00:01
Here in this question, level the alpha -beta carbons in each alkalahillite and draw all possible elimination products formed when each alkyl halide is treated with this odysium tertiary butoxid, yes.
00:25
So in the first case, the alpha carbon is actually that carbon which is directly attest to halogen.
00:33
So this should be alpha carbon.
00:39
Here in b, the carbon which is directly attached to halogen is this one.
00:46
So this central carbon is alpha carbon.
00:51
And in case of c, it should be this carbon.
01:00
And the beta carbon is next to it.
01:08
Next to alpha is beta.
01:10
So beta should be this one.
01:14
In a in c c it is 1 is this one 3 beta carbon second should be this one and third should be this and in b compound which should be next is this one next to alpha is here also so this is also beta carbon and this one so 3 beta carbon present in b, 3 in c and only 1 beta carbon is present in a.
02:17
Accordingly we can make the elimination product...