I like to screw around with chemistry when I have nothing else to do, but I got stuck on this problem. I was wondering if anybody could help me out. The products are what I'm having trouble with C17H21NO • HCl + H2O + NaHCO3 = NaCl + HCO + C17H21NO I'm just wondering if those products would be correct. Originally I had H2CO3 instead of bicarbonate, but it didn't look right, and didn't really make sense that I would get carbonic acid as a result. Edit: I don't think any of my products are correct haha, someone please help.
Um, just looking at it I would be inclined to agree with your gut. From the look of the equation (assuming you knew to an extent what you were doing), your organic molecule is simply going to catalyze the reaction, while the water will simply be a solvent in which the reaction will occur. Other than that, a simple substitution reaction would leave NaCl and, yes, carbonic acid. If you aren't sure how the organic molecule comes into play though, if you can give me a structure I can tell you with more certainty how it will be involved in the reaction process.
You are correct in assuming that I do somewhat know what I'm doing. In all likelihood if I didn't know what I was doing I probably would have gotten myself hurt instead of played with equations before playing with chems Not sure if that's what you mean by the structure.
like i said it depends on reaction conditions too. are any of those reagents meant to be a solvent, or meant to be inert or a catalyst? BE MORE SPECIFIC
Well the H2O would be the solvent in which the reaction would occur, I was thinking heat would be the catalyst, the solution being brought to 100 degrees celsius. My goal is to strip the HCl ion from the C17H21NO • HCl compound leaving me with freebase C17H21NO as one of my products. My goal with the original equation (which I managed to have lost, as it was written on a letter envelope) that the NaHCO3 would neutralize the HCl and leave me with a sodium compound, either a hydroxide or a salt, an acid and the free base C17H21NO.
When you dissolve the acid-salt in the water it will strip it. From there you'll want to raise the pH to keep the acid-salt from reforming when the water is gone. I would probably use NaOH though instead of your sodium bicarbonate: water is going to be easier to get rid of in the end (H+ + OH- = H2O) than cabonic acid when you get down to separating your product, and I don't see anywhere right off that the hydroxide would react with on your diphenhydramine. After raising your pH to ~7, you're going to want to look up solubility data for your molecules and separate the DPH and NaCl accordingly.
Carbonate would go straight to the aqueous layer, anyway. But it's a weaker base, and it's effervescent, so it would be a bitch to use in place of NaOH. Here's what I'd do... and don't ask me why in gods name you are trying to obtain the freebase of diphenydramine, but if you want it, then do this: Take up the HCl salt in water, drop conc. NaOH into the solution and stir it very generously as you go. Measure the pH regularly, until it is greater than 9 (the pKa of diphenhydramine) and not much more than 10. Then just extract it with dichloromethane or another easily obtainable non-water miscible solvent. Remember that halogenated solvents are more dense than water. You can do this in a ziplock bag, assuming you don't have a separating funnel lying around in your house. Do it outside. If you get an emulsion, drop saturated salt water (brine) into the bag. If that doesn't help, remove the water + emulsion, but save the organic layer.. Take the emulsion + aqueous phase, dump in a lot more water, or brine, and then extract again with DCM.. rinse and repeat. Pool together your organic extracts, and proceed to evaporate them in a pyrex dish, once again, assuming you don't have anything better on hand. But man... If I were you, I'd really recommend this is done purely for academic interest, and not for putting that free base in your system.
Note, my memory of henderson hasselbach equations and dealing with weak acids/weak bases is long forgotten. So don't read too much into my calculation there, but I can assure you that at pH nine you will mop up all of the ionized diphenhydramine, which is what you want at the end of the day. If you want, you could look it up and calculate the exact pH at which no more ionized species would exist, which is quite possibly lower than 9. But I am pretty sure that wouldn't be necessary, as pH 9 isn't going to be a problem for that molecule. You definitely don't need to boil it up, or mess around with chemical equations, it's a simple acid/base extraction. Good luck.