July 14, 2024, 09:52:52 PM
Forum Rules: Read This Before Posting

Topic: Polysaccharides modification  (Read 871 times)

0 Members and 1 Guest are viewing this topic.

Offline marsio95

  • New Member
  • **
  • Posts: 5
  • Mole Snacks: +0/-0
Polysaccharides modification
« on: February 01, 2022, 11:41:42 AM »
I'm trying to perform the following synthesis that I found on a paper : O-palmitoyl
"pullulan (OPP) is prepared by a method described
earlier [24] by Sunamoto et al. (1992). Briefly, 1g
pullulan is dissolved in 11 ml of dry dimethyl
formamide at 60°C. To the resulting solution, 1 ml
dry pyridine and 0.1 g palmitoyl chloride, dissolved
in 0.24 ml dry dimethyl formamide are added. The
mixture is stirred at 60°C for 2 hours followed by
1hatroom temperature. This mixture is then slowly
poured into 70ml absolute ethanol under stirring.
The precipitate so formed is collected and ished
with 80 ml absolute ethanol and 60 ml dry diethyl
ether. The white solid material obtained is dried in
vacuum at 50°C for 2 h".
I tried the same reaction several times (pure DMF not dry and lauryl chloride instead of palmitoyl chloride), adjusting the concentration of the polymer, or mixing it for more than 3 h (I actually went for 8, 16 and 24 h) but every time I got the same result that is two fractions: one soluble and another which does not dissolve in water and I don't really understand why. I thought that the concentration of the polymer was too high giving a viscous solution which in turns caused a poor diffusion of the acyl chloride but even under more dilute conditions nothing changes. Have you got any idea?

Offline wildfyr

  • Global Moderator
  • Sr. Member
  • ***
  • Posts: 1774
  • Mole Snacks: +203/-10
Re: Polysaccharides modification
« Reply #1 on: February 01, 2022, 06:23:31 PM »
When instructions say 3 times about dry reagents and you don't use a dry solvent, thats usually a problem. Acyl chlorides are quite water sensitive.

Offline marsio95

  • New Member
  • **
  • Posts: 5
  • Mole Snacks: +0/-0
Re: Polysaccharides modification
« Reply #2 on: February 03, 2022, 09:08:59 AM »
I tried this morning the same procedure with all dry solvents and again the same result. Two fractions one soluble another insoluble. NMR suggests that the insoluble fraction has a high content of lauryl groups while the soluble one has  a small content (the one I am expecting to find based on the reactants molar ratio). Any tips?  ??? ??? ??? ???

Sponsored Links