00:01
Hello students, in this question we have to predict the product of this reaction.
00:04
First of all, the given acid reacts with the sodium ethoxide in the presence of ethanol reflux.
00:10
So, this sodium will abstract the chloride ion and the ethoxide ion will attack to this benzene ring and this would result in the formation of a ethoxide compound that can be written as o -et and c double bond o -oh.
00:26
This is the product of this reaction.
00:28
And coming to the next one, the alkane is reacted with a sodium ethoxide in the presence of ethanol.
00:35
The same here the sodium will abstract the br bromide ion so that it will be eliminated as nabr and the ethoxide ion will attach to this compound.
00:45
This would result in the formation of a so here will be a o -et group and here will be the hydrogen group and here will be the deuterium.
00:57
And coming to the next one, so here a cyclohexane, methyl cyclohexane is reacted with the bromine in the presence of dichloromethane.
01:06
So, this would result in the dibromination of this compound.
01:10
Therefore, the resulting compound will be substituted with the 2 bromine group along the double bond.
01:18
Here will be the bethene group and here will be the bromine group.
01:21
And coming to the next one, so 1 methyl cyclohexane is reacted with the bromine in the presence of etoh that is ethanol.
01:30
And this would result in the formation of a monobromination of this compound...