00:02
Hello students, hope you are doing great.
00:04
So here we are given 1 methyl cyclohexin which reacts with h2so4 in presence of methanol, chlorine gas, bromine gas in presence of water and bh3.
00:15
Let us see all the four reactions one by one.
00:18
So first, it reacts with h2s of 4 in presence of methanol.
00:28
So here we have cyclohex.
00:31
Is 1 methyl cyclohexene.
00:34
It reacts with h2s of 4.
00:38
From h2so4 we get a proton.
00:42
Here the electrons of the pi bond have an affinity for the proton which will attack on proton and we get ch3 and plus charge here.
00:56
So here preferentially this carbocatin only be formed because this is a tertiary carbocatin which is the most stable carboketine.
01:03
Now it will it will react with methanol.
01:06
And the ch3 -o -minus from this methanol will attack on the positive charge and we here will get the two products.
01:14
And the two products will be like this.
01:20
O -c -h -3 below, ch3 above the plane.
01:25
And the next product will be o -c -h -3 present above the plane and ch -3 present below the plane.
01:36
So this will be the two products that we will get.
01:39
Let us see the next reaction.
01:41
In next reaction we again have methyl cyclohexene which reacts with chlorine gas.
01:52
Here the electrons of the pi bond will attack on the chlorine and here cl minus will leave.
02:00
Doing so we get a three -membered ring.
02:04
Let us assume that the electrons are attacking from below the plane.
02:08
So if they are attacking from below the plane, the ch3 will automatically move to above the plane.
02:15
So here we have cl plus charge and three member drink.
02:20
Now here cl minus left we will attack which will attack here from above the plane because the cl plus ring with the with the with the same.
02:31
With the ben cyclohex and ring is present below the plane.
02:35
So we will get the final product like this...