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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.
Solution :
1. Chemical Context and Regioselectivity
The addition of hydrogen bromide ($HBr$) to an unsymmetrical alkene like propene ($CH_3-CH=CH_2$) follows two distinct mechanistic pathways depending on the reaction conditions. The formation of 2-bromopropane represents Markownikoff's Rule, while the formation of 1-bromopropane represents the Peroxide Effect (Kharasch effect or anti-Markownikoff addition).
2. Mechanism 1: Electrophilic Addition (Markownikoff's Rule)
In the absence of peroxides, the reaction proceeds via an electrophilic mechanism [Per the Mechanism of Electrophilic Addition to Alkenes].
Step 1: Protonation of the double bond. The $H^+$ ion (electrophile) from $HBr$ attacks the double bond to form the most stable carbocation.
$$CH_3-CH=CH_2 + H^+ \rightarrow CH_3-CH^+-CH_3 \text{ (secondary carbocation)}$$[Justification: The secondary carbocation ($2^\circ$) is more stable than the primary carbocation ($1^\circ$) due to inductive effect and hyperconjugation.]
Step 2: Nucleophilic Attack. The bromide ion ($Br^-$) attacks the carbocation.
$$CH_3-CH^+-CH_3 + Br^- \rightarrow CH_3-CH(Br)-CH_3$$Result: 2-bromopropane.
3. Mechanism 2: Free Radical Addition (Peroxide Effect)
In the presence of benzoyl peroxide ($(C_6H_5COO)_2$), the reaction follows a free-radical chain mechanism.
Step 1: Initiation. Benzoyl peroxide undergoes homolytic cleavage to generate phenyl radicals, which eventually generate bromine radicals ($Br^\bullet$).
$$(C_6H_5COO)_2 \rightarrow 2C_6H_5COO^\bullet \rightarrow 2C_6H_5^\bullet + 2CO_2$$ $$C_6H_5^\bullet + HBr \rightarrow C_6H_6 + Br^\bullet$$Step 2: Propagation. The $Br^\bullet$ attacks the alkene to form the more stable alkyl radical.
$$CH_3-CH=CH_2 + Br^\bullet \rightarrow CH_3-\dot{C}H-CH_2Br \text{ (secondary radical)}$$[Justification: A secondary free radical is more stable than a primary free radical due to greater alkyl group substitution.]
Step 3: Abstraction. The secondary radical abstracts a hydrogen atom from $HBr$.
$$CH_3-\dot{C}H-CH_2Br + HBr \rightarrow CH_3-CH_2-CH_2Br + Br^\bullet$$Result: 1-bromopropane.
4. Visual Representation of Reaction Pathway
5. Comparative Summary
| Feature | Markownikoff Addition | Anti-Markownikoff (Peroxide) |
|---|---|---|
| Mechanism Type | Electrophilic Addition | Free Radical Addition |
| Reactive Intermediate | $2^\circ$ Carbocation | $2^\circ$ Free Radical |
| Major Product | 2-bromopropane | 1-bromopropane |
Final Solution: The addition of HBr to propene in the absence of peroxides follows the electrophilic addition mechanism yielding 2-bromopropane (Markownikoff's rule), whereas in the presence of benzoyl peroxide, it follows a free-radical chain mechanism yielding 1-bromopropane (Anti-Markownikoff addition/Peroxide effect).
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?
- Q9.17: 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.
- Q9.19: Why does benzene undergo electrophilic substitution reactions easily and nucleophilic substitutions with difficulty?
- 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?
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