Sciencemadness Discussion Board - Synthesis of longer chain tertiary alcohols

After reduction of the double bond a possible substitute for tBuOH is obtained, which is sterically more demanding and has a higher boiling point. The double bond, if not reduced, may lead to decomposition/polymerization. Maybe it's worth a trial using alpha-terpineol directly and look what happens.

All of that is discussed higher up, including using neat alpha-pinene and hydrogenation of the relevant carbinol. Worth taking a gander, IMO. Certainly our 'blueprints' were sound but execution met with numerous problems.

Personally I'm still very interested in developing a much better catalyst for the reduction of KOH with Mg in paraffinic media. And for NaOH/Mg, of course!

Any assistance remains appreciated.

[Edited on 27-8-2016 by blogfast25]

blogfast25 - 26-8-2016 at 17:19

The pinene related stuff starts getting interesting from about here:

http://www.sciencemadness.org/talk/viewthread.php?tid=15171&...

[Edited on 27-8-2016 by blogfast25]

aga - 28-8-2016 at 07:21

Today's 'progress' was yet another failure to create chloral as a precursor to trichloroacetic acid, which supposedly catalyses the pinene reaction.

20 hours and 260 litres of chlorine later ...

The resulting 2-phase liquid was heated to 80C on a water bath (to drive off any remaining ethanol & dissolved chlorine), then had an equal volume of conc sulphuric acid added.

The liquid remained biphasic, so they were separated, then the upper layer was weighed and had a stoichiometric amount of water added.

This just made it wet.

No crystals, nothing.

blogfast25 - 28-8-2016 at 15:57

@aga:

Welcome back!

Let's recap before we carry on.

Soon!

Alice - 28-8-2016 at 17:28

I had a look at a paper mentioned previously:

"Synthesis of terpineol from α-pinene by homogeneous acid catalysis"

http://www.sciencedirect.com/science/article/pii/S0920586105...

In the abstract chloroacetic acid and HCl are mentionend, but in the paper acetic acid, and oxalic acid as well.

Reaction conditions for oxalic acid:

4 h, 70 °C, 0.6 mol water, 0.25 mol alpha-pinene, 0.43 mol oxalic acid:

Conversion: 39.8 %

Selectivity: 60.8 %

=> 24% yield

My intuition says the numbers have an error, because there is a strong trend of increasing conversion for 0.03, 0.11, 0.22, and for 0.43 mol oxalic acid conversion is reported lower than for 0.22 which seems unlikely.

[Edited on 29-8-2016 by Alice]

aga - 29-8-2016 at 08:26

Getting the reaction mixture anywhere near homogenous was really hard.

At one point i even added a few drops of detergent to try to break the oil up.

blogfast25 - 29-8-2016 at 09:36

Can you summarise that last run?

aga - 29-8-2016 at 09:39

I'll go and dig thru the notes.

Will take a little time.

aga - 29-8-2016 at 12:20

OK. Trawled thru a pile of random notebooks and cannot find as much info as there is in this thread about what i did.

Reading it all (again) i think i simply ballsed up the whole thing due to inexperience.

Most likely errors appear to be :-

Lack of detailed notes (!)
Poor Mixing technique.
Inadequate Temperature control.
Impure reagents.
Poor Lab hygiene.
Careless Handling.

By sheer luck there were some (very few) crystals after the first attempt.

I suggest we go back to the start with a slightly more experienced operator and try that procedure again.

If Alice (and anyone else interested) would also please try it, then we can compare notes.

[Edited on 29-8-2016 by aga]

blogfast25 - 29-8-2016 at 15:19

@aga:

Do we have a plan?

Alice - 30-8-2016 at 03:23

Aga, can you remember how you achieved crystallization?

Considering the reported tiny amount of water being very near the molar range of oxalic acid, this might have a considerable impact on the reaction. After having had a closer look the example I was wondering about, lower conversion but more oxalic acid than the following shows something else:

4 h, 70 °C, 0.6 mol water, 0.25 mol alpha-pinene, 0.22 mol oxalic acid:

Conversion: 53.4 %

Selectivity: 64 %

What I've ignored is the fact that the difference isn't just the oxalic acid amount but the altered oxalic acid : water ratio. 1 : 1.4 vs. 1 : 2.7. Additionally the authors mention, the reaction to proceed mainly at the pinene/water interface due to bad solubility. So, why not enlarging the interface by using more of the more favorable ratio, i.e. 0.43 mol oxalic acid with 1.16 mol H2O? It is strange, that the reaction stops at 53.4% conversion isn't it? More interesting, the product doesn't really decompose after a prolonged heating time.

Anyhow, the reaction goes probably better the finer oxalic acid is ground.

My suggestion is:

2 h, 70 °C, 0.3 mol water, 0.25 mol alpha-pinene, 0.43 mol oxalic acid dihydrate.

How is the plan for reduction of the double bond?

[Edited on 30-8-2016 by Alice]

aga - 30-8-2016 at 05:59

Aga, can you remember how you achieved crystallization?

The 'product' was just left to stand overnight and there were a few crystals the next day.

blogfast25 - 30-8-2016 at 10:19

How is the plan for reduction of the double bond?

Assuming the unsaturated carbinol (2,2-dimethyl something unsaturated-2-ol), pictured in the pic of this post of the thread:

http://www.sciencemadness.org/talk/viewthread.php?tid=15171&...

... can be obtained in sufficient quantity then I believe that hydrogenation with Pd/C catalyst and ammonium formate must be possible and quite easy to perform.

The right hand side carbinol is the one to be hydrogenated:

[Edited on 30-8-2016 by blogfast25]

aga - 30-8-2016 at 11:27

Not got any Pd/C catalyst, but then, not got past just having a bottle of turps at the moment.

Running a Cl₂ generator for many hours to get nothing useful was a big waste of time (not to mention the cleanup time !).

No oxalic acid here, so are there any other possible catalysts that i might have ?

If not, maybe better to just attempt the synth on page 9 again (starting with the fractional distillation on page 8, of course ).

aga - 30-8-2016 at 12:06

Astonishing !

In a book published around 140 years ago, oxalic acid was produced by heating pine sawdust with KOH/NaOH, citing an earlier ref to 'Thorn' (no date given).

1:2 ratio pine:KOH, heat to 250 C with stirring, lixiviate etc.

Page 199, Manual of chemical technology, Rudolf von Wagner, Johannes Rudolf, 1822-1880.

https://archive.org/details/manualofchemical00wagnuoft

Found a few methods on utoob using sugar and HNO₃, but the sheer volume of NO_x produced makes them a LOT less attractive.

Pine and KOH/NaOH i have, so time for some chainsaw action.

(maybe the bench drill would be better so no lube oil gets in the sawdust)

OK, so may as well give Alice's method a whirl.

Alice - 30-8-2016 at 14:42

Oxalic acid is available in hardware stores for rust removal and as optical brightener for wood.

Pd/C is unavailable for me too as well as any kind of platinum catalysts, raney nickel. What seems available is urushiba nickel, but I'm afraid it's way too weak.
Seems like a deadlock, even if it would be performed by someone it remains difficult to access.

I think I'll go back to camphor with some oranozinc reagent and hope for less reduction product.

@aga, I like these old methods!

[Edited on 30-8-2016 by Alice]

blogfast25 - 30-8-2016 at 14:51

Pd/C is unavailable for me too as well as any kind of platinum catalysts, raney nickel. What seems available is urushiba nickel, but I'm afraid it's way too weak.
Seems like a deadlock, even if it would be performed by someone it remains difficult to access.

Nah. Get it from eBay, Amazon or prepare it yourself. Not hard at all.

aga - 30-8-2016 at 15:06

Erm, given that is an Amateur attempt, and that we got some (very few) crystals once, i'd suggest we do that procedure again, more carefully.

What 'carefully' means is also unknown, so any suggestions welcome.

Presumably Alice will also be conducting the experiment, which will be very useful in determining the best way to go.

It's still 37 C here in the day, which is why i suspect temperature control in all OC synths is one of the Key factors to 'getting it right'.

(cite: multiple chloral synths in summer failing and one when it was cooler getting at least a solid)

CuReUS - 30-8-2016 at 23:33

Why not get limonene instead of terpineol and then add two moles of H₂O to it to get a diol
the more OH on the molecule,the better

how to extract limonene from orange peels - https://www.youtube.com/watch?v=o4CBXkfVHDc

aga - 30-8-2016 at 23:52

Quote: Originally posted by CuReUS

Why not get limonene instead of terpineol and then add two moles of H₂O to it to get a diol
the more OH on the molecule,the better

Limonene extraction looks easy enough.

Any references for 'adding mols of water to limonene to get a diol' ?

Alice - 31-8-2016 at 00:43

@CuReUS

If double hydration on limonene would work, it would work on terpineol as well, right?

I'm not convinced about "the more OH the better" because the goal is good solubility of the corresponding alkoxide in alkanes. The only way I can imagine a diol might work is one that has chelating properties, like 1,4-diethyl-1,4-dihydroxycyclohexane which might form a soluble dimer or something similar as an alkoxide.

Alice - 31-8-2016 at 03:43

Quote:

Nah. Get it from eBay, Amazon or prepare it yourself. Not hard at all.

@blogfast, I don't have what could be called a lab, a fume hood, inert atmosphere, and so on. Unavailable means, I had to buy all the stuff and equipement too including a house with a cellar, a garage, a garden or whatever and that's simply too much.

[Edited on 31-8-2016 by Alice]

blogfast25 - 31-8-2016 at 06:53

@CuReUS:

We don't need a diol. We need a 2,2-dimethyl something-2-ol.

@aga:

Limonene has been considered. Buy it pure on eBay, cheap as chips. Just about every toilet cleaner/oven cleaner/bathroom cleaner/etc contains it, it gives these concoctions its 'natural' (LOL) citrussy odour.

Will revisit the possibility of turning this into a 2,2-dimethyl something-2-ol. Hydration of limonene leads to a mixture of terpineol isomer though. Hard to separate.

Memory lane:

Here's the original post in which I proposed an OC research project for aga:

http://www.sciencemadness.org/talk/viewthread.php?tid=62973&...

Halcyon days...
<hr>

Having said all that I'm now inclined to 'cut out the middleman' and purchase the alpha-terpineol:

https://shop.perfumersapprentice.com/p-6228-terpineol-alpha.... (thanks Alice!)

And play with that.

* Characterise
* Attempt to hydrogenate (Pd/C + ammonium formate)
* Attempt to isomerise to tetrahydro myrcenol (2,6-dimethyloctan-2-ol)

[Edited on 31-8-2016 by blogfast25]

Texium - 31-8-2016 at 07:21

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.

Not sure if this is still relevant as I haven't read every page of this thread, but I have the materials to make such an alcohol, 2-methyl-2-octanol, by the Grignard reaction of 2-octanone (which I made previously) and methylmagnesium iodide. Longer chain options that aren't 2-methyl-2-ols but could be prepared just as easily include 3-methyl-3-nonanol (substitute ethylmagnesium iodide) and 4-methyl-4-decanol (substitute n-propylmagnesium bromide).

aga - 31-8-2016 at 07:29

Oh, still very relevant - we've failed to make anything so far !

Please, dive into the synth immediately and see what you get.

It'd be great if you could post a link to a ref, or even an outline of the process.

I'll try to copy it if i have the reagents.

blogfast25 - 31-8-2016 at 07:32

I've since come back a little bit from the C10 requirement, which I fear may be a little too long.

Understand the function of the carbinol catalyst in the reduction of KOH by Mg (powder, filings). The t-alcohol side of things is the reactive side, forming the K alkanoate (KOR). The dangling alkyl group R is supposed to provide solubility in the alkane solvent. So far, so simple.

I fear that too long alkyl groups may reduce reaction speed due to steric hindrance. For that reason, my ideal candidate would be around C6, in my current thinking.

The simple truth is that this is all guestimated speculation and that we simply will never know without testing the candidate catalyst in a real experiment.

Who knows. Maybe a C10 is the answer to all problems? Test, test, test!

[Edited on 31-8-2016 by blogfast25]

Texium - 31-8-2016 at 07:37

@aga: I might try it this weekend when I'm home. I don't know yet, as there's a lot of other stuff that I want to do, but it would make a good video.

In that case, I think that C8 would be a good place to start, especially since it would be in the 2-methyl-2-ol form. I could prepare the other two that I mentioned and compare, though I would probably have to prepare more 2-octanone to do this, and it would take considerably longer because of that and the fact that my lab time is very limited now. This weekend if everything goes to plan, I'll prepare methyl iodide and do the Grignard. I'll record it and edit the footage over the next week.

[Edited on 8-31-2016 by zts16]

Alice - 31-8-2016 at 08:40

Quote:

Having said all that I'm now inclined to 'cut out the middleman' and purchase the alpha-terpineol (thanks Alice!)

You're welcome, but before you order have a look at those two:

http://shop.perfumersapprentice.com/p-6061-dimetol-g.aspx

http://shop.perfumersapprentice.com/p-6058-dimethyl-phenyl-e...

[Edited on 31-8-2016 by Alice]

CuReUS - 1-9-2016 at 04:32

@CuReUS:
We don't need a diol. We need a 2,2-dimethyl something-2-ol.

I don't mean to be rude but even after going through the whole thread I still don't know what you need.There are talks about turps,TCCA,chloral,banana oil,camphor,limonene,cat repellent.And no one had been able to prepare any of the above mentioned compounds to test whether they work at all !!!
so saying that "we don't need a diol" would be going a bit far,don't you think ? ,when you don't even know if the tert-OH would work or not ?

Quote:

Having said all that I'm now inclined to 'cut out the middleman' and purchase the alpha-terpineol:
* Attempt to hydrogenate (Pd/C + ammonium formate)

do you have a reference for that ?
I found two references for the hydrogenation of terpineol. Both the papers are by the same author but both the articles are in german.

http://onlinelibrary.wiley.com/doi/10.1002/jlac.19113810105/... - "colloidal Pd" on pg 69
http://onlinelibrary.wiley.com/doi/10.1002/jlac.19144030107/... - "Pd catalyst " on pg 101
I had another question. Suppose you started with limonene,you could put the OH on the side chain to make terpineol and reduce the cyclic double bond or you could reduce the alkene side chain and put the OH on the ring.
would doing the latter give any advantage over the former ?

blogfast25 - 1-9-2016 at 05:10

Quote: Originally posted by CuReUS

I. [...] ,when you don't even know if the tert-OH would work or not ?

Quote:

Having said all that I'm now inclined to 'cut out the middleman' and purchase the alpha-terpineol:
* Attempt to hydrogenate (Pd/C + ammonium formate)

do you have a reference for that ?
I found two references for the hydrogenation of terpineol. Both the papers are by the same author but both the articles are in german.

http://onlinelibrary.wiley.com/doi/10.1002/jlac.19113810105/... - "colloidal Pd" on pg 69
http://onlinelibrary.wiley.com/doi/10.1002/jlac.19144030107/... - "Pd catalyst " on pg 101

II. I had another question. Suppose you started with limonene,you could put the OH on the side chain to make terpineol and reduce the cyclic double bond or you could reduce the alkene side chain and put the OH on the ring.
would doing the latter give any advantage over the former ?

I. We have two t-alcohols that work as catalysts in KOH/Mg reactions, both tried and tested: t-butanol and 2-methyl-butan-2-ol. We also know why these work. So we logically assume that longer chain alcohols with the same type of reactive site will work too.

Primary diols BTW wouldn't last in that cauldron, just like primary and secondary alcohols don't. Only t-alcohols are stable enough, in these high temp. alkaline conditions.

II. Hydration of limonene mostly puts the OH in the side-group. The remaining double bound (in the ring) could be hydrogenated or cracked open to give a longer t-alcohol.

Thanks for the hydrogenation links!

I have plenty references for Pd/C+ammonium formate hydrogenations of simple molecules, it's a very effective method.

[Edited on 1-9-2016 by blogfast25]

CuReUS - 1-9-2016 at 22:10

Quote: Originally posted by CuReUS

Why not get limonene instead of terpineol and then add two moles of H₂O to it to get a diol

Primary diols BTW wouldn't last in that cauldron, just like primary and secondary alcohols don't. Only t-alcohols are stable enough, in these high temp. alkaline conditions.
II. Hydration of limonene mostly puts the OH in the side-group

I understand the confusion now.By diol I had meant a molecule with two OH groups attached to it,not two OH groups adjacent to each other like a glycol
what I meant by double hydration of limonene was putting two molecules of water in limonene,one on the side chain double bond and the other on the ring double bond to get 2 t-butyl OH groups !!!
the name of the compound formed after double hydration of limonene is 1,8-terpin ( CAS no-80-53-5) http://www.chemblink.com/products/80-53-5.htm
I have found a reference to do just that but I can't find the article,so pls find it someone

ref - Comptes Rendus Hebdomadaires des Seances de l'Academie des Sciences, , vol. 89, p. 361

[Edited on 2-9-2016 by CuReUS]

Alice - 2-9-2016 at 00:11

@CuReUS.

Comptes rendus hebdomadaires des séances de l'Académie des Sciences:

http://gallica.bnf.fr/ark:/12148/cb343481087/date

Texium - 4-9-2016 at 13:29

I made methyl iodide today, so I'm all set for the Grignard, and could do it tomorrow. However, the humidity in my lab right now is about 70%, so it may not be prudent.

aga - 4-9-2016 at 14:14

Prudence is for Business or Politics.

If the yield is low, you already got a reason why.

Go for it zts16 !

Texium - 4-9-2016 at 14:21

But also, if the yield is low I'll end up wasting two valuable reagents that I had to make myself, and then I'll have a lot of work to do before I have a chance to try again. I want to have enough 2-methyl-2-octanol to allow for a couple of tries at the sodium reaction.

blogfast25 - 4-9-2016 at 15:07

But also, if the yield is low I'll end up wasting two valuable reagents that I had to make myself, and then I'll have a lot of work to do before I have a chance to try again. I want to have enough 2-methyl-2-octanol to allow for a couple of tries at the sodium reaction.

Unless you have experience with the KOH/Mg reduction with the standard catalyst (t-butanol) it would be prudent to evaluate the 'new' carbinol back-to-back with t-butanol. The reduction has defied some of our best experimeters, notably garagechemist who had several fails before getting K. A 'negative' with the new catalyst may thus not necessarily be meaningful.

Texium - 4-9-2016 at 15:54

I totally agree. In fact, yesterday I went on eBay and ordered some t-butanol so that I could try the potassium version of the reaction myself and become comfortable with it before I begin playing with sodium.

blogfast25 - 4-9-2016 at 16:28

I totally agree. In fact, yesterday I went on eBay and ordered some t-butanol so that I could try the potassium version of the reaction myself and become comfortable with it before I begin playing with sodium.

Excellent. Thanks for getting this thread moving again.

If your longer carbinol works for K then it certainly would be worth trying for NaOH/Mg.

[Edited on 5-9-2016 by blogfast25]

Texium - 24-9-2016 at 13:11

I'm happy to say that today I finally got around to running the Grignard, and have successfully prepared almost 5 mL of what should be 2-methyl-2-octanol.

I also recently received a liter of t-butanol from eBay, so now I can try the control reaction that we're all familiar with to make sure that I'm able to get it to work before I start experimenting with the new alcohol.

aga - 24-9-2016 at 14:40

Coool !

Photos ! Write-up !

Any difficulties with it or was it do-able for most of us ?

Texium - 24-9-2016 at 14:48

As I previously said I would, I recorded the reaction. A video will be posted to YouTube once the editing is completed, probably later this week.

It was actually a lot easier than I had anticipated.

aga - 24-9-2016 at 15:18

So, el Gringo is actually not the big bad hard-ass it's Rep says it is ?

C'mon. Details dude !

OK. A proper write-up would be better, but that takes time.

If you could maybe dangle a dollop of hope to us Grignard-virgins, that'd be good.

Texium - 24-9-2016 at 15:29

Well, it was 76% humidity today, I didn't flame dry the glassware, and I used ether that I distilled directly from starting fluid without any additional drying. If I was to assume that my crude product was actually pure 2-methyl-2-octanol, I would have gotten a 79% yield. But I know that there are impurities because it is slightly yellow and smells slightly like 2-octanone. I don't want to try and remove these impurities because I have such a small volume that it may not leave me with enough to use. If I find out that I'm allowed to, I might take a tiny sample of it to school with me and see if I can run it in the GC/MS.

blogfast25 - 6-10-2016 at 12:57