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ammonium isocyanate
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Since water isn't a concern you would be using hydrated AlCl3, which is really easy to prepare so you might as well, as IIRC it's the method used in
industry and so probably better in some way.
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Paddywhacker
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Quote: Originally posted by JohnWW ![](images/xpblue/lastpost.gif) | Just as a matter of interest, how would that ratio of B-pinene to A-pinene and other terpenes in New Zealand pine needle oil, which I presume is from
the very common Pinus Radiata, or Monterey Pine, as distinct from other species, compare with that obtained from trees of the same species in
their home areas of Monterey, California, and Isla dos Cedros, México? .... |
Yes, it is Pinus radiata in New Zealand, but the Wiki does not mention that species as being used elsewhere. I have read somewhere that the local turpentine has more beta than alpha.
Thanks for the effort with the links, but they don't add any information to that provided by the link that I supplied. The reaction I asked about involves the opening of a four membered ring and the formation of a five membered ring. Involking
Markownikoff does not do it justice.
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Nicodem
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@Paddywhacker: It is just a normal carbocation rearrangement. The electrophilic addition of a proton on either alpha- or beta-pinene gives the same
product due to thermodynamic reasons (lower energy carbocation forms; or what used to be called the Markownikoff rule in the old, premechanistic
times). The carbocation then rearranges via a 1,2-shift* to the bornyl carbocation which is then quenched by the nucleophile (Cl(-), Br(-), H2O, etc.)
to give bornyl chloride/bromide/alcohol/etc. This 1,2-shifts in carbocations originating from terpenes is commonly called the Wagner-Meerwein
rearrangement** (or the Nametkin rearrangement when only a methyl group shifts). There are numerous such transformations of terpenes and terpenoids
known. For example, the pinenes first rerarange to bornyls which further rearrange to camphenes. The electrophilic addition of electrophilies other
than a proton can also induce the same rearrangement.
*
http://en.wikipedia.org/wiki/1,2-rearrangement
**
http://en.wikipedia.org/wiki/Wagner-Meerwein_rearrangement
https://themerckindex.cambridgesoft.com/themerckindex/NameRe...
@Formula409: Which one of the two 3-chlorobut-2-en-1-ols? I will not even bother looking for information unless you start to specify which information
and about which compound. You have already been told not to name or draw compounds in an ambiguous way.
@Sedit: Nucleophilic chlorination of alcohols with HCl proceeds way slower than the analogous bromination. It would be a waste of time to try the same
reaction with HCl, at least without an autoclave. Besides, in case you need 1,2-dibromoethane for SN2 based reactions, you should know that
1,2-dichloroethane is quite unreactive in comparison. It is also much more volatile, thus making isolation and storage more annoying (but less toxic).
…there is a human touch of the cultist “believer” in every theorist that he must struggle against as being
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ketel-one
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I was pleasantly surprised today. I put some vodka in a glass in the freezer. I was expecting it not to freeze at all since ethanol is kind of an
antifreeze. But it made a sort of clean ice slurry with soft white ice in ethanol. Good way to make ethanol?
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Jor
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I am going to buy oxalyl chloride and Rhodamine B to make chemiluminescence using diphenyloxalate. Do I really need triethylamine to scavenge the
formed HCl when reacting oxalyl chloride wiith phenol? I'd rather not buy it. And I don't want to use my pyridine for that purpose, as I only have
25mL of that.
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Klute
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I think you can use excess anhydrous K2CO3 as a substitute, i've done this for acylations and alkylations of amines with great results, whereas the
original procedure called for an double equivalent of amine or Et3N. But oxalyl chloride might be reactive enough to directly react with the
carbonate.. Considering the price of this reagent it's best to avoid lossing some with side reactions..
If you use excess phenol, and apply slight vacuum, you could remove the formed HCl and drive the reaction forward. But I think that even without
removing the formed HCl the reaction will be more or less complete anyway.
But triethylamine is dirt cheap, and very usefull, why not get just 100-250ml of it? It's a very versatil product in org. chem.. You can also prepare
PTC's from it, etc
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woelen
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Probably ethylene diamine or diethyl amine can be used equally well for taking away the HCl. Just any amine, which can form salts with HCl will do the
job.
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Klute
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Hum, no I don't think it would be wise to use a primary or secondary amine here, as it will get acylated. Amines are more nucleophilic than phenols at
neutral pH..
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crazyboy
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If a text says "dissolve X grams of formaldehyde in methanol" that would mean adding enough 37% soln. to methanol to equal X grams correct? It doesn't
mean dissolve the gas in methanol somehow?
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Sedit
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@Crazy
If it calls for X grams just make sure the concentration equals 37% how ever you may get there. Just adjust for it if your using a more dilute
solution of formaldahyde.
OK I just a quick question that arises from just playing around and for once it appears to be going how I assumed atlest to some extent.
The results of Mixing NaOH + NaNO3 + Al produced ammonia if the temperature was allowed to go to high (expected).
Is my assumption that below 60C that Nitrites should be the product correct?
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fall of the ruling country, the people who think they can do whatever they want without anybody else's consent. I've seen this story
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crazyboy
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Quote: Originally posted by Sedit ![](images/xpblue/lastpost.gif) | @Crazy
If it calls for X grams just make sure the concentration equals 37% how ever you may get there. Just adjust for it if your using a more dilute
solution of formaldahyde.
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Well the text didn't specify any concentration it just said add X grams. 37% just happens to be the concentration I can get formaldehyde in.
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Klute
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Hum, I think that refers to anhydrous formaldehyde, not the aqueous solution, overwise it would have be specified IMHO, as formol, formaldehyde
solution, 37% formaldehyde, etc. The term "dissolve" refers to bubbling the gas in methanol IMHO.
If you don't have paraformaldehyde, to generate dry formaldehyde gas, you can make it by vacuum distilling the water out of 37% formol, under take off
nearly ceases. Then you get a viscous colorless oil, that upon cooling whitens and solififies to paraformaldehyde. Be prepared to pour the still hot
residu on a cold surface and break it the rubbery lumps before they solidify to much, as it forms very hard blocks, nearly impossible to break up or
grind when hard. But in your case you can just add a few drops of H2SO4 to the residu in your flask, and then heat that up to depolymerise the
paraformaldehyde and generate dry formaldehyde gas, dry over P205, and dissolve in your methanol.
Hey, maybe even heating paraformaldehyde in methanol with a drop of H2SO4 could depolymerise it.. I think bases are more efficient at depolyerising
paraformaldehyde, so a catalytic amount of alkali methoxide in your methanol could depolymerise the paraformaldehyde. Have a look on the net to see if
you find any info on this..
EDIT: I'm even asking myself if the term formaldehyde doesn't refer to paraformaldehyde here, perhaps simply adding paraformaldehyde to methanol can
depolymerise it, forming the hemiacetal, or perhaps acetal, of formaldehyde and methanol..
EDIT2: found this:
"In the presence of small amounts triethylamine, paraformaldehyde dissolves in methanol already at 45° to 50° C. At reflux conditions, one obtains a
complete solution in 5 to 10 minutes."
http://www.freepatentsonline.com/4237065.html
surely it's depolymerising here.
and this:
"It dissolves readily in water or alcohol by hydrolysis or depolymerization to yield free formaldehyde"
http://www.wipo.int/pctdb/en/wo.jsp?IA=US1997004396&DISP...
[Edited on 17-9-2009 by Klute]
[Edited on 17-9-2009 by Klute]
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entropy51
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I don't know what syn you are using the HCHO in, but paraformaldehyde is often used directly instead of bubbling in anhydrous HCHO as a gas. I have
used paraformaldehyde directly in Blanc chloromethylations and as a solution in THF for reaction with Grignards.
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Jor
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I am soon attempting the luminol synthesis. However, for the last step you need sodium dithionite, to reduce the nitro-group to an amino-group. I
don't have this reagent. Are there alternative reducing agents? Like Sn/HCl?
All procedures describe reacting hydrazine with 3-nitrophthalic acid in high boiling (250C IIRC) triethylene glycol. Would ethylene diglycol also be
suitable? Or preferably another solvent?
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JohnWW
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Quote: Originally posted by ketel-one ![](images/xpblue/lastpost.gif) | I was pleasantly surprised today. I put some vodka in a glass in the freezer. I was expecting it not to freeze at all since ethanol is kind of an
antifreeze. But it made a sort of clean ice slurry with soft white ice in ethanol. Good way to make ethanol? | Yes, I have heard of cheap rot-gut spirits being made from the liquor of a fermented mix by freezing the water out, instead of
distilling the ethanol off and collecting it. However, the spirits obtained by freezing are liable to contain an excessive amount of fusel oil, i.e.
higher alcohols and esters, which in the much smaller amounts usually distilled over with the ethanol in the conventional manner impart spirits
(especially whiskeys and brandys) with their characteristic tastes and odors. The large amount of fusel oil remaining with alcohol after freezing out
the water is, unfortunately, toxic in large amounts.
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mr.crow
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First post time This is really basic because I am just starting out.
I have a large amount of Sodium Bromide I recrystallized from some pool solution from Wal-Mart. The crystals are beautiful plates, defiantly not salt.
My question is, how do I dry it off? Leave it in a buchner funnel, leave it in the air, or put it in an oven?
Also it appears it forms a dihydrate. Wikipedia says it melts at 34 degrees, so could I heat it in a beaker until it fuses solid? Or just keep it as a
dihydrate.
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Picric-A
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Sodium bromide melts at 747 deg C, its dihydrate dehydrates at 34deg C so if you wanted the anhydrous powder keeping it in a 200deg C oven for an hour
would leave you with the anhydrous product.
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Paddywhacker
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Quote: Originally posted by Jor ![](images/xpblue/lastpost.gif) | I am soon attempting the luminol synthesis. However, for the last step you need sodium dithionite, to reduce the nitro-group to an amino-group. I
don't have this reagent. Are there alternative reducing agents? Like Sn/HCl?
All procedures describe reacting hydrazine with 3-nitrophthalic acid in high boiling (250C IIRC) triethylene glycol. Would ethylene diglycol also be
suitable? Or preferably another solvent? |
Boiling points
diethylene glycol 244
triethylene glycol 285
glycerol 290 (with decomposition?)
so, no.
[Edited on 17-9-2009 by Paddywhacker]
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Picric-A
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Quote: Originally posted by Paddywhacker ![](images/xpblue/lastpost.gif) | Quote: Originally posted by Jor ![](images/xpblue/lastpost.gif) | I am soon attempting the luminol synthesis. However, for the last step you need sodium dithionite, to reduce the nitro-group to an amino-group. I
don't have this reagent. Are there alternative reducing agents? Like Sn/HCl?
All procedures describe reacting hydrazine with 3-nitrophthalic acid in high boiling (250C IIRC) triethylene glycol. Would ethylene diglycol also be
suitable? Or preferably another solvent?[/rquot
Boiling points
diethylene glycol 244
triethylene glycol 285
glycerol 290 (with decomposition?)
so, no.
[Edited on 17-9-2009 by Paddywhacker] |
Is this is question? if so, yes glycerol works, i have used it successfully before.
As for the reducing agent i tryed Zn/HCl method but it failed to produce any luminol. I also have no access to Na2S2O4
[Edited on 17-9-2009 by Picric-A] |
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entropy51
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There is an OTC product which consists mainly of sodium dithionite and carbonate. I don't know about the luminol synthesis, but it works for a number
of other purposes.
Rit Color Remover http://www.ritdye.com/Fabric+Treatments.28.51.7.49.lasso
Dithionite can also be prepared by reducing SO2 with sodium formate in aqueous methanol. There is a reference posted around here somewhere on this
method.
[Edited on 17-9-2009 by entropy51]
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1281371269
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I just received a joblot of quickfit stuff won on ebay. So that I can work out which bits I need to get, adaptors and so forth, I need to be able to
measure the joint sizes where the writing has worn off. I have some tools for woodwork that measure diameters in mm - is the 42/22 rating thing the
top of the joint width over the bottom of the joint width in mm?
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entropy51
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Google is your friend! http://en.wikipedia.org/wiki/Ground_glass_joint
I think you mean "diameter" rather than "width".
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watson.fawkes
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Yes, but the description there is slightly lacking, as
it has no table of standard sizes. See http://www.wilmad-labglass.com/pdf/taper_ground_joint_dims.pdf for a solid drawing with tables standard dimensions. The modern standard is ASTM
E676, which costs to acquire. Original dimensions are from NIST CS21-58, now withdrawn.
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crazyboy
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Quote: Originally posted by Klute ![](images/xpblue/lastpost.gif) | Hum, I think that refers to anhydrous formaldehyde, not the aqueous solution, overwise it would have be specified IMHO, as formol, formaldehyde
solution, 37% formaldehyde, etc. The term "dissolve" refers to bubbling the gas in methanol IMHO.
If you don't have paraformaldehyde, to generate dry formaldehyde gas, you can make it by vacuum distilling the water out of 37% formol, under take off
nearly ceases. Then you get a viscous colorless oil, that upon cooling whitens and solififies to paraformaldehyde. Be prepared to pour the still hot
residu on a cold surface and break it the rubbery lumps before they solidify to much, as it forms very hard blocks, nearly impossible to break up or
grind when hard. But in your case you can just add a few drops of H2SO4 to the residu in your flask, and then heat that up to depolymerise the
paraformaldehyde and generate dry formaldehyde gas, dry over P205, and dissolve in your methanol.
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Thanks, you're absolutely right a few hours after I posted that it hit me it has to be anhydrous because the next step is sodium borohydride
reduction.
So once I get paraformaldehyde how can I generate formaldehyde gas?
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kclo4
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paraformaldehyde can be decomposed by an acid or heating IIRC.
http://en.wikipedia.org/wiki/Paraformaldehyde#Reactions
Wiki agrees
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