00:01
We have given a benzene ring, let me draw that and the conjugated double bonds.
00:18
Now, we have a substitution on it.
00:22
We have two oxygen atoms.
00:27
One oxygen carries hydrogen atom, okay? and another oxygen atom carries nothing but the lone pair of electrons like another oxygen atom.
00:40
But here i need a correction, okay? we need a correction.
00:44
The correction is we have given in the question, in the figure, two lone pairs of electron, okay, but oxygen is from group 16, six valence electrons are there, right.
00:57
So, now, this means to stabilize it, we should have a double bond here actually, okay.
01:08
Otherwise, if there is single bond, then there should be three lone pair of electron and a negative charge, okay.
01:16
Okay, this is my concern is, okay, either it should have a pair of electron and a negative charge, then this compound is stabilized or either it should have another bond also.
01:37
So, now it is stabilized.
01:38
Okay, so now i am considering that we have a pair of electron and a negative charge here.
01:47
Okay now moving ahead so we have given na positive and we have given oh negative okay so this is oh negative why because the negative charge is on oxygen okay so actually this is naoh sodium hydroxide okay but since this is cation this is anion so they are shown separately okay now so so what happens that this oxygen is electron rich, okay, this is electron negative and this is electron rich.
02:25
Being electron rich, it will act as a nucleophile, okay, it will donate electron.
02:35
So this oh negative is here nucleophile.
02:41
And now it will be in search of a atom or species which is electron deficient.
02:49
Now consider this hydrogen atom.
02:50
So, this is bonded to electronegative oxygen atom being electronegative it has tendency to withdraw the bond pair of electron towards itself.
02:58
So, it will have partial negative charge and this hydrogen will have partial positive charge all its electron density is being withdrawn and this is bonded to electronegative atom ok.
03:09
So, that's why this part will act as a electrophile this hydrogen ok with partial positive charge.
03:19
So that's why the approach takes place okay this nucleophilic oh negative approaches this hydrogen and the bond breaking takes place okay and that's why now we have a product so this is benzene ring then single bond oxygen lone pair of electrons negative charge another oxygen atom so we have the two lone pairs okay and then since the bond is being broken okay so now i have a pair of electron and negative charge and we have here this h2o molecule okay so this can be represented as like this okay so now of course here i have this na na positive ion.
04:33
Okay.
04:34
So, this na positive ion which i am talking about.
04:37
So, actually this being having positive charge...