Sciencemadness Discussion Board

The short questions thread (1)

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Nicodem - 16-2-2009 at 00:59

Only few imines of the most electrophylic carbonyl compounds (formaldehyde, some benzaldehydes, etc.), or the ones that are additionally stabilized by hydrogen bonds, or endocyclic imines, or certain groups at the amine part, are stable enough for isolations involving extractions from aq. phases. Others hydrolyse very rapidly even in neutral pH, and about immediately in acidic aq. media.

PS1: I don't want to see any more drug cooks acronyms like P2P, MDP2P or similar! This is a forum for amateur science, so use chemical names.

PS2: The condensation Intergalactic_Captain was inquiring was the Knoevenagel condensation between substituted benzaldehydes and nitromethane. He is not asking about any Henry reaction, just about the applicability of that method at Org. Synth. for substituted benzaldehydes.

hector2000 - 16-2-2009 at 04:20

what is d-tartaric acid?
(L-(+)-tartaric acid) or(D-( − )-tartaric acid)?

[Edited on 16-2-2009 by hector2000]

Nicodem - 16-2-2009 at 04:51

http://en.wikipedia.org/wiki/Tartaric_acid

hector2000 - 16-2-2009 at 05:54

yes before i read it but i am confused
always d isomer is + and l isomer is - but tartaric acid has diffrent
d is - and l is +
d-tartaric acd is dextro tartaric acid or is d(-)tartarc acid?

sparkgap - 16-2-2009 at 07:07

The Small-type Capitalized D and L and the small d and l are two different things. See this.

Another reason to be careful with capitalization!

sparky (~_~)

hector2000 - 16-2-2009 at 11:02

yes you say true
my question is this:
d-tartaric acd is dextro tartaric acid or is D(-)tartarc acid?
i want to Resolution of Racemates

Nicodem - 17-2-2009 at 00:09

The prefix (+)- means dextrorotatory, (-)- is levorotatory. Read the wikipedia entry I linked above and the one Sparkgap provided to figure out which configuration of tartaric acid corresponds to which optical activity. Nobody here can answer your question but yourself since you do not use the prefixes correctly. For example, your name "d(-)tartarc acid" makes no sense whatsoever since "d-" is the prefix for dextrorotatory while "(-)-" is the prefix for levorotatory (you can not have a "dextrolevorotatory" compound :o ). Make up your mind or correct your capitalisation. It is quite probable you confused D- with d- and L- with l-, two things that have nothing in common.

Baphomet - 17-2-2009 at 03:02

I think he just got the dash '-' mixed up with the negative sign '-'

Indeed on some keyboards these are even the same key :D

hector2000 - 17-2-2009 at 05:38

@nicodem i know that and also i am confused about that
for example see these componet:
L(+)-tartaric acid merck number:100804
D(-)-tartaric acid merck number:800799
really which one is d-tartaric acid(dextrotartaric acid)

sparkgap - 17-2-2009 at 05:40

The "+" in L(+)-tartaric acid would imply that that is the dextrorotatory enantiomer.

sparky (~_~)

497 - 17-2-2009 at 19:52

Extra short and easy question for you:

If you monomethylate (say with MeOH/H2SO4) an alkyl hydroquinone, will it methylate the OH ortho to the alkyl group or meta to the alkyl group? For example, would it form 2-ethyl-4-methoxyphenol or 3-ethyl-4-methoxyphenol or, god forbid, a mixture of the two?

Thanks.

[Edited on 17-2-2009 by 497]

chemrox - 17-2-2009 at 22:33

alkyl are o,p directing so I would say the major product would be ortho but you'd get 3-5% of the meta as well

497 - 17-2-2009 at 23:22

But it's not actually methylating on the ring, so does that still apply?

not_important - 18-2-2009 at 00:22

You'll get a mix, more para than ortho if the alkyl group crowds the ortho HO- enough. You may find you get some C-methylation as well, primarily at the C-O positions but possibly at the C-alkyl as well; for most substrates this happens to only a small degree if at all.

Nicodem - 18-2-2009 at 00:24

Quote:
Originally posted by chemrox
alkyl are o,p directing so I would say the major product would be ortho but you'd get 3-5% of the meta as well

497 is asking about O-methylation, I guess with dimethyl sulfate or methyl hydrogensulfate, because "MeOH/H2SO4" does not makes much sense, unless he is talking about the benzoquinone catalysed monoalkylation of hydroquinones with alcohols which is however a different thing altogether... In any case electrophilic aromatic substitution has nothing to do with this.
Quote:
Originally posted by 497
But it's not actually methylating on the ring, so does that still apply?

With a methylating reagent, a mixture of all mono- and dimethylation products would be obtained, with the ratio depending on the reaction conditions, but never selectively only one product.
With the benzoquinone catalysed alkylation with MeOH you would obtain a mixture of both possible monomethylated products in a ratio approaching 1 : 1, but not exactly such. Electronic effects should in my opinion somewhat favour the formation of 2-ethyl-4-methoxyphenol due to the higher basicity of the carbonyl group distant from the alkyl in the benzoquinoid intermediate. But then again, this is hard to predict unless you use the DFT theory and do some computational chemistry.

To not_important: Concurrent C-alkylation of the phenoxides generally occurs <5% extent in most phenols, with some electrophiles more than others (allyl bromide is one such, but still ~5%), at least in my experience. Yet there are phenols that will give a much larger proportion of C-alkylation, for example resorcinol and some others. This is a topic that interest me quite a lot. If you happen to have a reference or a review paper discussing the topic, I would appreciate if you could share it.

[Edited on 18/2/2009 by Nicodem]

497 - 18-2-2009 at 00:46

Thank you Nicodem, that was exactly what I wanted to know. Although not the answer I was hoping to hear :P. I think I've found a way around that problem anyhow.

Ebao-lu - 18-2-2009 at 23:19

Did any one come accross a picture of JWH-018 binding to the CB receptor(of any kind), or any other 1-alkyl-3-aroylindole binding? I mean, which atoms are relevant for binding, and the alkyl chain disposition while binding. If possible, gei me a link, or tell where can i find a picture(i want just in brief for JWH018, not to study all the cannabinoid family)..

[Edited on 18-2-2009 by Ebao-lu]

Nicodem - 19-2-2009 at 00:35

If by "picture" you mean crystallographic data then the answer is negative since both CB receptors are transmembrane receptors and thus have not been crystallized (with or without a ligand) yet (I think rhodopsin is the only transmembrane protein ever crystalized up to now and of which the crystallographic data is available).
On the other hand, if by "picture" you mean a hypothetical model of interaction, then you better start looking for review papers on the topic as surely some model was described (if not graphically, at least by describing the interactions of the pharmacophore).

Ebao-lu - 19-2-2009 at 04:11

Yes, of course i mean a model of interaction. My main interest is the position of alkyl "tail" there - is it directed to fill the space between N and naphtoyl, or it is directed perpendicularry to the indole ring, or any other way...

Intergalactic_Captain - 19-2-2009 at 13:44

Nicodem - Thanks for pointing that out. I should have recognized the confusion earlier on, but I figured the members here with experience in the matter would know what I was asking.

Another unrelated question - I need a small amount of n-pentanol, and it doesn't appear to be a particularly common ingredient in anything. Hydrolysis of amyl acetate first came to mind, but the stuff's expensive and most likely iso-amyl acetate (the more commonly used banana/"juicy fruit" flavoring). Right now I'm thinking this - Baeyer-villiger oxidation of cyclopentanone (from pvc solvents) to caprolactone, hydrolysis to 6-HO-caproic acid, and then decarboxylation to the one-carbon shorter n-pentanol.

Anyone ever tried to decarboxylate an aliphatic substance? Would the hydroxyl group interfere in any way in a simple benzoic acid style carboxylation (add a base and heat)? The reason I ask is related to biodiesel - It seems that if decarboxylation of fatty acids was particularly easy, the market would be flooded with plant-derived gasolines rather than methyl-esters. Once again, maybe I'm not being clear in illustrating my confusion, but that's the parallel that's got me wondering.

...and, completely unrelated once again, anyone have any experience with organocadmium reagents? I found a couple papers on their ketone-forming and other reactions and have been completely fascinated for a few days now.

superman1451 - 19-2-2009 at 18:22

Question,
when decomposing ammonium dichromate why does it seem like there is way more product than reactant after the reaction has finished?

DJF90 - 19-2-2009 at 18:35

The chromium (III) oxide is produced is very voluminous compared to the ammonium dichromate.

not_important - 20-2-2009 at 00:57

Quote:
Originally posted by Intergalactic_Captain

Anyone ever tried to decarboxylate an aliphatic substance?

Most don't decarboxylate very well, it takes electron withdrawing groups nearby, preferably alpha. Thus malonic acid, acetoacetic acid, Cl3CCO2H, and so on, all lose CO2 fairly readily.
Quote:
Would the hydroxyl group interfere in any way in a simple benzoic acid style carboxylation (add a base and heat)?

I suggest reading the history of polymer chemistry, in particular the work of Wallace Carothers, and you may wish to reflect on the term polyester.
Quote:
The reason I ask is related to biodiesel - It seems that if decarboxylation of fatty acids was particularly easy, the market would be flooded with plant-derived gasolines rather than methyl-esters.

You got that right, babe. There have been a number of schemes for thermal decomposition of waste biological materials, many of which were to produce significant quantities of alkanes. To date none have been particularly successful, see Wiki on "thermal depolymerization".
Quote:
...and, completely unrelated once again, anyone have any experience with organocadmium reagents? I found a couple papers on their ketone-forming and other reactions and have been completely fascinated for a few days now.

Yes. Sensitive to moisture, oxygen, light, sometimes to the phase of the Moon, and occasionally to the existence of space-time. Toxic, if ignited have the delightful property of forming finely divided CdO smoke which is known to be a human carcinogen and has other interesting effects on health.

To some extent replaced by organocopper compounds, but still interesting chemistry. To get some useful details on the use of organocadmium compounds, you may wish to read this http://pubs.acs.org/doi/abs/10.1021/cr60315a001

497 - 21-2-2009 at 02:45

Quote:
Originally posted by Ullmann
And about the sodium methoxide : it can easily be done from a methanolic NaOH solution (3 molar) by addition of one equivalent of 3A molecular sieves, which can be obtained OTC for conservation of artcraft... also Na2CO3 in MeOH can be use to form it...


Is this true? I'm skeptical because I've never seen it mentioned anywhere else, always that you have to have Na metal to make NaOMe...

manimal - 21-2-2009 at 03:58

Quote:
Originally posted by Formula409
Actually, I was planning on exploring whether thiourea dioxide (http://www.sciencemadness.org/talk/viewthread.php?tid=11785) is suitable for performing the reduction, naturally I will need some sort of way of quantitatively measuring yield, so the reduction procedures detailed on Rhodium are unsuitable for research.

Formula409.

[Edited on 16-2-2009 by Formula409]


Rhodium has a procedure for the reduction of imines with sodium dithionite. It calls for forming the imine first in DMF by combining the ketone and amine, then adding the dithionite, Na2CO3 and water.

An adaptation using thiourea dioxide should adhere to the same general outline of forming the imine in the usual way and then adding to it the reducer, a base and water. Note the funky cosolvent used, which I think should be replaced with isopropanol or ethanol.

See here:
Quote:
To a solution of the imine (30 mmol), either preformed or prepared in situ from the amine (30 mmol) and the carbonyl compound (30 mmol), in dimethy1formamide (70 ml) at 110ｰC under nitrogen was added solid sodium hydrogen carbonate (120 mmol). The mixture was stirred vigorously; solid sodium dithionite (60 mmol) was added, followed immediately by water (30 ml). Gas evolution took place some minutes after the addition of the water. Stirring was continued at 110ｰC for 30 min; the reaction-mixture was allowed to cool to room temperature and then poured into water (300 ml). The aqueous solution was extracted with ether (4x75 ml) which was in turn washed with water (4x50 ml) and saturated brine (50 ml). The ethereal extract was dried and evaporated to give the amine. The product was purified by distillation or through the hydrochloride.

The amine hydrochlorides were prepared by adding a slight excess of 5 M hydrochloric acid to the neat amine. The mixture was stirred and the solid product was collected by filtration.

http://www.erowid.org/archive/rhodium/chemistry/redamin.dith...

[Edited on 21-2-2009 by manimal]

Formula409 - 21-2-2009 at 07:03

Thank you manimal. I also found this procedure which calls for the imine to be formed before adding the reducing agent. It doesn't need nitrogen, however and still gets quite a good yield.

http://www.erowid.org/archive/rhodium/chemistry/redamin.aque...
Quote:

To a solution of 1-(2,4-dimethoxyphenyl)-2-propanone (48.56g, 250 mmol) in 300 ml IPA was added a solution of methylamine hydrochloride (25.3g, 375 mmol) in 30 ml water followed by dropwise addition of a solution of NaOH (15g, 375 mmol) in 40 ml water during 10 minutes while stirring the mixture violently. When the addition was complete the mixture was stirred for another hour at room temperature.

A solution of sodium borohydride (5.5g, 145 mmol) in 20 ml water containing 25 mg NaOH (to prevent decomposition) was then added dropwise over 30 minutes while the mixture was stirred violently. When addition was complete the stirring was continued for two hours. The residual borohydride was destroyed by addition of 2M hydrochloric acid (1:5 37% HCl:H2O) until gas evolution ceased and pH 3 was reached. The alcohol was removed by distillation in a rotovap and the aqueous solution diluted with 100 ml water, extracted once with 50 ml toluene, made strongly alkaline with 25% aq. NaOH and then extracted with 2x50 ml toluene. The combined alkaline extracts was dried over MgSO4 and the solvent removed by distillation. The residual oil was dissolved in 200 ml EtOAc and 5N HCl/IPA was added in portions until pH 5 was reached. Several times the acid addition had to be stopped and the formed crystals removed by filtration. The salt was then recrystallised in IPA.

Yield: 48.5g 2,4-Dimethoxymethamphetamine HCl (79%).

One would imagine that one would just substitute in an equivalent molar amount of TUD to test it.

What would be an easy substrate to test this on? I was thinking along the lines of Fructose, as it is quite a readily available ketone.

Formula409.

497 - 21-2-2009 at 12:52

I don't know if TUD could directly replace NaBH4 in that writeup. TUD requires a strongly basic environment to work, while NaBH4 does not AFAIK. So I would add more NaOH..

UnintentionalChaos - 21-2-2009 at 15:44

Quote:

What would be an easy substrate to test this on? I was thinking along the lines of Fructose, as it is quite a readily available ketone.

Formula409.


Cyclohexanone? MEK?

kclo4 - 21-2-2009 at 16:16

Would fructose even work in that reaction as the ketone? It would fascinate me if it did.

smuv - 21-2-2009 at 18:13

Fructose is a ketone...at least something like 1% of the time. Fructose exists as an equilibrium between a linear ketone and a cyclic hemiketal. This is why fructose is considered a ketose.

Formula409 - 21-2-2009 at 18:50

Quote:
Originally posted by smuv
Fructose is a ketone...at least something like 1% of the time. Fructose exists as an equilibrium between a linear ketone and a cyclic hemiketal. This is why fructose is considered a ketose.

Damn. Looks like I'll try my hand at the amination with MEK unless anyone can think of a suitable substrate that isn't going to turn into a phenethylamine/amphetamine and get me vanned :P

Formula409.

Aubrey - 21-2-2009 at 19:14

"...A solution is made by stirring 73.5 grams (0.5 mole) of finely powdered
phthalimide with 250 ce. water and 125 grams of ice and then
adding 55 cc. sodium hydroxide solution (40° Be)."

rom Dye chemstry. There is an accent over the Be, what does this mean?

sparkgap - 21-2-2009 at 20:01

@Aubrey: see this.

sparky (~_~)

smuv - 21-2-2009 at 21:42

Under strongly basic conditions, imines don't form, except with really nucleophilic amines such as hydroxylamine and hydrazine (draw out the mechanism, see for yourself; the hemiaminal needs to be dehydrated, and in basic conditions this is not favorable).

To do this, with thiourea dioxide (which I am not familiar enough with to say if it would even work) you would need to generate the imine then treat it with base and reducing agent; you can't just substitute it directly into those borohydride procedures.

Formula409 - 21-2-2009 at 23:17

Quote:
Originally posted by smuv
Under strongly basic conditions, imines don't form

Are they stable in strongly basic conditions (pH 13.5-14), however?
Quote:

To do this, with thiourea dioxide (which I am not familiar enough with to say if it would even work) you would need to generate the imine then treat it with base and reducing agent; you can't just substitute it directly into those borohydride procedures.

Understood. A solution of pH 13.82 (0.66M) seems to be used in this reference (http://www.sciencemadness.org/talk/viewthread.php?action=att...).

Formula409.

[Edited on 21-2-2009 by Formula409]

not_important - 21-2-2009 at 23:28

Quote:
Originally posted by Formula409
... Looks like I'll try my hand at the amination with MEK unless anyone can think of a suitable substrate that isn't going to turn into a phenethylamine/amphetamine and get me vanned :P


Carvone (caraway), camphor, methyl isobutyl ketone (much less water soluble than MEK and used in some OTC products).

manimal - 22-2-2009 at 04:22

What is a kewl?

Formula409 - 22-2-2009 at 04:24

Quote:
Originally posted by manimal
What is a kewl?

In what context was the term used? Normally in chemistry/pyro circles it refers to one who comes onto a forum with a primary interest in creating a drug/explosion in the easiest way possible with no concern for the reaction mechanism or safety consideration.

Formula409.

sparkgap - 22-2-2009 at 05:30

@manimal: This may be enlightening.

sparky (~_~)

Aubrey - 22-2-2009 at 10:34

If i want to preprepare specific concentrations of NaOH / dH20, will these keep or do they decompose? Its the kind of thing that seems pretty useful for lots of experiments and it can be a pain when you spill litle granules o it and it make the newspaper orange. noob question i know ;)

smuv - 22-2-2009 at 11:24

@Formula: Imines will hydrolyze under strongly basic conditions, how quickly is hard for me to say; for that matter they hydrolize under acidic conditions as well. To avoid hydrolysis one could add the imine dropwisely to a solution of thioureadioxide and the base; hopefully this would reduce the bulk of the imine faster than it hydrolized. Also if you could acomplish this reaction in an appreciably dry solvent (say methanol) hydrolysis would be minimal.

@Aubrey: Sodium Hydroxide solutions are stable so long as they are tightly capped. If they are not tightly capped they absorb CO2 from the air to contaminate the solution with a lot of carbonates. Store concentrated NaOH solutions in HDPE or PMP plastic bottles as concentrated solutions will (slowly) attack glass.

Ebao-lu - 22-2-2009 at 13:54

I'd like to add, never use usual plastic bottles (like coca cola, sprite, soda water etc) for alkaline solutions storage(and ammonia solution). The bottom will be destroyer after some time, and the solution will leave the bottle

As for imines reduction with TUD, there is a problem of solubility also. If you want to do this in water, then imine should be stable there, and soluble(i'm not sure the base is a good catalyst fot hydrolysis, but it will surely proceed without any catalyst if the imine is not stanle enough). And if it is some fatty-like alkylarylketone, i.m sure its solubility in water is an issue.
Carying out the reduction in organic solvents is also problematic because TUD is having a slight or a poor solubility in most organic solvents, and its deprotonated form - even worse. (still, maybe it is possible, but should take much time, or a special method to evelate the deprotonated TUD solubility there). And if the imine is comperatively stable in water and alkaline, but not enough soluble, you can try to use a kind of biphasal system - IPA layer with imine and water layer with NaOH/TUD. The main problem is the eqiulibrium concentration of a ketone in comparison to the rate of reduction by TUD - you can get only alcohol if the imine is unstable.


[Edited on 22-2-2009 by Ebao-lu]

[Edited on 22-2-2009 by Ebao-lu]

hellfire23 - 22-2-2009 at 15:57

I am planning on doing a nitric acid synth using the KN03 and H2SO4 route, but I have recently acquired 2lbs of sodium bisulfate I was wondering if I could use a sodium bisulfate solution as a substitute for the sulfuric acid, as that acid is somewhat precious to me.

DJF90 - 22-2-2009 at 16:19

I'm pretty sure this has been covered before here. I'm not 100% but I think you can heat the solid bisulfate with the nitrate to yield nitric acid. UTFSE.

http://www.sciencemadness.org/talk/viewthread.php?tid=2823&a...

Edited to add link.

[Edited on 23-2-2009 by DJF90]

Disposal of sodium sulfide

pantone159 - 22-2-2009 at 19:45

I have a bottle of sodium sulfide, that I got about 2 years ago. I did not open it until now, and it seems the bottle was not well sealed - there is a solution inside, along with some crystals.

I figure the stuff is no good now, and I plan to dispose of it. Is there a convenient way to react it into something less hazardous?

Nicodem - 23-2-2009 at 00:43

Quote:
Originally posted by pantone159
I have a bottle of sodium sulfide, that I got about 2 years ago. I did not open it until now, and it seems the bottle was not well sealed - there is a solution inside, along with some crystals.

That is a quite normal situation for sodium sulfide nonahydrate. It is still useful for most things you would want to use it, just use the solid and decant the liquid.
You can destroy the sulfide with most oxidants or you can form an insoluble sulfide, but pouring small amounts into the sink will do no harm to the ambient. There are plenty of organisms feeding on H2S and sulfides and none of those who would be harmed by it live in the sewage system.

dann2 - 23-2-2009 at 08:03

Hello,

Can anyone direct me in the direction of a graph or table showing saturated Sodium Chloride solution densities, or a table/graph telling me grams or moles per liter of saturated solutions. (I have lots of info. grams/moles per 100grams water).
I obtained a paper on Sodium Chloride densities (thanks to noe_c) but I never realized something as simple as NaCl solutions could be made so complicated!!!!
I could work it out with pencil (like the constipated Accountant) but a table would be nice.
TIA,
Dann2

pantone159 - 23-2-2009 at 23:19

Quote:
Originally posted by NicodemYou can destroy the sulfide with most oxidants

Bleach (5-6% NaOCl) would be fine, presumably? Since you say that it is still good, I'll keep the Na2S-9H2O, but I do want to know how it might be treated, I might practice with some of it. There actually seems to be some mold growing in there, maybe I'll scrape it out, pour off the liquid, and treat that with bleach as practice. Nice point about the sewer system - that is kind of where sulfides go normally, so a little Na2S will fit right in!

Formula409 - 25-2-2009 at 03:08

Could someone please provide me with some data regarding n-methylcamphoramine/n-methylcamphorimine (BP, solubility, etc). Google is turning up nothing, so my naming nomenclature must be incorrect.

Thanks!

Formula409.

sparkgap - 25-2-2009 at 04:35

"my naming nomenclature must be incorrect"

Apparently; your "n", if it does imply that it's attached to the nitrogen atom should be capitalized. Anyway, how about drawing the structure so we all have a better idea of what to look for?

sparky (~_~)

Aubrey - 25-2-2009 at 13:32

When pouring liquid Phtalimide after reacting Phtalic Anyhdride / Ammonia into a dish does anyone have a trick for preventing the fine light silvery dust diffusing into the air around? It solidifies immediately on contact with a room temperature container. The last time i attempted this my product was off-white, i'm wondering if there was ammonia residue left.

Formula409 - 26-2-2009 at 02:05

Quote:
Originally posted by sparkgap
Apparently; your "n", if it does imply that it's attached to the nitrogen atom should be capitalized. Anyway, how about drawing the structure so we all have a better idea of what to look for?

sparky (~_~)

Wiki says that Camphor is 1,7,7-trimethylbicyclo [2.2.1]heptan-2-one, so I am assuming that the methyl amine of it is 1,7,7-trimethylbicyclo [2.2.1]heptan-2-N-methylamine?


Formula409.

[Edited on 26-2-2009 by Formula409]

Nicodem - 26-2-2009 at 02:31

Which isomer? Endo or exo at the amine group? Why don't you give a CAS number or correct name. It is pointless to search for information about a compound only to latter find that I wasted time with the wrong one.

Formula409 - 26-2-2009 at 02:57

Quote:
Originally posted by Nicodem
Which isomer? Endo or exo at the amine group?

d-Camphor and Exo at the amine please.

Formula409.

[Edited on 26-2-2009 by Formula409]

manimal - 26-2-2009 at 17:10

Methylamination of camphor gives d,l N methylbornylamine. The isomeric bornylamines melt at 163C and 180C and boil at 189C and 199C respectively, so you can expect the N methyl values to be even higher.

It would be typical of hydrocarbon amines in that it would dissolve in organic solvents, would be insoluble in H2O and would form H2O-soluble hydrochloride salts.

[Edited on 26-2-2009 by manimal]

Nicodem - 27-2-2009 at 04:33

(1R)-N-Methylisobornylamin: bp 89-90°C at 10 torr; Alpha<sub>D</sub> = -73.2° (EtOH, 20-23°C)
Derivatives: hydrochloride mp >320°C, Alpha<sub>D</sub> = -59.7° (EtOH); hydroiodide mp 203-205°C; picrate mp 171.5-172.5°C.
Reference: Collection of Czechoslovak Chemical Communications, 26 (1961) 2602-2605

(1R)-N-Methylbornylamin: bp 205°C at 759 torr, 95-110°C at 22 torr, 82-83°C at 10 torr; Alpha<sub>D</sub> = 77.3°C (EtOH, 20°C)
Derivatives: hydrochloride mp 338-339°C; picrate mp 178-179°C, 178.5-179°C
Reference: Collection of Czechoslovak Chemical Communications, 26 (1961) 2602-2605;
Journal of the Chemical Society, 75 (1899) 1152-1153;
Bulletin of the Chemical Society of Japan, 49 (1976) 1897-1898.

Picric-A - 27-2-2009 at 06:35

I had a biology lesson today in which we were testing for sugars and starch with iodine (+KI) and benedicts solution.
The results of the tests, as im sure many of you know, is the benedicts forms a Cu2O precipitate in the presence of a reducing sugar, and the iodine solution goes deep blue in the presence of starch.
After completing a test for sugar with benedicts and getting the Cu2O precipitate I was curious so I added some iodine solution to this precipitated mix.
The results were unexpected and even my teacher couldn’t explain it- At first the blue color appeared showing starch was indeed present, and masking the color of the Cu2O at the same time. After a few seconds the deep blue color slowly faded and turned back to the normal Cu2O color. This can be repeated many times.
Can somebody explain what is going on here? Is the Cu2O slowly reducing the I2 complex?
Thanks in advance,

panziandi - 27-2-2009 at 12:38

Was the Benedicts solution (blue) reduced fully to the orange/red cuprous oxide? What sugar were you testing and was it pure? Also Benedicts solution is a complex organic using citrate complexed-Cu2+, repeat the experiment using Fehlings solution and add the iodine again. Also you could just test to see what is observed when I2 is added to neat Benedicts, to a suspension of pure Cu2O, and to a solution of the sugar.

My guess is that excess Cu2+ reacted with the I2 and you precipitated white CuI admixed with the orange Cu2O, the blue colour may have been due to excess Cu2+ contrasting with the white CuI.

Picric-A - 27-2-2009 at 12:59

I am not sure as to what the sugar was, we were basically given 4 beaker with unknown contents of proteins, sugars and starches. The solution I am referring to in this did contain starch.
I am not sure if the benedicts had been fully reduced however I am pretty sure it had, it was completely orange in color.
I have doubts as to whether CuI is formed because the solution would of become noticeably lighter with the white ppt. but it didn’t.
Also why the time delay between dark blue (I2/Starch) back to the orange/red ?

panziandi - 27-2-2009 at 13:08

I'm a little confused by what you are saying.

Let me see if my understanding is correct:

You had 4 beakers of unknowns, the one which tested +ve for reducing sugar, you then added iodine to AFTER you had added and reacted Benedicts. So you basically had a tube with a bit of sugar solution, Benedicts solution, copper (I) oxide and then added some Iodine solution and you saw a blue colour followed by a chaneg back to orange?

Well in my experience you tend to use excess Benedicts, so you will likely still have Cu2+, only way to be sure is to allow the Cu2O to ppt and settle and look at the colour of the supernatent, also you could add excess sugar.

It could be that the solution of sugar was contaminated with some starch, since the starch-iodide is a sensitive test.

Certainly if you started with starch and hydrolysed it to glucose then rana Benedicts then added iodine I wouldn't be at all surprised if you got +ve result, but that doesn't sound like what happen.

I suggest repeating it in a test tube changing the variables as in my first post.

Also just remembered that Benedicts solution is alkaline IIRC so you will be removing I2 as IO- and I- you likely have a complex set of redox reactions occuring.

[Edited on 27-2-2009 by panziandi]

Ebao-lu - 27-2-2009 at 15:18

The starch is colored by I2 via the formation of via I3- anion, that is favorable to occupy some special vacant place in the starch. So if there is no I3-, the colour should dissapear. I3- is formed by an equilibrium reaction I2 + I- = I3-, and it can break down as well if there is some good I- consumer (like Cu(+) that forms CuI, Cu2+ that can (not for sure) oxydise I- to give I2). So probably, first starch was coloured with iodine, and then I3- lost its I- to give an inert CuI(or oxydised by Cu2+), and the colour dissapeared. And due to there is very few I- in comparison to I2(if KI was not specially added), it took very few Cu+(or Cu2+), thus the experiment can be repeated many times.
All of the written is IMHO

[Edited on 27-2-2009 by Ebao-lu]

sparkgap - 27-2-2009 at 20:21

My memory seems to be failing me today... :(

So I wanted to ask if there are any halomethylation procedures for aromatic rings that won't destroy ester or acetal groups. I'm pretty sure concentrated HCl is not nice to those functionalities.

sparky (@_@)

not_important - 27-2-2009 at 21:07

Quote:
Originally posted by sparkgap
So I wanted to ask if there are any halomethylation procedures for aromatic rings that won't destroy ester or acetal groups. I'm pretty sure concentrated HCl is not nice to those functionalities.

sparky (@_@)


Might start by looking at these

http://pubs.acs.org/doi/abs/10.1021/ja00178a064


http://www3.interscience.wiley.com/journal/114126991/abstrac...
http://www3.interscience.wiley.com/journal/104056097/abstrac...

where the halo-ether is then reacted with the aromatic group using a transition metal halide as catalyst. This might transesterfy, the full articles might give an answer.

querjek - 1-3-2009 at 18:54

I have a solution with a few mixed quaternary ammonium compounds in water. Is it safe to store this in PETE for a week or two?

Sedit - 1-3-2009 at 23:08

I just seen on How its Made of them using some chemical to etch the stainless steel with a picture of the maker, Any idea of what was used? He put the chemical on a sponge and just pressed if for a few seconds and it etched the SS. Im thinking acid but im woundering if anyone has a more definitive answer.

Jor - 2-3-2009 at 02:30

A mixture of nitric and hydrofluoric acid is typically used in industry...
But this is a rather dangerous acid. Did the user were thick gloves?

not_important - 2-3-2009 at 04:10

Quote:
Originally posted by querjek
I have a solution with a few mixed quaternary ammonium compounds in water. Is it safe to store this in PETE for a week or two?

Depends on concentrations and pH. PETE is a (poly)ester, anything that speeds ester hydrolysis can trash PETE. Polyolefins are safer in most cases.

Sedit - 2-3-2009 at 08:52

Quote:
Originally posted by Jor
A mixture of nitric and hydrofluoric acid is typically used in industry...
But this is a rather dangerous acid. Did the user were thick gloves?


You know honestly I cant remember, He was making a sword and had gloves on most of the time so im pretty sure he had them on. He diped a sponge in the chemicals and pressed it on a stencial and it etched the SS in a split second.

Jor - 2-3-2009 at 09:03

I would never ever store my chemicals in PETE. Not a good plastic for chemicals storage.

I would recommend everyone to store their chemicals in PE (HDPE or LDPE), PP , glass, aluminium (solvents) or PTFE. As far as I know these are about the only plastics sold as UN-approved containers.

[Edited on 2-3-2009 by Jor]

Sedit - 2-3-2009 at 18:19

Will Phosphoric acid reduce with SiO2 and Carbon like calcium phosphate will?

Paddywhacker - 2-3-2009 at 20:53

Quote:
Originally posted by Sedit
Will Phosphoric acid reduce with SiO2 and Carbon like calcium phosphate will?


Unlikely, unless it is contained in a pressure vessel. Otherwise it will vaporise long before the required temperature is reached.

kclo4 - 2-3-2009 at 21:05

I think it may to some degree, but what point would the SiO2 serve? the phosphoric acid certainly isn't going anywhere when it is heated. I'd imagine that those temperatures carbon would be plenty reactive and a reaction could happen.

It has been noted that when making pyrophosphoric acid from phosphoric acid, the glass gets eaten away slightly.

Ebao-lu - 5-3-2009 at 11:14

Does anyone know about vanillin diethylacetal (or other, maybe cyclic acetal) synthesis? Can it be made from vanillin/alcohol + acidic catalyst(like with benzaldehyde, piperonal, anisic aldehyde)? Or it is less reactive and requires ethyl ortho-formate?
Second question is wether this acetal is resistant to base or not, and water-containing base particularly? The problem is, that theoretically phenoxide anion may break down into EtO- and quinoethoxymethide intermediate(under basic conditions), or at normal pH under acid/base catalysis the acetal can break directly into quinoethoxymethide and EtOH. Or these reactions are unlikely to proceed?

Paddywhacker - 5-3-2009 at 11:45

And, unless you protect the aldehyde, you will also get the Cannizaro reaction. Can't answer your question, but I'd suggest you get a good book on protection reactions.

DJF90 - 5-3-2009 at 12:00

Acetals are hydrolysed by acid only. If you want a carbonyl protecting group thats unreactive towards acid also use a thioacetal. I would use ethane-1,2-diol or 1-2-dithiol to form the cyclic acetal/thioacetal, as this is general practice as I understand.

Acetal formation is only acid catalysed, so using acid catalyst and an alcohol with a carbonyl will yield the acetal. The whole reaction is in equilibrium so to make the acetal use a dean stark and toluene solvent, to break the acetal use excess aqueous acid (well lots of water, the acid is just the catalyst). Making the thioacetal a lewis acid like BF3 is generally used as the catalyst. So best stick to the acetal (use ethane-1,2-diol and an acid catalyst, I think sulfuric will work fine but p-toluenesulfonic acid would be better soluble with the toluene solvent).

I'm not sure if the acetal you are forming is special or not but I would say it should be unaffected by base just as other acetals are, unless there are other functionalities present that are base intolerant.

Paddywhacker: You are right that the cannizzaro reaction can occur with basic conditions but the mechanism requires the formation of two negative charges on the molecule, which is supposedly only going to happen in forcing conditions (conc NaOH, maybe hot also I can't remember).

len1 seemed to find that even a "weak" base like sodium carbonate can cause the reaction come to think of it (see his thread on making benzaldehyde in prepublication, I think it is via the route of benzyl and benzal chloride).

Edited to add cannizzaro mechanism.

[Edited on 5-3-2009 by DJF90]

[Edited on 5-3-2009 by DJF90]

Cannizzaro.jpg - 26kB

Ebao-lu - 5-3-2009 at 12:56

Vanillin indeed seems to be quite an unusual aldehyde. Thus, it does not undergo Cannizzarro at all, because of OH group in p-position. I dont know why it happens, probably because OH is deprotonated by base first, and the phenoxide anion is stabilized via delocalization and conjugation with CHO group, giving a structure like O=C6H3(OCH3)=C(O-)H, that can't further be attacked by OH- like aldehyde at carbonyl. I donk know wether it can undergo 1,6 michael OH- addition into same position, but either because of electrostatic repulsion it can't, or because the resulting product is present at very low concentrations in equilibrium, or because it further can't be deprotonated because of there are already 2 negative charges in the molecule.
And anisic aldehyde unlike vanillin can undergo cannizzarro, because there is no OH

As for acetal formation, on the one hand this problem should be deminished in acidic conditions, but on the other - OH group is still much better electron donor to be conjugated with CHO(then for example OCH3), because H is more flexible. This difference can also be critical to form an acetal with acid/EtOH. The case is not all carbonyl compounds can give acetals (at appropriate rate of reaction, or equilibrium constant) with using ethanol and acid, thus ethyl ortho-formate is used for making acetals of arylalkylketones. Maybe vanilin is also a problematic carbonyl conpound in this regard

kclo4 - 5-3-2009 at 16:38

Does anyone know how to determine if a chemical will pass the blood brain barrier?

sparkgap - 5-3-2009 at 17:06

Usually, you can reasonably suspect that a drug possessing predominantly lipophilic (a.k.a. nonpolar) groups, or can be metabolized to something like that will pass the blood-brain barrier.

sparky (~_~)

Sedit - 5-3-2009 at 19:49

Im starting to create some lead electrodes for an electro chemical cell.
Now I have to admit that I am scared of this.
Im a plumber by trade and have used the material Im making the electrods out of many many times,(old plumbing pipes), yet working with them the right way scares the hell out of me(go figure).

How can I minimize any risk I might have and what may they be?
Please dont say lead poisoning because yes I understand that but what is the biggest concern with this? Salts seem pretty scary due to there solubility.

What other concerns does one have and has most worrys been "over rated"?

querjek - 5-3-2009 at 19:51

Quote:
Originally posted by kclo4
Does anyone know how to determine if a chemical will pass the blood brain barrier?

I believe I heard once that its MW had to be below 600 or 550ish or something, but don't quote me on any of that...

chemrox - 5-3-2009 at 19:54

I tried registering at thevespiary and was blocked by yet another silly riddle. This time one had to know something about a movie I would probably never see. It's too bad becaue it seemed like an interesting site but the honchos seem to be making sure all their members share common esthetic values. Thanks all the same.

Sedit - 5-3-2009 at 20:01

? Iv never seen any riddle chemrox, are you clicking the link provided here?

Paddywhacker - 5-3-2009 at 22:51

Quote:
Originally posted by Ebao-lu
Does anyone know about vanillin diethylacetal (or other, maybe cyclic acetal) synthesis? Can it be made from vanillin/alcohol + acidic catalyst(like with benzaldehyde, piperonal, anisic aldehyde)? Or it is less reactive and requires ethyl ortho-formate?
...

Acetal formation can be catalyzed by H2SO4 or by toluene sulphonic acid, but perhaps the nicest catalyst is acid ionic resin, especially if you wash the junk out of it with methanol first.

But acetal formation is very sensitive to water. Sometimes, even a Dean & Stark will not help because the amount of water that can dissolve in the hot toluene is enough to scupper the equilibrium.

One of the best procedures is to use a zeolite molecular seive in a soxhlet apparatus to trap the water from refluxing toluene. Unfortunately you don't get a visual confirmation of the reaction the way that you do by watching water accumulate in a Dean & Stark.

But another way to make difficult acetals is to trans-acetal from excess of a lower molecular weight acetal. For example, an acid catalyst plus excess of the ethylene glycol acetal of acetone would probably work. The acetone would come off at it's boiling point, driving the reaction forward. But you'd still need to exclude water with a drying tube, etc.

Nicodem - 6-3-2009 at 00:16

Quote:
Originally posted by kclo4
Does anyone know how to determine if a chemical will pass the blood brain barrier?

You might want to read about the Lipinski's Rule of Five (and the links therein). The part most pertaining the absorption trough various lipid bilayer membranes, like the blood-brain barrier and similar, is the rule about the logP upper and lower limits. If the logP is too high (the compound being too lypophilic) the drug gets absorbed in the lipids. If it is too low (the compound being too hydrophilic) then it can not pass trough the hydrophobic blood-brain barrier and is rapidly excreted by the kidneys. Molecular size is less important, but must be whiting the limits of the rule just like all the other parameters. Of course, the prerequisite is metabolic stability, or else the compound never makes it to the target organ.

Ebao-lu - 6-3-2009 at 00:43

Quote:
But another way to make difficult acetals is to trans-acetal from excess of a lower molecular weight acetal. For example, an acid catalyst plus excess of the ethylene glycol acetal of acetone would probably work. The acetone would come off at it's boiling point, driving the reaction forward. But you'd still need to exclude water with a drying tube, etc.

Rather a nice idea. Distilling off acetone will drive the equilibrium to yield more high MW acetal.
By excluding of water, you mean there would be another haemorrhoidsm making the aceton ethylene acetal itself?(because water-toluene azeotrope is bouling at higher t then acetone). Is it possible to use petroleum ether instead (as its aseotrope should be boiling at lower point then acetone)? As for the traces of water while reaction of acetone ethyleneacetal + vanillin should not play a big role afaik, because of acetone is distilled off and water is generally consumed (either by direct hydrolysis of acetone ethyleneketal, or while hydrolysinig of the resulting vanillin ethyleneacetal).
As for acid catalysts, i dont yet know what will be used. I've planned boric acid but found that it is not used to make acetals (though it is a nice catalyst for esters). Sometimes NH4Cl is used as a catalyst, but in those procedures the RM should stay for several days, so this catalyst is not quite good. I dont have toluenesulfonic acid also, and have only a dilute solution of H2SO4. Maybe NaHSO4 is usable?

Ebao-lu - 6-3-2009 at 01:48

2 Sedit: Pb is toxic as a vapour (if molten), and as Pb2+ (if ingested), you need only to enquire if Pb2+ can pass through the skin. I dont see any other potential hazard with Pb. Just protect yourself from physical contact with the solution(even if it does not pass through skin), and from ingesting it. If the solution from electrochemical cell would be further used, you can analyse the Pb2+ content there and reduce it using some reagents like Na2S. The same you can do with a waste solution. If there are oxides like PbO, Pb3O4 etc - dont utilize the waste solution in a back yard, find some special place(and convert Pb to the most inert salt like PbS)

Sorry but do want to compose _anode_ from Pb? You will get plenty of Pb2+ in solution. Resistant anodes are made from PbO2. Myabe there is some way to "passivate" Pb surface making it PbO2, but i don't remember if it is possible(probably yes).

[Edited on 6-3-2009 by Ebao-lu]

[Edited on 6-3-2009 by Ebao-lu]

kclo4 - 7-3-2009 at 13:37

All right, so I am wanting to make some Fatty acid amides.
I was planning on making them by mixing the fatty acid, with urea and heating. This idea was inspired by ScienceSquirrel in his post:

"ScienceSquirrel - 7-9-2008 at 10:36

One of the nicest preparations of acetamide is heating acetic acid with urea. The byproducts are ammonia and carbon dioxide and the acetamide is distilled using an air condenser."

I don't have the free fatty acids, they are in trigylcerides. I don't really want to bother with making the fatty acid since I don't have much free time for this and I am running low on sodium hydroxide. Would it be possible for me to react Urea with the triglyceride to form the amides and have glycerol or something related left behind?

It would be nice if it worked, but I'm not to sure I can see how that reaction would happen.

Paddywhacker - 7-3-2009 at 22:34

Re: Urea plus triglycerides

Something happens, that's for sure.

When heating about 1g urea and 5 ml olive oil in a boiling tube the urea initially dissolves.

On further heating there is a boiling action with the evolution of volatiles. No noticable smell. Non-inflammable.

On strong heating there are whitish fumes given off and a white solid lump develops in the liquid.

There is scope here for somebody to investigate further and report your observations.

Good luck.

kclo4 - 7-3-2009 at 23:29

you did this? Thank you very much I find this useful and very interesting!

Think this reaction produces fatty acid amides?

Paddywhacker - 7-3-2009 at 23:50

Quote:
Originally posted by kclo4
you did this? Thank you very much I find this useful and very interesting!

Think this reaction produces fatty acid amides?


No problem. Everything was to hand, so I had to give it a go.

I have no idea if the amide was formed.

After posting, the mixture had cooled. I added about 10 ml isopropyl alcohol and heated to boiling. The oil and alcohol misced together, but the white lump remained apparently unaffected.

I poured off the alcohol/oil mixture and washed the white lump again with alcohol. The combined alcohol/oil extracts were cooled and the oil settled out as a separate layer. No crystals or other solide were seen.

The white lump was broken up and boiled with isopropyl alcohol. It seemed not to be very soluble. The condensed white material from the fumes of heating behaved the same as the white lump.

But I'm losing interest. This is your experiment. Pick up thy test tube and run.

kclo4 - 8-3-2009 at 00:14

yeah I will when I can haha I don't get to do chemistry much. Interesting though that urea dissolves in oil. Possibly the weirdest thing I've heard in a while.


Update for those who care.. I guess: heating urea with olive oil. urea melted and let off ammonia gas, and eventually formed solid white clumps that did not dissolve easily in water - I suspect it was some form of melamine or something else that is a byproduct of urea decomposition. The oil seem unaffected.

[Edited on 8-3-2009 by kclo4]

dann2 - 8-3-2009 at 18:25

Hello,

Can I make K (or Na) Chromate from K (or Na) Dichromate by adding base (KOH or NaOH) to a solution of the Dichromate?
I am trying to do a Chloride titration using Silver Nitrate. It calls for Chromate as indicator. Have none. Perhaps I could add some K Dichromate and then add base (to the solution that I am going to titrate) and that would do for the Chromate indicator?

TIA,
Dann2

Paddywhacker - 8-3-2009 at 22:56

Quote:
Originally posted by dann2
Hello,

Can I make K (or Na) Chromate from K (or Na) Dichromate by adding base (KOH or NaOH) to a solution of the Dichromate?
I am trying to do a Chloride titration using Silver Nitrate. It calls for Chromate as indicator. Have none. Perhaps I could add some K Dichromate and then add base (to the solution that I am going to titrate) and that would do for the Chromate indicator?

TIA,
Dann2


Yes that will work. You know how to work out the stoichiometry?

But it is probably unnecessary. Dilute dichromate will have plenty of chromate in equilibrium.

dann2 - 9-3-2009 at 05:43

Thanks Paddywhaker,

I don't know the Stoichiometry. I was simply going to take a (not concentrated as you suggested) solution of Na Dichromate , add some NaOH solution and use a few drops of that for the indicator. Would that be OK?
This is where I got the suggestion in the first place. http://jchemed.chem.wisc.edu/JCESOFT/CCA/CCA8/MAIN/8/08/32/m...
I have lots of Dichromate.
Will this work as an indicator for a Chloride titration in a solution of Chlorate and/or Perchlorate using silver Nitrate?.
I guess I could try.

Thanks,
Dann2

querjek - 9-3-2009 at 08:59

I need some potassium hydroxide for a reaction, but can't find any. I can, however, find potassium bicarbonate and calcium oxide.

Would the following work:
KHCO3 + CaO + H2O -> KOH(aq) + CaCO3(solid) + H2O

Then filtering the resultant CaCO3 and boiling away excess water?

Thanks!

EDIT: Well, I found some potash, which is a mixture of mostly KCl + K2CO3. Any way to precipitate out those chloride ions?

[Edited on 9-3-2009 by querjek]

Picric-A - 9-3-2009 at 11:05

Potassium hydroxide used to be manufactured industrially by boiling an aquous soloution of wood ash (K2CO3) with slaked lime (Ca(OH)2.
Boiling KHCO3 with CaO will first produce CaHCO3 which decomposes to CaCO3. Filtering is going to be a bummer though, that CaCO3 formed is very fine, takes ages. May help if you Pre-decompose the KHCO3…

kclo4 - 9-3-2009 at 14:55

In order to aid in filtering the calcium carbonate formed you can reflux it to encourage the particles to become larger. I hear that works with calcium sulfate, so it should work with calcium carbonate assuming it is at least slightly soluble.

Sedit - 9-3-2009 at 17:45

Iv heard many ways of making (aq) solutions of NaOH and KOH from differnt materials but if one needed dry(as possible) caustic flakes how would one go about evaportating these solutions with out it reacting with the CO2 in the air forming carbonates?

IE if I had (aq)solution of NaOH how can one produce pure NaOH crystals from said solution?

[Edit]
Picric-A Iv come to the conclusion that if a substance is so fine that it quickly forms that unfilterable mud then leaving it to set for how ever long and decanting into a filter is the best method of operation. When they are that fine they seem to easyly form a solid clump on setting to make decantment easy and high yeilding.

[Edited on 9-3-2009 by Sedit]

kclo4 - 9-3-2009 at 18:54

just boil it away. I don't think it would be able to react with hardly any CO2 when the water vapor is coming out of the solution fairly rapidly and displacing the air around it.
If you were to boil away the water with a condenser above it, I bet it would make it even harder for CO2 to react with it. .. and obviously you wouldn't have the condenser being cooled or anything.

What is the percentage of CO2 in the air anyways? I bet it doesn't effect it noticeably unless it is left in a container for hours, and even then I doubt it would effect it that much.

barbs09 - 10-3-2009 at 00:52

Hi, This will probably be an easy one for some one.. For the purposes of extracting a crop of NaClO3 from a cell containing NaCl and NaClO3, two methods are generally proposed 1, concentrating he solution untill the chlorate will crystallise out and 2, salting the chlorate out by the addition of a salutation solution of NaCl.

My question is: since NaClO3 is over twice the solubility of NaCl for the same volume of water, why does the more soluble of the two salts precipitate out?? I would have thought the least soluble salt (NaCl) would have crystallised out first. I cannot find a satisfactory answer anywhere.

Thanks in advance
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