What paper is this ?
For the first one, an OH- group attacks the carbon attached to the bromine, the bonding pair of electrons between c-br is donated to the br making br- ion, and simultaneously a lone pair of electrons from the OH- is donated to the carbon making a carbocation
This is heterolytic fission
Search sn1 / sn2 reactions on youtube, look for the khan academy videos
Last edited by s.aley; 22-05-2012 at 01:55.
(Original post by sabre2th1)
For the first reactive intermediate, the markscheme gives the following answer:
I understand what you are trying to say, but I am confused as to how I would draw the intermediate
Regarding the second intermediate, how many methyl groups have to be present in order to prevent an attack on a Carbon atom? There are 3 methyl groups in this example, but what if there were 2? Would they still prevent an attack (and therefore lead to a carbocation forming first?)
> Draw the carbon with the bromine atom as the central atom(I'll call it the central atom for ease of understanding, however, strictly speaking it is not the central atom).
> Draw all the other CH2 or 3 groups on one side of the central C atom and all the hydrogen connected to the central C atom on the other side of it
> Draw the Br on another side of the Central C atom and the OH group on the opposite side to it.
For example, when 1-bromopropane reacts with OH- ions the following reactive intermediate forms before propan-1-al is formed:
As for the second question as to how many methyl groups need to be present to prevent attack from nucleophile:
> If it is a tertiary halogenoalkane (3 methyl/ethyl etc groups around the central C atom) then it will always be SN1 mechanism
> If it is a secondary halogenoalkane (2 methyl/ethyl etc groups around the central C atom) then it might form either SN1 or SN2 based on its structure and the conditions such as the solvent used.
-For example if it is a larger molecule it would prefer SN1 over SN2 since there is a greater repulsion from the larger chain on carbon atoms on one side.
-If a polar solvent such as water or ethanol is used SN1 mechanism is favored.
Why is that so?
- The other reason why the formation of carbocation may be favored is because alkyl groups bonded to the central carbon atom tends to stabilize the charge on the carbocation, so the greater number of alkyl groups, the greater the stability of the carbocation, thus the stability is tertiary>secondary>primary
- Polar solvents also help stabilize the charge on the carbocation, thus its formation is favored...
If there is any other confusion then please don't hesitate to ask.
Last edited by Waseef; 22-05-2012 at 14:28.