00:01
Here in this question you have to decide the products formed by these nucleophilic substitution reactions and decide that whether the reaction is following as in one path or as in two path.
00:17
So first we are deciding it.
00:20
So in case of a, just see it is the tertiary alky alkyl halite.
00:25
In case of tertiary alkylchalite, tertiary carbonium ion is more stable.
00:30
That is why likely mechanism should be sn1.
00:39
Here it is primary alkali halide and stronger nucleophile.
00:47
When it is primary alkyl halide and stronger nucleophile, that means it should be sn2 mechanism.
00:59
Here this is again secondary alkyl halite but stronger nucleophile.
01:08
Secondary alkali halide and stronger nucleophile so it should be sn2 mechanism and here it is again secondary alkaliate but our nucleophilic reagent ch3oh is weaker and that is why it should be sn1 mechanism so accordingly we are deciding the product formation in the first case, sn1 mechanism that means the first step is carbocation is formed with the removal of this cl minus i.
02:04
So here it should be trigonal planar structure of the carbonium ion, even positive charge and cl minus is left.
02:25
Once this is formed in the second step products are formed.
02:35
Methyl alcohol can attack from both the side from here or chances of attack from the down is equal and by that the product which is formed is first product should be if the methyl group is coming here then it is och3 here and hydrogen is lost as proto.
03:16
It is the first product and the second product should be och3 downwards and two methyl will remain as such on the product also.
03:41
These are mirror images of each other and they are in ancientness.
03:54
In case of b where the reaction is taking place by s &2 mechanism because it is primary alkaline halide...