Synthesis of Benzimidazole involves reaction of NH2 group of O Phenylenediamine and COOH group of the other compound, I am not an organic chemist and want to know why some people have used 4N Hcl in the start of the reaction and why others have used 10% NaOH?
I have tried the same method given in some of the papers but i am not getting any precipitate after reaction..dose means my product has not formed? ( I have a different reactants like o phenylenediamine and glycine)
Sorry to ask a basic question but I am really interested to learn organic chemistry.
Some of the papers i have referred:
1) synthesis and characterization of some new shiff bases
2) synthesis and analgesic activity of novel derivativesof 1,2-substitutedbenzimidazole
3) synthesis and characterization of novel benzimidazole derivatives
In fact, benzimidazole cyclization was built from 2 steps reaction, at first, the NH2 group was reacted with COOH group to form an amide, CONH2 group. Then, the second NH2 group form imine with the carbonyl of CONH2. Both cyclizations can be occurred either in acidic or alkaline solution, depending on the solubility of your starting material, the possibility for undesired reaction from other functional groups and so on.
Some organic compounds were not spontaneously precipitated after the reaction, so it doesn't mean that you don't get your product. Another treatment is required, such as evaporation, neutralization, recrystallization, etc. depends on the physicochemical properties of the product. Good luck.
I imagine you are using aqueous solution of NaOH or HCl, therefore after reaction is complete you can simply extract your product with dichloromethane or chloroform (benzimidazoles are usually nicely soluble in chlorinated solvents), then evaporate your solvent and recrystalize your crude product from, for instance, diethyl ether (benzimidazoles are usually poorly soluble in ethers but it also depends on substituents) to avoid preparative chromatography.
Remember you can follow your reaction very easily by TLC, generally speaking it is a very easy and necessary to learn technique if you want to be sure if reaction is over, especially if you don't have access to NMR or GC. Good luck!
In this type of reaction either an acid or a base can act as a catalyst
The key step is attack of the amine, RNH2, at the carbonyl carbon, RXC=O, to form and amide, RN(H)-C(O)R, ending with loss of HX (such as H-OH if you are starting with a carboxylic acid RC(O)X, where X=OH).
In the case of addition of an acid:
The carbonyl can become protonated (RXC=O+-H) and this cationic species is more easily attacked by the amine.
Essentially the protonated transition state RN+H2-(RX)C-OH is more stable than RN+H2-(RX)C-O-
In the case of addition of a base:
The amine can also be deprotonated (RN-H) and this anionic species more easily attacks the carbonyl compound. (Or an intermediate can be deprotonated lowering the energy of the transition state en route to the amide)
The decision to choose between these (as mentioned above) can be to do with solubility, and the particular reagents involved in the reaction.
i my point of view, it is not basic question this reaction may have tricky point specifically in how to get rid of solvent correctly and how to purify the final product. for the reaction between glycin and Ortho phenylene diamine (as basic example) in 4 N HCL solution with heating for 2-3 h at 70-80 C, the mixture neutralized using alkaline solution of NaOH or even NH4OH til neutralization , then the reaction concentrated using Rotary evaporator ,the product obtained as buff ppt with melting point (not corrected) near 230 C .