Mechanism of Some Important Reactions of Alkenes

MECHANISM OF SOME IMPORTANT REACTIONS OF ALKENES

1. Mechanism of halogen addition:

The mechanism proposed is an ionic mechanism.

In the first step the exposed electrons of the $\pi$- bond of the alkene attacks the halogen in the following way:

As $\pi -e^{\prime }s$of the alkene approach the bromine molecules, the electrons of bromine – bromine bond drift to make bromine molecule polarised. The more distant bromine develops a partial negative charge and nearer bromine becomes partially positive. Polarization weakens the bond and cleaves it heterolytically.

In second step, one of the bromide ions predicted in step I attacks one of the carbon atoms of the bromonium ion. The nucleophilic attack results in the formation of a vicinal dibromide by opening the three – membered ring.

On reaction of cyclopentane with bromine in $CC{l}_4,$ anti – addition occurs and the products of the reaction are trans – 1, 2 dibromocyclopentane enantiomers (as a race`mate)

Addition of bromine to cis – 2 – butene gives racemic form of 2, 3 – dibromobutane.

Bromine adds to trans – 2 – butene to form meso compound, thus the reaction is stereospecific in nature.

2. Mechanism of halohydrin formation

It can be explained by the following mechanism:

If the alkene is unsymmetrical, the halogen ends up on the carbon atom with greater number of hydrogen atoms.

3. Syn - hydroxylation

Hydroxylation with permanganate is carried out by reaction at room temperature. It's a good method for the synthesis of 1, 2 – diols.

Mechanism in both cases involves formation of cyclic intermediates, then in several steps, the cleavage at oxygen – metal bond takes place producing glycol and $Mn{O}_2$or Os metal.

Cis – 2 – butene when treated with cold alkaline $KMn{O}_4$gives meso glycol and trans – 2 – butene gives racemate.

4. Oxidation reactions of alkenes

(i) With cold dilute $KMn{O}_4$(Baeyer's reagent) alkenes give 1, 2 – glycols.

$C{H}_3-CH=C{H}_2+H_{2}O+O\to C{H}_3-CHOH-C{H}_{2}OH$

Propene From $KMn{O}_4$ Propylene glycol

(ii) With hot alkaline $KMn{O}_4$

Cleavage of C = C bond takes place leading to formation of carboxylic acids, ketones and $C{O}_2+H_{2}O$depending upon structure of alkene.

Hence by identifying the products formed during alkaline $KMn{O}_4$oxidation, it is possible to determine the position of the double bond in an alkene molecule.

(iii) With ozone alkenes first give ozonides which upon reductive cleavage with Zn dust and $H_{2}Oor{H}_2/Pd$ gives aldehydes / ketones or a mixture of these depending upon the structure of alkene.

This two step – conversion of alkene into ozonide followed by decomposition with $Zn/H_{2}O$to give aldehydes / ketones or a mixture of these is called reductive ozonolysis.

If however ozonide is decomposed with $H_{2}O,$ the initially formed aldehydes are further oxidized to the corresponding acids by $H_2{O}_2$produced in the reaction. This is called oxidation ozonolysis.

The oxidation reactions of alkenes are summed up as follows:

Illustration 1. C₅H₁₀ represents three isomeric alkenes A₁, A₂ and A₃. Each on hydrogentation gives 2 – methyl butane. A₁ and A₂ on oxymercuration – demercuration gives the same 3⁰ alcohol. A₂ and A₃ on hydroboration oxidation gives different 1⁰ alcohol. Assign structures to A₁, A₂ and A₃ and explain the reactions.

Solution:

This shows that each of A₁, A₂ and A₃ have same C – chain only position of

C = C is to be decided

$A_1,A_2\underset{demerocuration}{\overset{oxymercuration}{\to }}{3}^{0}Alcohol$

hence A_1­ and A₂­ have

$A_{1}and{A}_3\stackrel{hydroborationoxidation}{\to }{1}^{0}Alcohol$

This indicates presence of (CH₂ =) grouping at the terminal. Hence