Questions & explanations
1. Arrange the following carbocations in decreasing order of stability: (CH3)3C+, CH3-CH=CH-CH2+, PhCH2+, (CH3)2CH-CH2+, Ph2CH+.
- CH3-CH=CH-CH2+ > PhCH2+ > (CH3)3C+ > (CH3)2CH-CH2+ > Ph2CH+
- (CH3)3C+ > PhCH2+ > CH3-CH=CH-CH2+ > Ph2CH+ > (CH3)2CH-CH2+
- PhCH2+ > Ph2CH+ > (CH3)3C+ > CH3-CH=CH-CH2+ > (CH3)2CH-CH2+
- Ph2CH+ > PhCH2+ ≈ CH3-CH=CH-CH2+ > (CH3)3C+ > (CH3)2CH-CH2+
Answer: Ph2CH+ > PhCH2+ ≈ CH3-CH=CH-CH2+ > (CH3)3C+ > (CH3)2CH-CH2+
Ph2CH+ has two phenyl rings, giving extensive resonance stabilisation, making it most stable. PhCH2+ (benzyl) and CH3-CH=CH-CH2+ (allyl) have resonance with one pi system; they are similarly stable. (CH3)3C+ is tertiary, stabilised by hyperconjugation and +I, but less than resonance-stabilised ones. (CH3)2CH-CH2+ is primary, least stable, and rearranges via 1,2-hydride shift to (CH3)3C+. Thus order: Ph2CH+ > PhCH2+ ≈ allyl > (CH3)3C+ > (CH3)2CH-CH2+.
2. Arrange the following in decreasing order of basicity: p-methoxyaniline, aniline, m-nitroaniline, p-nitroaniline.
- p-methoxyaniline > aniline > p-nitroaniline > m-nitroaniline
- aniline > p-methoxyaniline > m-nitroaniline > p-nitroaniline
- p-methoxyaniline > aniline > m-nitroaniline > p-nitroaniline
- p-nitroaniline > m-nitroaniline > aniline > p-methoxyaniline
Answer: p-methoxyaniline > aniline > m-nitroaniline > p-nitroaniline
p-Methoxyaniline has +M -OCH3 group, increasing electron density on nitrogen, making it most basic. Aniline is next. m-Nitroaniline has -I and -M nitro group, but meta position prevents direct resonance with the N lone pair, so less basic than aniline. p-Nitroaniline has -M nitro group at para, which directly delocalises the N lone pair into the ring, making it least basic. Thus order: p-methoxyaniline > aniline > m-nitroaniline > p-nitroaniline.
3. Arrange in decreasing order of acidity: ethanol, acetic acid, phenol, 2,4-dinitrophenol, acetone, 2,4-pentanedione.
- acetic acid > 2,4-dinitrophenol > phenol > 2,4-pentanedione > ethanol > acetone
- 2,4-dinitrophenol > acetic acid > 2,4-pentanedione > phenol > ethanol > acetone
- 2,4-dinitrophenol > acetic acid > phenol > 2,4-pentanedione > ethanol > acetone
- 2,4-dinitrophenol > acetic acid > 2,4-pentanedione > phenol > acetone > ethanol
Answer: 2,4-dinitrophenol > acetic acid > 2,4-pentanedione > phenol > ethanol > acetone
Acidity depends on conjugate base stability. 2,4-dinitrophenol (pKa 4.1) has two -NO2 groups withdrawing electrons by -M and -I, making it strongest. Acetic acid (pKa 4.76) has resonance-stabilized conjugate base. 2,4-pentanedione (pKa 8.9) has enolate resonance. Phenol (pKa 10) has weaker resonance. Ethanol (pKa 16) has no resonance. Acetone (pKa 20) has enolate but less stabilized than ethanol's alkoxide due to charge on carbon.
4. Chlorobenzene undergoes electrophilic substitution slower than benzene, yet the incoming group goes to ortho/para positions. Why?
- Chlorine is activating due to +M and directs ortho/para.
- Chlorine is deactivating due to -I, but in the transition state +M stabilises ortho/para attack.
- Chlorine is deactivating due to -M and directs meta.
- Chlorine is activating due to +I and directs ortho/para.
Answer: Chlorine is deactivating due to -I, but in the transition state +M stabilises ortho/para attack.
Chlorine has a strong -I effect (withdraws electrons through sigma bonds) making the ring electron-poor and deactivating. However, its lone pairs provide a +M effect that can stabilise the positive charge in the sigma complex only when the electrophile attacks ortho or para. Hence, chlorobenzene is deactivating but ortho/para directing.
5. Which C-H bond has the lowest bond dissociation energy?
- CH3-H in methane
- (CH3)2CH-H in propane
- CH2=CH-CH2-H in propene
- CH3CH2-H in ethane
Answer: CH2=CH-CH2-H in propene
The allylic C-H bond in propene has the lowest bond dissociation energy (about 88 kcal/mol) because the resulting allyl radical is stabilised by resonance delocalisation of the unpaired electron over two carbon atoms. In contrast, methane (98 kcal/mol), ethane (98 kcal/mol), and isopropyl (95 kcal/mol) lack such resonance stabilisation.
6. Which of the following benzoic acids is the strongest acid?
- p-nitrobenzoic acid
- m-methoxybenzoic acid
- p-methoxybenzoic acid
- benzoic acid
Answer: p-nitrobenzoic acid
p-nitrobenzoic acid is strongest because the nitro group is strongly electron-withdrawing by both -I and -M effects, stabilising the carboxylate anion. m-methoxybenzoic acid is stronger than p-methoxybenzoic acid because at meta, the +M of methoxy cannot reach the COOH, so only -I operates, making it more acidic than the para isomer.
7. Which statement correctly describes the electromeric effect in ethene when attacked by H+?
- Permanent transfer of pi electrons to the more electronegative carbon
- Delocalisation of pi electrons through resonance structures
- Partial polarisation of sigma bonds along the carbon chain
- Temporary complete transfer of pi electrons towards the attacking electrophile
Answer: Temporary complete transfer of pi electrons towards the attacking electrophile
The electromeric effect is a temporary, complete transfer of a pi-bond electron pair to one atom in the presence of an attacking reagent. For ethene + H+, the pi electrons move completely towards the terminal carbon (towards the electrophile), which is a +E effect. This is distinct from permanent inductive or mesomeric effects.
8. Which carbocation is most stable?
- CH3CH2+
- (CH3)3C+
- CH2=CH-CH2+
- PhCH2+
Answer: PhCH2+
Benzyl carbocation is most stable due to resonance delocalization of the positive charge into the aromatic ring, providing extensive stabilization. Although tertiary carbocation has hyperconjugation, the resonance effect in benzyl carbocation is stronger, making it more stable than tertiary, allyl, or primary carbocations.
9. Arrange in decreasing order of acidity: (a) CH3COOH, (b) C2H5OH, (c) C6H5OH, (d) CH≡CH.
- a > c > b > d
- a > c > d > b
- c > a > d > b
- a > d > c > b
Answer: a > c > b > d
Acidity depends on conjugate base stability. Carboxylic acids (pKa ~5) are strongest due to resonance-stabilised carboxylate. Phenols (pKa ~10) are next due to resonance in phenoxide. Alcohols (pKa ~16) are weaker than terminal alkynes (pKa ~25) because alkoxide is less stable than acetylide. Thus order: a > c > b > d.
10. Bromination of 1-methoxy-3-nitrobenzene gives which major product?
- 2-bromo-1-methoxy-3-nitrobenzene
- 4-bromo-1-methoxy-3-nitrobenzene
- 5-bromo-1-methoxy-3-nitrobenzene
- 6-bromo-1-methoxy-3-nitrobenzene
Answer: 2-bromo-1-methoxy-3-nitrobenzene
Methoxy is ortho/para directing; nitro is meta directing. In 1-methoxy-3-nitrobenzene, position 2 is ortho to methoxy and meta to nitro, making it activated. Position 4 is para to methoxy but ortho to nitro, causing deactivation. Position 2 is less hindered and more activated, so bromination occurs at position 2.
11. Arrange in decreasing basicity: (a) p-methoxyaniline, (b) aniline, (c) p-nitroaniline, (d) m-nitroaniline.
- b > a > d > c
- a > b > c > d
- a > b > d > c
- a > d > b > c
Answer: a > b > d > c
Electron-donating groups increase basicity by making lone pair more available. p-Methoxyaniline (pKb 8.7) is most basic due to +M of -OCH3. Aniline (pKb 9.4) is next. m-Nitroaniline (pKb 11.5) is more basic than p-nitroaniline (pKb 13.0) because -NO2 at para exerts stronger -M effect. Thus order: a > b > d > c.
12. Which of the following correctly describes the hybridisation of carbon in methane, ethene, and ethyne?
- sp3, sp2, sp
- sp2, sp3, sp
- sp, sp2, sp3
- sp3, sp, sp2
Answer: sp3, sp2, sp
Carbon in methane (CH4) forms four sigma bonds, so it is sp3 hybridised. In ethene (C2H4), each carbon forms three sigma bonds and one pi bond, so it is sp2 hybridised. In ethyne (C2H2), each carbon forms two sigma bonds and two pi bonds, so it is sp hybridised. Thus, the correct order is sp3, sp2, sp.