00:01
Alkyl halides in the presence of strong bases undergo elimination reactions.
00:05
Consider one bromohexane.
00:12
Consider one bromohetane whose structure is in the presence of sodium ethoxide and ethyl alcohol.
00:29
It forms one hexene as a major product where it undergoes e2 elimination because of the strong base, that is sodium methoxy.
00:48
Consider two chlorohexene in the presence of sodium methoxy and methanol to produce two hexene.
01:05
This is the major product where it undergoes elimination because of the presence of sodium metoxide strong base.
01:14
The product formed alkin is formed at the more substituted carbon.
01:19
So this is the e1 elimination.
01:22
Consider 2 -thyl -butane in the presence of sodium ethyl -o -ethyl and ethyl alcohol we obtain a product which is 2 -ethyl to beotene.
01:49
This is the major product.
01:52
The base sodium ethyl is also a strong base...