I wonder if its possible to remove BzOH from BzCl without distillation?
I remember that CaCl will "absorb" (lower?) alcohols like water. Maybe this could work with BzOH?not_important - 25-6-2010 at 20:37
A) Short questions thread
B) What relative proportions? Makes a difference on what might work.
C) With a boiling point difference of 25 C, distillation is about the best way to separate them. Reduced pressure, Vigruex column - easiest for
moderate vacuum distillations and sufficient to give lean separation Good dry atmosphere practices help reduce the formation of the alcohol.
D) Treat it with HCl gas, which converts the alcohol to the chloride, then add hexane to force separation of the hydrochloric acid (HCl aq) formed,
remove hexane by distilling (oops) it off.
Alpha Centauri - 30-6-2010 at 20:09
It´s a reation Product from BzOH and aq. HCl, 33%. It contains 80% BzCl and probably 20% BzOH.
I know distillation is the best way but maybe the is another way to convert the remaining BzOH to BzCl - about 95% BzCl would be enough.smuv - 1-7-2010 at 00:26
Do you have any way of determining what the actual ratios are?
Just by looking at the suitabilities of the two in water, you can probably remove a good amount of the BnOH by dissolving the mixture in any available
hydrocarbon and washing repeatedly with water.Nicodem - 1-7-2010 at 01:46
It´s a reation Product from BzOH and aq. HCl, 33%. It contains 80% BzCl and probably 20% BzOH.
You can not get benzoyl chloride (BzCl) from benzoic acid (BzOH) this way! Surelly you mean benzyl chloride (BnCl) from benzyl alcohol (BnOH). Avoid
chemical shorthands if you don't know what they stand for.
If you got only an 80% conversion by using that method (how did you analyse it?), then you made something wrong. If using the method from Vogel the
conversion should be complete. Reuse that mixture in the same process just like not_important already suggested you and distil the product. Washing a
petroleum ether solution with water, kind of what Smuv suggested, will do little to nothing as the partition coefficient of benzyl alcohol does not
allow for much partition into the aqueous phase (logP for octanol/water is about 1.20).Alpha Centauri - 2-7-2010 at 15:57
The benzyl chloride content was determined by mixing the dry reaction product from BnOH / HCl with an ethanolic solution of sodium ethoxide. After a
little heating you will see clouds of NaCl. Filter, dry and weigh the NaCl and you can calculate the yield...
A 5x excess of 33% HCl was used and the yield was about 80% like reported at different sources. I have never read from a 100% conversation in this
reaction. Maybe with HCl (gas) or fuming HCl, 37% and ZnCl as booster the yields are close to 100%.
chucknorris - 20-10-2012 at 00:56
Hey, can you help me with this one?
I have product that contains about 50% benzyl alcohol and bunch of all kind of random shit and I need to get pure benzyl alcohol. Can I just vacuum
distill the fractionates and collect the calculated vacuum BP of benzyl alcohol and get pure, non-corrupted product?
