Sciencemadness Discussion Board

Synthesis of longer chain tertiary alcohols

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blogfast25 - 29-12-2010 at 07:58

1. Introduction:

I’m opening this thread with permission and on behalf of Nicodem and on request by me.

For those still unfamiliar with the currently hottest topic in the ‘Chemistry in General’ section, the small scale production of potassium metal by non-electrochemical means, I would suggest to read this thread here:

http://www.sciencemadness.org/talk/viewthread.php?tid=14970

And in particular woelen’s (moderator of the ‘Chemistry in general’ section) excellent ‘executive summary’ which sums up current state of the art of this highly interesting and useful chemical reaction:

http://woelen.homescience.net/science/chem/exps/synthesis_K/...

The overall reaction is KOH(s) + Mg(s) === > K(l) + MgO(s) + ½ H2(g) @ about 200C, in an inert paraffinic/naphtenic solvent. Equipment? A pyrex flask and a refluxer, essentially…

The natural kinetic hindrance to this reaction is overcome by means of a catalyst, namely t-butanol (2-methyl-2-propanol) which reacts with KOH in the given circumstances:

t-BuOH + KOH === > t-BuOK + H2O

The ‘miracle’ of potassium production is then achieved by a previously unknown or little known redox reaction:

2 t-BuOK(dissolved) + Mg(s) === > (t-BuO)2Mg(dissolved?) + 2 K(l)

In the mean time successful production of potassium using 2-methyl-2-butanol (t-amyl alcohol or t-pentanol) has also been achieved…

2. Production of sodium:

It rather follows logically that if potassium can be produced in this way, possibly other alkali metals can too. And indeed the US patent on which our success was based does also give an example of the production of sodium metal using the same principle: reduction of a sodium alkoxide with solid magnesium in an inert solvent at about 200C. Unfortunately the method prescribes inordinately long ‘cooking’ times of up to 13 hours, hardly practical for the home chemist.

Among those who’ve been most proactive in bringing about successful replication of the ‘potassium patent’ it’s widely believed that reduced solubility of the equivalent sodium alkoxides in paraffinic/naphtenic solvents, possibly due to the higher lattice energy associated with the smaller Na+ ion (compared to K+), is the cause of the slower reaction rate.

If so, the solution to reduce reaction times in the case of sodium would possibly be the use of longer chain tertiary alcohols. Of potential interest here would be:

• 2-methyl-2-pentanol
• 2-propyl-2-butanol
• 2-ethyl-2-pentanol
• 2-propyl-2-pentanol and such like
• ‘tertiary fatty alcohols’

An interest has thus arisen in the synthesis of such alcohols with a very practical purpose.

Nicodem has kindly pledged (time allowing) to put up a review of the synthesis of tertiary alcohols and I would like to appeal to all organic wiz kids in this section to contribute with whatever relevant knowledge on the subject that you might have. May this become a very sticky thread!

Thank you.

rrkss - 29-12-2010 at 09:33

I've successfully synthesized 3-methyl-3-pentanol from ethyl magnesium bromide and ethyl acetate. Both precursor chemicals are easy to obtain otc.

I'm going to quote my procedure from a previous post of mine which can be found in this thread.

http://www.sciencemadness.org/talk/viewthread.php?tid=13844#...

Quote: Originally posted by rrkss  
Well I am done with the grignard. It was my first grignard reaction and was very interesting. I started it by adding 1 mL of ethyl bromide to 60 mL of ether in a 250 mL roundbottom with 13g of magnesium. I calculated a 0.7g excess of magnesium from what I needed.

The reaction took off and was very exothermic forcing me to put the flask in an icebath right away since my condenser was not catching all the boiling ether. Once things calmed down I added the ethyl bromide dropwise over the next 2 hours. The grignard reagent was a nice dark gray almost purplish color.

After the grignard was prepared, I added the ethyl acetate dropwise. The ethyl acetate reacting with the grignard was even more exothermic than the preparation of the grignard and took a ton of work to get the addition rate just right.

Once the reaction was done, I quenched with acid. I added the 2.0 M H2SO4 solution dropwise to the flask. The addition was highly exothermic and formed a white jellylike precipitate. I soon ran out of room and was forced to transfer this mess into a larger flask and do multiple acid rinses. Being strapped for time I ended up probably using way to much acid to quench the reaction (500 mL of 2 M H2SO4 solution) but eventually all the magnesium reacted and the precipitate dissolved. I then transferred my organic layer to a new roundbottom and dried it with anhydrous MgSO4.

I lost about 50 mL of organic layer during the workup. Probably due to ether's slight solubility in water and my hastly workup. Hopefully after distillation, did not lose too much of my alcohol which supposedly is slightly soluble in water as well. Tommorow when I distill the ether away and get my product, I will finally know.

Things I learned.

1. Transfer the finished reaction mixture to a different and larger flask before doing the acid quench.

2. Do the quench when you are not strapped for time as it takes much longer than expected.


The alcohol decomposes upon distillation so it will need to be purified under reduced pressure.

[Edited on 12-29-10 by rrkss]

blogfast25 - 29-12-2010 at 09:47

Thanks!

Quote: Originally posted by rrkss  
I've successfully synthesized 3-methyl-3-pentanol from ethyl magnesium bromide and ethyl acetate. Both precursor chemicals are easy to obtain otc.

The alcohol decomposes upon distillation so it will need to be purified under reduced pressure.

[Edited on 12-29-10 by rrkss]


Does this mean that 3-methyl-3-pentanol would not survive the 200C needed for our alkali alkoxide reductions?

I seem to understand the general reaction path, so let me see if I can come up with some creative (but not necessarily possible!) molecule rearranging. Bit rusty on the old organic side, you see :o

According to that scheme, 2-ethyl-2-pentanol (for instance) could be synthesized from methyl propanoate and propyl magnesium bromide...

[Edited on 29-12-2010 by blogfast25]

Jor - 29-12-2010 at 09:54

AFAIK, you have to wash the ether layer with some aqeous sodium carbonate solution, as traces of acid decompose the alcohol by a elimination reaction (E1 mechanism).
During heating I think it just decomposed, giving you water and a alkene.

NurdRage - 29-12-2010 at 10:01

I'm doing a bit of long-chain t-alcohol synthesis on the side. Currently i'm starting with 1-octadecene and trying to couple that with acetone (with several steps in between) to get a C21 tertiary alcohol.

As for amateur approaches, a problem i see is finding suitable domestically available long-chains to start with. The most readily available activated long-chain carbon sources i know are 2-butanone, ethyl acetate, acetone, ethanol and methanol.

Any others that i missed?

[Edited on 29-12-2010 by NurdRage]

blogfast25 - 29-12-2010 at 10:11

Quote: Originally posted by NurdRage  
I'm doing a bit of long-chain t-alcohol synthesis on the side. Currently i'm starting with 1-octadecene and trying to couple that with acetone (with several steps in between) to get a C21 tertiary alcohol.

As for amateur approaches, a problem i see is finding suitable domestically available long-chains to start with. The most readily available activated long-chain carbon sources i know are 2-butanone, ethyl acetate, acetone, ethanol and methanol.

Any others that i missed?

[Edited on 29-12-2010 by NurdRage]


The esters are relatively easy to synthesise though, are they not?



[Edited on 29-12-2010 by blogfast25]

NurdRage - 29-12-2010 at 10:29

In a grignard, the alcohol side of the ester is lost.

Nonetheless they play a "chain doubler" role in that they add two grignard reagents to the acid side. so you get C2n+m where n is grignard carbon length and m is the carboxylic acid carbon length.

As carbon sources themselves we'd need a long-chain carboxylic acid, i don't know of any other than acetic acid.

I suppose though with esters the problem simplifies to that if we can get a good long chain alcohol we could convert it to a grignard and pull a chain-doubler trick to essentially get twice the chain length (plus acid length) we started with.


Ozone - 29-12-2010 at 10:56

The decomposition is dehydration to yield the olefin. This is especially pronounced for the aforementioned tertiary alcohol. 2° and 1° alcohols will not be as easy to dehydrate, but in the presence of sulfuric acid, particularly if you are concentrating it while the distillation takes place, most will react (at least to some extent, at the expense of yield).

The aqueous solubility of the product is poor (45 g/L), so I would recommend salting-out the aqueous phase and extracting it into a more volatile organic solvent.

The BP is 122.4 °C, BTW.

Cheers,

O3

3-methyl-3-pentanol_E1_small.jpg - 44kB

rrkss - 29-12-2010 at 11:05

Quote: Originally posted by blogfast25  
Does this mean that 3-methyl-3-pentanol would not survive the 200C needed for our alkali alkoxide reductions?
[Edited on 29-12-2010 by blogfast25]


It should survive. The decomposition was more charring on the glass but it did distill over. This alcohol has a high boiling point so there was lots of localized heating causing the charring. When I distilled under vacuum the charring problem was eliminated.

blogfast25 - 29-12-2010 at 11:50

Quote: Originally posted by NurdRage  
In a grignard, the alcohol side of the ester is lost.

Nonetheless they play a "chain doubler" role in that they add two grignard reagents to the acid side. so you get C2n+m where n is grignard carbon length and m is the carboxylic acid carbon length.

As carbon sources themselves we'd need a long-chain carboxylic acid, i don't know of any other than acetic acid.

I suppose though with esters the problem simplifies to that if we can get a good long chain alcohol we could convert it to a grignard and pull a chain-doubler trick to essentially get twice the chain length (plus acid length) we started with.



What about butyric acid? Should be quite easy to make... And the other, longer fatty acids? Nicodem mentioned something about margerine...

[Edited on 29-12-2010 by blogfast25]

bahamuth - 29-12-2010 at 12:25

Octadecanoic acid (stearic acid) from candles would be easy enough to estrify with e.g. methanol and further react that FAME (Fatty Acid Methyl Ester) with ethyl magnesium bromide yielding 1,1-diethyloctadecan-1-ol if I understand rrkss's reaction scheme right and NurdRage's explanation correct.

Please correct me if i am wrong.

Pic of suggested compound:


1,1-diethyloctadecan-1-ol.png - 7kB

blogfast25 - 29-12-2010 at 12:48

Quote: Originally posted by bahamuth  
Octadecanoic acid (stearic acid) from candles would be easy enough to estrify with e.g. methanol and further react that FAME (Fatty Acid Methyl Ester) with ethyl magnesium bromide yielding 1,1-diethyloctadecan-1-ol if I understand rrkss's reaction scheme right and NurdRage's explanation correct.

Please correct me if i am wrong.

Pic of suggested compound:




Actually going by the picture and assuming you’ve got the reaction right (I’m pretty lost for the moment! Must read up on Grignards…) you’ve got 3-ethyl-3-icosanol there: the longest chain is 20, not 18. But that’s definitely a long chain tertiary alcohol!

blogfast25 - 29-12-2010 at 12:52

This was Nicodem's remark over at the K thread with regards to Grignarding margerine...

Quote: Originally posted by Nicodem  
IUsing a longer chain t-alcohol as catalyst might prove useful in the synthesis of sodium where t-BuOH is unlikely to work given the practical insolubility of t-BuONa in alkanes. A crude way to "fatty t-alcohols" can be the reaction of a large excess of MeMgI or EtMgI(Br) with pure margarine in refluxing THF or other ethers. The preparation of grignard reagents require Mg and dry solvents. Purifying the products would require some knowledge of organic chemistry, but already with extractions it might be possible to get them pure enough for this reaction. I doubt it requires a very pure catalyst, as all the impurities would be destroyed by the reaction conditions (hot KOH!).


blogfast25 - 29-12-2010 at 13:26

Oh, I get it now. Ester + 2 RMgX causes the alcohol of the ester to split off and then the second R group is added on and the =O converted to –OH. If R = methyl you get a 2-methyl-2-… ol, with R = ethyl a 3-ethyl-3-… ol, with R = propyl a 4-propyl-4-… ol. All tertiary (and all ‘in theory’!)

Caproic (hexanoic) acid esters must be quite OTC and with methyl magnesium halides would give 2-methyl-2-octanol. Already quite a long chain really…

Methyl butanoate is a food additive.

[Edited on 29-12-2010 by blogfast25]

watson.fawkes - 29-12-2010 at 13:36

Quote: Originally posted by NurdRage  
As carbon sources themselves we'd need a long-chain carboxylic acid, i don't know of any other than acetic acid.
Alkali salts of propanoic acid (CH3CH2COOH) are readily available as food preservatives, where they are more commonly known as propionates.

bahamuth - 29-12-2010 at 13:54

Quote: Originally posted by blogfast25  
Quote: Originally posted by bahamuth  
Octadecanoic acid (stearic acid) from candles would be easy enough to estrify with e.g. methanol and further react that FAME (Fatty Acid Methyl Ester) with ethyl magnesium bromide yielding 1,1-diethyloctadecan-1-ol if I understand rrkss's reaction scheme right and NurdRage's explanation correct.

Please correct me if i am wrong.

Pic of suggested compound:




Actually going by the picture and assuming you’ve got the reaction right (I’m pretty lost for the moment! Must read up on Grignards…) you’ve got 3-ethyl-3-icosanol there: the longest chain is 20, not 18. But that’s definitely a long chain tertiary alcohol!


You are indeed correct. Never heard of or used the alkane name Icosanol before though.

According to this: http://chemistry2.csudh.edu/rpendarvis/carb-enolate.html#est... the suggested product of stearic acid and EtMgBr would be 3-ethyl-3-icosanol by reaction:
AcORRMgX.GIF - 5kB

High boiling point would be expected, perhaps over 250 degrees C, and "fatty" enough, a bit bulky though.

Stearic acid is OTC for hand lotion and candle hobby production everywhere in pure enough form.



Quote: Originally posted by blogfast25  
Oh, I get it now. Ester + 2 RMgX causes the alcohol of the ester to split off and then the second R group is added on and the =O converted to –OH. If R = methyl you get a 2-methyl-2-… ol, with R = ethyl a 3-ethyl-3-… ol, with R = propyl a 4-propyl-4-… ol. All tertiary (and all ‘in theory’!)

Caproic (hexanoic) acid esters must be quite OTC and with methyl magnesium halides would give 2-methyl-2-octanol. Already quite a long chain really…

Methyl butanoate is a food additive.

[Edited on 29-12-2010 by blogfast25]


In theory yes, not all who struggle to get sodium metal can get their hand on or synthesize Grignard reagents and/or get hold of suitable solvents.

Anyways, I was intrigued by the synthesis as I have yet to do a Grignard at all.



[Edited on 29-12-2010 by bahamuth]

blogfast25 - 29-12-2010 at 14:00

Thanks bahamuth! The name for the C20 n-alkane is icosane, its alcohols would be icosanols...

Methyl hexanoate 1 kg from sigma-aldrich for £42:

http://www.sigmaaldrich.com/catalog/ProductDetail.do?D7=0&am...

Sample (unspecified size) £15.


[Edited on 29-12-2010 by blogfast25]

[Edited on 29-12-2010 by blogfast25]

entropy51 - 29-12-2010 at 14:38

Quote: Originally posted by NurdRage  
As carbon sources themselves we'd need a long-chain carboxylic acid, i don't know of any other than acetic acid.
Grignard reagents can also be used to build up carboxylic acids by treatment with CO2 as dry ice:

R-MgX + CO2 --> --> R-COOH

Ozone - 29-12-2010 at 18:52

If you want stearic acid, saponify lard (manteca). You can't get much cheaper than that. Palm oil triacylglyceride is primarly n-hexadecanoic (palmitic).

A mixture of C20-C30 n-alkanols is known a policosanol. This is frequently derived from sugar cane cuticle wax. An excellent substitute for carnauba wax, it shares similar issues with solubility. n-C28OH, octacosanol (the dietary suppliment), for example, is only very sparingly soluble in DCM, or better, chloroform...and not much else (or so it appears to me, having isolated it from sugarcane process mud).

They are quite crystalline and difficult to re-dissolve once isolated.


NurdRage - 29-12-2010 at 19:00

Yeah.... if we go TOO long it won't be soluble... which kinda defeats the purpose of making this in the first place; to make sodium ions soluble in aliphatic solvents.

coming in from a completely different direction: could we make crown ethers by amateur means? t-butanol and a crown ether (say 15-crown-5) might do the trick and solvate sodium ions... assuming the ether remains stable under these harsh conditions.


blogfast25 - 30-12-2010 at 07:29

Personally I very much like the idea of a 2-methyl-2-alkanol with total chain length around C10. These could be Grignarded with MeMgBr on linear long chain alkanoates, with ‘alka ≈ C10’. They would have the same ‘active site’ (2-methyl-2ol) as t-butanol but a medium-long paraffinic tail for solubility in alkanes/napthenics.

Fairly OTC precursors appear to be:

Ethyl octanoate: C2H5OOCC7H15, food additive (pineapple, apple like aroma).

Decanoic acid: C9H17COOH, used in perfumery, lubricants, greases, rubber, dyes, plastics, food additives and pharmaceuticals (Wiki).

Methyl nonyl ketone: CH3COC9H19, used in pest control, as cat repellant.

It’s hard to believe that, say sodium 2-methyl-2-octanoate, would not be quite soluble in Shellsol or equivalent.

Much longer chain alkoxides would need much higher catalyst/hydroxide ratios. For KOH the ratio is currently roughly 0.1 ml of t-butanol / 1 g of KOH) (for NaOH that would already be higher to account for the lower MW of NaOH). For a t-alcohol with twice the chain length as t-butanol that ratio would approx. double and for, say C20, it would have to be multiplied by about 5! The catalyst cost then becomes a real factor… And these octyl-nonyl-decyl precursors might not come very cheap!

watson.fawkes - 30-12-2010 at 10:13

Quote: Originally posted by blogfast25  
Much longer chain alkoxides would need much higher catalyst/hydroxide ratios.
Perhaps. If the solubility increases and subsequently transfer mobility does likewise, you'll certainly need less catalyst per mole. You might also need less per gram.

blogfast25 - 31-12-2010 at 07:45

@Watson:

Probably. The reaction rate in the case of a successful 2-methyl-2-alkanol will depend mainly on:

• Volatility: as pointed out by you early on in the K thread, volatile alcohols will concentrate more in the vapour phase. Higher boiling t-alcohols have the edge there.

• The equilibrium constants of the initiation, propagation and termination steps are all likely to be a function of n, the length of the alkane backbone.

I wonder also if it would be worth revisiting primary or secondary alcohols but long chain ones: octanol, nonanol or decanol for instance. So far no one has explained why the alcohol has to be a tertiary one and it is possible the patent authors chose t-butanol because at least it’s slightly less volatile than short chain primary alcohols.

Or how about 2-methyl-2-ols derived from the naphthenic acids 'isomer' mix:

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



[Edited on 31-12-2010 by blogfast25]

blogfast25 - 31-12-2010 at 07:48

Some sources of potentially interesting precursors:

Berje (US):

Methyl heptylate (?? Methyl heptanoate ??)
Methyl-2-nonenoate (??)
Methyl nonyl ketone
Nonenal-cis-6
Nonenol-cis-6
3-octanol

Web: http://www.berjeinc.com/chemmp.html

Oxford Chemicals (UK):

Methyl octanoate

http://www.oxfordchemicals.com/oxford/ocweb.nsf/0/18510EF22B...

Grau aromatics (Germ.):

Capric acid (decanoic acid)
Caproic acid (hexanoic acid)
Capylic acid (octanoic acid)
Heptanol
Valeric acid (pentanoic acid)
Lauryl alcohol (dodecanol)
Myristic acid (tetradecanoic acid)
Octanol
Pelargonic acid (nonanoic acid)

http://www.grau-aromatics.de/fileadmin/dokumente/flavourfrag...

Penta group:

Probably the largest inventory, includes 1-octanol, 2-octanol, 3-octanol

http://www.pentamfg.com/query.asp?page=193

SAFC global:

Methyl octanoate 99 %

http://www.safcglobal.com/catalog/ProductDetail.do?N4=W27280...


Ungerer (global):

ethyl hexanoate, octanoate, decanoate

http://www.ungererandcompany.com/content/view/23/33/

Source page:

http://www.thegoodscentscompany.com/data/rw1008791.html

blogfast25 - 31-12-2010 at 09:16

Or isopropyl myristate – isopropyl tetradecanoate: the non-aqueous component of the two-phase mouthwash, Dentyl pH.

Ready for double Grignarding with CH3MgBr to 2-methyl-2-pentadecanol? It doesn’t get more OTC than that!

[Edited on 31-12-2010 by blogfast25]

bbartlog - 31-12-2010 at 10:42

Quote:
So far no one has explained why the alcohol has to be a tertiary one


Nicodem offered what sounded like a pretty good explanation somewhere in this thread.

(edit): actually it was in the potassium thread, not this one.

[Edited on 31-12-2010 by bbartlog]

not_important - 31-12-2010 at 12:07

Another source of middling length carboxylic acids is 'green' weed control products. Some of these use C8-C12 straight chain acids, with n-nonanoic (pelargonic) acid being the most effective and thus favored in products. Generally the acid(s) is mixed with some hydrocarbon solvent, add NaHCO3 or Na2CO3 solution to tie up all the acid as the salt, then steam distilling off the hydrocarbons (the free acids are pretty goat-like smelly, keeping them as the salt until needed is not a bad idea).


blogfast25 - 31-12-2010 at 12:23

Quote: Originally posted by not_important  
Another source of middling length carboxylic acids is 'green' weed control products. Some of these use C8-C12 straight chain acids, with n-nonanoic (pelargonic) acid being the most effective and thus favored in products. Generally the acid(s) is mixed with some hydrocarbon solvent, add NaHCO3 or Na2CO3 solution to tie up all the acid as the salt, then steam distilling off the hydrocarbons (the free acids are pretty goat-like smelly, keeping them as the salt until needed is not a bad idea).



Any trademarks spring to mind? These would be about the right length IMHO and could be easily esterified back to methyl or ethyl esters.

[Edited on 31-12-2010 by blogfast25]

smuv - 31-12-2010 at 12:46

While experimentation with sodium/potassium production isn't really my thing; I would not mind making these tertiary alcohols for others to experiment with (i'd actually enjoy making a product for a specific purpose).

I think grignard is not a good way to do this, because, its a pretty expensive and time consuming process, when you consider large volumes of ether, dry solvents, preparing alkyl halides etc. I have thought of some practical alternatives.

1. BHT. While BHT is not a tertiary alcohol, it is a phenol, so no posibility of alpha hydride elimination. Hopefully it will not be metallated at any position other than the OH group. Once the OH group is deprotanated the protons of the ring become pretty basic. BHT is available for less than $15 a pound and should be very fat soluble.

2. Pinnacol coupling of a long chain ketone. This will form a diol composed of 2 tertiary alcohols.

3. Isomerize a branched readily available secondary alcohol to a tertiary alcohol.

4. Oxidize a branched paraffin to an alcohol. This can be done with KMnO4 under certain conditions (do a patent, search isobutane to t-butanol, look at earlier patents)

I think the ones to focus on are 1 and 3. Option 1 because BHT is directly OTC and might work. Option 3 because I have an easy and cheap method in mind starting from menthol. In theory, 2 is good and easy but the chain lengths will need to be very long, because per molecule you will have two alcohols that will be deprotonated. 4 seems good on paper, but I just can't think of a pure source of branched parafins. Also, like the pinnacol problem, multiple branches would lead to polyols, reducing solubility.

Therefore, I propose BHT be researched and also 1-isopropyl-4-methyl-cyclohexanol, which can be made in just a couple of steps from menthol. A scheme is attached below (I haven't bothered to show stereochemistry), I haven't worked out the best method for hydration of the olefin, but It is something I can definitely work out.

untitled.png - 5kB


[Edited on 12-31-2010 by smuv]

not_important - 31-12-2010 at 13:03

Search for "fatty acid weed control". Safer's Superfast and/or Topgun Weed Control, SCYTHE (almost 60% pelargonic) , and Weed-Aside, are ones that I remember but there I'm sure there are others.

not_important - 31-12-2010 at 13:07

Quote: Originally posted by smuv  
(big snip)...

4. Oxidize a branched paraffin to an alcohol. This can be done with KMnO4 under certain conditions (do a patent, search isobutane to t-butanol, look at earlier patents)

...


Also bromination of paraffins under fairly mild conditions where the hydrogen reactive will be tert > sec > primary. Do it under fairly mild conditions, treat the reaction mix with cold NaHCO3 solution (to reduce the amount of elimination) to get the alcohols.

blogfast25 - 1-1-2011 at 09:15

@not_important:

Thanks for both these posts. Read and noted… Petroleum jelly (Vaseline) is probably full of branched paraffins. Not sure how you go about brominating that though...

@smuv:

Agreed of course on the Grignard reaction as being difficult and fairly wasteful. At some point we’re going to have to look into catalyst recovery which would make that less important but for now we need to take into account that even small amounts of a target alcohol (which might then prove not to work!) are resourceful..

All of your ideas are appealing, especially as we may have to look at alcohols with cyclic/aromatic functionality. But as this is a ‘labour of love’ we’ll also need some ‘division of labour’…

Personally I’ll probably pursue the isopropyl myristate to 2-methyl-2-pentadecanol route because it will give me a ‘Grignard experience' and because the alcohol should work in highly paraffinic solvents and may be recoverable/reusable from it…

BTW: what software do you use to depict those wonderfully clear structural formulas?


[Edited on 1-1-2011 by blogfast25]

blogfast25 - 1-1-2011 at 09:48

Another possible and very OTC precursor may be naphthalene (moth balls). Nurdrage got very good results by using Tetralin, the naphthalene derivative, as a solvent for K production. This may be due to higher alcohol solubility in Tetralin as opposed to in more paraffinic solvents. This could possibly be taken advantage of by means of naphthalenic functionality in the t-alcohol.

smuv - 1-1-2011 at 10:50

I use chemdraw, actually IMO the image isn't that clear because I didn't put much effort into resizing the image with a photo editor.

I don't think naphthalene is a good idea because naphthalenes are generally easier to metallate and reduce than phenyl derivatives. While tetralin is a naphthalene derivative it has no naphthalene ring. Therefore in terms of acidity and ease of oxidation tetralin is more similar to a dialkyl benzene derivative.

About getting the best use per gram of (pre-)catalyst, I think instead of catalyst recovery, effort should be made to make a semi-continuous batch process. I see it going like this.

1. Dehydrate KOH by heating with Mg as in the first part of the patent. Either allow the solution to settle and siphon off the bulk of the solvent or vac filter the solution. Save this solvent, it will be reused as "dehydration solvent".

2. Add the dehydrated KOH and Mg to another solution, the "reduction solution". The reduction solution is a mixture of the solvent and pre-catalyst you are using. Heat this up as usual and reduce the potassium hydroxide. When the run is done, filter, to recover the "reduction solution" (and potassium).

3. Repeat the process with a new batch, re-using the recovered "dehydration solvent" and "reduction solution".

This of course will only work with an alcohol that is not appreciably lost at the reaction temps.





blogfast25 - 1-1-2011 at 12:50

On the naphthalene I was simply thinking it might be a cheap and easy precursor. Surely you wiz kids can crack up one of the resonance rings and graft a 2-methyl-2-ol on there? ;-)

Regarding recovery/reuse of the solvent, assuming indeed the t-alcohol is the catalyst and that none of it gets lost through degradation or evaporation, then at the end of the reaction when all Mg has been used up the alcohol should be in the dissolved potassium alkoxide form. Separate the liquid from the solids and it should strictly speaking be reusable as such: adding fresh KOH (with water contained in it) and Mg should hydrolyse the K alkoxide wholly or partly to KOH + alcohol, the drying process then continues via Mg + H2O and after that the reduction reaction should proceed. No one has yet attempted to recycle the solvent + catalyst in that way.

--------------------------

Looking at the availability of longer chain precursors for 2-methyl-2-alkanols by Grignard, the choice is basically between 2-ketones and methyl or ethyl alkanoates, the latter requiring twice the amount of methyl magnesium iodide (a serious cost factor because of the iodine).

Of the equivalent 2-ketones and methyl or ethyl alkanoates the ketones are more expensive. And the cost of any of these precursor (at least going by Sigma-Aldrich [SA] and similar) rises almost exponentially with chain length. From about C6 the price starts getting expressed in gram amounts, rather than litre amounts.

The most affordable are:

2-pentanone – methyl propyl ketone – MPK (SA: £23.50/L),
2-hexanone – methyl butyl ketone – MBK (SA: £30.50/50 gram and
4-methyl-2-pentanone – methyl isobutyl ketone MIBK (SA: £32.80/L)
2-octanone – methyl hexyl ketone – MHK (SA: £22.10/500 gram)

Higher ketones become seriously expensive: £29.60/100 gram from MPbio.com for methyl nonyl ketone(MNK) – 2-undecanone. This MNK is also an active ingredient in cat/dog repellent products like ‘Get off my garden’ sprays and granules but the active ingredient appears to be about 2.5 g/500 g. 2-octanone being a slight exception to the rule: cheaper than 2-hexanone (?!?)

MIBK would be a precursor for 2,4-dimethyl-2-pentanol with an estimated BP of about 140 – 150C. But it’s not much longer than the working 2-methyl-2-butanol (t-amyl alcohol, BP ≈ 100C).

MHK would be a precursor for 2-methyl-2-octanol, about the right length, methinks. (1-) octanol has a BP of 195C, so we should also be in the right ball park on volatility…


[Edited on 2-1-2011 by blogfast25]

blogfast25 - 3-1-2011 at 08:05

Another seriously OTC candidate for a precursor of longer chain 2-methyl-2-alkanols is… fractionated coconut oil!

Acc. Wiki:

Fractionated coconut oil is a fraction of the whole oil, in which the long-chain fatty acids are removed so that only medium chain saturated fatty acids remain. Lauric acid, a 12 carbon chain fatty acid, is often removed as well because of its high value for industrial and medical purposes. Fractionated coconut oil may also be referred to as caprylic/capric triglyceride oil or medium chain triglyceride (MCT) oil because it is primarily the medium chain caprylic (8 carbons) and capric (10 carbons) acids that make up the bulk of the oil.

Being glycerol esters this oil could be Grignarded straight to the t-alkanols or hydrolysed to the acids for work up (caproic (C6) and caprylic acid (C8) are both soluble in methanol, 7.98 and 6.31 M, capric acid is soluble in alcohol and ether).

Nicodem - 3-1-2011 at 13:02

I guess the most practical t-alcohol to chose as a target would be 2-methylhexan-2-ol which can be made from n-butyl bromide and acetone via grignard addition:


JACS, 51, 1483 (1929) (attached + a later related paper)

n-Butyl bromide can be made from the fairly easily obtainable n-butanol with the NaBr/H2SO4 approach as described on this forum elsewhere on various lower alcohols.

I could not find any preparative synthesis of fatty t-alcohols from triglycerides, but there are a few articles where there is described that the treatment of triglycerides with alkylmagnesium gives the corresponding t-alcohols. For example, here is the CA abstract of one such paper where the use of the reaction for analytical purposes is discussed:
Quote:
DOI: 10.1007/BF02534006
Rapid analysis of jojoba wax fatty acids and alcohols after derivatization using Grignard reagents.
Pina, M.; Pioch, D.; Graille, J. Dep. Oleagineux, Cent. Coop. Int. Rech. Agron. Dev., Montpellier, Fr.
Lipids (1987), 22(5), 358-61. CODEN: LPDSAP ISSN: 0024-4201. Journal written in English. CAN 107:42039 AN 1987:442039 CAPLUS

Abstract: The title method, based on the action of Grignard reagents, differed from methods previously described, which were essentially based on wax alcoholysis. Grignard reagents, esp. MgEtBr, reacted on ester functions to turn the wax constituents into primary and tertiary alcs., the latter being the fatty acid derivs. The mixt. of these alcs. was analyzed by a single gas chromatog. injection. The overall time, .apprx.1.5 h, made this method suitable for routine anal. It could be also considered for analyzing low-carbon-condensation org. acid esters.


Attachment: t-alcohols via grignard reactions.pdf (352kB)
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Attachment: t-alcohols via grignard addition on acetone.pdf (574kB)
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Eclectic - 3-1-2011 at 20:01

OK. Has anyone here ever heard anything about Barbier-Gringnard reactions? Alkyl halide, zinc dust, water as solvent? It's not exactly rocket science....


Also kind of a hot topic for "Green" chemistry.

blogfast25 - 4-1-2011 at 08:06

Nicodem:

Interesting how it would seem possible to directly Grignard the triglycerides. But it does cost about 6 mol of GR per mol of triglyceride and needs the expensive CH3I as a source of alkyl.

The reactions using acetone as a substrate are far more attractive in that respect, thanks for both *.pdfs. Acetone is dirt cheap and easy to dry. For the 2-methyl-2-alkanols, the precursors would be respectively 1-chloro(butane, pentane, hexane, heptane, octane etc) and these are relatively affordable [cough!], going by Sigma-Aldrich:

Per 100 ml: C4 = £26.8, C6 = £19.00, C7 = £29.20, C8 = £17.30.

The butyl chloride could be made cheaper by chlorinating n-butanol (£15.30/500 ML) but I think the resulting alcohol may still be a bit short (a C6). Also 1-hexanol (£8.20/100 ML) isn’t too bad and the resulting C8 alcohol would be in the right BP area.

As ‘natural’, OTC precursors for C6-C8 ols, some essential oils may be considered: citronella oil contains two primary alcohols, geraniol and citronellol, both of which could be chlorinated quite easily but then may be difficult to completely separate from the rest of the oil. And OTC essential oils are very expensive and probably best left as a posh person’s belief system…

Quote: Originally posted by Eclectic  
OK. Has anyone here ever heard anything about Barbier-Gringnard reactions? Alkyl halide, zinc dust, water as solvent? It's not exactly rocket science....


Also kind of a hot topic for "Green" chemistry.


Ecclectic, so do you believe the ‘one pot reaction’ (as it’s described in Wiki):

Octyl chloride + acetone + Zn powder (all in water) === > 2-methyl-2-decanol + Zn(OH)Cl

… would effectively proceed? Octyl choride is of course not miscible with water to begin with… I somehow doubt that.


[Edited on 4-1-2011 by blogfast25]

ScienceSquirrel - 4-1-2011 at 08:58

Quote: Originally posted by Eclectic  
OK. Has anyone here ever heard anything about Barbier-Grignard reactions? Alkyl halide, zinc dust, water as solvent? It's not exactly rocket science....


Also kind of a hot topic for "Green" chemistry.


It should be noted that the Barbier-Grignard reaction is specific for propargyl or allyl halides that are highly activated.
Plain vanilla halides like propyl bromide will not work as far as I know.

blogfast25 - 4-1-2011 at 09:56

How do the reactivities of Grignard reagents compare for Cl:Br:I (all other things being equal)? I see a lot of references to bromides being used, rather than chlorides. Sigma offer 1-bromooctane for £19.20/100 g.

Edit: reactivity seems to go I > Br > Cl

And what are good procedures to sufficiently dry the reagents?


[Edited on 4-1-2011 by blogfast25]

garage chemist - 5-1-2011 at 12:27

Bromide grignards are easier to start and react more smoothly than the chlorides.
Also, the alkyl bromides are very easy to make from the alcohol, KBr and H2SO4. 1-Butanol is a cheap precursor.
The chlorides are less easy to make, as HCl doesn't readily react with primary alcohols unless a catalyst like anhydrous ZnCl2 is present.

blogfast25 - 5-1-2011 at 12:52

I see.

Sigma’s offer of £6.60 / 100 gram (0.606 mol) of 1-bromohexane is attractive. At assumed conversion rate of 90 % with acetone it would make about 80 g of 2-methyl-2-octanol.

From that same source, 1-bromobutane seems expensive: £5.70 for 25 gram, almost quadruple of the 1-bromohexane! Of course you can brominate 1-butanol but HBr isn’t cheap either…

DJF90 - 5-1-2011 at 13:01

Quote:
Of course you can brominate 1-butanol but HBr isn’t cheap either…

Hence why you use KBr and H2SO4. An excess of H2SO4 aids the reaction. NaBr will also substitute.

NurdRage - 5-1-2011 at 13:37

I was thinking, before we go off and start finding ways to make tertiary alcohols, it might be more prudent that we verify a particular long-chain alcohol works before we try to make it.

Alfa Aesar has a huge array of alcohols for purchase, a fair number are tertiary.

Just to name a few:
4-methyl-4-nonanol (C10)
2,6-Dimethyl-2-heptanol (C9)
3-ethyl-3-hexanol (C8)

While they are not cheap, if one of them works then we know we're in the right area and can refocus our attention on making that particular alcohol or its isomer. I think this would be a more efficient approach overall because It would be rather wasteful if we spent so much time and money trying to make an alcohol that turns out not to work.

blogfast25 - 5-1-2011 at 14:09

Quote: Originally posted by DJF90  
Hence why you use KBr and H2SO4. An excess of H2SO4 aids the reaction. NaBr will also substitute.


Yes but the REAL cost is the Br, in whatever form you use it… Using NaBr/KBr + conc. H2SO4 isn't going to be substantially cheaper. Not to mention yield and work up concerns, as opposed to simply buying the n-alkyl bromide. I challenge anyone to match SA's n-hexyl bromide with a homebrew product in total cost.

Quote: Originally posted by NurdRage  
I was thinking, before we go off and start finding ways to make tertiary alcohols, it might be more prudent that we verify a particular long-chain alcohol works before we try to make it.

Alfa Aesar has a huge array of alcohols for purchase, a fair number are tertiary.

Just to name a few:
4-methyl-4-nonanol (C10)
2,6-Dimethyl-2-heptanol (C9)
3-ethyl-3-hexanol (C8)

While they are not cheap, if one of them works then we know we're in the right area and can refocus our attention on making that particular alcohol or its isomer. I think this would be a more efficient approach overall because It would be rather wasteful if we spent so much time and money trying to make an alcohol that turns out not to work.


Broadly speaking agreed: it would be worthwhile checking whether 2-methyl-2-ol functionality is the only thing that works or whether tertiary 3-ols or 4-ols do the magic too… I’m gonna have a look at these products from Alfa Aesar, Nurdrage.

Edit:

That 2,6-Dimethyl-2-heptanol, for instance:

http://www.alfa.com/en/GP100W.pgm?DSSTK=B25691&rnd=72439...

£14.50 / 50 gram (BP 180C) is a bit of a steal, IYAM! The others you mentioned seem a little outside of my league on price...


[Edited on 5-1-2011 by blogfast25]

Nicodem - 5-1-2011 at 14:25

Tetrahydromyrcenol (2,6-dimethyloctan-2-ol, 18479-57-7) seems to be the cheapest saturated long chain t-alcohol available via chemical vendors. Si**a sells it for 54 EUR/kg which is in the price range of solvents. All the other long chain t-alcohols are quite expensive, 2-methylhexan-2-ol included.

NaBr, KBr or NH4Br can be used for the bromination of lower primary and secondary alcohols. I never applied this method on n-butanol, but on isobutyl alcohol (Me2CHCH2OH) it gave me a 33% yield of isobutyl bromide. However, I'm sure the reaction can be easily optimized to give yields above 60% on isobutanol, n-butanol and possibly also on amyl alcohols and other alcohols immiscible in aqueous H2SO4. Otherwise, butyl bromide is very cheap, but making it is fun and of good pedagogic value. Like I said, this method is not very suitable for higher alcohols, but these can be made into bromides by refluxing them with 48% HBr(aq). This way I made 3-phenylpropyl bromide in good yields from 3-phenylpropan-1-ol. The reaction is extremely slow at room temperature (only traces of bromide form), but only takes a couple of hours at reflux with good stirring (biphasic mixture).

blogfast25 - 6-1-2011 at 03:45

@Nicodem: Excellent find on the tetrahydro myrcenol (2,6-dimethyloctan-2-ol, 18479-57-7)! SA UK = £38 / kg, £ 15 for a(n unspecified) sample. BP ≈ 200C. This is probably a derivative from a natural product, hence the lower price and the trivial name.

@All: I’ve rearranged the proposed reaction mechanism a little bit, to try and focus the mind, here (the K thread):

http://www.sciencemadness.org/talk/viewthread.php?tid=14970&...

The four crucial reactions are:

2 KOH(s) + 2 ROH(sol) < == > 2 KOR(sol) + 2 H2O(sol) (I)

2 KOR(sol) + Mg(s) < == > 2 K(l) + Mg(OR)2(sol) (II)

Mg(OR)2(sol) + H2O(sol) < == > MgO(s) + 2 ROH(sol) (III)

K(l) + 2 ROH(sol) < == > 2 KOR(sol) + H2(g) (IV)

The third one is almost certainly the thermodynamic driver but probably not the rate determining step (RDS). Probably either the K alkoxide formation or the K alkoxide reduction is the RDS.

Assuming I’m seeing this right, from there it might be possible to predict a ‘winning alcohol’. Which kind of boils down to some relatively simple choices [cough!]:

1. tertiary alcohol function in 2 position or other position?
2. methyl, ethyl or other functions on the tertiary C-OH?

Personally I would have thought that longer R groups in a tertiary alcohol R-OH would render the -OH group more active because the R group is electron rich, increasing the partial charge δ- on the electrophile oxygen atom. So that would favour ethyl functionality over methyl functionality. By this reasoning 3-ethyl alka-3-ols would be more ‘active’ than 2-methyl alkan-2-ols, at least with respect to reactions (I) and (IV)…

[Edited on 6-1-2011 by blogfast25]

garage chemist - 9-1-2011 at 14:50

Yesterday and today I made a batch of 1-bromobutane.
I used:

50g 1-butanol
70ml conc. H2SO4
40ml water
100g KBr

Butanol and H2SO4 were mixed, then the water added (strong exotherm) and the mix poured upon the powdered KBr.
It was refluxed for about an hour and then slowly distilled over a free flame over the course of 2 hours.
The lower layer of the biphasic distillate was separated, washed twice with 10ml conc. H2SO4, then with water, Na2CO3 solution and water again, shaken with CaCl2 until clear, and distilled over phosphorus pentoxide.
It came over completely at 100-101°C.
I obtained 79g, an astoundingly high 85% yield! I never had that kind of yield with the ethyl bromide prep.
In this 1-bromobutane prep, butanol is the limiting reagent and KBr is present in slight excess, while in the EtBr prep I generally use an excess of ethanol.

The 1-bromobutane thus prepared has a slightly pungent smell component and fumes a slight bit in air. I wonder how this comes. I know this phenomenon from purified chloroform that's been distilled over P2O5- it also fumes slightly, which I have no plausible explanation for.

blogfast25 - 10-1-2011 at 05:30

Hey, that’s fantastic GC, great yield too! Hopefully the fuming won’t interfere with the Grignard reaction the 1-bromobutane is intended for.

Nicodem - 10-1-2011 at 11:10

Good work GC! I always suspected that with some reflux the bromide/H2SO4(aq) system would work the same as HBr(aq) also for longer alcohols. Nice to see it does, as bromides are much easier to come by than HBr. The yield is good probably only because of lower losses due to lower volatility. My yield of 1-bromo-3-phenylpropane was also thereabout, even after vacuum distillation and completely pure by 1H NMR (no alcohol, alkene, ether or isomers). So refluxing in in highly acidic media does no damage in this reaction. Though, the fuming of your product indicates perhaps there is still some HBr inside. In my opinion, this should not prevent the magnesium insertion initiation. The HBr should even etch the Mg surface for better reaction with n-BuBr.
Please post your results in Prepublication section if you ever manage to make the whole preparation till 2-methyl-2-hexanol.

garage chemist - 10-1-2011 at 11:25

I did not take pictures of the BuBr synthesis- there is nothing really special to it, it went completely according to textbook.

When I make the grignard I will take pictures since you showed interest.

The lower volatility of BuBr is certainly an important factor that gives improved yields compared to EtBr. Every distillation of EtBr entails significant losses to the atmosphere.
But I also felt that the reaction itself went much faster with BuBr since the reaction temperature was higher.
EtBr forms only slowly in the boiling reaction mixture, requiring slow distillation over 6 hours or more.
With BuBr the alcohol has a higher boiling point and allows a higher reaction temperature to be reached which seems to be beneficial.

blogfast25 - 16-1-2011 at 14:10

I have a cheap precursor to 1-dodecene, namely lauryldimethylamine oxide:

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

… which upon mere heating snips off the amine base to yield 1-dodecene (Cope reaction).

Now I’m wondering if that can be converted to 1-bromodecane fairly easily (or is this a case of 'close but no cigar'?) I’ve been reading up on anti-markovnikov hydrobrominations (requiring peroxide (!) catalysis) to force the halogen into 1- position. I'm guessing that by 'peroxide' here has to be understood the real deal, i.e. organic peroxides.

Anybody here have some sound ideas? SM searches on ‘markovnikov’ or ‘anti-markovnikov’ yielded little of any use.

I can get a litre of lauryldimethylamine oxide for 4 quid, sounds too good to pass up...

[Edited on 16-1-2011 by blogfast25]

garage chemist - 16-1-2011 at 14:22

Today I tried to make Na/K alloy in Dioxane.
The potassium coalesces perfectly in the hot solvent, but the Na/K alloy stubbornly refuses to do the same! It just forms shiny drops that divide into smaller drops as the vial is swirled, but the drops never unite to bigger ones at any temperature, not even when squeezed together with a spatula. Very annoying.
It might have something to do with the fact that I used dioxane that is probably saturated with KOH. The next time I will use fresh solvent.
The synthesis of tert-heptanol has to wait until I have made Na/K alloy and dried my ether with it.

blogfast25 - 19-1-2011 at 08:49

Methyl nonyl ketone (2-undecanone) and to lesser extent methyl heptyl ketone (2-nonanone) are the prime ingredients (>90 %) of the essential oil ‘Rue oil’ (aka Ruda or Ruta oil, an essential oil derived from Ruta graveolens). Very OTC but not very cheap…

Precursors to 2-methyl undecan-2-ol and 2-methyl nonan-2-ol.

garage chemist - 19-1-2011 at 14:52

I now made the Na/K alloy in fresh dioxane, but found that dioxane is not very suitable for coalescing sodium or Na/K alloy. Molten sodium does not lose its crust in dioxane and I had to hold the solid sodium into the molten potassium for over a minute until it started to dissolve. I did not swirl the flask in order to not divide the liquid alloy into smaller droplets, and sucked the alloy into a 2ml syringe for easy dispensing into ether.
So from now on I will make the alloy under kerosene or Shellsol, perhaps with some t-BuOH added.


blogfast25 - 20-1-2011 at 14:04

Very strange how it works for K but not Na or NaK... A bit worrying too: I was hoping to use dioxane (assuming I can get my hands on some!) for any 'chemical' Na ('pok's thread'). You are certain it works well for K, yes?

blogfast25 - 22-1-2011 at 09:12

I’m having considerable trouble getting hold of most of the materials I identified above. As usual the Big Boys aren’t willing to sell to t’Little Guy. No tetrahydro myrcenol, n-octanol or n-hexanol, nor derivatives neither. Of these alcohols many are used in food flavourings/fragrances as esters (often acetates) but are hard to obtain as pure substances. The only one I’ve found from a supplier that sells OTC is (quite expensive) amyl (pentyl) acetate from Kremer Pigments (G.) That can be converted to n-pentanol, then n-chloropentane, then used as Grignard on acetone to yield 2-methyl-2-heptanol.

Kremer also sells cyclohexanol which after bromination could be used as Grignard to yield 2-cyclohexyl-2-propanol. Not sure whether the sodium alkoxide would be well soluble in Shellsol D70 or similar.

Then there’s ‘essential oil of rue’ (OTC but not cheap), supposedly containing mainly methyl octyl ketone and methyl heptyl ketone. These should be purifiable with steam distillation and MgSO4 drying, and Grignarded with MeI to 2-methyl-2-alkanol or with EtBr to 3-ethyl-3-alkanol.

And isopropyl myristate as the organic phase of ‘Dentyl’ mouth wash can be double Grignarded (that becomes expensive) with MeI to 2-methyl-2-alkanol or with EtBr to 3-ethyl-3-alkanol. Very long chain alcohols…

Fractionated coconut oil is a triglyceride of mainly C8 and C10 fatty acids, which could be double Gignarded with MeI to 2-methyl-2-alkanols or with EtBr to 3-ethyl-3-alkanols. Alternatively, saponify, isolate the fatty acids, re-esterify to methyl esters and double Grignard these.


Also, pentyl magnesium chloride as an alkylating agent used on MEK would yield 3-methyl octan-3-ol, a fairly long chain tertiary 3-methyl-3-alkanol...

[Edited on 23-1-2011 by blogfast25]

blogfast25 - 26-2-2011 at 11:32

Thanks to ScienceSquirrel and IPN over at the potassium thread:

http://www.sciencemadness.org/talk/viewthread.php?tid=14970&...

… another useful idea has emerged: synth. of the tertiary alcohol α-terpineol: http://en.wikipedia.org/wiki/Terpineol

Acc. several literature references hydration of α-pinene in excess acetone in the presence of dilute sulphuric acid at 80-85 C oil bath leads to terpineol being formed. The excellent *.pdf unearthed by IPN (first link above) gives several examples. One example cited involves 2 g of turpentine oil (source of pinenes and VERY OTC), 4 ml of a 15 v/v% H2SO4 solution in an excess of 25 ml of acetone, with maximum conversion of pinenes to terpineols occurring at about 4 h at 80-85 C.

Ideas for work up (essentially elimination of the excess acetone) of the terpineol from organic wizzkids here would be much appreciated…

ScienceSquirrel - 27-2-2011 at 03:03

I would suggest;

1) add the reaction mixture to water / ice / diethyl ether

2) wash with sodium carbonate

3) dry and distill

Eclectic - 27-2-2011 at 05:47

Skip step 1? Just neutralize acid, remove alcohol or acetone solvent that was used to make turpentine miscible with water. The terpineol boils at about 220 C. Another approach might be to make terpin hydrate, then dehydrate that to terpineol in refluxing glacial acetic acid or dilute phosphoric. Terpin may be worth a try too.

blogfast25 - 27-2-2011 at 08:12

The folks from ‘The Good Scents Company’ tend to be quite useful for stuff that’s used in perfumery or as food additives (data on α-terpineol):

http://www.thegoodscentscompany.com/data/rw1011251.html

It would appear to be only slightly soluble in water. How about, after neutralisation, steam distillation to get rid of the acetone (is terpineol actually soluble in acetone?) and possible also other low boiling contaminants?

Some of the stuff that accompanies the α-terpineol seems harmless enough for the product’s purpose but it would be nice to get to at least > 90 % purity, if only to be sure how much you’re using when dispensing it. So how about maybe a few solvent extractions with low boiling solvents like DCM and maybe others to try and extract a maximum of the junk? Steam distil again to get rid of these solvents…

ScienceSquirrel: what’s the idea behind the ether?

Ecclectic: isn’t glacial acetic acid refluxing likely to lead to acetates of α-terpineol? The pinenes lead to acetates with glacial acetic acid (see Wiki on pinenes) which may be the other route: turps + glacial acetic acid -> terpineol acetate. De-esterify to alcohol.

Eclectic - 27-2-2011 at 13:20

I found a few routes to terpineol googleing terpineol with terpin hydrate.
At least one recommended refluxing terpin hydrate with glacial acetic acid, another with dilute phosphoric.
Terpin hydrate itself might work, and is easy to isolate from turpentine as crystals.

Things what can be made from turpentine seems almost as interesting as makin' potassium!

[Edited on 2-27-2011 by Eclectic]

UnintentionalChaos - 1-3-2011 at 16:24

I bring you a painfully cheap solution to your woes! Shipping is rather slow, I warn you.

http://www.newdirectionsaromatics.com/dimethyl-benzyl-carbin...

This stuff is cheap as dirt! For those not up on their old style names, it's 1,1-dimethyl-2-phenylethanol.

[Edited on 3-2-11 by UnintentionalChaos]

NurdRage - 1-3-2011 at 19:46

great find, i'm getting it.... like now!

UnintentionalChaos - 1-3-2011 at 23:07

http://www.thegoodscentscompany.com/msds/md101937.html lists the b.p. as 215.

For some reason the MSDS from new directions lists a much lower b.p. I suspect this was under reduced pressure (and they missed listing this). I'd find it very hard to believe 108C given that unsubstituted 2-phenylethanol boils at 219C.

Although it may not be worth the effort for everyone else, I am still interested in alpha-terpineol since I like scents and it can be made from limonene (almost all of orange essential oil, which you can also distill yourself from orange peels). Relevant paper attached

[Edited on 3-2-11 by UnintentionalChaos]

Attachment: op068012d.pdf (33kB)
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blogfast25 - 2-3-2011 at 09:10

UC: well done that man!

1,1-dimethyl-2-phenylethanol at lookchem:

http://www.lookchem.com/cas-100/100-86-7.html

This is indeed very similar to the 2-methyl-alkan-2-ol desired structure. And the phenol func. comes bearing promise of possible further functionalisation too. Now I need a source in Europe, this is too good to pass up. And yes, atm. BP must be > 200 C, which is great!

On the limonene -> alfa-terpineol route: trifluoroacetic acid is a bit of an obstacle.

You sure orange essential oil is mainly limonene?



[Edited on 2-3-2011 by blogfast25]

turd - 2-3-2011 at 09:47

Quote: Originally posted by UnintentionalChaos  
For some reason the MSDS from new directions lists a much lower b.p. I suspect this was under reduced pressure (and they missed listing this). I'd find it very hard to believe 108C given that unsubstituted 2-phenylethanol boils at 219C.

108°C is at ~13 mmHg. I wonder how easily the thing does elimination?

UnintentionalChaos - 2-3-2011 at 09:50

Quote: Originally posted by blogfast25  

You sure orange essential oil is mainly limonene?


http://www.newdirectionsaromatics.com/msds/orangesweetmsds.h...

http://www.newdirectionsaromatics.com/msds/orangesweetgc.htm

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

Attachment: isolating natural compounds.pdf (53kB)
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blogfast25 - 2-3-2011 at 13:18

@ UC:

Thanks for nuttin’: now I won’t be able to eat an orange w/o trying to imagine a way to convert the limonene to terpineol but using no CF3COOH! :D (come on organic wizzkids!)

Interesting *.pdf too…


That New Direction Aromatics shop sells other interesting stuff:

Neryl actetate: an unsaturated long chain acetate:

http://www.thegoodscentscompany.com/data/rw1033551.html

‘Octanyl’ acetate: they mean octyl acetate:

http://www.newdirectionsaromatics.com/msds/octanylacetate.ht...

Both are precursors to long alcohols, that could be chlorinated or brominated to Grignard reagents for alkylation of acetone to 2-methyl-2-ol type t-alcohols….

Even linalool is a t-alcohol that would be worth just testing for K-synth. if I could get my hands on some:

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

[Edited on 2-3-2011 by blogfast25]

Eclectic - 3-3-2011 at 07:51

Could someone with a reflux/distillation setup try refluxing equal parts turpentine and glacial acetic acid with 1 drop of sulfuric for 4-24 hours, followed by distilling off any excess acetic acid and neutralizing the residue with NaHCO3/water? Floral scent would indicate terpineol and/or it's acetate, citrus would mean it went all the way to limonene.

NurdRage - 3-3-2011 at 20:46

i tried ordering the carbinol, but they said it would be on back-order for 6+ weeks or so. Anyone else had any luck getting from them?

smuv - 3-3-2011 at 23:54

@Blogfast, use formic acid, it is the very old way of doing this. Chloroacetic acid also works. Oxalic acid may work as well.

blogfast25 - 4-3-2011 at 07:41

Quote: Originally posted by NurdRage  
i tried ordering the carbinol, but they said it would be on back-order for 6+ weeks or so. Anyone else had any luck getting from them?


In theory they ship to outside N.America but looking at the conditions it’s not worth doing. I’m looking for a similar business in Europe but no luck so far…

Quote: Originally posted by smuv  
@Blogfast, use formic acid, it is the very old way of doing this. Chloroacetic acid also works. Oxalic acid may work as well.


You have any references for that? It would be great if formic acid worked…

smuv - 4-3-2011 at 09:41

@b Unfortunately I don't have anything specific on-hand (I definitely would have posted it if I did ;)). I do know it works, it was an old way of determining where the double bond in natural products was located. Why not run a google books search and see what comes up; definitely confine your search to older literature.

Also, the reason why these acids work, and carboxylic acids like acetic acid don't, is that formic and chloroacetic acid are strong enough to protonate the alkene, giving a carbocation which is vulnerable for attack by the formed carboxylate. Trifluoroacetic acid works in the same way, but it is way overkill, acetic acid on the other hand, just is not acidic enough to protonate the alkene.

All this being said, it is possible that if you reflux the stuff in glacial acetic acid, with a catalytic amount of H2SO4, you may get addition to the double bond.

[Edited on 3-4-2011 by smuv]

blogfast25 - 4-3-2011 at 09:53

Quote: Originally posted by smuv  
Also, the reason why these acids work, and carboxylic acids like acetic acid don't, is that formic and chloroacetic acid are strong enough to protonate the alkene, giving a carbocation which is vulnerable for attack by the formed carboxylate. Trifluoroacetic acid works in the same way, but it is way overkill, acetic acid on the other hand, just is not acidic enough to protonate the alkene.

All this being said, it is possible that if you reflux the stuff in glacial acetic acid, with a catalytic amount of H2SO4, you may get addition to the double bond.

[Edited on 3-4-2011 by smuv]


The Wiki entry on alpha-pinene states explicitly that alpha-terpineol acetate is obtained from refluxing pinene with glacial acetic acid.

And a paper linked to above points to efficient conversion of pinene to terpineol with dilute sulphuric acid in an excess of acetone (refluxing).

Eclectic - 4-3-2011 at 10:01

That's from pinene though, with a strained ring. In
USA, pine turpentine is very cheap, and might be more reasonable starting material than limonene. If you want to start with limonene:

http://www.scielo.br/scielo.php?script=sci_arttext&pid=S...

smuv - 4-3-2011 at 10:04

The sulfuric acid method would be simpler, if it works. The problem is sometimes these things undergo dehydration even with very dilute acids, which is why often the ester is formed and then hydrolyzed. But since you have a reference for this specific case it is worth a shot.

About, the conversion of alpha-pinene to alpha-terpineol acetate, I can't comment about the wiki ref, but comparing the ring opening of a cyclobutane to the addition of a carboxylic acid to a double bond, is not exactly comparing apples to apples.

Anyways, I hope everything works out. Might be the first terpene synth on this forum.

blogfast25 - 4-3-2011 at 12:07

Eclectic and smuv:

Decisions, decisions, eh? Both pine turpentine and orange oil are highly OTC (although trichloroacetic acid is less so). I’m surprised no one on the organics board hasn’t tried these synths yet. There’s a first time for everything, I guess…

MrHomeScientist - 22-3-2011 at 13:31

Quote: Originally posted by blogfast25  
Thanks to ScienceSquirrel and IPN over at the potassium thread:

http://www.sciencemadness.org/talk/viewthread.php?tid=14970&...

… another useful idea has emerged: synth. of the tertiary alcohol α-terpineol: http://en.wikipedia.org/wiki/Terpineol

Acc. several literature references hydration of α-pinene in excess acetone in the presence of dilute sulphuric acid at 80-85 C oil bath leads to terpineol being formed. The excellent *.pdf unearthed by IPN (first link above) gives several examples. One example cited involves 2 g of turpentine oil (source of pinenes and VERY OTC), 4 ml of a 15 v/v% H2SO4 solution in an excess of 25 ml of acetone, with maximum conversion of pinenes to terpineols occurring at about 4 h at 80-85 C.

Ideas for work up (essentially elimination of the excess acetone) of the terpineol from organic wizzkids here would be much appreciated…


I'm working on a process to isolate alpha-terpineol fom turpentine, basically following the procedure laid out in the paper. This will be my first foray into organic chemistry, so sorry if the following questions are naieve:

1) I assume 'turpentine oil' is the same thing as standard turpentine? Specifically, "Klean-Strip 100% Pure Gum Spirits Turpentine" is what's available here and that's what I'd like to use.
2) At the end of the reaction, could you not eliminate the extra acetone by simple heating? Terpineol boils at over 200C, so the acetone and any water should boil off much sooner for easy separation.

Basically, the plan is to follow the procedure outlined above in the paper, neutralize the product solution with a base (baking soda or NaOH), and heat it to boil off the acetone.

Any help would be greatly appreciated!

blogfast25 - 23-3-2011 at 05:43

Thanks MrHomeChemist for wanting to help out here. I was going to run a test today using the same procedure but something’s popped up. Hopefully tomorrow, then we can ‘compare notes’?

Your questions:

1) I bought turpentine paint stripper/thinner and made sure it was ‘the real deal’ by smelling it. If it smells strongly of pine oil, it should contain about 70 %+ of alfa-pinene (acc. info on turpentine in google) and be fit for purpose.

2) As regards the work up of the final product, like you said, firstly neutralise the mix with weak NaOH. I’m assuming the reaction product mix will be a two phase system (two layers), one the alcohol (and other by-products), the other acetone/water and some sodium sulphate (and some other acetone soluble by-products). I believe the alcohol phase will be the bottom one (but won’t put my hand in the fire on that: acetone has a lower density than alfa-terpineol but the acetone phase also contains some (heavier) water). If I’m right on this, then separate the phases and wash the alcohol phase a few times with clean water, this will leach out any remaining acetone (which is highly soluble in water). Then dry the product with some crushed anhydrous MgSO4 or dry Na2CO3.

But, but, but. If no two separate phases are present then distilling off the acetone will be needed. Use steam distillation for this: it’s very safe, low temperature and because acetone is both very volatile and soluble in water it should get rid of the acetone real easily. Then dry the product with some crushed anhydrous MgSO4 or dry Na2CO3.


The end product should be a flagrance with an odour quite different from turpentine: alfa-terpineol is used in perfumery for instance.

You can also check the tertiary alcohol nature of the product by means of potassium dichromate: tertiary alcohols like alfa-terpineol CANNOT be oxidised by it, see here:

http://www.chemguide.co.uk/organicprops/alcohols/oxidation.h...

I think this is an interesting first organic experience for anyone...

MrHomeScientist - 23-3-2011 at 07:15

Quote: Originally posted by blogfast25  
Thanks MrHomeChemist for wanting to help out here. I was going to run a test today using the same procedure but something’s popped up. Hopefully tomorrow, then we can ‘compare notes’?


Definitely. I just ordered some 24/40 glassware from "DrBob" from over in the reagents acquisition section, so I'm hoping that will arrive early next week. In the meantime I can try to rig something up with my non-jointed liebig as a refluxer to do some preliminary runs.

Quote: Originally posted by blogfast25  
Your questions:

...(snip)

Thanks for answering my questions. If the product is two phase that would simplify things greatly. Your plan for the workup sounds good to me. I've got some CaCl2 as a drying agent too, as I'd have to dehydrate MgSO4 (Epsom salt) in my oven. I'm always wary of getting chemicals anywhere near the kitchen, even if they seem innocuous.

Quote: Originally posted by blogfast25  
But, but, but. If no two separate phases are present then distilling off the acetone will be needed. Use steam distillation for this: it’s very safe, low temperature and because acetone is both very volatile and soluble in water it should get rid of the acetone real easily. Then dry the product with some crushed anhydrous MgSO4 or dry Na2CO3.

Steam distillation is something I haven't done either, but I looked around a bit and it seems pretty straightforward. Any reason I couldn't use a standard distillation (i.e. oil bath or heating the flask directly)?

Quote: Originally posted by blogfast25  
You can also check the tertiary alcohol nature of the product by means of potassium dichromate: tertiary alcohols like alfa-terpineol CANNOT be oxidised by it, see here:

http://www.chemguide.co.uk/organicprops/alcohols/oxidation.h...

I think this is an interesting first organic experience for anyone...


Perfect, I have some dichromate already and I was wondering how I'd be able to check my results. I had also thought of the smell test, as you mentioned - turpentine has that pine scent, and alpha-terpineol "has a pleasant odor similar to lilac" according to the wiki.

Thanks for the link too, that site has a lot of really useful information. Looking forward to trying this!

blogfast25 - 23-3-2011 at 14:14

Well, direct distillation has the disadvantage that when the acetone content gets low, the BP of the stuff gets naturally close to that of the remaining component (the alfa-terpineol), so rather high. But it may be possible to get rid of most of the acetone by straight distillation, then switch to steam distillation when the boiler temperature is getting too high for comfort... It depends a bit on your set up.

By steam distillation here can simply be understood: add some water to the mix and some boiling stones (low boiling mixtures tend to boil rather bumpily, not w/o danger). Now you start distilling off gently and the steam that passes through the product mix carries with it most of the volatiles, including the acetone, while the alcohol is condensed in your column. When little water is left, separate the phases and dry the product unless you can still smell acetone in there.

I will have a first run tomorrow.

blogfast25 - 24-3-2011 at 09:57

Well, that first run wasn’t successful because my refluxer couldn’t take the flow and reagent mixture was spitting out at the top. Using a demoted graduated pipette (top and bottom clipped off) the new refluxer now works fine. This is the setup: 100 ml round flask (62 ml total charge), electrical submersion heater (but manual temp. control, ON - OFF), and refluxer:



Turps and the mixture of acetone/dilute H2SO4 form a two phase system with turps floating on top:



All ready for a run tomorrow…

Polverone - 24-3-2011 at 16:47

I came across US Patent 2432556 in an unrelated search, which claims that terpene alcohols are produced in superior yield from terpenes using sulfamic acid instead of sulfuric acid as catalyst. The improvement is not vast but sulfamic acid is also easier to handle and may be more readily available in some places than sulfuric acid.

blogfast25 - 25-3-2011 at 05:51

Thank you, Polverone!

MrHomeScientist - 25-3-2011 at 05:55

Very interesting find Polverone! Thanks for passing it along. I'm going to try to replicate the paper's method first, and if I can get that to work then sulfamic acid is definitely something to try for comparison. They let their reaction run at a lower temperature but for a much longer period, though. Four hours is plenty long enough for me :P

I rigged up an acceptable reflux apparatus last night, so hopefully I'll be able to do a run tonight. I'll start by only going for 2 hours just to see if it works, and this should give me a decent yield according to the paper.

edit: @Blogfast: So that second picture was before you did any reaction, correct? That's promising that it already shows two layers.

[Edited on 3-25-2011 by MrHomeScientist]

blogfast25 - 25-3-2011 at 09:39

Quote: Originally posted by MrHomeScientist  
@Blogfast: So that second picture was before you did any reaction, correct? That's promising that it already shows two layers.



Yes, it was before refluxing. The alcohol may be better soluble in acetone than pinene itself though: only one way to find out. alfa-terpineol is listed as soluble in ethanol, for instance.

That pinene floats on the acetone/H2So4 mix is good because the boiling bottom phase will constanty agitate the pinene!

Good luck with your run. I had to postpone again. Hopefully tomorrow :)

First Run

MrHomeScientist - 25-3-2011 at 21:48

I did my first run of this tonight, and it seems to have gone well. I haven't tested the product yet, but here's a photo of the results to whet your appetite:

Product after neutralizing.jpg - 194kB

This was after neutralizing with NaOH. There's two layers there, but it's hazy so I'm letting it settle overnight. The bottom layer is oily, and the top is aqueous. My starting solution also had two layers, but was completely clear. The color change was surprising.

Just wanted to post a quick update. It's late, so I'll finish up and post the full details tomorrow.

[Edited on 3-26-2011 by MrHomeScientist]

blogfast25 - 26-3-2011 at 05:55

Lookin' good!

MrHomeScientist - 26-3-2011 at 16:41

Here’s the writeup for my first run of the terpineol synthesis. I tried to follow the method outlined in the paper. It’s a bit long, and I’ve got even more photos if something isn’t clear.

A 150mL flask was used as the reaction vessel, and into this was mixed:
4g Crude Turpentine (4.6mL – this was chosen by assuming turpentine is 50% a-pinene)
1.6mL 4.4M Sulfuric Acid
25mL Acetone


The mix was biphasic with turpentine floating on top, as seen by blogfast25.

Reactants.jpg - 237kB

This was immersed in a water bath, and connected to a liebig condenser using Teflon tape to afford a better seal. Additional cooling was not used.

Apparatus.jpg - 256kB

The water was heated to the 80 – 85C range and held there for 2 hours. I only went for two hours because Figure 2 in the original paper shows that around 70% of the terpineol should be produced at this point. The maximum yield would be at 4 hours. Sometimes it creeped up to 90C, because my hotplate isn’t great, so tweaking was necessary every so often.

Shortly after reaching about 70C, the mix started changing color to yellowish and the volume began decreasing. This is likely the acetone boiling away and escaping the condenser due to insufficient cooling. Throughout the reaction, reflux was observed so at least some of the acetone survived. I’ll fix this in the next run.

During Reaction.jpg - 436kB

After 2 hours, heating was stopped and the apparatus disassembled. A strong scent of turpentine and acetone was detected. The mix had separated into two layers, both orange-yellow colored.

Initial Product.jpg - 239kB

I neutralized the acid with 0.5g NaOH in 3mL water, which evolved considerable heat. This was when the picture in my post above was taken. After letting it settle overnight, this is what it turned into:

Neutralized Settled Product.jpg - 265kB

The colors flipped! It almost looks like there are 3 layers there, but I never found a third in subsequent steps so I think that's an illusion. The layers were then separated, and the bottom layer turned out to be fairly clear while the top layer was orange. I assumed the bottom was the oil layer, so I tried to wash this with 2mL of water. Turns out, this was the aqueous layer! It was miscible with my wash water.

I then prepared a test solution of a few mg of potassium dichromate in 9mL water, acidified with about 10 drops of 4.4M sulfuric acid. This was split into 3 test tubes, and combined with 0.5mL of the following solutions (from left to right):
Test tube A: original, raw turpentine (the control)
Test tube B: lower layer
Test tube C: upper layer


A & C separated into layers, while B was miscible with the dichromate solution.

Dichromate Test Start.jpg - 308kB

These were placed in a warm water bath for a few moments, with no changes observed at the time. The heat was turned off, and I left for about an hour. When I came back, the tubes looked like this:

Dichromate Test Finish.jpg - 312kB

It’s a little hard to see, but A had become a light olive green, B had not changed, and C had become a darker green with the oil layer darkening significantly.

This is when I realized the original upper layer must have been the organic, so I took that and shook it with 2mL of water to remove soluble contaminants. I then repeated the dichromate test, and ended up with a bit darker green solution.

Upper Layer Test.jpg - 309kB
After heating:
Upper Layer Test Finish.jpg - 268kB

EDIT: After letting this sit while I wrote up this post, this solution has become an even darker green, which looks like a positive test for a tertiary alcohol!

Conclusions
It probably worked! The darker green color of the upper layer of my product seems to mean that some terpineol was produced, so more was available for the dichromate to reduce.

One source of error was insufficient cooling, which allowed almost all of the acetone solvent to evaporate and escape. In the next run I’ll use water cooling on my condenser, which should solve this problem.

If anyone has any suggestions on how to improve, please let me know. As I’ve said, this is my first experience with organic chemistry so there’s likely some basic things I’m unaware of. I really enjoyed doing it though!

[Edited on 3-27-2011 by MrHomeScientist]

blogfast25 - 27-3-2011 at 05:41

Good write up!

A few comments so far, maybe some more later on.

Cooling: the easiest way is to wrap your refluxer in kitchen towel (absorbent paper), wetted with iced water. My own refluxer which is very much the same size as yours is cooled that way and can handle (easily) twice the amount of boiler charge (I’ll be basing everything on 4 ml turps, not 2). Then occasionally (say every 10 minutes) you need to pipette some more iced water on your cooler…

There is one glaring error in your write up though: as I indicated above, potassium dichromate CANNOT oxidise t-alcohols, a ‘negative’ test result would thus indicate no primary or secondary alcohols are present in the end mix. Clearly there are but that’s probably quite normal: if you look at the article, final analysis shows it to be quite a witches’ brew! For the purposes of a catalyst for K synth. these minor constituents probably won't harm...

http://www.chemguide.co.uk/organicprops/alcohols/oxidation.h...

For your next run I’d suggest to use ethanol or methanol (or ‘methylated spirits’) as a control: that would allow you to compare intensity of colour and speed of oxidation.

Did you get to smell the organic phase at all?

Good work though!


MrHomeScientist - 27-3-2011 at 08:10

Whoops! You're right of course, I got my dichromate test results backwards. I looked at the paper again and there's definitely a lot going on in the products. Can the dichromate test even tell us anything then? All I can conclude from mine is that turpentine contains some reducible material, the aqueous product phase does not, and the organic product phase appears to contain more than the turpentine. The latter may be because I only went for 2 hours, and so the product ratios will be different.

Using methanol as a control is a great idea. I think because there's so many extra products hanging around, the speed of oxidation will be what we want to look at. The higher the concentration of a-terpineol, the slower it should change colors I would think.

For cooling, I found an old fish tank pump and I'll try to use the condenser as it was meant to be used - with ice water flowing through the jacket. The paper towel suggestion is good though, if this pump doesn't still work.

I did smell the products, but it was all a turpentine/acetone scent. I imagine the lilac smell of terpineol is easily overwhelmed by turpentine's strong pine scent. It might be more evident, had I gone the full 4 hours. That will be the best test I think, since the dichromate test is iffy.

Thanks for the encouragement, and I really appreciate the comments and advice! I'll see if I can do another run today.

blogfast25 - 27-3-2011 at 10:02

Regards the cooler, sorry, I didn’t see it was a proper Liebig with cooling mantle. Don’t pump too much or it will affect the temperature of the refluxed liquid.

One of the reaction products mentioned is fenchyl alcohol (fenchol), that’s a seconday alcohol and would be affected by dichromate (I think). To get a clear negative would probably only work with pure terpineol, which is why a comparison with a readily oxidisable substance is more appropriate here…

Your set up should also easily be capable of dealing with 2 - 3 the paper recipe quantities.

I’ve just distilled off some acetone today and my first run will definitely be tomorrow.

blogfast25 - 28-3-2011 at 08:37

Sigh… another teething problem occurred: refluxer flooding. This is where the upward vapour flow exceeds the downward liquid flow and liquid starts building up in the refluxer column. Once initiated it just gets worse and worse, with more vapour condensing in the held-up liquid: I ended up with a column full of liquid!

The main cause here was too narrow a diameter of the ‘bottom’ of the modified pipette (the column), but thankfully nothing taking off another bit of glass off that part couldn’t solve. By then it was too late to achieve a full run of 4 h, so instead I ran some useful tests. Latest set up:



The BP of the reagent mix was about 62 - 63 C, in line with an acetone phase laced with a bit of dilute H2SO4. In principle that means that under total reflux the water bath temperature should be of no importance: the temp. of the reboiler being always it’s constant BP. In practice the bath temp. does affect the boiling rate, as the heat per unit of time transferred from bath to reboiler is directly proportional to the temp. difference between the bath and the reboiler. The boiling rate could then be calculated by dividing the transferred heat (per unit of time) by the molar heat of evaporation of the boiler liquid (acetone, basically). The difference in reboiling rates was palpably visible while toggling manually between 80 and 85 C bath temperature.

But if the run tomorrow is successful and yields a K-synth. useful product, I’ll be making the product using my trusted acetone still (of much larger capacity) and a plain steam bath at 100 C bath temp.

The power of the submersion heating coil is about 280 W (measured) and about right for the job of manually regulating bath temp. The ‘paper towel cooling mantle’ also worked fine, with no acetone vapours escaping at the top but it needs regular addition of iced water.

After about an hour or so of putting the thing through its paces the turpentine phase (top) had turned a light yellow, as did MrHomeScientist’s:





The acetone/H2SO4 phase was turbid. I could have sworn the mix also smelled differently after the test but that may just be wishful thinking…





[Edited on 28-3-2011 by blogfast25]

MrHomeScientist - 29-3-2011 at 08:35

Nice! Your upper layer looks exactly like the color I found at about that time in the reaction. Nice to get confirmation :) Sorry to hear about your refluxer problems though, at least you got some preliminary results.

How much acetone are you using? It looks like much more than what I have. That could be due to different container sizes though.

I've noticed the same thing with boiling rate as you have. I ran a second test Sunday night, this time with cooling water running through the liebig. Everything was going great until my hotplate started to get away from me. I saw no change in volume until about 1.5 hours in, when my hotplate creeped up around 90C. It seems that as long as bath temperature stays in the 80-85C range (like the paper uses) things are fine, but above that vapors begin to escape my system. I let it run to 2.5 hours, but by then practically all of the acetone was gone and I was down to about the same volume as my first run. Darn. I really need to babysit my system apparantly.

I was busy yesterday so I didn't test my product yet, but it does have a somewhat lighter color. Still smells like turpentine/acetone though. I'll try and get to that tonight or tomorrow, and post more photos.

blogfast25 - 29-3-2011 at 12:34

MrHomeScientist:

I used twice the amount of the recipe, so about 62 ml. During refluxing, I could see vapour/liquid getting dangerously close to the top of the refluxer when bath temp. was 85 C. Our refluxers must be operating similarly!

After about 2 ½ h, I noticed that the boiling behaviour started to change: it seemed to reboil less… And after 3 ½ hours the liquid refused to boil at all, so I stopped.

The cause was soon found: the content of the flask was about 86 g at the end of the run, presumably because of cooling water infiltration! How this affects the conversion is of course unknown but of course it increases the BP! The next, much larger test will be done with my acetone still.

The two phase system was then neutralised with 5 % NaOH which made the acetone/water phase cloud over. The two phases are now being allowed to separate to the fullest extent possible, overnight:



One thing is striking: the transformation of the odour. The treated product mix has a flagrance that is almost completely different from the starting mix. The new flagrance is extremely pleasant, very ‘clean’ and ‘citrussy’, reminiscent of some lemon flavoured toilet cleaners, with perhaps overtones of pine. Something has been brewed in there. If only I had access to FT-IR!

Update:

After standing overnight the cloudiness had coalesced into the kind of dross you find in your bath water! Draining off the watery/acetone phase and replacing it with clean water still didn’t get rid of it. But the oily phase filtered well to a clear, light yellow and very flagrant liquid (1 - 2 ml):



For now I’ll just call it hydrated pine oil. I’ll desist from drying and further attempts at characterising it (only IR could conclusively tell what it is anyway) because I’m running another batch, this time 4 times the recipe quantities (and no cooling water infiltration!!). But the ‘finished’ product smells like I described it: so nice I’d gladly put some of it behind my ears!


[Edited on 30-3-2011 by blogfast25]

blogfast25 - 31-3-2011 at 12:24

Today a run with 4 times the quantities the paper called for, using a home made still that I use to recycle acetone with. Above the water cooled still column head is paper cooled corrugated flexible hose (not shown), normally used for cooling and directing the distillate into a distillate flask. On this occasion it was clamped into vertical position to provide the cooler:




Yesterday I distilled some 125 g of acetone (it came over at 56.1 - 56.3 C) w/o problems but today the damn thing had decided to leak a little from the joint to holds the column and still head together. But only a little and it was remedied by topping up the acetone level through the cooler a couple of times. Total run time just over 3 hours.

Two phases existed in the boiler liquid at all times. The cooled reaction mix was then neutralised and spilt into three phases:



You don’t see that everyday! But the mystery of the third, bottom phase was soon revealed when on standing crystals began to appear: these are Na2SO4 which salted out the acetone-water phase into Na2SO4-water and acetone (mainly).

The top, ‘oily’ phase was then decanted off and washed with some clean water. Tomorrow it will be filtered and dried. It looks amberish and smells like yesterday’s product.

$$^&%^**_)_)+

During the three hours of babysitting yesterday’s pine oil hydrate was subjected to a K2Cr2O7 oxidation test.

100 ml of 0.1 M K2Cr2O7 solution in 1 M H2SO4 was prepared. Here’s 10 ml of it:



To it was added an excess of methylated spirits (about 0.5 ml) and this was heated om steam bath. Predictably it turned blue-green (from Cr3+). This is a control.

To another 10 ml of the solution was then added drop by drop some methylated spirits, heating in between additions, to achieve the same colour as the control above. It took 8 drops to get colour matching.

To two test tubes both with the dichromate was added respectfully 8 drops of the hydrated pine oil (HPO) obtained yesterday and 8 drops of turpentine. In both cases the dichromate oxidised something: the HPO t5ube went dark to brownish but not green/blue. The turpentine control also reacted slightly, darkening the dichromate solution.


[Edited on 1-4-2011 by blogfast25]

MrHomeScientist - 31-3-2011 at 13:07

Very interesting results! The smell you describe is the most striking thing. My products have so far all smelled of turpentine/acetone, but somewhat less strong than the original turpentine. Perhaps I need to run it longer to further get rid of it. Like I said before, I imagine that pine scent is strong enough to overwhelm the more pleasent scent of the real product.

I've never seen 3 phases in mine before! You said the top was the oil and bottom was water/acetone - what was the middle, milky layer?

I've unfortunately been busy this week every day after work, so I haven't had time to work on my own runs. I've neutralized my product and let it sit, and it's settled out to be very similar to my first run. I'm betting since most of my solvent ended up evaporating away this batch will be nearly the same as the first batch. I might have to add more acetone, like you did. I've just been trying to stick to the paper's formula to make sure I can replicate it before I started making any changes, but your results really look promising. I'll try and do more work this weekend, and maybe get to try out my fancy new glassware! Keep it up!

blogfast25 - 1-4-2011 at 05:04

Quote: Originally posted by MrHomeScientist  
I've never seen 3 phases in mine before! You said the top was the oil and bottom was water/acetone - what was the middle, milky layer?



Hi!

The three phases appeared after neutralising the acid, the bottom one of them splitting into two. Turned out that the bottom shown in the photo was water + Na2SO4 (which crystallised and made a general nuisance of itself in the outlet of the separating funnel). The milky middle phase was mainly acetone: the sodium sulphate had salted it out of the water phase. Top was the ‘oily’ phase, presumably holding the alcohol(s).

The characteristic odour is what kind of gives me hope that there’s alfa terpineol in there. Unfortunately I’ve no real means of assaying the product. But if this really is ‘hydrated pine oil’ then I can see why some would want to produce and trade it…

The real test will be to try and use it in the K-synth!


[Edited on 1-4-2011 by blogfast25]

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