Question

NH ОН

Imagine these two products are dissolved in CH2Cl2. The solution is treated with 2-propanol/HCl until pH is approximately 7, resulting in the formation of crystals. The crystals are isolated using vacuum filtration. The sequence of actions described above results in the separation of the two products A and B. Why? How? Write a concise discussion.

Answer 1

OH Basic in nature Neutral at in nature When treated with protic acid, compound A undergoes protonation at benzylic nitrogen forming the salt as shown below. HCI salt of A is more polar than neutral coumpound A. The A.HCI is therefore insoluble in hydrophobic medium and separates out as an crystaline form. HCl in propanol | NHẠ*CH A.HCI Note that to maintain the hydrophobicity of the rection solution, the "HCl solution in 2-propanol" is used as a solvent instead of using the "HCl in methanol". The addition of methanolic HCl solution to the compound A in DCM may lead to solubilize the salt A.HCI The compound B is a neutral compound and therefore at pH 7 it does not undergo the protonation of any of the functional group (both the ester and alcohol). Therefore, Compound A remains in neutral form and soluble in the rection mixtute. HCl in propanol OH remains in neutral and soluble form Therefore, on filtration the A.HCl are collected as cystal and and filtrate on concentration gives the neutral B as the pure product