00:01
Hello students, hope you are doing great.
00:04
In this question we are given few reactions and we have to find out the conjugate acid and base formed in this particular reaction.
00:11
The first reaction we are given here is ch₃oh reacts with oh-, this oh - takes proton from the alcohol and there is formation of ch₃o - plus h₂o.
00:27
Since this compound gave proton, so this is acting as a acid over here, therefore here it will form the conjugate base and water is acting as a conjugate acid here and we can see that this group, the conjugate base is a very strong conjugate base, therefore it will readily take proton from the water molecule and it will again form the alcohol.
00:52
Therefore, the backward reaction is favored here.
00:55
Therefore, here we can say that equilibrium shifts to left.
01:05
The next reaction we are given here is this.
01:10
Here we have a ketone, cyclic ketone which reacts with nah, this h - will take a proton from here as this is an acidic proton as it is present right next, it is bonded to the carbon which is right next to the carbonyl group and carbonyl group is an electron withdrawing group, therefore this proton is the acidic.
01:34
Since it is an acidic proton, therefore it will form a conjugate base, so here we have the conjugate base and na plus acts as a conjugate acid over here...