In the previous section, we completed a discussion on some properties of alkenes and Markovnikov's rule. In this section, we will see anti Markovnikov's rule.
Some basics can be written in 7 steps:
1. In the previous section, we have seen the addition reaction between HBr and an unsymmetrical alkene. This reaction follows the Markovnikov's rule.
2. But if peroxide is present in the reaction mixture, the reaction will take place contrary to the Markovnikov’s rule.
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This effect was discovered in 1933 by M.S. Kharash and F.R. Mayo at the university of Chicago.
♦ This effect is known as Peroxide effect.
♦ This effect is also known as Kharash effect.
♦ This effect is also known as addition reaction anti to Markovnikov’s rule.
3. Consider the reaction between HBr and prop-1-ene.
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Based on the discussions in the previous section, we know that, the product will be 2-Bromopropane.
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But if benzoyl peroxide is present, the product will be 1-Bromopropane.
4. The benzoyl peroxide molecule consists of two C6H5-C=O groups attached together by a -O-O- group.
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The peroxide undergoes homolysis as shown in fig.13.74 (a) below:
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| Fig.13.74 |
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We see that, as a result of the homolysis, two free radicals are obtained as products. Each is a free radical because, the O atom has an unpaired electron.
5. This free radical will again undergo homolysis.
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The homolysis takes place at the bond between the C6H5 group and the CO2 group. This is shown in fig.13.74(b) above.
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We see that, the product is again a free radical ($\rm{\dot{C}_6 H_5}$) and a molecule of CO2.
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We have seen the Lewis structure of CO2 [See fig.4.11 in section 4.1]. It is clear that, the homolysis in fig.b enables the CO2 group to acquire an extra electron needed for the new double bond.
6. The $\rm{\dot{C}_6 H_5}$ radical attacks the HBr molecule.
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The HBr undergoes homolysis. This is shown in fig.13.74(c) above.
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The products are $\rm{\dot{H}}$ radical and $\rm{\dot{Br}}$ radical.
7. The $\rm{\dot{H}}$ radical combines with $\rm{\dot{C}_6 H_5}$ radical to form C6H6. This is shown in fig.13.74(d) above.
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We want the $\rm{\dot{Br}}$ radical. It is required for our further discussions.
◼ The $\rm{\dot{Br}}$ radical attacks the molecule of prop-1-ene, resulting in the formation of 1-Bromopropane or 2-Bromopropane. We want to know which one is the major product.
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Let us first see the mechanism of the reaction which produces 1-Bromopropane. It can be written in 3 steps:
1. First the $\rm{\dot{Br}}$ radical attacks the prop-1-ene molecule.
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The radical gets attached to the first C atom. This is shown in fig.13.75(a) below:
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| Fig.13.75 |
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Note that, the $\rm{\dot{Br}}$ radical brought one electron required for the bond. The other electron on the bond is obtained by breaking the double bond of the prop-1-ene.
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When the double bond is broken in this way, the second C atom has now an unpaired electron.
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A new radical is formed in this way. The newly formed radical is a secondary free radical. This is because, the "C atom with the unpaired electron" is attached to two other C atoms.
2. The secondary free radical now attacks a HBr molecule. It results in the homolysis of HBr. This is shown in fig.13.75(b) above. Thus we get a $\rm{\dot{H}}$ radical and a new $\rm{\dot{Br}}$ free radical.
• We must note a point here. It can be written in 3 steps:
(i) The $\rm{\dot{Br}}$ free radical in fig.13.75(a) is obtained by the action of the peroxide.
(ii) But the $\rm{\dot{Br}}$ free radical in fig.13.75(b) is obtained by the action of the secondary free radical. This radical can attack another prop-1-ene molecule.
(iii) So a chain reaction will be set up.
3. The $\rm{\dot{H}}$ free radical attacks the secondary free radical. This is shown in fig.c above.
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The $\rm{\dot{H}}$ free radical gets attached to the second C atom.
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The $\rm{\dot{H}}$ free radical brought one electron required for the bond. The second electron is already present as an unpaired electron.
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Thus we get 1-Bromopropane.
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Next, let us see the mechanism of the reaction which produces 2-Bromopropane. It can be written in 3 steps:
1. First the $\rm{\dot{Br}}$ radical attacks the prop-1-ene molecule.
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The radical gets attached to the second C atom. This is shown in fig.13.76(a) below:
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| Fig.13.76 |
• Note that, the $\rm{\dot{Br}}$ radical brought one electron required for the bond. The other electron on the bond is obtained by breaking the double bond of the prop-1-ene.
• When the double bond is broken in this way, the first C atom has now an unpaired electron.
• A new radical is formed in this way. The newly formed radical is a primary free radical. This is because, the "C atom with the unpaired electron" is attached to only one other C atom.
2. The primary free radical now attacks a HBr molecule. It results in the homolysis of HBr. This is shown in fig.13.76(b) above. Thus we get a H radical and a new $\rm{\dot{Br}}$ free radical.
• As before, we must note a point here. It can be written in 3 steps:
(i) The $\rm{\dot{Br}}$ free radical in fig.13.76(a) is obtained by the action of the peroxide.
(ii) But the $\rm{\dot{Br}}$ free radical in fig.13.76(b) is obtained by the action of the primary free radical. This radical can attack another prop-1-ene molecule.
(iii) So a chain reaction will be set up.
3. The $\rm{\dot{H}}$ free radical attacks the primary free radical. This is shown in fig.c above.
• The $\rm{\dot{H}}$ free radical gets attached to the first C atom.
• The $\rm{\dot{H}}$ free radical brought one electron required for the bond. The second electron is already present as an unpaired electron.
• Thus we get 2-Bromopropane.
◼ We want to know this:
In an actual reaction, when peroxide is present, what will be the major product? 1-Bromopropane or 2-Bromopropane?
• We can now write the answer. It can be written in 3 steps:
1. We have seen the two possibilities. Let us compare them.
• In fig.13.75, we see that, the prop-1-ene changes into a secondary free radical.
• In fig.13.76, we see that, the prop-1-ene changes into a primary free radical.
2. Secondary free radicals are more stable than primary free radicals.
• So in the reaction mixture, there will be a greater quantity of secondary free radicals.
• Consequently, the $\rm{\dot{H}}$ free radical will be reacting more with secondary free radicals. This will result in a greater quantity of 1-Bromopropane.
3. So the answer is:
In an actual reaction, when peroxide is present, the principal product will be 1-Bromopropane.
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This result is contrary to Markovnikov’s rule. If we apply Markovnikov’s rule, we would be expecting 2-Bromopropane. It is the presence of peroxide which causes the contradicting result.
• Based on the discussions so far, we can write a comparison between two cases: Case I and Case II. The comparison can be written in 5 steps:
1. Which are the reactants in each case?
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Case I is the reaction between prop-1-ene and HBr.
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Case II is also the reaction between the same prop-1-ene and HBr. But benzoylperoxide is also present.
2. Which is the attacking reagent?
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In case I, the H+ attacks the prop-1-ene molecule.
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In case II, the $\rm{\dot{Br}}$ free radical attacks the prop-1-ene molecule.
3. What is the result of the attack?
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As a result of the attack,
♦ In case I, prop-1-ene changes to primary/secondary carbocation.
♦ In case II, prop-1-ene changes to primary/secondary free radical.
4. What is the end product?
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In case I, the end product is 2-Bromopropene.
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In case II, the end product is 1-Bromopropene.
5. How is Markovnikov’s rule applicable to the two cases?
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Markovnikov’s rule states that:
The negative part of the addendum gets attached to that carbon atom which possesses lesser number of hydrogen atoms.
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In case I, the Br- indeed gets attached to the second C atom which has only one H atom.
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In case II, the Br- gets attached to the first C atom which has two H atoms. This is contrary to the Markovnikov’s rule.
• The anti-Markovnikov’s effect is not observed if we use HCl or HI in the place of HBr.
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First we will discuss about HCl. It can be written in 2 steps:
1. In fig.13.74(c) above, we saw that the free radicals formed from the peroxide, causes the homolysis of HBr.
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But such a homolysis is not possible in the case of HCl
2. The reason can be understood if we compare the bond strengths.
♦ H-Cl bond has a bond strength of 430.5 kJ mol-1.
♦ H-Br bond has a bond strength of 363.7 kJ mol-1.
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Obviously, H-Cl bond is much stronger. It cannot be subjected to homolysis easily.
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Next we will discuss about HI. It can be written in 2 steps:
1. The H-I bond strength (296.8 kJ mol-1) is comparatively low.
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So The H-I bond can be easily subjected to homolysis by the radicals formed from the peroxide.
2. But the $\rm{\dot{I}}$ free radicals so formed will combine together to form I2 molecules. Those $\rm{\dot{I}}$ free radicals will not attack the alkene.
Now we will see a solved example
Solved example 13.12
Write IUPAC names of products obtained by addition reaction of HBr to hex-1-ene.
(i) in the absence of peroxide.
(ii) in the presence of peroxide.
Solution:
1. The condensed formula of hex-1-ene is shown in fig.13.77(a) below:
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| Fig.13.77 |
2. The H atom will get attached to either the first C atom or the second C atom.
• Similarly, the Br atom will get attached to either the first C atom or the second C atom.
• This is because, those two C atoms are the ones which form the double bond.
3. When peroxide is absent, we can apply the Markovnikov’s rule.
• So Br- will get attached to that C atom with the lesser number of H atoms. That is the second C atom.
• So the H atom gets attached to the first C atom.
• Thus we get the product: 2-Bromohexane. This is shown in fig.b.
4. When peroxide is present, we can apply the anti-Markovnikov’s rule.
• So Br- will get attached to that C atom with the greater number of H atoms. That is the first C atom.
• So the H atom gets attached to the second C atom.
• Thus we get the product: 1-Bromohexane. This is shown in fig.c
We have completed a discussion on anti-Markovnikov's rule. In the next section we will see a few more chemical properties of alkenes.
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