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Author: Subject: The Short Questions Thread (4)
Sulaiman
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[*] posted on 21-6-2022 at 22:03


Quote: Originally posted by mayko  
I tried to grow some rochelle's salt crystals.

The salt is deliquescent so I GUESS that some sort of dessicator is required.

Edit : just checked.. <30% RH to dessicate.

Also, decomposes above 55C

[Edited on 22-6-2022 by Sulaiman]




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[*] posted on 6-7-2022 at 08:16


Quote: Originally posted by AvBaeyer  
Dry, very dry.

AvB


thank you , but tell me this in many of the articles read and videos viewed , the topic of the reaction being anhydrous only shows up when using NaH as a base but otherewise nothing in the use of the bases methoxide or ethoxide....why is that? Is it something one should know from the start .....I had my sample of 4-amina acetophenone analysed , and it turns out to be benzocaine and benzoic acid.....is it that the amine reacted with the enol or the base....quite confused ....I'm currently on test #13 and the same results even though i introduce the ethyl ethoxide last slowly with an excess of benzocaine ....but the results are the same once i add the HCl to keep the reaction to the right.....then heated to decarboxylate.......any hints , and yes i've run the reaction with anhydrous reagents, but the results seems to be the same.......solo




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mayko
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[*] posted on 6-7-2022 at 20:30


Quote: Originally posted by Sulaiman  
Quote: Originally posted by mayko  
I tried to grow some rochelle's salt crystals.

The salt is deliquescent so I GUESS that some sort of dessicator is required.

Edit : just checked.. <30% RH to dessicate.

Also, decomposes above 55C

[Edited on 22-6-2022 by Sulaiman]


Where are you getting this? I've never heard of any special drying apparatus being necessary. Wikipedia gives a melting point of 75C, a dehydration point of 130C, and a decomposition point of 220C so I don't feel too bad about heating the aqueous solutions.

I poked and prodded the syrup, dripped ethanol and acetone on the surface, smeared a thin layer on the side of the beaker, left a toothpick in... nothing. I would have thrown it out except I got lazy cleaning and then after many weeks of sitting I thought I saw mold in it... no, the syrup was crystallizing in a couple points! Within a day or so it was solid throughout. I saved a bit, redissolved the crystallized mass, and then used the residue as a seed. I now have a jumble of fat needles, not really the blocks I expect but I'm going to use them to reseed another round and see how that goes.




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clearly_not_atara
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[*] posted on 7-7-2022 at 07:46


It's on Wikipedia.

Quote:
Rochelle salt crystals will begin to dehydrate when the relative humidity drops to about 30 per cent and will begin to dissolve at relative humidities above 84 per cent.




[Edited on 04-20-1969 by clearly_not_atara]
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mayko
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[*] posted on 7-7-2022 at 11:04


"dehydrate" in that passage refers to the loss of water of crystallization, not crystallization from solution.



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[*] posted on 8-7-2022 at 19:09


The benzylic bromination of 2-methylnaphthalene, using NBS, is ideally done in presence of a radical initiator, benzoyl peroxide is usually used for that purpose.
With such a radical initiator, I could use a number of solvents, as the benzylic bromination will in that case be the main- and not side reaction.

Tried it once without, solventless, 20h in a melt, and the yield was(side reaction aromatic bromination) just a third of the theoretical yield.

So, here is the question: can I use the 2,4-dichlorobenzoylperoxide(50% under some silicon oil, because explosive :o) as alternative radical initiator?
I expect the answer to be "of course", but I figured it won't hurt to ask.




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[*] posted on 9-7-2022 at 09:56


I'm making 5-formyl vanillyl and after reacting vanillyl and hexamine, and adding HCl I dumped the reaction on ice. The reaction contains 55 grams water, 55 grams acetic acid, 50 ml 30% HCl, and about 50 grams organics. I dumped this hot mixture on 400 grams of ice of approximately 0 degrees.

Now what I didn't expect was this mixture to go below 0 degrees, but it went all the way down to -7 degrees. Why is that? What endothermic process is going on here?
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[*] posted on 9-7-2022 at 10:20


Quote: Originally posted by Tsjerk  
I'm making 5-formyl vanillyl and after reacting vanillyl and hexamine, and adding HCl I dumped the reaction on ice. The reaction contains 55 grams water, 55 grams acetic acid, 50 ml 30% HCl, and about 50 grams organics. I dumped this hot mixture on 400 grams of ice of approximately 0 degrees.

Now what I didn't expect was this mixture to go below 0 degrees, but it went all the way down to -7 degrees. Why is that? What endothermic process is going on here?

Ice melting is endothermic. Because you've got a lot of acid, that lowers the melting point just like salt would.




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[*] posted on 28-7-2022 at 20:06


when nitrosyl chloride reacts with limonene, why does it add only to the cyclohexene double bond, and not to the 2-propene side chain?



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[*] posted on 28-7-2022 at 22:55


Quote: Originally posted by Tsjerk  
... adding HCl I dumped the reaction on ice... what I didn't expect was this mixture to go below 0 degrees, but it went all the way down to -7 degrees. Why is that? What endothermic process is going on here?
IIRC the mixture of ice and concentrated HCl solution produces quite low temperatures, you diluted HCl in other reagents but anyway you still observed the endothermic process



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[*] posted on 28-7-2022 at 23:01


Quote: Originally posted by mayko  
when nitrosyl chloride reacts with limonene, why does it add only to the cyclohexene double bond, and not to the 2-propene side chain?
Mayko that's true, but I do not know why. I'm not too much skilled in theory, I like practical chemistry more. Anyway it is nice experiment https://www.sciencemadness.org/whisper/viewthread.php?tid=15...



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[*] posted on 29-7-2022 at 05:46


Yep, I've been wanting to try the limonene to carvone conversion for a while, finally getting set up and want to understand more about the mechanisms



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[*] posted on 18-8-2022 at 05:39


i want to nitrate this compound....being that the group is large and its an o,p directional i will get the p position.....my concern is if there will be any reaction with my side chain exposed to the sulfuric and nitric acid used to place the nitro group....i.e esterification of the alcohol function ...and it might affect the direction of the nitro placement...?........solo

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[*] posted on 18-8-2022 at 09:14


I guess you will eliminate water forming an alkene before forming an ester. I wouldn't worry about forming an ester as it would probably not effect direction of the nitration too much and it would be reversible.

[Edited on 18-8-2022 by Tsjerk]
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[*] posted on 18-8-2022 at 10:14


......thanks i though of making the nitronium ion first with sulfuric acid and nitric acid then add to my compound with some cooling...solo




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[*] posted on 19-8-2022 at 00:36


Can BiCl3(aq) be made easily from Bi2O3? Do I need concentrated HCl or will dilute HCl do? I supposed I need an excess of the acid to prevent the hydrolysis of the BiCl3 in solution.

Bi2O3(s) + 6HCl(aq) = 2BiCl3(aq) + 3H2O

BiCl3 is difficult/expensive to get here, but Bi2O3 is cheap and easily available.

Thanks.
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[*] posted on 19-8-2022 at 03:53


I have never done said reaction but I know bismuth has a tendency to form oxychlorides. A cursory looks seems to indicate it's not as bad as zinc where the isolation of the anhydrous chloride itself is next to impossible but seems it might be difficult to do without that showing up as a contaminant.



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[*] posted on 19-8-2022 at 23:31


Quote: Originally posted by BromicAcid  
I have never done said reaction but I know bismuth has a tendency to form oxychlorides. A cursory looks seems to indicate it's not as bad as zinc where the isolation of the anhydrous chloride itself is next to impossible but seems it might be difficult to do without that showing up as a contaminant.


From what I read, the hydrolysis to form oxychlorides can be reversed/prevented by using an excess of acid.
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[*] posted on 27-8-2022 at 13:24


Quote: Originally posted by artemov  
Quote: Originally posted by BromicAcid  
I have never done said reaction but I know bismuth has a tendency to form oxychlorides. A cursory looks seems to indicate it's not as bad as zinc where the isolation of the anhydrous chloride itself is next to impossible but seems it might be difficult to do without that showing up as a contaminant.


From what I read, the hydrolysis to form oxychlorides can be reversed/prevented by using an excess of acid.


Yeah, that's correct. Dissolving in 1+1 acid is okay. Maybe even 1+2 would be good.

[Edited on 27-8-2022 by Bedlasky]
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[*] posted on 27-8-2022 at 14:51


The Bismuth question is related to its iodo-complexes discussed here.
I'm also interested in these complexes, but I only have Bi metal. I tried to dissolve it in ~16% HCl, but it did not dissolve at all, neither when boiled.
I then tried to dissolve the same piece in 53% nitric acid and it did dissolve, giving off NO as orange-brown fumes. The solution was neutralised with ammonia solution until pH paper showed the pH was > 9. A fine, slightly off-white precipitate was formed (would do a good job as a warm white pigment).
The precipitate was suction filtered on a piece of Whatman white band filter paper put on a coarse fritte. Approx. 10% of the precipitate ran through during the first filtration, but virtually all was filtered when the filtrate was run through the same filter paper again. The residue was rinsed by pouring water on it after the initial solution has run though (still under suction).

The residue was dissolved in the earlier used 16% HCl solution. Another 0.5ml 36% had to be added in order to dissolve all of the residue. Heating was needed.
Around 10% was spilled and when the spilled drops combined with drops of water in the sink, a milky liquid immediately formed. It is evident that in order to keep Bi in solution, an excess of strong acid is required.


[Edited on 27-8-2022 by Bezaleel]
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[*] posted on 27-8-2022 at 23:19


Quote: Originally posted by Bezaleel  

The residue was dissolved in the earlier used 16% HCl solution. Another 0.5ml 36% had to be added in order to dissolve all of the residue. Heating was needed.
Around 10% was spilled and when the spilled drops combined with drops of water in the sink, a milky liquid immediately formed. It is evident that in order to keep Bi in solution, an excess of strong acid is required.

[Edited on 27-8-2022 by Bezaleel]


Thanks Bedlasky.

Bezaleel:
Yes I am trying to make the red salt cesium pentaiodobismuthate.
This is the only color missing from my rainbow salt series. Hope it really turns out red! lol

Would just adding more 16% HCl dissolve all of the ppt, or is a strong acid (>30% HCL?) absolutely necessary?
I have some <20% HCl, but not >30%. Guess I might need to make a small amount.

I bought some cesium salt (expensive!) and Bi2O3. Waiting patiently for their delivery :)

Oh, I managed to make nickel chloride and lead acetate from nickel coins and lead shots using hydrogen peroxide and the respective acid (dilute HCl/freeze distilled vinegar). Not sure if it will work with bismuth. I dun have nitric acid and dun really want to deal with NO2.


[Edited on 28-8-2022 by artemov]
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[*] posted on 21-9-2022 at 16:44


Quote: Originally posted by artemov  
Would just adding more 16% HCl dissolve all of the ppt, or is a strong acid (>30% HCL?) absolutely necessary?
I have some <20% HCl, but not >30%. Guess I might need to make a small amount.

16% solution should be fine, but an excess of acid is necessary to keep Bi dissolved.
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[*] posted on 27-10-2022 at 20:11


....what can i substitute Sulfolane with? DMSO?........solo



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[*] posted on 11-11-2022 at 10:58


What is the easiest way to use Cl2 gas given off by an electrolysis cell? Would love to turn it into HCl



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[*] posted on 2-12-2022 at 05:07


What is a safe but difficult fractional distillation?

Where the 2 components are almost identical.
I've read that vinegar is difficult but has almost an 18c difference in bp and no azeotrope it doesn't seem that it should be hard. Running a liter now.
Edit:
when looking at the vapor liquid equilibrium i see the difficulty
Should be a good test

[Edited on 3-12-2022 by Rainwater]




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