00:01
Hello students here given by question in which we have the reactant which is having the carbonyl compounds that is ketone and it is reacting with the ns2 -ns2 -cho -h.
00:11
So this will reduce the ketone and ketone will convert it into the hydrocarbons.
00:20
So now we have to find the product.
00:23
So the product is we have this six -famber drink than another six -fibon.
00:30
Member ring is used here and then again there is a six member ring which is used and then the five member ring fuse with this six member ring then we have the double bond and then is 3 and on this carbon we have o t double bond o then then is 3 so this is the final product of the first part of the question now we have the part of the question in which we have to find the product that is having the h plus ion.
01:06
So now in the presence of the h plus ion the product form will be the five fambard ether ring and it is attached here with the tho so this is the product here then we move to the next part of the question so in this we have the anzaldehyde which is reacting with the secondary amine.
01:33
So now this secondary amine will have the bone pair of the electrons on the nitrogen which will attack on the carbonyl center to form the product.
01:45
So now after attacking the nucleophile on the carbonyl center then the product will be we have the benzene ring and then c h double bond and then five member ring is there and then there is a positive charge.
02:07
So this is the final product of the reaction.
02:11
Then we move to the next part of the question.
02:16
So in this we have this ethyl bromide which is reacting with these reagents and we have to find the product.
02:25
So when firstly the ethyl bromide reacts with the alcoholic chaos then there is a formation...