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Q9.19:
Why does benzene undergo electrophilic substitution reactions easily and nucleophilic substitutions with difficulty?
Solution :
Step 1: Electronic Structure and Resonance Hybridization
Benzene ($C_6H_6$) is a planar, cyclic molecule where each carbon atom is $sp^2$ hybridized. The remaining unhybridized $p_z$ orbital on each carbon overlaps laterally to form a delocalized $\pi$-electron cloud above and below the plane of the ring [Per Hückel’s Rule for Aromaticity].
This $\pi$-electron cloud creates a region of high electron density, making the benzene ring inherently electron-rich. This makes it an ideal target for electrophiles (electron-deficient species) and extremely repulsive to nucleophiles (electron-rich species).
Step 2: Analysis of Electrophilic Substitution ($S_EAr$)
Because the benzene ring is electron-dense, electrophiles ($E^+$) are naturally attracted to it. The reaction mechanism follows two distinct phases:
- Formation of Arenium Ion: The electrophile attacks the $\pi$-cloud, forming a resonance-stabilized carbocation known as the $\sigma$-complex or Wheland intermediate.
- Restoration of Aromaticity: A base removes a proton from the $sp^3$ hybridized carbon, restoring the stable aromatic sextet.
The transition state is stabilized by the resonance of the delocalized electrons, lowering the activation energy for electrophilic attack [Per Transition State Theory].
Step 3: Obstacles to Nucleophilic Substitution ($S_NAr$)
Nucleophiles ($Nu^-$) are negatively charged or electron-rich. When a nucleophile approaches benzene, the following barriers prevent reaction:
- Electronic Repulsion: The incoming nucleophile encounters massive electrostatic repulsion from the electron-rich $\pi$-cloud of the benzene ring.
- Nature of the C-H Bond: The $C-H$ bonds in benzene are relatively strong and non-polar, offering no leaving group capability under standard conditions.
- Instability of Intermediate: A nucleophilic attack would result in a carbanion intermediate. Unlike the carbocation intermediate in $S_EAr$, this carbanion is highly unstable because it is localized and lacks the stabilization provided by the electron-delocalization of the ring.
Visual Representation of Electronic Environment
Comparative Summary
| Feature | Electrophilic Substitution | Nucleophilic Substitution |
|---|---|---|
| Reagent | Electrophile ($E^+$) | Nucleophile ($Nu^-$) |
| Ring Interaction | Attractive (Electrostatic) | Repulsive (Electrostatic) |
| Intermediate | Resonance-stabilized Carbocation | High-energy/Unstable Carbanion |
Final Solution: Benzene undergoes electrophilic substitution easily because its electron-rich $\pi$-cloud acts as a donor to electron-deficient electrophiles, forming a stable intermediate. Conversely, it resists nucleophilic substitution due to strong electrostatic repulsion between the incoming nucleophile and the $\pi$-electron system, combined with the lack of an energetically favorable pathway to form a carbanion.
More Questions from Class 11 Chemistry Hydrocarbons EXERCISES
- Q9.1: How do you account for the formation of ethane during chlorination of methane ?
- Q9.10: Why is benzene extra ordinarily stable though it contains three double bonds?
- Q9.11: What are the necessary conditions for any system to be aromatic?
- Q9.12(i): Explain why the following systems are not aromatic? (i)
- Q9.12(ii): Explain why the following systems are not aromatic? (ii)
- Q9.12(iii): Explain why the following systems are not aromatic? (iii)
- Q9.13(i): How will you convert benzene into (i) $p$-nitrobromobenzene
- Q9.13(ii): How will you convert benzene into (ii) $m$-nitrochlorobenzene
- Q9.13(iii): How will you convert benzene into (iii) $p$-nitrotoluene
- Q9.13(iv): How will you convert benzene into (iv) acetophenone?
- Q9.14: In the alkane $H_3C-CH_2-C(CH_3)_2-CH_2-CH(CH_3)_2$, identify $1^{\circ},2^{\circ},3^{\circ}$ carbon atoms and give the number of H atoms bonded to each one of these.
- Q9.15: What effect does branching of an alkane chain has on its boiling point?
- Q9.16: Addition of HBr to propene yields 2-bromopropane, while in the presence of benzoyl peroxide, the same reaction yields 1-bromopropane. Explain and give mechanism.
- Q9.17: Write down the products of ozonolysis of 1,2-dimethylbenzene ($o$-xylene). How does the result support Kekulé structure for benzene?
- Q9.18: Arrange benzene, $n$-hexane and ethyne in decreasing order of acidic behaviour. Also give reason for this behaviour.
- Q9.2(a): Write IUPAC names of the following compounds : (a) $CH_3CH=C(CH_3)_2$
- Q9.2(b): Write IUPAC names of the following compounds : (b) $CH_2=CH-C \equiv C-CH_3$
- Q9.2(c): Write IUPAC names of the following compounds : (c)
- Q9.2(d): Write IUPAC names of the following compounds : (d)
- Q9.2(e): Write IUPAC names of the following compounds : (e)
- Q9.2(f): Write IUPAC names of the following compounds : (f)
- Q9.2(g): Write IUPAC names of the following compounds : (g)
- Q9.20(i): How would you convert the following compounds into benzene? (i) Ethyne
- Q9.20(ii): How would you convert the following compounds into benzene? (ii) Ethene
- Q9.20(iii): How would you convert the following compounds into benzene? (iii) Hexane
- Q9.21: Write structures of all the alkenes which on hydrogenation give 2-methylbutane.
- Q9.22(a): Arrange the following set of compounds in order of their decreasing relative reactivity with an electrophile, $E^+$ (a) Chlorobenzene, 2,4-dinitrochlorobenzene, $p$-nitrochlorobenzene
- Q9.22(b): Arrange the following set of compounds in order of their decreasing relative reactivity with an electrophile, $E^+$ (b) Toluene, $p-H_3C-C_6H_4-NO_2$, $p-O_2N-C_6H_4-NO_2$.
- Q9.23: Out of benzene, $m$-dinitrobenzene and toluene which will undergo nitration most easily and why?
- Q9.24: Suggest the name of a Lewis acid other than anhydrous aluminium chloride which can be used during ethylation of benzene.
- Q9.25: Why is Wurtz reaction not preferred for the preparation of alkanes containing odd number of carbon atoms? Illustrate your answer by taking one example.
- Q9.3(a): For the following compounds, write structural formulas and IUPAC names for all possible isomers having the number of double or triple bond as indicated : (a) $C_4H_8$ (one double bond)
- Q9.3(b): For the following compounds, write structural formulas and IUPAC names for all possible isomers having the number of double or triple bond as indicated : (b) $C_5H_8$ (one triple bond)
- Q9.4(i): Write IUPAC names of the products obtained by the ozonolysis of the following compounds : (i) Pent-2-ene
- Q9.4(ii): Write IUPAC names of the products obtained by the ozonolysis of the following compounds : (ii) 3,4-Dimethylhept-3-ene
- Q9.4(iii): Write IUPAC names of the products obtained by the ozonolysis of the following compounds : (iii) 2-Ethylbut-1-ene
- Q9.4(iv): Write IUPAC names of the products obtained by the ozonolysis of the following compounds : (iv) 1-Phenylbut-1-ene
- Q9.5: An alkene ‘$A$’ on ozonolysis gives a mixture of ethanal and pentan-3-one. Write structure and IUPAC name of ‘$A$’.
- Q9.6: An alkene ‘$A$’ contains three $C-C$, eight $C-H$ $\sigma$ bonds and one $C-C$ $\pi$ bond. ‘$A$’ on ozonolysis gives two moles of an aldehyde of molar mass 44 u. Write IUPAC name of ‘$A$’.
- Q9.7: Propanal and pentan-3-one are the ozonolysis products of an alkene? What is the structural formula of the alkene?
- Q9.8(i): Write chemical equations for combustion reaction of the following hydrocarbons: (i) Butane
- Q9.8(ii): Write chemical equations for combustion reaction of the following hydrocarbons: (ii) Pentene
- Q9.8(iii): Write chemical equations for combustion reaction of the following hydrocarbons: (iii) Hexyne
- Q9.8(iv): Write chemical equations for combustion reaction of the following hydrocarbons: (iv) Toluene
- Q9.9: Draw the cis and trans structures of hex-2-ene. Which isomer will have higher b.p. and why?
CBSE Solutions for Class 11 Chemistry Hydrocarbons
Chapters in CBSE - Class 11 Chemistry
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