00:01
Hello students, we have to complete some chemical reactions in this question.
00:06
We have been given a pantane molecule which we have to treat with bromine.
00:13
So, this will undergo a free radical substitution reaction.
00:18
Hence, will cause a mono -bromination.
00:21
Now, bromination in this molecule can take place at three different positions.
00:27
First, bromination can take place either on this carbon, on this carbon and on this carbon.
00:34
Based on the stability of free radicals that will be formed as an intermediate.
00:40
If bromination has taken place here, free radical must have been formed here.
00:45
Similarly, in the second product, free radical must have been formed here and in the third product, the free radical must have been formed on this carbon.
00:55
Now, if we look at the stability of free radicals, we can see that this free radical will have only two alpha hydrogens.
01:05
This free radical will have three alpha hydrogens here and two here.
01:10
This free radical will have two alpha hydrogens here and two alpha hydrogen here.
01:15
The most number of alpha hydrogens are present in this molecule.
01:20
So, this will be our major product which we will name as a.
01:24
This will be the major product and these other products will be termed as the minor products.
01:33
Now, this will be the minor product which we will name as c and this will be formed in a concentration which will be less than the minor and more than the major.
01:46
So, this will be named as b.
01:48
Now, let's move to the next part of question.
01:52
In this question, we have to treat our major product that was formed as a with alcoholic koh.
02:01
So, in this reaction, the koh will act as a base.
02:09
Now, base will cause a e2 type of elimination reaction.
02:15
We have two types of alpha hydrogen present in this molecule.
02:19
So, let's name them ha and hb.
02:23
If ha is removed, then alkene will be formed like this.
02:28
So, we will get a di -substituted alkene.
02:33
Now, if hb is removed, we will get a alkene formed here because reaction will take place like this.
02:45
Hence, we will get a mono -substituted alkene here.
02:51
Now, more substituted alkene that is known as the sex -f product is more stable.
02:59
So, this will be formed as a major product which we will name as e and this will be formed as a minor product that we will name as.
03:09
Now, let's move to the next part of question.
03:13
In next questions, we have to treat our minor products that were formed in the first reaction with alcoholic koh.
03:21
So, here we have this proton here.
03:25
So, this will again cause an e2 type of elimination reaction...