In the below reaction the first starting material have two amine groups. In that which amine group will easily react with aldehyde to form schiff base?
Reaction Concentration :- Eq.Mol Concentration
Both of the two NH2 groups can react with the carbonyl of the aldehyde. The F atom is a very small one so there will be an easy access to the ortho position . Had there been a much larger atom (e.g. Br), the para position will be an easier destination for the reaction.
I may add a 3rd product to what you expect: there is possibility of having 2 imine groups attached together to the aromatic ring . Therefore, you may need to isolate the 3 products in your research and then determine the % yield of each.
Both of the two NH2 groups can react with the carbonyl of the aldehyde. The F atom is a very small one so there will be an easy access to the ortho position . Had there been a much larger atom (e.g. Br), the para position will be an easier destination for the reaction.
I may add a 3rd product to what you expect: there is possibility of having 2 imine groups attached together to the aromatic ring . Therefore, you may need to isolate the 3 products in your research and then determine the % yield of each.
They are pretty similar. So like Nizar said above I suspect you will get a mixture. Its possible the one closer to the fluorine will be less reactive due to fluorines electron withdrawing, but I doubt it’ll have that great of an effect.
I did a similar reaction where I was trying to methylate one of two phenols on an aromatic aldehyde. I ended up getting 3 products: starting material, monomethylated (my desired product), and dimethylated. The separation was very easy as I suspect it will be for you.
You will obtain a mixture of both isomers. Product 1 may be a major one in small excess approx.58-65%1 and 35-42% of 2) bacause of the steric factors and orto sigma-withdrawing effect of F-atom. To sepatate them chromatography is the only way.
Two isomers are possible. You will have a difficulties to separate them, due to the similar chemical structures, but separation is not impossible.
I believe that product 1 will be by far major. and if the addition of aldheyde is slow, or its concentration is low when compared with the amine u may get even better selectivity.
Good luck
I also believe both the products are likely to be formed with minor variations in quantities. And formation of the di-imine in small quantity can also not be ruled out.
Please note that one of the two components, that is, the aldehyde is, in general, quite reactive.
Both options are likely to forme Product 1 may be a major one in small excess approx. Because the two NH2 groups can react with the carbonyl of the aldehyde
As Dr. Nizar mentioned,don't forget to consider the formation of a third product resulting from the double Shiff base reactions on both amino groups.
The amine group in P- position (1) is easier than O- position
because of high inductive effect on O- group by F therefor O- amine less basic
Fluorine has the higher electronegativity in the periodic table and withdraws the elections; this will make conjugation in the double bond. Finally the lone pair on the nitrogen of amine group No. 2 will be not free just like to the nitrogen of the amine group No. 1. Therefor the major product will be compound No. 1 and the byproduct is the compound No. 2
The amine group meta- w.r.t. fluorine group reacts with aldehyde to form Schiff base with faster rate.
Three product are possible but he major product will be compound No. 1
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