00:01
Hello students, hope you are doing great.
00:04
In this question, we will see bromination of trans -cinnamic acid.
00:08
Let us draw the structure of the trans -cinnamic acid first.
00:11
So here we have a double bond, here we have cooh group present away from us and hydrogen present towards us.
00:20
On the other carbon, we have the phenyl ring present towards us and hydrogen present away from us and this is trans -cinnamic acid.
00:35
Since this is undergoing bromination, this will react with the bromine molecule.
00:41
The pi electrons will attack on the bromine and here the bromide ion is removed.
00:46
Doing so, there is formation of a three -membered carbon ring like this.
00:51
So here we have br, br is a positive charge and the rest of the groups remain as it is.
00:58
So we have the carboxylic acid group away from us, hydrogen towards us, hydrogen away from us from the other carbon and the phenyl ring towards us.
01:10
Now in the next step, the bromide ion which left from here will act as a nucleophile and will attack on these two carbons.
01:19
If it attacks on this carbon, then we will have a different product and if it attacks on this carbon, then we will have different products.
01:28
Let this be number one and let this be number two.
01:32
Doing so, we will be synthesizing two different enantiomers.
01:37
Let us see the two enantiomers that are formed here.
01:40
So here we have br and br, rest of the groups are as it is...