This post is about a particular addition reaction mechanism found in alkene chemistry.
We've all seen the test for the alkene double bond.
Nice bit of chemistry as the alkene decolorises the bromine water.
What's exactly going on in that reaction?
Is it a straightforward as it seems ?
Well, as you might have guessed, I wouldn't be asking the question if it wasn't exactly a straight forward reaction,
The mechanisms are complex and there are two of them if you use bromine water (Br2) and propene (CH3CHCH2).
So in this post let's
just look at the reaction between bromine itself and propene and then in the next post on hydrocarbons (Hydrocarbons (11)) look at the influence of the water.
Overall the reaction goes like this:
CH3CH=CH2 + Br2 = CH3CHBrCH2Br
propene bromine 1,2-dibromopropane
Now the reaction has to go via collision between the bromine molecule and the propene molecule.
The discussion lies around how that collision happens and what sort of entity results from it.
You'll see from the diagram that the π bond is quite exposed as it is above and below the plane of the alkene molecule.
We also know that the π bond is weaker than the σ bond.
Collision between the bromine molecule and the alkene at the π bond is likely to result in a change especially if the collision is energetic enough.
The reaction happens at room temperature even in the dark so the collision is probably energetic enough.
But what will happen if the bromine molecule approaches the alkene double bond?
The π electrons could affect the bromine molecule and in fact we say they induce a temporary dipole in the bromine molecule.
You can see in the diagram to the left that a pair of electrons moves towards the bromine molecule.
The Bromine molecule is polarised in the presence of the π bond as shown by the δ+ and δ- charges.
The effect of the approach of the π bond is to push a pair of bonding electrons onto the δ- bromine atom as shown by the other double headed curly arrow.
Let's see how that looks in propene:
You can see that the initial product is an unstable cation because the triangular ring structure is very strained.
As ROC Norman (my organic chemistry hero by the way!!) states, and you'll hear more from him if you stick with this blog, "the intermediate carbocation is not able to undergo free rotation about the new C-C bond but is held rigid by interaction with the electrophile which has been added."
Hence the triangular structure on the right is rigid and the C-C bond in it cannot rotate.
It is this carbocation that is then attacked (very rapidly) from the opposite side to the bromine atom already attached.
Here is the overall mechanism of electrophilic addition of bromine in propene:
You'll see in the representation that I have left the intermediate as a simple carbocation and ignored the triangular structure (its called an epoxide ring structure`)
But note too how the curly arrow from the bromide ion moves towards the positive charge on the intermediate.
So this is called an electrophilic addition mechanism.
Electrophilic because the bromine molecule acts as in electrophile once a dipole is induced in it.
Addition because the two reactants from one product.
Check out the next post (Hydrocarbons(11)) for what happens when this reaction is carried out in the presence of water.
Pages on the "Mole" and "Using the Mole" in chemical calculations are here
We've all seen the test for the alkene double bond.
Nice bit of chemistry as the alkene decolorises the bromine water.
What's exactly going on in that reaction?
Is it a straightforward as it seems ?
Well, as you might have guessed, I wouldn't be asking the question if it wasn't exactly a straight forward reaction,
The mechanisms are complex and there are two of them if you use bromine water (Br2) and propene (CH3CHCH2).
So in this post let's
just look at the reaction between bromine itself and propene and then in the next post on hydrocarbons (Hydrocarbons (11)) look at the influence of the water.
Overall the reaction goes like this:
CH3CH=CH2 + Br2 = CH3CHBrCH2Br
propene bromine 1,2-dibromopropane
Now the reaction has to go via collision between the bromine molecule and the propene molecule.
The discussion lies around how that collision happens and what sort of entity results from it.
In our example then we have a double bond with π and σ bonds.
You'll see from the diagram that the π bond is quite exposed as it is above and below the plane of the alkene molecule.We also know that the π bond is weaker than the σ bond.
Collision between the bromine molecule and the alkene at the π bond is likely to result in a change especially if the collision is energetic enough.
The reaction happens at room temperature even in the dark so the collision is probably energetic enough.
But what will happen if the bromine molecule approaches the alkene double bond?
The π electrons could affect the bromine molecule and in fact we say they induce a temporary dipole in the bromine molecule.
The Bromine molecule is polarised in the presence of the π bond as shown by the δ+ and δ- charges.
The effect of the approach of the π bond is to push a pair of bonding electrons onto the δ- bromine atom as shown by the other double headed curly arrow.
Let's see how that looks in propene:
You can see that the initial product is an unstable cation because the triangular ring structure is very strained.
As ROC Norman (my organic chemistry hero by the way!!) states, and you'll hear more from him if you stick with this blog, "the intermediate carbocation is not able to undergo free rotation about the new C-C bond but is held rigid by interaction with the electrophile which has been added."
Hence the triangular structure on the right is rigid and the C-C bond in it cannot rotate.
It is this carbocation that is then attacked (very rapidly) from the opposite side to the bromine atom already attached.
Here is the overall mechanism of electrophilic addition of bromine in propene:
You'll see in the representation that I have left the intermediate as a simple carbocation and ignored the triangular structure (its called an epoxide ring structure`)
But note too how the curly arrow from the bromide ion moves towards the positive charge on the intermediate.
So this is called an electrophilic addition mechanism.
Electrophilic because the bromine molecule acts as in electrophile once a dipole is induced in it.
Addition because the two reactants from one product.
Check out the next post (Hydrocarbons(11)) for what happens when this reaction is carried out in the presence of water.
Pages on the "Mole" and "Using the Mole" in chemical calculations are here
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