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len
Harmless
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Registered: 17-7-2006
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I tried to steam distill C6H5NO2. Its a real b...., and was hard to get moving. The first thing that came over was a yellow oil, lighter than water,
also smelling of almonds. I have no idea what that is, but would love to know. I know it was present in the original mixture, and was not formed
during the steam distill.
The heavy stuff came over with vigorous boiling, but was unfortunately still orange. This not having had the effect I desired I decided to start again
with fractional distill of petrol this time attacking with KMnO4 prior to nitration to remove branched aromatics, alkenes and any other non-benzene
stuff that might be in there.
So far I have surpising results in that the sub 90C fraction turned out to be 400/750 = 53%, as opposed to 68% last time (last time almost nothing
came over in the 90C-105C range, this time the flow here was substantial). The vigreux is really unstable when fractioning close boiling point
mixtures with many components such as petrol. This affects the results substantially. I can maintain the temperature in the boiling flask below to an
accuracy of 10C only. Most of the time nothing is coming over, and the bulb temperature is sub 50C. When the mixture below reaches a temperature at
which a component present in substantial proportion is at bp at the top of the vigreux it comes over in a rush and the bulb temperature rises rapidly.
A small percentage component does not have the energy to penetrate the vigreux on its own. I might need a better T controller before I proceed
[Edited on 13-8-2006 by len]
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not_important
International Hazard
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You try putting insulation around the fractionating column? It's also tough unless there's a substantial amount of fraction in relation to the column
'size', otherwise acting as you describe. The insulation makes it easier to keep the column at the boiling point of the currrent fraction. Sometimes a
'hat' over the upper section of the flask helps as well.
There are series of test that could be run to get an idea what is in each product group. Or maybe you're thinking of building an FTIR and a NMR 8-)
The oder of the lighter layer makes me wonder if you got some co-distillation (besides water)
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len
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I finally got round to titrating the benzene fraction with KMnO4 to remove the alkenes and substituted aromatics. The idea being to separate the
benzene from the residual alkanes by nitration. I first did the procedure with pure toluene. Here are the results.
Toluene does not reduce KMnO4 to Mn2+ under any conditions. Even in almost 30% acid it yields a dark sediment of MnO2. I suggest that the authors of
a book stating 'clarification of solution' did not do this experiment.
This oxidation is extremely messy. It takes a lot of KMnO4, 3.2gms per ml of toluene. In addition it is not very soluble, 5gms/100gm H2O, the
reaction, although it starts at room temperature, completes only at the boiling point of toluene after 10mins, and it produces a sticky precipitate of
MnO2 which covers any clean flask surfaces obscuring observation of the reaction. This makes it difficult to tell when all the MnO4 has been used up
despite its intense colour. All glassware, has to be subsequently cleaned in conc HCL to remove the MnO2. To put this in perspective, to oxidise
40ml of toluene and thus make about 50gms of benzoic acid 130gms of KMnO4 are required, dissolved in 3l of water! The whole thing needs to be boiled
for 20mins.
Doing the oxidation on my benzene fraction took all day. I did not have a 3l flask, and so had to add the KMnO4 in stages. In the end I was simpling
adding KMnO4 in powder form to replace that consumed. I added 106gms to 40ml of the second fraction, and this still did not bring about an end point.
There was an unpleasant smell produced, somewhat like molten plastic, I even had to check I had no heating rubber anywhere. I read that benzoic acid
smells like maple syrop so I presume the smell was from lower carboxilic acids, resulting from oxidation of alkenes. I stopped the eperiment at the
end of the day. After about 10 additions of KMnO4 and heating and cooling cycles there would have hardly been any benzene left - despite my use of a
reflux followed by air condenser.
I presume most of my fraction was toluene even though it was collected below 100C, and I have to remove my earlier claim to making nitrobenzene - and
therefore proving the quantity of benzene in Australian petrol. I most likely made the almost identical nitrotoluene - seeing toluene can come over
before 100C. I looked up the vapour pressure of benzene, and its already half an atmosphere at 55C. This means that if the petrol cotains a fraction
x of it, it will be in a ratio of about x/3 in the vapour of petroleum ether at its boling point, ie it will be evaporating 1/3 as fast as the
petroleum ether. Seeing the ether forms about 50% of petrol, while benzene is expected in the range 1-5%, there will be hardly any benzene left by
the time all the ether had evaporated in a simple distillation. In a setup with N theoretical plates I can expect the x/3 will change roughly to
x/3^n. The vigreux collumn does not correspond to theoretical plates, and I have difficulty in estimating how many it is equivalent to. Probably
2-3, which still means all the benzene would have left the the pet ether. I hence have to modify my procedure, and repeat the experiment.
1) Collect the whole low boling point fraction, below the bp of benzene, 40-80C.
2) Titrate a small portion of this with permanganate to find amount required for complete fraction. To the latter add about 500ml of water,
sufficient acid to take it to MnO2 (I think H2SO4 can be counted diprotic in this regard), and slightly over the calculated amount of KMnO4.
3) Boil with reflux and shacking for 30mins. Decolorise the
solutions with NaHSO3. Only benzene and alkanes can now be left in the organic layer
4) Separate the organic layer and nitrate to separate the alkanes from benzene.
Does anyone have any comments on the new procedure
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len1
National Hazard
Posts: 595
Registered: 1-3-2007
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Well I have now been able to complete the first two steps above, the results are surprising.
Under slow distilation of petrol using a vigreaux there is a very wide regions of temperatures in which nothing comes over, between 75-109 degrees.
This allows easy separation into a low and high boiling point mix. From the bp of the later it is obvious it contains toluene. Any benzene present
will be contained in the low bp mix. Checking its vapour pressure, which is already 1/2 atmosphere at 50 degrees, it is obvious it will all be
carried away by the low boiling point fractions well before its own boiling point is reached. 500ml of petrol generated 190ml of the low bp mixture.
This was then titrated with KMnO4 as stipulated above, and this is the surprise. Several titrations showed that 3.6gms KMnO4 are fully reduced by the
low bp mixture to MnO2. Assuming this is due to alkenes with a single double bond, we can take dimethyl-butene as a representative with a suitable
boiling point. This is then converted into a 5-carbon carboxylic acid and carbon dioxide if the double bond is terminal, i.e. a gain of 10 electrons
per mole. The KMnO4 releases 3 electrons per mole, so 3.6 grams oxidize 3.6/158 * 3/10 moles of the alkene = 0.0068 mole, thats 0.56 gram.
In any case the low bp fraction is highly unsaturated which is a real surprise. This means the nitration products obtained in the first few
experiments were all aliphatic nitrates. Referring to the TNT book on this site, these are all quite explosive. Just as well I resisted the
temptation at purification by distillation.
The benzene content can not be established by this method due to the high level of unsaturation. Attempts at chromatography will be next. Dont know
if IR can help me get a quantitative result with such low concentrations?
[Edited on 17-4-2007 by len1]
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not_important
International Hazard
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Welcome back
What if you steam distilled the mix after oxidation, 'titrating' that near the end to avoid overshoot, and leaving it alkaline so there are no free
carboxylic acids. What comes over with the water should be any alkanes, benzene if any, and ketones if they were formed.
I think that some alkanes with tertiary hydrogens do get oxidised by alkaline KMnO4, in acid solutions that's a for sure; you didn't specify the
temperature you used, alkane oxidation would be unlikely in the cold. If you can make some bromine, you might try checking double bonds that way as
well; make a known concentration dilute solution of Br2 in an inert solvent, titrate in dim light.
[Edited on 2-3-2007 by not_important]
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len1
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Hello again Not Important, I see you have been busy.
Its possible what you say, I think steam distillation would be a way to separate the remnant hydrocarbons from the large amount of MnO2 they cling to.
For 20ml of starting hydrocarbon the result is about 160gms MnO2 in 1 l of water in which the non-oxidised product (less than 5ml) is immersed in.
Im not sure if I will recover much from this.
Your really interesting point is that alkanes also get oxidized by KMnO4. I did not know that. I have always been told that KMnO4 does not attack
alkanes. Checking the internet I see that is not quite so, although rates and temperatures are not quoted. Could the entire reaction I witnessed be
with alkanes? I dont think so. Although proving it will take me on another side tour. The reaction conditions were 100-120 and acidic, below that
there is no reaction, or its very slow. I thought it KMnO4 attacks 3 alkanes it might also attack benzene, but could find no references to this.
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