00:01
Hello students, we have been given two substrates in this reaction and we have to form the major product from them.
00:10
Now we have been given an naoh base.
00:17
So, this will liberate hydroxyl ions.
00:20
This proton is most acidic but this is hindered because of these functional groups.
00:26
This is also a methyl group which is hindering it.
00:29
So, proton will be abstracted from this methyl group.
00:32
Hence, we will get a negative charge generated on this carbon atom like this.
00:42
Now, this will attack on the molecule that contains these two carbonyls.
00:53
So, this will attack on this carbon leaving this double bond towards oxygen.
01:00
So, we will get a molecule which will contain one o - and a ch2 group on which there is a carbonyl and a group like this.
01:15
Now, we can show its hydrolysis because when base abstracted this proton it becomes water.
01:26
So, water will be available here which will donate its proton back to the o-.
01:31
Now, we will get a hydroxyl group here and here we have a ch2 c double bond o and this pi bond.
01:45
Now, since presence of base is still there and we are heating this, this will undergo a e2 type of elimination.
01:58
So, one proton will be abstracted from the alpha position and this hydroxyl group will be removed as a water molecule.
02:09
So, the molecule that will be formed will appear like this.
02:17
It will contain a double bond here and here we have our carbonyl on which there is an alpha beta double bond as well...