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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 Reaction Overview
The addition of hydrogen bromide ($\text{HBr}$) to propene ($\text{CH}_3\text{CH=CH}_2$) follows two distinct mechanistic pathways based on reaction conditions: the electrophilic addition (Markovnikov’s rule) and the free-radical addition (Anti-Markovnikov’s rule or Peroxide Effect/Kharasch effect).
2. Case I: Electrophilic Addition (Absence of Peroxides)
In the absence of peroxides, the reaction follows the Markovnikov rule: the negative part of the addendum ($\text{Br}^-$) attaches to the carbon atom with the fewer number of hydrogen atoms.
Mechanism:
- Step 1 (Protonation): The electrophile $\text{H}^+$ attacks the double bond to form the more stable carbocation intermediate.
$\text{CH}_3\text{CH=CH}_2 + \text{H}^+ \rightarrow \text{CH}_3\text{CH}^+\text{CH}_3$ [Formation of secondary carbocation, which is more stable than a primary carbocation per the Inductive Effect]. - Step 2 (Nucleophilic Attack): The bromide ion ($\text{Br}^-$) attacks the secondary carbocation.
$\text{CH}_3\text{CH}^+\text{CH}_3 + \text{Br}^- \rightarrow \text{CH}_3\text{CH(Br)CH}_3$ (2-bromopropane).
3. Case II: Peroxide Effect (Free Radical Mechanism)
In the presence of benzoyl peroxide, the reaction proceeds via a free-radical chain mechanism, overriding Markovnikov's rule. This occurs only with $\text{HBr}$.
Mechanism:
- Step 1 (Chain Initiation): Peroxide undergoes homolytic cleavage.
$(\text{C}_6\text{H}_5\text{COO})_2 \rightarrow 2\text{C}_6\text{H}_5\text{COO}^\bullet \rightarrow 2\text{C}_6\text{H}_5^\bullet + 2\text{CO}_2$.
The phenyl radical abstracts H from $\text{HBr}$ to form a bromine radical: $\text{C}_6\text{H}_5^\bullet + \text{HBr} \rightarrow \text{C}_6\text{H}_6 + \text{Br}^\bullet$. - Step 2 (Chain Propagation): The $\text{Br}^\bullet$ adds to the double bond to form the more stable carbon radical.
$\text{CH}_3\text{CH=CH}_2 + \text{Br}^\bullet \rightarrow \text{CH}_3\text{CH}^\bullet\text{CH}_2\text{Br}$ (Secondary free radical).
$\text{CH}_3\text{CH}^\bullet\text{CH}_2\text{Br} + \text{HBr} \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{Br} + \text{Br}^\bullet$ (1-bromopropane).
4. Comparative Analysis of Intermediates
| Reaction Condition | Intermediate Species | Major Product |
|---|---|---|
| No Peroxide | $2^\circ$ Carbocation | 2-Bromopropane |
| With Peroxide | $2^\circ$ Free Radical | 1-Bromopropane |
5. Visual Representation of Reaction Pathway
Final Solution: The addition of HBr to propene in the absence of peroxide follows the electrophilic Markovnikov addition to yield 2-bromopropane via a secondary carbocation intermediate. In the presence of benzoyl peroxide, a free-radical chain mechanism takes precedence, where the bromine radical attacks the terminal carbon to form a more stable secondary radical, ultimately yielding 1-bromopropane.
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) $CH_3(CH_2)_4 CH (CH_2)_3 CH_3$ with a $CH_2–CH(CH_3)_2$ group attached to the CH.
- Q9.2(g): Write IUPAC names of the following compounds : (g) $CH_3 – CH = CH – CH_2 – CH = CH – CH – CH_2 – CH = CH_2$ with a $C_2H_5$ group attached to the third CH from the left.
- 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
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