Okay, the infamous talking Barbie doll didn’t really say that. But she should have.
So I’m toodling along in chemistry lab, having a great time, making excellent banana oil and doing very nice crystallizations (and why, oh why, do caffeine crystals have to come out so damn fluffy? I spent half the lab chasing caffeine around), and last night opened my lab manual to find that tomorrow morning I will actually be required to think in chemistry lab. Who said there’d be thinking?
We will be taking some quantity of benzaldehyde, adding KOH, and refluxing it for an hour. While it’s heating, we will have the joy and delight of outlining exactly how we will separate and purify the mixture of benzoate anion and benzyl alcohol that will result. Then the lab TA will give us the procedure for how we’ll REALLY do it. Since I screwed the pooch on the midterm, I’d kind of like my outline to actually be correct.
Now if I can acidify the benzoate and wash it out from the mixture in a separatory funnel with some carbonate or other, I’m a happy woman. I can dry and recrystallize the acid from the aqueous layer (I love crystallization), and perhaps distill the alcohol from the organic layer, provided the benzyl alcohol will stay in the organic layer. Am I barking
up the right tree at all? Would I need to add an organic solvent? Would the benzyl alcohol mind its own business and stay the heck out of the aqueous layer?
Other suggestions are welcome. It appears that this is not a small enough quantity that I can just say “Oh, I’ll just separate it in the gas chromatograph. Have a good day, Xuebin.”
It appears we’ll be given “more information on the experiment” at some point in the lab, but it’s not clear whether that will be AFTER we write our silly outline. It’s that sort of lab course.
Hi Denise,
I would think that to make benzoic acid from benzaldehyde you’d need an oxidizing agent, not a base. Hmmm… On the other hand, to get an alcohol, wouldn’t you need a reducing agent? Anyway, if you give more details about the reaction, maybe I’ll recognize it better.
What’s the solvent going to be? I can’t tell from your post if you’re starting with an organic or water solvent. I don’t remember refluxing anything myself except in water. An organic solvent could be ether, I guess, but you wouldn’t want to heat that up. Plus, if the solvent is organic, how would you get the KOH to react? But if your original solvent is water, which I’m guessing it must be, then to do any kind of extraction you’d have to add an organic solvent and move one or both of the products into that layer. If you are dealing with benzoate then I think you have to add some acid to protonate the anion and make it organic soluble. If both the products get moved to the organic layer, you could add a base and then extract the benzoate with water. I don’t know about the benzyl alcohol staying out of the aqueous layer–for this technique to work it better, but it might be kind of acidic which would be a problem. Maybe look up the pKa to get an idea?
Don’t they have different boiling points? Could you distill off the alcohol first and then crystallize the benzoic acid?
This is all off the top of my head, so take it with a grain of salt. I could be completely wrong. But good luck!
Also, look up the two products’ solubilities in water. Don’t bother with the pKa. See if one’s way more soluble than the other.
The reaction is the Cannizaro reaction, using a stoichiometric amount of OH-. The OH- substitutes onto the carbonyl carbon. Hydride transfer takes place between the substituted carbonyl and another benzaldehyde, forming the carboxylic acid, and the deprotonated alcohol. Since the acid is a stronger acid (duh) than the alcohol, the acid is deprotonated by the alcohol. Thus, benzoate anion and alcohol.
Well, my thought was protonate the benzoate with a strong acid, then wash it out with the water. The benzoate is obviously more acidic than the alcohol, otherwise the alcohol would be the anion from between the two. I couldn’t figure out what the organic solvent would be.
Finally it occurred to me that my husband, who took this class a couple of years ago but can’t remember what happened, must have his lab report.
So I made this much outline:
1 Add acid to protonate benzoate.
2 Extract benzyl alchol with X organic solvent.
3 Wash out acid, dry in rotary evaporator, recrystallize in water to purify. Dry in Buchner funnel.
4 Alcohol should be in the organic layer. Extract by fractional distillation. (I’m now thinking this is a bad idea because of the low flash point, but then, that’s the whole point of a condenser, right?)
The result from the lab report? Extract with ethyl ether. Acid is in organic layer, dry and recrystallize (yes!)
Alcohol is in organic layer. Dry with rotary evaporater (?!) and remove additional ether by blowing compressed air across the product. This part I don’t quite get, because the melting point of benzyl alcohol is quite low (-15C)and I’m expecting both the alcohol and the ether would be liquid. Then the alcohol was blown dry some more with compressed air (?!) and analyzed in the gas chromatograph.
This gives me enough of a sense of purpose to write my skeleton instructions for lab tomorrow. (I always outline the lab myself so I can include labeled blanks for what observations to record in terms of appearance of reactants, products, masses along the way, etc.)
A-HAH! I looked up the boiling points of the two liquids in the organic layer, and ether is 34.6C and benzyl alcohol, as I had determined earlier, is more than 200C. So I’m assuming that there’s a point where the rotary evaporator is evaporating like nobody’s business, and then it’s not, and you say to yourself, “The rest of this stuff must be alcohol, then.”
Sounds good. I didn’t know this reaction before–it’s not even in my orgo book (the Wade book). Well, the hydride transfer sounds neat.
So when you evaporate off the ether, is that like some quick and dirty way of separating two liquids, without bothering to do a distillation? Seems like it would work since they have such different BPs. Wow–this benzyl alcohol stuff sounds pretty impressive, being a liquid from -15C to 200C.
The only thing I don’t get is, why would you intentionally protonate the benzoate, if what you’re trying to do is move it into the aqueous layer (which I guess you’re not going to do after all, judging from the lab report. But anyway…). Since the anion is charged, wouldn’t it move into the aqueous layer faster and more completely if you didn’t protonate it? I take back what I said about adding a base though–the alcohol IS the base and you wouldn’t want to deprotonate that too.
Well, sounds like you’ve got it under control. Have fun in lab tomorrow!
I got to the lab and the TA told us we could do our own outline if we wanted and then there were procedures in the other room we should follow. Most people didn’t bother to guess. I’m glad I did; it really made me think.
You’re right, protonation before separation is futile.
It turns out we used ether as the solvent, washed the organic layer with sodium bisulfite to remove the remaining benzaldehyde, removed most of the ether with rotary evaporation, and I get to remove the rest with distillation. and then analayze by gas chromatography. Whee.
The acid washed out in the acqueous layer, THEN we protonated it to precipitate the acid, and dried it in a Buchner funnel with a couple of ice-water rinses. Next week I get to recrystallize it and determine the melting point.
This turned out to be fun.