00:01
In this video, we are focused on writing the reactions as per the given information.
00:05
So, if we talk about the step one, then we can here say that it is talking about converting one bromobutane that will look like this.
00:15
So here we are going to have bromine that is we can say bromine substitute.
00:20
And then if it reacts with m .g in the presence of dry ether, then it will be here giving us basically grignard reagent which will look like this and here we are going to have mgbr.
00:36
So we can say that it is our grignard reagent.
00:40
So let us write it and it is it will look like this and we can put this entire reaction inside a box because this will be our answer for the first step and it will look like this.
00:54
So now let us see how we can find out the answer for the second step.
01:00
So let us write over here that in the step second, we are basically having that reaction of grignard reagent with the ketone to form alcoxide salt.
01:12
So here it will be ketone.
01:14
Firstly, we are basically drawing the ketone structure.
01:18
And here there will be charge development that is delta plus.
01:22
And here there will be charge development delta minus.
01:24
Now if this is our ketone, then it is here going to react with grignard reagent and grignard reagent will look like this as we have drawn over.
01:37
Now this grignard reagent is having the nucleophilic site and nucleophilic side is this one because the grignard reagent is a nucleophile, that is it is electron rich and now here this ketone is having basically we can say electrophilixite and this one is the electrophilixite.
02:00
Electrophile means electron deficient and here nucleophile means electron rich species.
02:07
Hence the reaction will look like this, that is it is here going to attack over here and there will be bond transfer like this and now it can be here drawn the final product which is basically going to be, let us draw over here, it will be here r and then it will be here c and then here it is o minus and then it will be here ch2.
02:33
Now it is here ch2 and now it is here ch2 and then it will be here ch3 and hence here we will be having one more r.
02:43
So this is basically our alcoxide salt.
02:48
So we can write it as as alcoxide and then here we can mention that this is our salt.
02:56
So now based on this very information, we can say that we have drawn or written the answer for the second part.
03:04
Now it is better if we leave it as such without putting it into our bracket because then it will create confusion.
03:10
So this is our answer for the step two and then let us proceed for the step three.
03:15
So in the step three we can say that we need to show the step two.
03:20
The protonation of the alcoxide salt using ammonium chloride.
03:25
So here we can say that r and then it will be here like this, that is o minus, and then here we are having r, and then here we are having this kind of structure, and then here it will be h, and then it is here n, and now it will be here h, and then it is here cl minus, and then there will be h and now it will be here one more h.
03:52
So here as we know that there is charge development of minus here and plus here and this will act as a proton source.
04:01
So here we can say that this will attack over here on this very proton and this bond will go like this and hence we can see that the final product in this very reaction is going to form.
04:15
So basically ammonium chloride here is acting as an acid and this alcoxide salt is acting as a base...