00:01
Some reaction will be given which reaction will be correct i will be identified here so i take first that is here two methyl group and here bromine group will be connected and hydrogen group will be here and given that o cs3 negative and here shown that arrow that will be show and this will here and hydrogen will be here.
00:35
In this reaction we know very well the ocs3 will a base so a waste only and only abstract proton so as you can see here this will abstract br and hydrogen as a hydride ion leave here so this reaction does not take place so i will write here no reaction will will be take this.
01:06
Now i take second example that is this is given and also here hydrogen and here here we are and also a nucleophile or base that will here and arrow will shown here this will be removed.
01:32
As you can see that here the base abstract acidic proton and it will put our electron in anti -bonding to bromine so here e2 type mechanism will be operate in e2 type mechanism here anti -wonding electron will be removed so we are as a living group it will remove from here and alkin formation will be take place so this is our final answer then option b it will will be our correct answer because here e2 type mechanism will be operate to antito to each other.
02:19
So this is our correct answer.
02:23
Now i take second question that is in which option here carbocatin will be rearrangement.
02:35
So first this question and second will be like that molecule.
02:43
Third option, that is here 2 methyl and here 1 chlorine will be present and 4...