Hydration of Alkenes:

Hydration of Alkenes:

• this reaction is the addition of water to a molecule with a double bond  ,another kind of electrophilic addition reaction is called hydration. As the name suggests,
• Hydration requires the presence of a strong acid catalyst.
• ( A catalyst is a substance that accelarates a reaction, but is not actually consumed in the reaction).
• Similar to  the addition of hydrogen halides to alkenes, the addition of water is regioselective .
• The first step is the attack of the electrons of the pi bond on a proton (usually shown on a hydronium ion).
•  In the next step of the reaction, a lone pair of electrons from the oxygen of water attack the carbocation to give another intermediate which bears a positive charge on oxygen, this is called an oxonium ion
• In the final step of the reaction, a proton is removed from the oxygen by water to regenerate the hydronium ion, and to give the alcohol product.
• Rearrangements are possible here as well
• Water, itself, is too weak an acid to transfer the proton to the alkene, so acid catalysis is essentialNote also that the product is an alcohol.
• Note that the Markovnikov Rule is followed for this reaction, i.e., the proton adds to the less substituted carbon of the alkene double bond (the methylene carbon, making it a methyl group), and the hydroxyl group adds to the more highly substituted carbon atom of the alkene double bond. So the regiochemistry of the hydration reaction is closely parallel to that previously discussed for the hydrochlorination reaction.

example  with rearrangement and ring expansion.

electrophillic additions and rearrangements

electrophillic additions and rearrangements

During   electrophillic     addition ,    carbocations are the intermediates .They can rearrange by a 1,2 HYDRIDE SFIFT or 1,2 ALKYL SHIFT .What is the thermodynamic driving force for this shifting process? If  the hydride shift results in the conversion of a secondary (2^o) carbocation to a (more stable) tertiary (3^o) carbocation or from less satble carbocation to more stable carbocation then rearrangement takes place.  It is thus a thermodynamically downhill reaction, and, as it turns out, the energy barrier to the shift is relatively low, meaning that it can occur quite rapidly .

The following is an example of rearrangement reaction.

In most examples of carbocation rearrangements  ,that you are likely to encounter, the shifting species is a hydride or methyl group.  However,  any alkyl group is capable of shifting.  Sometimes, the entire side of a ring will shift over in a ring-expanding rearrangement.  Consider the following electrophilic addition of HBr to an the following alkene: 

.

Markovnikov’s Rule

Markovnikov’s Rule

what is it ?

While adding , an , unsymmetrical reagent such as HCl,Where H+  is  the electrophile,and Cl- is the nucleophile , to an unsymmetrical  double bond ,  

 the major product in each case is always the one where the electrophile (hydrogen) adds to the carbon in the double bond  with the most hydrogens, and the nucleophile(chlorine) adds to the carbon with the fewest hydrogens.

In other words, this reaction is REGIOSELECTIVE.  

you can see that in each case, we’re breaking C-C (π) and forming a new C-H and C-Cl bond. But ,

The major product in each case is always the one where the hydrogen adds to the carbon with the most hydrogens, and the chlorine adds to the carbon with the fewest hydrogens.

In other words, this reaction is  REGIOSELECTIVE

To describe this, the term “most substituted” is often thrown around a lot, so here is a graphical explanation:

REGIOSELECTIVE ???????? 

If more than one reaction could occur between a set of 

reactants under the same conditions giving products that

 are constitutional isomers  and if one product forms in greater 

amounts(MAJOR PRODUCT) than the others(MINOR 

PRODUCT), the overall reaction is said to be regioselective.

This empirical observation was first pointed out in 1870 by one Vladimir Markovnikov and this pattern of regioselectivity has become known as “Markovnikov’s rule”:

“when an unsymmetrical alkene reacts with a hydrogen halide to give an alkyl halide, the hydrogen adds to the carbon that has the greater number of hydrogen substituents, and the halogen to the carbon having the fewer number of hydrogen substituents”

electrophillic addition ----1,4 and 1,2 addition reactions

• In conjugated dienes, the π bonds are separated by exactly one single bond.

• In isolated dienes, the π bonds are separated by two or more single bonds.

Conjugated dienes represent a special category, because they contain one continuous system of overlapping p orbitals—that is, the two π bonds together comprise one functional group

composed of four overlapping p orbitals. 

There fore , conjugated dienes exhibit special properties

and reactivity.

and 1,3-Butadiene is a conjugated diene

Electrophilic Addition

Addition of HX Across 1,3-Butadiene .

When butadiene is treated with HBr, electrophillic addition takes place,instead of one

two products are formed .

These two products can be explained with a  two-step mechanism:

protonation to form a carbocation followed by nucleophilic attack. which is :

first  electrophile attacks leading to formation of allylic carbocation that 

undergoes rearrangement .These two carbocations attract the nucleophile that leeds to formation of the final product .

Thermodynamic Control vs. Kinetic Control

The expected addition product from reactions of this kind is the result of 1,2-addition, i.e. bonding to the adjacent 

carbons of a double bond  and   the unexpected product 

comes from 1,4-addition

 Product compositions are  temperature dependent, as the 

addition of HBr to 1,3-butadiene demonstrates.

CH2=CH-CH=CH2 + HBr   reaction temperature	CH3CHBr-CH=CH2 + 1,2 addition yield	CH3CH=CHCH2Br  1,4 addition yield
0 ºC 40 ºC	70% 15%	30% 85%

Bonding of an electrophilic atom or group to one of the end 

carbon atoms of a conjugated diene (#1) generates an allyl 

cation intermediate.

Such cations are stabilized by charge delocalization, and it is

 this delocalization that accounts for the 1,4-addition product

 produced in such addition reactions.

Nucleophillic attack to first acrbocation leads to 1,2 product.

Nucleophillic attack to second  acrbocation leads to 1,4 product.

An explanation for the temperature influence is shown in the

 following energy diagram for the addition of HBr to 1,3-

butadiene. 1,2 product. is termed kinetic product.. At higher 

temperatures, equilibrium is established between the 

products, and the thermodynamically favored 1,4 product. 

dominates.1,4 product  is termed  thermodynamic product.

All said and done 1,2 product ,  termed kinetic product , requires lower activation energy and is formed at loewr temperatures as major product.1,4 product is the minor product.

At higher temperatures ,as per the requirements of
reaction, 1,4 product domminates ,since energy requirements in the form of higher activation energy is achieved

ELECTROPHILLIC ADDITION REACTION - AN EXAMPLE

When 3-methyl pent-1-ene is treated with HBr, H+ attacks first as an electrophile. This creates a secondary carbocation .

This is followed by an hydride shift to generate a more stable tertiary carbocation.

Each  carbocation can get attacked by bromine nucleophile to give a final addition product.

In doing so , in secondary carbocation , a new chiral center is created at C -2.This gives to two configurations at this carbon .

why ? 

Since a carbocation is planar ,nucleophile can attack from the top as well as bottom giving raise to two new configurations.

However in the given case two chiral centers are present.One chiral centre has a fixed configuration and doesent change during reaction but the other changes to give diastereomers,

The MAJOR product does not have chiral center and therofore ACHIRAL .

Electrophilic Addition to Alkenes Mechanism

Two characteristics of the double bond help us understand why these addition reactions occur:

  1. An addition reaction results in the conversion of one p bond and one s bond .

2. The electrons of the p bond are exposed. Because the pi bond results from overlapping of p orbitals, the p electrons lie above and below the plane of the double bond:

The first step of the reaction is a relatively slow addition of

the electrophilic proton to the nucleophilic alkene to form a carbocation

intermediate.

In the second step, the positively charged carbocation intermediate

(an electrophile) reacts rapidly with the negatively charged bromide

ion (a nucleophile).

Electrophiles include proton donors such as Brønsted-Lowry acids, neutral reagents such as bromine (because it can be polarized so that one end is positive), and Lewis acids such as BH₃, BF₃, and AlCl₃. Metal ions that contain vacant orbitals—the silver ion (Ag^_), the mercuric ion (Hg²⁺), and the platinum ion (Pt²⁺),  for example—also act as electrophiles .

Nucleophiles
Nucleophile means "nucleus loving" which describes the tendency of an electron rich species to be attracted to the positive nuclear charge of an electron poor species, the electrophile .
The nucleophilicity expresses the ability of the nucleophile to react in this fashion.
In general terms this can be appreciated by considering the availability of the electrons in the nucleophile. The more available the electrons, the more nucleophilic the system.
Hence the first step should be to locate the nucleophilic center.  At this point we will be considering Nu that contain lone pairs and may be anionic, however the high electron density of a  C=C is also a nucleophile. 

When H–Br reacts as an electrophile, it is attacked at H, losing Br–. 

So electrophilic addition of a proton (which is what this is) to an alkene gives aproduct best represented as a carbocation. 

This carbocation rapidly reacts with the bromide ion just

formed. Overall, H–Br adds across the alkene.

 This is a useful way of making simple alkyl

bromides.

The energetic s of this  reaction are shown below . Proton attack leads to formation of a transition state1. This is higher in energy.After a sigma bond is formed , a carbocation is formed which is lower in energy. Energetically it is higher then  either the reactants or the products. In step 2 nucleophillic attack forms transition state 2 silghtly higher in energy.Agter forming the final product,an alkyl halide its energy is lowered .

 an example to show electrophillic addition :

the bromine atom ends up on the more substituted carbon, why ?

Protonation at one end gives a stabilized, benzylic cation .

while protonation at the other would give a highly unstable primary cation, and therefore does not take place.

 The benzylic cation   gives the benzylic alkyl bromide.

Markovnikov’s rule

‘The hydrogen ends up attached to the carbon of the double bond that had more hydrogens to start with.’ 

I don’t suggest you learn this rule, though  you may hear it referred to. As with all ‘rules’ it is much more important to understand the reason behind it.

One way to state Markovnikov’s rule is to say that in the addition of HX to an  

alkene, the hydrogen atom adds to the carbon atom of the double bond that 

already has the greater number of hydrogen atoms.*

or

MARKOVNIKOV'S RULE

The hydrogen of the HX is added to the carbon which has the greater number of hydrogens and the halogen is added to the carbon which has the lesser number of hydrogens.


a example to show reaction of hydrogen iodide with an alkene.

Electrophilic addition to dienes

Conjugated dienes  , in the presence of proton (electrophile ) and nucleophile ,gives two  

carbocations .

Each gives a different addition product.

One is 1,2 addition product .if addition is taking place
between neighboring atoms .If addition takes place between
1st and 4th atoms then 1,4 addition product. 

E 1 cb reaction -------- a ------------------ video

please find here video for e1cb reaction.


you tube url : http://youtu.be/GE8kw2yRdEs

good luck

PSK CHAKRAVARTHY