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Author: Subject: The Short Questions Thread (4)
Etaoin Shrdlu
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[*] posted on 4-5-2014 at 19:26


Quote: Originally posted by Cheddite Cheese  
I know the Barbier reaction is usually used for allylation. Can it also work for benzylation?

[Edited on 5-5-2014 by Cheddite Cheese]

Apparently. http://www.arkat-usa.org/get-file/34424/
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[*] posted on 5-5-2014 at 13:14


Quote: Originally posted by Brain&Force  
Anyone know any strontium/lithium or any other red-burning salt that is soluble in methanol?


I'll answer my own question: lithium nitrate. Though I'm not sure if it'll burn nicely; I haven't tried it yet.

[edit] it does!

[Edited on 2014-5-6 by Töilet Plünger]




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[*] posted on 7-5-2014 at 21:43


Solubility Data: (NH4)2SO4 in HNO3 / H2SO4?
Exactly what it says - is ammonium sulfate soluble to any appreciable degree in either of these acids, at low temperatures (assume 0 to 20 degrees C)?




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[*] posted on 8-5-2014 at 04:15


I imagine they would be soluble to a limited degree. Nitrating baths can be made of NH4NO3 and H2SO4, and the NH4NO3 dissolves into the sulfuric acid to an appreciable extent to create nitric acid. This is a very similar situation.
I don't know the precisely how soluble they would be, but I imagine you should be able to get a 5-10% soln fairly easily




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[*] posted on 8-5-2014 at 08:32
Soxhlet extractor question


I´m buying soxlet extractor with volume of 150 ml and I was looking for extraction thimbles.. They are not cheap at all and what I found is package of 25 pieces with size 34mm in external diameter and lenght 100mm for 35 pounds with shipping.. my question is if it would fit in and work ? Does anyone know? Or shall I rather look for 22x80 thimbles? :o



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Etaoin Shrdlu
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[*] posted on 8-5-2014 at 10:00


Give the supplier your dimensions and ask them.
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[*] posted on 8-5-2014 at 10:21


Quote: Originally posted by elementcollector1  
Solubility Data: (NH4)2SO4 in HNO3 / H2SO4?
Exactly what it says - is ammonium sulfate soluble to any appreciable degree in either of these acids, at low temperatures (assume 0 to 20 degrees C)?

http://www.uea.ac.uk/~e770/aim/model2/descr2.htm




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[*] posted on 9-5-2014 at 05:39


.......when halogenating an aldehyde with the alpha carbon without any free hydrogens, if there is al alkene adjacent to the alpha carbon will the halogenation favor the aldehyde or the alkene or both.....solo


Note- March 5th edition page 914 sectsion 14-3, talks about the halogenation of aldehydes with the alpha carbon abscent of any hydrogens

[Edited on 9-5-2014 by solo]




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[*] posted on 9-5-2014 at 11:12


If there are no alpha hydrogens, halogenation will not occur (what atom would the halogen replace?). I think the halogen would just add across the double bond.



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[*] posted on 9-5-2014 at 12:39


.......the hydrogen on the aldehyde...as stated on the reference provided......will get halogenated, but my question is will the alkene be halogenate as well or just the aldehyde site.....solo

[Edited on 9-5-2014 by solo]




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[*] posted on 9-5-2014 at 15:02


So the reaction forms an acyl halide? Very interesting . . .



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[*] posted on 9-5-2014 at 15:22


.....well, if it halogenates an aldehyde with no alpha hydrogen it would be great...if only the alkene attached to the alpha carbon is not halogenated with this particular type of aldehyde.....one can do many things with that .....hence the reason for my inquiry....solo

[Edited on 10-5-2014 by solo]




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[*] posted on 10-5-2014 at 13:53


Quote: Originally posted by solo  

Note- March 5th edition page 914 section 14-3


edition of what?

Also, what would prevent the alkene from being halogenated?

Can you be more specific about conditions, substrate, etc.?




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[*] posted on 13-5-2014 at 19:33


I mixed potassium nitrate in methanol. It doesn't appear to dissolve, but the methanol turns blue. I've also had this happen with a sample of methanol sitting in a low-oxygen environment with just a match in it (there weren't any nitrates in the methanol). What causes this?



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[*] posted on 13-5-2014 at 20:27


I have recently made what I believe to be copper nitrite, by the addition of copper sulfate to a small excess sodium nitrite. The solution is dark green, similar to concentrated copper chloride. I wish to isolate the copper nitrite, but am unsure about how to go about it. Does anyone have any information on copper nitrite? Or any ideas on how to separate it?



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[*] posted on 13-5-2014 at 20:48


You just have a mixture of copper, sodium, sulfate, and nitrite ions now.
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[*] posted on 13-5-2014 at 20:56


Yes I know. However this doesn't answer any of my questions. I have looked in the Merck index and in the CRC, and neither contain anything on copper nitrite.



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[*] posted on 14-5-2014 at 10:12


Sorry, let me try again. You haven't "made" copper nitrite because none of the compounds changed states, you have made a mixture of copper, sodium, sulfate, and nitrite ions.

I'm almost 100% certain you can't isolate copper nitrite without it decomposing. Maybe with highly specialized equipment and conditions. Sodium sulfate looks marginally less soluble than copper sulfate or sodium nitrite at cool temperatures, perhaps it is possible to crystallize that out...

[Edited on 5-14-2014 by Etaoin Shrdlu]
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[*] posted on 15-5-2014 at 13:05


So earlier today a mixed solutions of povidone iodine that was neutralized with excess sodium thiosulfate. This instantly precipitated an orangish yellow precipitate. What happened here?
EDIT: it starts out yellow and turns orange quickly.

[Edited on 15-5-2014 by bismuthate]




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[*] posted on 15-5-2014 at 15:29


The reaction should produce tetrathionate. Could this possibly be decomposing to sulfur?



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[*] posted on 15-5-2014 at 15:33


The yellow is much brighter that sulfur.My theory is that the tetrathionate decomposes to the sulfide.



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[*] posted on 18-5-2014 at 11:35


More flame questions:

First of all, is it possible to color the flame of a fireplace or some other wood fire (in other words, a vapor-diffusion flame)? I've tried just dumping copper sulfate on wood fires, but with very little change.

Second, when it is said that the bluest part of a flame is the hottest, isn't that technically inaccurate? Isn't the bluest part the best-mixed part of the flame, where there is enough oxygen to burn completely? And isn't the color caused by the emission spectrum of the burning substance?




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[*] posted on 18-5-2014 at 12:15


Quote: Originally posted by Brain&Force  
More flame questions:

First of all, is it possible to color the flame of a fireplace or some other wood fire (in other words, a vapor-diffusion flame)? I've tried just dumping copper sulfate on wood fires, but with very little change.

Copper perchlorate works wonderfully well for this.




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[*] posted on 18-5-2014 at 17:53


Quote: Originally posted by Brain&Force  


Second, when it is said that the bluest part of a flame is the hottest, isn't that technically inaccurate? Isn't the bluest part the best-mixed part of the flame, where there is enough oxygen to burn completely? And isn't the color caused by the emission spectrum of the burning substance?


How wonderful, I am studying that right now in physics. Every particle that moves (has heat) emits electromagnetic radiations (visible light, infrared, etc.). Particles around our temperature (20°C) emits infrared light. When it gets hotter, the energy emited is stronger, the frequency gets higher and the wavelenght lower : we start to get to the visible spectrum, things are red (like heating ovens). When it gets a bit hotter, it continues to go up the spectrum, it gets yellow. Once it gets hotter than that, it is around the middle of the visible spectrum, it emits a good quantity of light of all the wavelengths of the spectrum : all colors together = white. When the energy is higher, the light emited is at the top of the visible spectrum (blue) and UV. That's why UV protection is used by those who bind metal together ( I don't know the word for it in english). Finaly, when it is hotter than that, it is 100% UV or even stronger. The equation to know the principal wavelength emited is "0,0029/T " where T is the temperature in Kelvins. For exemple, the wire in a light bulb is around 3000K, its principal wavelenght is 967 nanometers, which is infrared. It also emits light near that number. 700 being red, that's why light bulbs with wires are red/yellow. It also means that most of its power is lost in infrared that we cannot see, that's why these bulbs efficiency is so bad. The spectrums look like this : (http://faculty.virginia.edu/consciousness/images/black%20bod...)

That being said, light is also created not only by heat, but also when an electron absorbs energy, get to an higher level of energy, then release the energy absorbed as a photon (light) to get back to normal. Depending of the electronic configuration of the atoms, it can only acces to precise levels of energy, so it will always let go only precise amounts of energy corresponding to these levels. That is why compounds let go different colours. The addition of the wavelengts they can release makes their color.. (That is how fluo light bulb work, the electrons of a gas that emit wavelengths in the visible spectrum are excited by an electrical current, then release their energy as light. That is why these bulbs are a lot more efficient than the others which release a lot of invisible particules of light). These spectrums looks like this : http://utahscience.oremjr.alpine.k12.ut.us/sciber06/9th/stan...

[Edited on 19-5-2014 by alexleyenda]
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[*] posted on 18-5-2014 at 18:18


I thought about it, I didn't really answer cleary. To answer more directly to your question, the bluest part really is the hottest, except if you put a compound in it with an emission spectrum that releases blue.
Yes, it is usually the best mixed part where it burns completely, because the more it burns, the more energy is released, the hotter it gets, the frequency gets higher and the wavelenght (in that case) gets more blue.
The color may be caused by the emission spectrum of the burning substance, but most of the time, it is not. Most molecules and atoms require extreme excitation to give off visible light, a flame is not enough to triger that, except in the case of some compounds such as copper, lithium and all.

[Edited on 19-5-2014 by alexleyenda]
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