So I've been at it again, benzaldehyde from benzyl alcohol and dilute (around 15% this time) nitric acid. If anyone's interested in knowing, I got 70%
conversion with almost no side products from heating to 80C and gently shaking over the course of 48 hours.
But to the main point of this post, I've converted about 80g altogether of benzaldehyde to its bisulfite adduct by repeated shaking
with saturated sodium bisulfite solution, and washed it thoroughly with methanol and DCM to remove benzyl alcohol/benzyl nitrite. When I went to
regenerate free benzaldehyde from this sulfonic acid adduct, I only got full dissolution of the adduct and a light orange solution resulted.
I added my adduct to a solution of 90g of sodium bicarbonate dissolved in 700mL of water. This is an excess of sodium bicarbonate if we only need to
deprotonate the bisulfite ion in equilibrium, but does anybody know what exact conditions are necessary to free up the benzaldehyde and get a second
layer? I don't want to heat to drive off CO2 and raise the pH that way, since heat is unlikely to benefit my benzaldehyde's purity. I am hesitant to
increase the liquid volume further by adding something like saturated NaOH solution (I also don't have NaOH right now), but if necessary I can make
that happen. Any advice from the experts here?
aldehyde recovery
starlight_ - 8-11-2018 at 08:47
Suitable proportions of carbonate and water to adduct can be found in here:
Also watch out for strong alkalis like NaOH / solutions that are too concentrated or you get Cannizzaro reaction
[Edited on 8-11-2018 by starlight_]happyfooddance - 8-11-2018 at 09:30
700 ml is a lot of water to use for that amount of BzH (80g), you would be better off with a solution that is closer to saturated, and of Na2CO3, not
NaHCO3. You can convert a fair of the bicarbonate to carbonate by boiling the solution heavily, but unless you are using a reflux condenser you can
expect to lose a few grams of BzH to the atmosphere as it steam distills quite easily.
The adduct is fairly soluble in water, that is what you still have, the adduct dissolved in water. NaHCO3 is not a strong enough base to deprotonate
the adduct in any reasonable quantity.Amos - 8-11-2018 at 12:09
My proportions are almost exactly the same as the ones in the link starlight_ provided. I suppose I'll reflux the whole mixture and get back to you. happyfooddance - 8-11-2018 at 21:30
My proportions are almost exactly the same as the ones in the link starlight_ provided. I suppose I'll reflux the whole mixture and get back to you.
As for proportion I alway use about a 2x stoichiometric excess of a saturated solution of base. But in your case I believe the problem is that you
used the wrong base. In the link posted len1 uses sodium carbonate, not sodium bicarbonate as you said that you used.
Another tip that helps me significantly w/ BzH because of its close density to the water/salts mixture, is to use distilled water. Tap water with
calcium can form fine precipitates especially with carbonates. Shaking this in a funnel has always resulted in an emulsion forming for me.
Edit: You can alternatively add a stoichiometric eq. (in respect to NaHCO3) of NaOH to your mixture. The NaHCO3 will neutralize the NaOH and you will
be left with Na2CO3.
[Edited on 11-9-2018 by happyfooddance]SHADYCHASE54 - 9-5-2019 at 22:08
I have a quick question, would it be possible to calculate the relative purity of benzaldehyde through the formation of its bisufite adduct? If this
is possible and or has been covered in a previous thread would someone be so kind as to point me in the right direction. Or simply offer your
experience pertaining to this query. Thanks.
