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AndersHoveland
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Quote: Originally posted by jwarr  | | I'd like to remove carbonyl adulterants from my denatured ethanol. Would mixing it with sodium hydroxide and then distilling be sufficient?
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Yes, not aware of any reason it would problematic. Just be aware that the sodium acetate will easily melt at higher temperatures, and you would want
to stop distillation as soon as all the alcohol is out, otherwise some of the sodium acetate may decompose, giving off acetone vapor.
I'm not saying let's go kill all the stupid people...I'm just saying lets remove all the warning labels and let the problem sort itself out.
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hkparker
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I do believe ethanol reacts with NaOH to make sodium ethoxide...
http://www.sciencemadness.org/talk/viewthread.php?tid=2656
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"Nothing is too wonderful to be true if it be consistent with the laws of nature." -Michael Faraday
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AndersHoveland
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Only if the ethanol is not diluted with water, and solid NaOH is used. Obviously, sodium ethoxide reacts with water.
I'm not saying let's go kill all the stupid people...I'm just saying lets remove all the warning labels and let the problem sort itself out.
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hkparker
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oh course, it reacts to reform the reactants...
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"Nothing is too wonderful to be true if it be consistent with the laws of nature." -Michael Faraday
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solo
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In the enclosed study it the TEA they use mean triethylamine or the triethanolamine.....i'm confused since they sometimes use the same abbreviation
for both.....but i feel it may be the triethylamine since it may be removing the hydrogen chloride from the reaction....any thoughts.....solo
Attachment: An Insight of the Reactions of Amines with Trichloroisocyanuric Acid.pdf (160kB)
This file has been downloaded 3111 times
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Polverone
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| Quote: Originally posted by solo | | In the enclosed study it the TEA they use mean triethylamine or the triethanolamine.....i'm confused since they sometimes use the same abbreviation
for both.....but i feel it may be the triethylamine since it may be removing the hydrogen chloride from the reaction....any thoughts.....solo
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It is triethylamine according to the diagram of reaction scheme 2.
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Panache
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does anyone know what is added to automatic dishwashing liquids (these are rarer, as opposed to powders and tablets) to stabilise the carbonate,
bicarbonate solutions, or is it something completely different?
Also downward distillations is when the condenser is lower than the still head.
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Lambda-Eyde
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Is the -CH<sub>2</sub>OH substituent activating or deactivating? I.e. will benzyl alcohol, on treatment with an alkyl chloride and
FeCl<sub>3</sub> give the meta or ortho/para product?
I'm guessing meta, since benzaldehyde and benzoic acid both have deactivated rings, but I'm asking since I didn't find the particular substituent
(-CH<sub>2</sub>OH) in my organic chemistry book. Google didn't turn up anything relevant either.
If it is deactivating, i.e. meta-directing, is there any way to reversibly modify the -CH<sub>2</sub>OH substituent reversibly to an
activating, ortho/para-directing one? Would converting the alcohol to a benzyl ether change the reactivity of the ring? Also, would this reaction be a
competing side reaction in a Friedel-Crafts alkylation? The Williamson ether synthesis is conducted in a basic environment, to what extent does this
reaction occur in the conditions of a F-C alkylation?
This just in: 95,5 % of the world population lives outside the USA
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smuv
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Hardness of Fe++
Why is Fe(II) considered a borderline acid?
It has lower Electron affinity, Ionization energy, electronegativity and chemical hardness than both Cd(II) and Hg(II), both of which are weak acids.
Also anyone have any quantitative hardness data for thioethers? At this point I am just extrapolating from H2S.
--Thanks
p.s. data: http://pubs.acs.org/doi/abs/10.1021/ja00364a005
@ lambda, somewhat deactivating. The electronegativity of oxygen pulls electron density from the ring. There is also no way for the oxygen to
contribute its lone pairs to the ring. So slightly meta directing. I have no data for product distributions but I would expect a lot of para product
as well.
"Titanium tetrachloride…You sly temptress." --Walter Bishop
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fledarmus
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| Quote: Originally posted by Lambda-Eyde | Is the -CH<sub>2</sub>OH substituent activating or deactivating? I.e. will benzyl alcohol, on treatment with an alkyl chloride and
FeCl<sub>3</sub> give the meta or ortho/para product?
I'm guessing meta, since benzaldehyde and benzoic acid both have deactivated rings, but I'm asking since I didn't find the particular substituent
(-CH<sub>2</sub>OH) in my organic chemistry book. Google didn't turn up anything relevant either.
If it is deactivating, i.e. meta-directing, is there any way to reversibly modify the -CH<sub>2</sub>OH substituent reversibly to an
activating, ortho/para-directing one? Would converting the alcohol to a benzyl ether change the reactivity of the ring? Also, would this reaction be a
competing side reaction in a Friedel-Crafts alkylation? The Williamson ether synthesis is conducted in a basic environment, to what extent does this
reaction occur in the conditions of a F-C alkylation?
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This is a slightly activating, ortho-para director. It acts very much as a similar alkyl substituent would act (eg n-propyl). Because the oxygen can't
actually contribute electrons directly into the ring, it only acts as an electron withdrawing group on what would otherwise be a methyl group.
A Japanese patent (JP08176049) measured the products of iodination on the methyl ether and obtained a 1:0.5:2.6 ratio of ortho, meta, and para
substituted compounds.
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smuv
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Generally it is considered a moderately deactivating meta director. I cannot read the patent you post, but it may be a special case.
"Titanium tetrachloride…You sly temptress." --Walter Bishop
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DDTea
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When taking an object out of a muffle furnace and allowing it to cool in a desiccator, should the vacuum be applied as it cools?
I don't think that it should. Objects cool by two mechanisms: conduction and blackbody radiation. By removing the air from the desiccator, much of
the conductive medium is being removed. So the object can only conduct heat away via the desiccator platform or radiate heat (less efficient).
My colleague believes that the vacuum increases the rate of cooling. I feel like doing an experiment to prove my point would be a dick move, and I
may have forgotten some thermodynamics. Who is right?
"In the end the proud scientist or philosopher who cannot be bothered to make his thought accessible has no choice but to retire to the heights in
which dwell the Great Misunderstood and the Great Ignored, there to rail in Olympic superiority at the folly of mankind." - Reginald Kapp.
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Mixell
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Well, If the object is indeed a solid that would not sublimate under vacuum, applying more vacuum means less contact with particles that could take
heat away from your object, thus lowering the rate of cooling. Although if it does sublimate/evaporate, then the vacuum should increase cooling rate
by promoting some of the more agitated particles to escape the object (or liquid) and take away a good amount of heat with them (by lowering the force
applied by the air pressure, thus lowering the boiling point/increasing the sublimation rate).
[Edited on 5-8-2011 by Mixell]
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watson.fawkes
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| Quote: Originally posted by DDTea | | When taking an object out of a muffle furnace and allowing it to cool in a desiccator, should the vacuum be applied as it cools?
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"Should" is relative to your goal. If your only goal is speed of cooling, then eliminating a fluid cooling
medium (gas or liquid) will make it cool more slowly. That's not nearly the whole story, though. Since thermal emissivity goes up as the fourth power
of temperature, radiation completely dominates thermal loss at high temperatures. Fluid cooling will add something, but it may not be worth it.
The best reason to keep it under vacuum is to prevent from reacting with the atmosphere while it's hot. For some materials, this isn't a concern. If
you've got materials that will oxidize, or more rarely nitridize, vacuum is perfectly warranted.
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Mixell
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But on a more practical note, I doubt very much that if the object does not sublimate, that the air particles in a closed vessel could relieve much
heat from it. The lion share of the heat will be transfered to the desiccator platform. Although the air particles might transfer the heat to the
glass casing of the desiccator, allowing some non negligible heat transfer.
Could be quite interesting to check this out.
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DDTea
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| Quote: Originally posted by Mixell | ...The lion share of the heat will be transfered to the desiccator platform. Although the air particles might transfer the heat to the glass casing of
the desiccator, allowing some non negligible heat transfer.
Could be quite interesting to check this out. |
That was my reasoning: the air would transfer energy to the glass casing which, in turn, would heat the surroundings and result in a net loss energy
(i.e., cooling) in the system (desiccator and its contents). However, watson.fawkes brought up an interesting point about the importance of radiative
energy loss at higher temperatures.
In this procedure, we're using the muffle furnace to burn off residual organics on some weigh boats (they appear to be sand--they're used in LECO C/S
analyzers) at 1000*C--i.e., raging orange glow.
If, at this temperature, blackbody radiation is the primary means of cooling, then the vacuum issue may be moot--at least until the weigh boats cool
below a certain temperature.
"In the end the proud scientist or philosopher who cannot be bothered to make his thought accessible has no choice but to retire to the heights in
which dwell the Great Misunderstood and the Great Ignored, there to rail in Olympic superiority at the folly of mankind." - Reginald Kapp.
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watson.fawkes
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| Quote: Originally posted by DDTea | | If, at this temperature, blackbody radiation is the primary means of cooling, then the vacuum issue may be moot--at least until the weigh boats cool
below a certain temperature. |
I'll add only to keep them out of air drafts while cooling. This can thermally
shock ceramics and glass enough to break them.
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AndersHoveland
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| Quote: Originally posted by Lambda-Eyde | Is the -CH<sub>2</sub>OH substituent activating or deactivating? I.e. will benzyl alcohol, on treatment with an alkyl chloride and
FeCl<sub>3</sub> give the meta or ortho/para product?
I'm guessing meta, since benzaldehyde and benzoic acid both have deactivated rings... |
Although chlorination of nitrobenzene with FeCl3/Cl2 leads to the meta isomer, methyl groups are said to be 2,4-directing. So while nitration of
nitrobenzene lead mostly to the 3-isomer of dinitrobenzene, when toluene is nitrated only about 5% of the product is 3-nitromethylbenzene.
| Quote: Originally posted by fledarmus |
This is a slightly activating, ortho-para director. It acts very much as a similar alkyl substituent would act (eg n-propyl). Because the oxygen can't
actually contribute electrons directly into the ring, it only acts as an electron withdrawing group on what would otherwise be a methyl group.
A Japanese patent (JP08176049) measured the products of iodination on the methyl ether and obtained a 1:0.5:2.6 ratio of ortho, meta, and para
substituted compounds. |
The above answer would seem true.
I'm not saying let's go kill all the stupid people...I'm just saying lets remove all the warning labels and let the problem sort itself out.
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Picric-A
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Dibenzyl Ketone
Dry distillation of calcium phenylacetate with a current of dry N2 yields dibenzyl ketone in good yields.
I am having problems with what i thought would be a simple procedure of making the calcium salt. Here is what a did;
-1 mole of cacium hydroxide was suspended in 200ml of hot water.
-To the suspension, 2 moles of phenylacetic acid was added and the solution was boiled for 5 mins
-the water was then evaporated. This left me with what seemed to be the phenylacetic acid crystals (smelt like it) and calcium hydroxide powder.
Does anybody know why the two didnt react? What am i doing wrong? I thought it would be siple acid-base chemistry :s
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smaerd
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edit - nevermind I see now 
[Edited on 11-8-2011 by smaerd]
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redox
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| Quote: Originally posted by Picric-A | Dry distillation of calcium phenylacetate with a current of dry N2 yields dibenzyl ketone in good yields.
I am having problems with what i thought would be a simple procedure of making the calcium salt. Here is what a did;
-1 mole of cacium hydroxide was suspended in 200ml of hot water.
-To the suspension, 2 moles of phenylacetic acid was added and the solution was boiled for 5 mins
-the water was then evaporated. This left me with what seemed to be the phenylacetic acid crystals (smelt like it) and calcium hydroxide powder.
Does anybody know why the two didnt react? What am i doing wrong? I thought it would be siple acid-base chemistry :s |
In a different thread, Vogelzang made calcium phenylacetate from phenylacetic acid and calcium oxide instead of the hydroxide. Maybe you can try this.
My quite small but growing Youtube Channel: http://www.youtube.com/user/RealChemLabs
Newest video: Synthesis of Chloroform
The difference between chemists and chemical engineers: Chemists use test tubes, chemical engineers use buckets. |
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Mixell
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Does iodination of hexane is possible at RT? and if yes, how fast the reaction is and what is the favorable species?
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theflickkk
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I don't think that hexane would under iodination at RT. Free-radical sub is the mechanism by which it would probably go by. However, with I2, this is
likely to be practically non-existent at RT.
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Mixell
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Bromination proceeds quite fast at RT as I've seen on youtube, I'm afraid that iodination proceeds similarly (the iodine will be dissolved in hexane
and will stay like that for a few weeks, so I need to be sure that it won't react at all).
[Edited on 13-8-2011 by Mixell]
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theflickkk
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Bond energy of :
C-H bond approx. 397KJ/mol
Br-Br bond approx. 193KJ/mol
I-I bond approx. 151KJ/mol
C-Br bond approx. 280KJ/mol
C-I bond approx. 240KJ/mol
H-Br bond approx. 366KJ/mol
H-I bond approx. 299KJ/mol
Calculating energies of bonds broken + bonds formed:
For bromination= 397 + 193 - 280 - 366 = -56KJ/mol
For iodination = 397 + 151 - 240 - 299 = +9KJ/mol
Since bromination is exothermic, its likely to be energetically favorable while the opposite would likely apply to iodination.
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