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Presentation on theme: "MARKOVNIKOV’S RULE."— Presentation transcript:


2 The compounds which contains only carbon and hydrogen are called hydrocarbons. Hydrocarbons are classified into three categories according to c-c bond present. Hydrocarbon Saturated Unsaturated Aromatic alkane alkene alkyne Symmetrical Unsymmetrical CH2=CH CH3-CH=CH2 Ethene Propene

3 In case of symmetrical alkene when HBr is react with the product will be the same.
ie H H H H H H C=C HBr H C C H OR H C C H H H H Br Br H I-Bromoethane I-Bromoethane Similarly- H H H CH3-CH(Br)-CH2-CH3 H C C = C CH3 + HBr Bromobutane CH3-CH2-CH(Br)CH H Bromobutane

4 CH3-CH=CH2+HBr I-Bromopropane CH3-CH-CH3 Br
But when Hbr is added to the unsymmetrical alkene then we will get two different product i.e. CH3-CH2-CH2-Br CH3-CH=CH2+HBr I-Bromopropane CH3-CH-CH3 Br In this case the addition governed by Markovnikov Rule who was a Russian chemist. The rules goes like this During the addition acroos unsymmetrical multiple bond the negatively part attacking molecule joins the carbon atom which carries small number of hydrogen atoms while the positive part goes to carbon atom with more hydrogen atoms. So in this above case 2-Bromopropoane is the major product i.e. CH3-CH=CH2+HBr CH3-CH-CH3+ CH3-CH2-CH2 Br Br (90%) (10%)

5 In order to understand Markovnikov’s rule.
Mechanism- In order to understand Markovnikov’s rule. Let consider the attack of electrophile(H+) on the unsymmetrical alkene molecule (Propene). Here two intermediate carbocation are formed. (+) CH3-CH-CH3 CH3-CH=CH2+ H Degree(carbocation) CH3-CH2-CH2 1 Degree(carbo cation)

6 CH3-CH-CH3 or H-C-CH-C-H CH3-CH=CH
As we know that carbocation is stabilises by no-bond resonance structure. More the resonating structure more in the stable. (+) H (+) H H CH3-CH-CH or H-C-CH-C-H CH3-CH=CH 2 Degree(carbonation) H H H H H H H + C=CH-CH CH3-CH=C H CH3-CH=CH H H H + H H + H-C=CH-CH H-C=CH-CH3 H H Total no: 6(six) Where as in case of 1degree carbon cation it has 3 resonating structure. So 2 dgree carbon cation is more stable than 1 degree carbon cation. This 2 degree carbon cation further react with Br- to form the addition product.

7 (+) (-) CH3-CH-CH3+Br CH3-CH-CH3 Br
(+) (-) CH3-CH-CH3+Br CH3-CH-CH3 Br So we can describe Markovnikov ‘s Rule as follows. The electrophilic addition to an unsymmetrical alkene always occur through the formation of more stable carbo cation intermediate.

8 Thus the complete mechanism for the addition of HBr to propane as follow.
HBr H+ + Br- CH3-CH=CH2+H CH3-CH-CH3 2◦ Carbocation CH3-CH-CH3+Br CH3-CH-CH3 Br Anti Markovnikov Rule or peroxide effect Observed by Kharasch that when HBr is added to alkene in presence of O O organic peroxides provide (C6H5-C-O-O-C-C6H5) the reaction takes place opposite to Markovnikov rule. This is known as anti Markovnikov ‘srule or peroxide effect. According to this rule the Br atom will join to that carbon atom which carries more hydrogen atoms. While the hydrogen (H)-atom go to the other C- atom.

9 Thus CH3-CH=CH2+HBr CH3-CH2CH2Br
1-Bromopropane Mechanism/ Explanation for anti- Markovnikov‘s Rule:- The addition of HBr to alkene in the presence of Peroxide occurs by free radical mechanism. Here the first two steps result into generation of free radical Br. Then it adds to double bond in such a way to give more stable radical. Then this free radical abstracts a H – free radical form to complete the addition. Peroxide

10 The various steps of mechanism are:-- O O O homolysis .
C6H5----C-O-O---C-C6H C6H5----C----O C6H5+2CO2 Free radical (phenyl free radical). 2. C6H5+H----Br C6H6+Br (Free radical). . CH3CH-CH2Br 3. Br+CH3-CH CH degree free radical(more stable)(I) CH3-CH-CH2 (less stable) (II) Br 1 degree free radical 2 degree free radical is more stable then 1 degree free radical. It will be preferably formed. 4. CH3-CH-CH2Br + H---Br CH3-CH2-CH2Br + Br

11 Exceptional behaviour of HBr :- It is interesting to note that peroxide effect is applicable only to HBr and not to HF,HCl or HI. This can be understand on tha basis of energy changes in the prpogation step using different halogen atom. . Step iii X+CH3-CH=CH2 CH3-CH-CH2X X=F ΔH= -209 KJ/MOL X=Cl ΔH= -101 KJ/MOL X=Br ΔH= -42 KJ/MOL X=I ΔH= +12 KJ/MOL . . Step iv H-X+CH3-CH-CH2X CH3-CH2-CH2X+X X=F ΔH= +159 KJ/MOL X=Cl ΔH= +27 KJ/MOL X=Br ΔH= -37 KJ/MOL X=I ΔH= KJ/MOL

12 From the above data it has been observed that only with HBr both steps are exothermic where as others do not posses both steps as exothermic. In other words homolysis of HF or HCl does not take place. In case of HI, the homolysis of HI is possible as evident from step 4 but the attack of I to propene does not take place (step III is exothermic). So the free radical I will combine with similar free radical from I2 molecule. Thus peroxide effect is observed incase of HBr.

13 Thank You

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