​R−CH2​−R′+N2​+H2​O
  • Key Point: Suitable for compounds sensitive to acid (e.g., with acid-sensitive functional groups like ester).

10. Stephen Reduction

  • Reactants: Nitriles (R−C≡N).
  • Reagents/Conditions: SnCl2​/HCl, followed by hydrolysis.
  • Products: Aldehydes.
  • Equation: R−C≡N+SnCl2​+2HCl→R−CH=NH⋅HClH2​O/H+​R−CHO
  • Key Point: Converts nitriles to aldehydes.

11. Rosenmund Reduction

  • Reactants: Acyl chlorides (RCOCl).
  • Reagents/Conditions: H2​ in presence of Pd supported on BaSO4​ (poisoned catalyst), heated.
  • Products: Aldehydes.
  • Equation: RCOCl+H2​Pd/BaSO4​ (poisoned)​RCHO+HCl
  • Key Point: BaSO4​ acts as a poison to prevent further reduction of aldehyde to alcohol.

12. Gattermann Reaction (Formylation of Benzene, not Diazonium)

  • Reactants: Benzene.
  • Reagents/Conditions: HCN and HCl in presence of anhydrous AlCl3​.
  • Products: Benzaldehyde.
  • Equation: C6​H6​+HCN+HClAnhydrous AlCl3​​IntermediateH2​O​C6​H5​CHO
  • Key Point: Similar to Gattermann-Koch but uses HCN.

13. Gattermann-Koch Reaction

  • Reactants: Benzene.
  • Reagents/Conditions: CO and HCl in presence of anhydrous AlCl3​ and CuCl.
  • Products: Benzaldehyde.
  • Equation: C6​H6​+CO+HClAnhydrous AlCl3​/CuCl​C6​H5​CHO
  • Key Point: Direct formylation of benzene.

14. Aldol Condensation

  • Reactants: Aldehydes or ketones having at least one α-hydrogen.
  • Reagents/Conditions: Dilute alkali (NaOH, KOH, Ba(OH)2​).
  • Products: β-Hydroxy aldehyde (aldol) or β-hydroxy ketone (ketol), which upon heating undergo dehydration to form α,β-unsaturated carbonyl compounds.
  • Equation: 2CH3​CHODil. NaOH​CH3​CH(OH)CH2​CHOHeat​CH3​CH=CHCHO+H2​O
  • Key Point: Intramolecular aldol is also possible.

15. Cannizzaro Reaction

  • Reactants: Aldehydes that do NOT have an α-hydrogen.
  • Reagents/Conditions: Concentrated alkali (NaOH, KOH).
  • Products: Alcohol and carboxylic acid salt (disproportionation reaction – one molecule is oxidized, another is reduced).
  • Equation: 2HCHOConc. NaOH​CH3​OH+HCOONa
  • Key Point: Formaldehyde, benzaldehyde are common examples.

16. Perkin Reaction

  • Reactants: Aromatic aldehyde and acetic anhydride (or other acid anhydride with α-hydrogens).
  • Reagents/Conditions: Sodium salt of the corresponding acid (e.g., sodium acetate) and heating.
  • Products: α,β-unsaturated aromatic acid.
  • Equation: Aromatic aldehyde+(CH3​CO)2​OCH3​COONa,Heat​Cinnamic acid+CH3​COOH
  • Key Point: Used to synthesize cinnamic acid derivatives.

17. Cross Aldol Condensation

  • Reactants: Two different aldehydes and/or ketones, at least one of which has α-hydrogen.
  • Reagents/Conditions: Dilute alkali.
  • Products: Mixture of up to four products. If one reactant has no α-hydrogen, products are mainly from its condensation with the other.

18. Cross Cannizzaro Reaction

  • Reactants: Two different aldehydes, both lacking α-hydrogen.
  • Reagents/Conditions: Concentrated alkali.
  • Products: Mixture of alcohol and acid salt. Generally, the more reactive aldehyde (e.g., formaldehyde) gets oxidized, and the other aldehyde gets reduced.

19. Gabriel Phthalimide Synthesis

  • Reactants: Phthalimide, ethanolic KOH, alkyl halide (primary).
  • Reagents/Conditions: Followed by hydrolysis (NaOH or NH2​NH2​).
  • Products: Pure primary aliphatic amines.
  • Key Point: Not for aromatic primary amines (aryl halides don’t undergo SN2 with phthalimide).

20. Hoffmann Bromamide Degradation

  • Reactants: Primary amide (RCONH2​).
  • Reagents/Conditions: Br2​ and KOH/NaOH (aqueous or ethanolic).
  • Products: Primary amine with one carbon less than the amide.
  • Equation: RCONH2​+Br2​+4KOH→RNH2​+K2​CO3​+2KBr+2H2​O
  • Key Point: Step-down reaction.

21. Carbylamine Reaction (Isocyanide Test)

  • Reactants: Primary amine (aliphatic or aromatic), chloroform (CHCl3​), and alcoholic KOH.
  • Reagents/Conditions: Heat.
  • Products: Offensive smelling isocyanide (carbylamine).
  • Equation: RNH2​+CHCl3​+3KOHHeat​RNC+3KCl+3H2​O
  • Key Point: Test for primary amines only.

22. Hinsberg’s Test

  • Reagents: Benzene sulphonyl chloride (C6​H5​SO2​Cl).
  • Purpose: To distinguish between primary, secondary, and tertiary amines.
    • 1° Amine: Forms N-alkylbenzenesulphonamide, which is soluble in KOH.
    • 2° Amine: Forms N,N-dialkylbenzenesulphonamide, which is insoluble in KOH.
    • 3° Amine: Does not react.

23. Sandmeyer Reaction

  • Reactants: Arenediazonium salt (ArN2+​X−).
  • Reagents/Conditions: Cu2​Cl2​/HCl, Cu2​Br2​/HBr, or Cu2​(CN)2​/KCN.
  • Products: Aryl chloride, aryl bromide, or aryl cyanide, with evolution of N2​.
  • Equation: ArN2+​Cl−Cu2​Cl2​/HCl​ArCl+N2​
  • Key Point: Replacement of diazonium group.

24. Gattermann Reaction (Diazonium)

  • Reactants: Arenediazonium salt (ArN2+​X−).
  • Reagents/Conditions: Cu powder/HCl or Cu powder/HBr.
  • Products: Aryl chloride or aryl bromide, with evolution of N2​.
  • Equation: ArN2+​Cl−Cu/HCl​ArCl+N2​+CuCl
  • Key Point: Similar to Sandmeyer but uses copper powder directly.

25. Balz-Schiemann Reaction

  • Reactants: Arenediazonium salt (ArN2+​X−).
  • Reagents/Conditions: HBF4​ (fluoroboric acid), then heat.
  • Products: Aryl fluoride.
  • Equation: ArN2+​Cl−HBF4​​ArN2+​BF4−​Heat​Ar−F+BF3​+N2​
  • Key Point: Only method to prepare aryl fluorides.

26. Coupling Reactions (Diazonium Salts)

  • Reactants: Arenediazonium salt and phenol (weakly alkaline) or aromatic amine (weakly acidic).
  • Products: Colored azo dyes (Ar−N=N−Ar′).
  • Equation: C6​H5​N2+​Cl−+C6​H5​OHH+​C6​H5​−N=N−C6​H4​−OH(p)+HCl

27. Schotten-Baumann Reaction

  • Reactants: Phenol or amine.
  • Reagents/Conditions: Benzoyl chloride (C6​H5​COCl) in aqueous NaOH.
  • Products: Benzoate ester (from phenol) or benzamide (from amine).
  • Equation: Ar−OH+C6​H5​COClNaOH/H2​O​Ar−OCO−C6​H5​+HCl

II. Other Important Reactions & Conversions

1. Oxidation of Alcohols

  • Primary Alcohol to Aldehyde: PCC (Pyridinium Chlorochromate) in CH2​Cl2​. RCH2​OHPCC​RCHO
  • Primary Alcohol to Carboxylic Acid: Strong oxidizing agents like KMnO4​/H+, K2​Cr2​O7​/H+, or Jones reagent. RCH2​OHKMnO4​/H+​RCOOH
  • Secondary Alcohol to Ketone: PCC, Jones reagent, K2​Cr2​O7​/H+. R2​CHOHK2​Cr2​O7​/H+​R2​CO
  • Tertiary Alcohols: Resistant to oxidation under normal conditions; undergo dehydration with strong oxidizing agents.

2. Reduction of Carbonyl Compounds

  • Aldehyde/Ketone to Alcohol:
    • LiAlH4​ (Lithium Aluminium Hydride) or NaBH4​ (Sodium Borohydride).
    • Catalytic Hydrogenation (H2​/Ni, Pt, or Pd). RCHOLiAlH4​​RCH2​OH (Primary alcohol) R2​COLiAlH4​​R2​CHOH (Secondary alcohol)

3. Reduction of Nitro Compounds to Amines

  • R−NO2​Sn/HCl or Fe/HCl or H2​/Pd​R−NH2​

4. Ammonolysis of Alkyl Halides

  • R−X+NH3​→R−NH2​ (mixture of 1°, 2°, 3° amines, and quaternary salt unless excess NH3​ is used)

5. Hydroboration-Oxidation

  • Reactants: Alkene.
  • Reagents: 1) BH3​⋅THF, 2) H2​O2​/OH−.
  • Products: Alcohol (Anti-Markovnikov addition of water, syn-addition).
  • Equation: R−CH=CH2​1)BH3​⋅THF,2)H2​O2​/OH−​R−CH2​−CH2​OH

6. Ozonolysis

  • Reactants: Alkene or Alkyne.
  • Reagents: 1) O3​, 2) Zn/H2​O (reductive ozonolysis) or H2​O2​ (oxidative ozonolysis).
  • Products: Aldehydes and/or ketones (from alkenes), carboxylic acids (from alkynes, or aldehydes/ketones from terminal alkynes).
  • Equation (Alkene): R2​C=CR2​1)O3​,2)Zn/H2​O​R2​CO+R2​CO

7. Hydration of Alkenes/Alkynes

  • Alkenes:
    • Acid-catalyzed hydration (H2​SO4​/H2​O): Markovnikov addition, carbocation intermediate.
    • Hydroboration-Oxidation: Anti-Markovnikov addition (see above).
  • Alkynes:
    • HgSO4​/H2​SO4​: Markovnikov addition of water, forms enol which tautomerizes to ketone (or acetaldehyde for ethyne). CH≡CHHgSO4​/H2​SO4​​CH3​CHO

8. Dehydration of Alcohols

  • Reagents: Conc. H2​SO4​ (heat), Al2​O3​ (heat).
  • Products: Alkenes.
  • Equation: R−CH2​−CH2​OHConc.H2​SO4​,Heat​R−CH=CH2​+H2​O
  • Key Point: Follows Saytzeff’s rule for major product (more substituted alkene).

9. Reduction of Carboxylic Acids and Esters

  • Carboxylic Acids: Only LiAlH4​ reduces carboxylic acids to primary alcohols. RCOOHLiAlH4​​RCH2​OH
  • Esters: LiAlH4​ or H2​/Ni. RCOOR′LiAlH4​​RCH2​OH+R′OH

10. Decarboxylation of Carboxylic Acids

  • Reagents: Soda lime (NaOH + CaO) and heat.
  • Products: Alkane with one carbon less than the acid.
  • Equation: RCOONa+NaOHCaO,Heat​R−H+Na2​CO3​
  • Key Point: Commonly used for sodium salts of carboxylic acids.

11. Haloform Reaction (Iodoform Test)

  • Reactants: Compounds containing CH3​CO− group (methyl ketones) or CH3​CH(OH)− group (secondary alcohols with methyl group at α-carbon).
  • Reagents: X2​ (typically I2​) and NaOH.
  • Products: Haloform (CHX3​) and sodium salt of carboxylic acid.
  • Equation: CH3​COR+3I2​+4NaOH→CHI3​↓+RCOONa+3NaI+3H2​O
  • Key Point: Yellow precipitate of iodoform (CHI3​) is a positive test.

12. Grignard Reagent Reactions

  • Formation: R−X+MgDry Ether​R−Mg−X
  • Reactions:
    • With aldehydes (except HCHO) → Secondary alcohols
    • With formaldehyde (HCHO) → Primary alcohols
    • With ketones → Tertiary alcohols
    • With CO2​ followed by hydrolysis → Carboxylic acids
    • With nitriles followed by hydrolysis → Ketones

13. Diazotisation

  • Reactants: Primary aromatic amine (e.g., aniline).
  • Reagents: NaNO2​/HCl.
  • Conditions: 0−5∘C (273-278 K).
  • Products: Arenediazonium salt.
  • Equation: Ar−NH2​+NaNO2​+2HCl273−278K​Ar−N2+​Cl−+NaCl+2H2​O

III. Important Topics/Concepts

1. Isomerism

  • Structural Isomerism: Same molecular formula, different structural formula.
    • Chain, Position, Functional, Metamerism, Tautomerism.
  • Stereoisomerism: Same molecular formula and sequence of bonded atoms, but different spatial arrangement.
    • Geometric (Cis-Trans) Isomerism: Restricted rotation around a double bond or in cyclic compounds.
    • Optical Isomerism: Presence of chiral center (asymmetric carbon) and non-superimposable mirror images (enantiomers). D-L configuration, R-S nomenclature.

2. Acidity and Basicity

  • Acidity: Ease of donation of H+. Stability of conjugate base.
    • Carboxylic Acids: Stronger acids than phenols and alcohols due to resonance stabilization of carboxylate ion.
    • Phenols: More acidic than alcohols due to resonance stabilization of phenoxide ion.
    • Alcohols: Weakly acidic.
    • Factors Affecting Acidity:
      • Electron-withdrawing groups (EWG) increase acidity (stabilize conjugate base).
      • Electron-donating groups (EDG) decrease acidity (destabilize conjugate base).
      • Hybridization: sp > sp$^2$ > sp$^3$ (for C-H bonds, e.g., alkynes are more acidic than alkenes/alkanes).
  • Basicity: Ease of acceptance of H+ (or donation of lone pair).
    • Amines:
      • Gaseous Phase: Tertiary > Secondary > Primary > Ammonia (purely +I effect).
      • Aqueous Solution: (depends on +I effect, steric hindrance, solvation)
        • Methylamines: Secondary > Primary > Tertiary > Ammonia
        • Ethylamines: Secondary > Tertiary > Primary > Ammonia
      • Aromatic vs. Aliphatic: Aliphatic amines > Ammonia > Aromatic amines.
        • Aromatic amines are weaker due to resonance delocalization of lone pair.
        • EWG on aromatic ring decreases basicity.
        • EDG on aromatic ring increases basicity.

3. Electronic Displacement Effects

  • Inductive Effect (+I and -I): Permanent displacement of σ electrons along a carbon chain due to electronegativity difference.
    • +I groups: Alkyl groups (electron-donating).
    • −I groups: Halogens, NO2​, COOH, CN, etc. (electron-withdrawing).
  • Resonance Effect (+R and -R) / Mesomeric Effect: Delocalization of π electrons or lone pair of electrons within a conjugated system.
    • +R groups: Groups that donate electrons to the conjugated system (e.g., −OH,−OR,−NH2​,−Cl).
    • −R groups: Groups that withdraw electrons from the conjugated system (e.g., −NO2​,−CN,−CHO,−COOH).
  • Hyperconjugation: Delocalization of σ electrons of C-H bond with an adjacent π bond or a vacant p-orbital. Explains stability of carbocations, free radicals, and alkenes. (“No bond resonance”)

4. Reaction Mechanisms (Brief Overview)

  • Electrophilic Addition (EA): Alkenes, Alkynes (e.g., addition of HX, H2​O, X2​). Markovnikov’s rule.
  • Electrophilic Substitution (ES): Aromatic compounds (e.g., nitration, halogenation, sulphonation, Friedel-Crafts). Role of activating/deactivating groups, ortho/para/meta directing.
  • Nucleophilic Substitution (SN1 & SN2): Alkyl halides.
    • SN1: 2-step, carbocation intermediate, 3° > 2° > 1°, racemic mixture, polar protic solvent.
    • SN2: 1-step, transition state, 1° > 2° > 3°, inversion of configuration, polar aprotic solvent.
  • Nucleophilic Addition (NA): Aldehydes and Ketones.
    • Reactivity: Aldehydes > Ketones (steric and electronic factors).
  • Elimination Reactions (E1 & E2): Formation of alkenes from alkyl halides or alcohols.
    • E1: 2-step, carbocation intermediate, 3° > 2° > 1°, favored by polar protic solvent, heat.
    • E2: 1-step, strong base, favored by polar aprotic solvent, heat, 3° > 2° > 1°.
    • Saytzeff’s Rule: Major product is the most substituted alkene.
    • Hofmann Elimination: Less substituted alkene is major product (bulky base).

IV. Practice Questions (MCQs)

1. Question: Which of the following reagents is used for the Clemmensen reduction? a) LiAlH4​ b) H2​/Pd−BaSO4​ c) Zn(Hg)/Conc.HCl d) NH2​NH2​/KOH

2. Question: Aniline on reaction with bromine water gives a white precipitate of: a) o-Bromoaniline b) p-Bromoaniline c) 2,4,6-Tribromoaniline d) m-Bromoaniline

3. Question: Which of the following compounds will not undergo Aldol condensation? a) Ethanal b) Propanal c) Methanal d) Propanone

4. Question: The reaction used to prepare pure primary aliphatic amines is: a) Hoffmann bromamide degradation b) Gabriel phthalimide synthesis c) Ammonolysis of alkyl halides d) Reduction of nitro compounds

5. Question: Which of the following is the strongest base in aqueous solution? a) Ammonia b) Trimethylamine c) Methylamine d) Dimethylamine

6. Question: The product of Stephen reduction of CH3​CN is: a) CH3​CH2​NH2​ b) CH3​CHO c) CH3​COOH d) CH3​CH3​

7. Question: Williamson’s synthesis is used to prepare: a) Aldehydes b) Ketones c) Ethers d) Alcohols

8. Question: Which of the following will give a positive iodoform test? a) Ethanol b) Propan-1-ol c) Pentan-3-one d) Methanol

9. Question: The correct order of acidity for phenol, ethanol, and water is: a) Phenol > Water > Ethanol b) Ethanol > Water > Phenol c) Water > Phenol > Ethanol d) Phenol > Ethanol > Water

10. Question: Benzene diazonium chloride on reaction with H3​PO2​ (hypophosphorous acid) and water gives: a) Phenol b) Benzene c) Chlorobenzene d) Aniline

11. Question: In the Reimer-Tiemann reaction, the electrophile involved is: a) Dichlorocarbene b) Chlorocarbene c) Chloroform molecule d) Dichloromethyl cation

12. Question: Which of the following functional groups acts as a meta-director and deactivating group in electrophilic aromatic substitution? a) −OH b) −NH2​ c) −NO2​ d) −OCH3​

13. Question: The Hoffmann bromamide degradation reaction is a significant reaction because: a) It converts a primary amine to a secondary amine. b) It converts an amide to an amine with one less carbon atom. c) It produces a mixture of amines. d) It is used to synthesize tertiary amines.

14. Question: Which of the following compounds is most reactive towards nucleophilic addition reactions? a) Propanal b) Propanone c) Ethanal d) Benzaldehyde

15. Question: Aniline, when treated with concentrated sulphuric acid, gives sulphanilic acid. The product exists as: a) Anion b) Cation c) Zwitterion d) Neutral molecule

16. Question: The reaction of an alkyl halide with sodium in dry ether to form a higher alkane is known as: a) Wurtz reaction b) Friedel-Crafts reaction c) Kolbe’s electrolysis d) Williamson synthesis

17. Question: Which reagent is used to convert an acyl chloride to an aldehyde? a) LiAlH4​ b) NaBH4​ c) Pd/BaSO4​ (Rosenmund catalyst) d) Zn(Hg)/HCl

18. Question: Which of the following is an example of α,β-unsaturated carboxylic acid formed in Perkin’s reaction? a) Benzoic acid b) Cinnamic acid c) Salicylic acid d) Acetic acid

19. Question: The decreasing order of acidity for carboxylic acids, phenols, and alcohols is: a) Carboxylic acid > Phenol > Alcohol b) Phenol > Carboxylic acid > Alcohol c) Alcohol > Phenol > Carboxylic acid d) Carboxylic acid > Alcohol > Phenol

20. Question: A primary amine on reaction with Hinsberg’s reagent forms a product which is: a) Soluble in aqueous KOH b) Insoluble in aqueous KOH c) Soluble in dilute acid d) Does not react

V. Answers and Explanations

I. Multiple Choice Questions (MCQs) – Answers

1. Answer: c) Zn(Hg)/Conc.HCl Explanation: Clemmensen reduction uses zinc amalgam and concentrated hydrochloric acid to reduce carbonyl compounds (aldehydes and ketones) to alkanes.

2. Answer: c) 2,4,6-Tribromoaniline Explanation: The amino group (−NH2​) in aniline is a strong activating group. It activates the benzene ring so strongly towards electrophilic substitution that bromine water reacts to substitute bromine at all ortho and para positions, leading to the formation of a white precipitate of 2,4,6-tribromoaniline.

3. Answer: c) Methanal Explanation: Aldol condensation requires the presence of at least one α-hydrogen atom in the aldehyde or ketone. Methanal (formaldehyde, HCHO) does not have any α-hydrogen atoms. Ethanal (CH3​CHO), Propanal (CH3​CH2​CHO), and Propanone ((CH3​)2​CO) all have α-hydrogens.

4. Answer: b) Gabriel phthalimide synthesis Explanation: Gabriel phthalimide synthesis is specifically known for producing pure primary aliphatic amines, avoiding the formation of mixtures of primary, secondary, and tertiary amines that typically result from ammonolysis of alkyl halides. Hoffmann bromamide also produces pure primary amines, but it is a step-down reaction (loses a carbon).

5. Answer: d) Dimethylamine Explanation: In aqueous solution, the basicity of amines is influenced by the +I effect, solvation effects, and steric hindrance. For methylamines, the observed order of basicity is dimethylamine > methylamine > trimethylamine > ammonia. This is due to an optimal balance of electron-donating effects and solvation stability for the secondary amine.

6. Answer: b) CH3​CHO Explanation: Stephen reduction converts nitriles into aldehydes. CH3​CN (ethanenitrile) is reduced to CH3​CHO (ethanal) through an imine intermediate.

7. Answer: c) Ethers Explanation: Williamson’s synthesis is a well-known method for preparing ethers by the reaction of an alkyl halide (preferably primary) with a sodium alkoxide.

8. Answer: a) Ethanol Explanation: The iodoform test is positive for compounds containing a CH3​CO− group (methyl ketones) or a CH3​CH(OH)− group (secondary alcohols with a methyl group on the carbon bearing the hydroxyl group). Ethanol (CH3​CH2​OH) contains the CH3​CH(OH)− structural unit, hence it gives a positive iodoform test. Propan-1-ol and Methanol do not have the required structures. Pentan-3-one does not have a CH3​CO− group.

9. Answer: a) Phenol > Water > Ethanol Explanation: Acidity is determined by the stability of the conjugate base.

  • Phenoxide ion (C6​H5​O−) is stabilized by resonance with the benzene ring.
  • Hydroxide ion (OH−) is less stable than phenoxide.
  • Alkoxide ion (C2​H5​O−) is destabilized by the +I effect of the ethyl group. Therefore, the order of acidity is Phenol > Water > Ethanol.

10. Answer: b) Benzene Explanation: Benzenediazonium chloride reacts with hypophosphorous acid (H3​PO2​) in the presence of water to reduce the diazonium group to a hydrogen, yielding benzene. This is a method for replacing the diazonium group with H.

11. Answer: a) Dichlorocarbene Explanation: In the Reimer-Tiemann reaction, chloroform (CHCl3​) reacts with a strong base (like KOH) to generate dichlorocarbene (:CCl2​), which is the electrophile that attacks the activated phenolic ring.

12. Answer: c) −NO2​ Explanation: Electron-withdrawing groups (EWG) like −NO2​, −CN, −CHO, −COOH, etc., are meta-directing and deactivating in electrophilic aromatic substitution reactions. Activating groups like −OH, −NH2​, −OCH3​ are ortho-para directors.

13. Answer: b) It converts an amide to an amine with one less carbon atom. Explanation: The Hoffmann bromamide degradation reaction is specifically used to step-down a carbon chain. It converts a primary amide into a primary amine containing one carbon atom fewer than the original amide.

14. Answer: c) Ethanal Explanation: Reactivity towards nucleophilic addition in aldehydes and ketones is influenced by steric hindrance and the magnitude of the positive charge on the carbonyl carbon.

  • Aldehydes are generally more reactive than ketones due to less steric hindrance and greater positive charge.
  • Among aldehydes, smaller aldehydes are more reactive. Ethanal (CH3​CHO) is more reactive than propanal (CH3​CH2​CHO) due to less steric hindrance.
  • Benzaldehyde’s carbonyl carbon is less positive due to resonance with the benzene ring, making it less reactive than aliphatic aldehydes. Thus, ethanal is the most reactive.

15. Answer: c) Zwitterion Explanation: Sulphanilic acid (p-aminobenzenesulphonic acid) has both an acidic group (sulphonic acid, −SO3​H) and a basic group (amino, −NH2​). In the same molecule, the acidic group donates a proton to the basic group, forming an internal salt structure with both positive and negative charges. This dipolar ionic form is called a zwitterion.

16. Answer: a) Wurtz reaction Explanation: The Wurtz reaction involves the coupling of two alkyl halides using sodium metal in dry ether to form a symmetrical alkane with double the number of carbon atoms.

17. Answer: c) Pd/BaSO4​ (Rosenmund catalyst) Explanation: Rosenmund reduction is a specific catalytic hydrogenation reaction using hydrogen gas in the presence of palladium supported on barium sulfate (poisoned catalyst) to reduce acyl chlorides to aldehydes. LiAlH4​ and NaBH4​ reduce acyl chlorides to alcohols. Zn(Hg)/HCl is Clemmensen reduction for carbonyl to alkane.

18. Answer: b) Cinnamic acid Explanation: Perkin’s reaction involves the condensation of an aromatic aldehyde with an acetic anhydride (or other acid anhydride with α-hydrogens) in the presence of the corresponding sodium salt to form an α,β-unsaturated aromatic acid. Cinnamic acid is a prime example formed from benzaldehyde and acetic anhydride.

19. Answer: a) Carboxylic acid > Phenol > Alcohol Explanation: The decreasing order of acidity is based on the stability of their conjugate bases. Carboxylate ions are stabilized by two equivalent resonance structures, phenoxide ions by multiple resonance structures (but not equivalent), and alkoxide ions are destabilized by +I effect and lack resonance.

20. Answer: a) Soluble in aqueous KOH Explanation: In Hinsberg’s test, a primary amine reacts with benzene sulphonyl chloride to form an N-alkylbenzenesulphonamide. This sulphonamide has an acidic hydrogen atom attached to the nitrogen, which can be removed by a strong base like KOH, forming a soluble potassium salt.

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