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Hello students welcome back.
00:02
In this question i am given in the following instruction procedure choose the number where nicotine will be found.
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So the reaction given is this given this compound this is nephtholene okay and this is reacted with a compound known as nicotine so this is the compound here we have a nitrogen and a double compound here and here we have nitrogen and nitrogen.
00:44
So these are reacted in the presence of diethyl ether and then the product is reacted in the presence of 10 % hl and h2.
01:02
Then we have an aquas layer and we have an aqueous layer and we have an ether layer.
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Okay, now this ether layer, now we take, now we proceed the reaction to the adverse layer.
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Okay, so this adverse layer, there is no space where i'm writing it here.
01:26
So the adverse layer is done reactable 20 % of noh and h2o, okay? and this further produces an adverse layer and an ether layer.
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Okay so now we need to tell where the you know we need to tell where the nicotine would be reduced choose the number where the nicotine would be found so you know nicotine will be found on the first number plus the third number okay so if we check this is the if we check the numbers okay this is given to be one and this is this one evaporation gives us 2 and this on an evaporation give us 3...