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Explain in detail about the addition of hydrogen halide to an unsymmetrical alkene. |
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Answer» Solution :Addition `HBr` to unsymmetrical alkene : In the addition of hydrogen halide to an unsymmetrical alkene, two products are obtained.  Mechanism : Consider addition of `HBr` to propene Step `1` : Formation of electrophile : In `H-Br`, `Br` is more electronegative than `H`. When bonded electron moves toward `Br`, POLARITY is developed and creates an electrophile `H^(+)` which attaches the double bond to form carbocation , as shown below.  Step `2` : Secondary carbocation is more stable than primary carbocation and it predominates over a the primary carbocation. Step `3` : The `Br`-ION attack the `2^(@)` carbocation to from `2-` Bromobutane, the major product. Consider addition of `HBr` to `3`-methyl -`1`- butene. Here the expected product according to Markovnikoff's rule is `2`-BROMO-`3`-methyl butane but the actual major product is `2`-Bromo- `2`-methyl butane. This is because, the secondary carbocation formed during the reaction rearranged to more stable tertiary carbocation. attack of `Br`-on this tertiary carbocation gives the major product `2`-bromo-`2` methyl butane.  Carbocation reaarangement  Anti-Markovnkoff's Rule (Or) Peroxide Effect (Or)Kharasch Addition : The addition of `HBr` to an alkene in the presence of organic peroxide, gives the anti Markovniko's product. This effect is called peroxide effect. `underset("PROPANE")(CH_(3)-CH=CH_(2))+HBr`  Mechanism : The reaction proceeds via free radical mechanism. Step `1` : The weak `O-O` single bond linkages of peroxides undergoes homolytic cleavage to generate free radical.  Step `2` : The radicals abstracts a hydrogen from `HBr` thus generating bromine radical. `overset(.)(C_(6))H_(5)+HBrtoC_(6)H_(6)+overset(.)Br` Step `3` : The bromine radical adds to the double bond in the way to form more stable alkyl free radical.  Step `4` : Addition of `HBr` to secondary free radical  The `H-Cl` bond is stronger `(430.5 kJ mol-1)` than `H-Br` bond `(363.7kJmol^(-1))`, thus `H-Cl` is not cleaved by the free radical. The `H-I` bond is weaker `(296.8kJmol-1)` , than `H-Cl` bond. Thus `H-I` bond breaks easily but iodine free radicals combine to form iodine molecules instead of adding to the double bond and hence peroxide effect is not observed in `HCl` & `HI`. |
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