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Author: Subject: Green Synthesis of 3-methyl-heptan-3-ol from common materials
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[*] posted on 2-4-2016 at 09:52


Great news guys!!!

I was up until 12 last night in the lab refluxing and distilling, and I woke up at 6 today to do workup. I redistilled also.

Doing the reaction starting with 82.4 mL of BuOH, I finished with 64.8 mL of BuBr, corresponding to a 67% yield!

To accomplish this, I did it my own way which makes the most sense. First I mixed water, NaBr, and sulfuric acid, THEN added the BuOH dropwise whilst heating up. It refluxed for an hour before I started to do downward distillation, which took a few hours at this scale.

I will certainly write the whole procedure in my paper... it was labor intensive but well worth it.

According to Av's paper, I should expect to see 86% yield, therefore there is a 19% difference in the two procedures, however the following must be remembered:
a) Workup. Lionel Joseph et. al. does not specify HOW he made his BuBr for 86% yield. He could've done it like Pavia, who omits downward distillation, or he could've distilled and not did workup. Again, there is no 'standard procedure' for preparing this, so we don't know what he did or didn't do.
b) Purity. Did they dry it? Distill it? I did both of these multiple times, and it is possible that each time I may have lost a small amount of product. Purity is, however, necessary if I plan to use this for Grignard reactions.
c) Laboratory. As home chemists, we are limited by the quality of our equipment, materials, and our time. Working at home in and of itself often sacrifices a bit of our yield for some reason.

According to Macroscale and Microscale Organic Experiments, a typical yield is 10 g from 10 mL BuOH, corresponding to a 67% yield.

Considering these facts, I would say I did pretty damn good this time around!




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[*] posted on 2-4-2016 at 10:59


Nice job. I suspect that there is more going on here however which is responsible for the lowered yields. With 1,3-propanediol, I got an 83% yield on a scale using ~40ml of the alcohol. The dibrominated compound is significantly less volatile than butyl bromide, so it's possible that this is another source of loss as well. Even still, a 19% loss of yield solely due to product lost to vapor seems quite high to me, so there might be a difference in the reaction which is hurting the yield here, but not with the diol. One thing which stands out to me immediately is that BuBr has a lower boiling point than water, while 1,3-dibromopropane has a higher boiling point than water. Perhaps prefilling the receiver with a bit of ice water could help improve the yields. Regardless, a 67% yield is not a bad yield, especially considering that none of the reagents are particularly expensive. Also, just a note from my runs of this reaction, the distillation to remove the product from the reaction mixture can be pushed very quickly, and if you're source of heat is powerful enough it can be finished in under an hour.



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[*] posted on 2-4-2016 at 12:00


The best thing about this is that ScienceHideout has gone from a 31% yield to a 67% yield! A lot of learning has happened.



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[*] posted on 2-4-2016 at 13:09


Quote: Originally posted by ScienceHideout  
I was up until 12 last night in the lab refluxing and distilling, and I woke up at 6 today to do workup. I redistilled also.

Doing the reaction starting with 82.4 mL of BuOH, I finished with 64.8 mL of BuBr, corresponding to a 67% yield!

Impressive commitment and a well deserved result. Congratulations. One point - Yields should be calculated and reported on a mass basis (its alot easier to weigh product (tared flask) than to measure the volume without losses/error). I'd personally also charge the principal input material by mass also, as its more accurate than doing so volumetrically, especially if you have a balance with 0.01 g resolution or better.

Quote: Originally posted by ScienceHideout  
To accomplish this, I did it my own way which makes the most sense. First I mixed water, NaBr, and sulfuric acid, THEN added the BuOH dropwise whilst heating up. It refluxed for an hour before I started to do downward distillation, which took a few hours at this scale.

I'm not sure how much of an issue the order of addition is in this case. Alkene and ether formation are really only an issue at elevated temperature with primary alcohols, so as long as the components are brought together at eg. room temperature it shouldn't make much difference.

When I prepared ethyl bromide, the ethanol and water were combined, cooled in a water-ice bath, and sulfuric acid added. This was followed by the addition of sodium bromide in one portion and sufficient heating to provide a slow rate of distillation. The crude yield for this reaction was ca 80 % th, not bad considering the additional problem of volatility encountered with ethyl bromide (b.p. 38 *C). The order of addition specified here was used to minimise formation of hydrogen bromide, something I have had issue with previously when attempting to preform the hydrobromic acid in situ before adding alcohol as you have done. Did you notice any evolution of fumes as you added the sulfuric acid?

Quote: Originally posted by ScienceHideout  
a) Workup. Lionel Joseph et. al. does not specify HOW he made his BuBr for 86% yield. He could've done it like Pavia, who omits downward distillation, or he could've distilled and not did workup. Again, there is no 'standard procedure' for preparing this, so we don't know what he did or didn't do.

A reference is given to the preparative method. The author of the paper appears to have varied only the reaction duration to determine the effect on the yield of this process. Unfortunately I do not have the referenced text - perhaps another member may have it in their own private collection?

Quote: Originally posted by ScienceHideout  
b) Purity. Did they dry it? Distill it? I did both of these multiple
times, and it is possible that each time I may have lost a small amount
of product. Purity is, however, necessary if I plan to use this for
Grignard reactions.

I'm not sure why you did both operations multiple times? You should have distilled to isolate the crude product, done your washes, dried, and re-distilled. That should have been sufficient.

Quote: Originally posted by ScienceHideout  
c) Laboratory. As home chemists, we are limited by the quality of our equipment, materials, and our time. Working at home in and of itself often sacrifices a bit of our yield for some reason.

I'm not sure I agree with you on that one. If you're concerned about the quality of your starting materials, you should purify them before use. Equipment usually falls down to having the right (sized) apparatus for the job, particularly when distillation is involved. The only real problem I've encountered is the lack of controlled "room temperature", particularly if your lab is located outside in a shed or garage.

Quote: Originally posted by gdflp  
Nice job. I suspect that there is more going on here however which is responsible for the lowered yields.

@ScienceHideout - Have you attempted to quantify losses during the workup? Mass before and after drying? That would help pinpoint the problem.

Quote: Originally posted by gdflp  
Perhaps prefilling the receiver with a bit of ice water could help improve the yields.

Good idea. Collecting volatile alkyl halides under water is an old trick to minimise cooling requirements and improve yields. I did this when I made isopropyl bromide, actually. A plain receiver bend reaching to the bottom of a measuring cylinder filled with a little cold water to submerge the end works nicely.

Quote: Originally posted by Magpie  
The best thing about this is that ScienceHideout has gone from a 31% yield to a 67% yield! A lot of learning has happened.

Agreed. Lets hope the success continues through the later stages of the synthesis also!

[Edited on 2-4-2016 by DJF90]
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[*] posted on 2-4-2016 at 19:51


gdflp- Interesting facts with the diol, It probably is partially because of the volatility, but I would think that it is an easier reaction. If you add one bromine, you make an electron withdrawing group... wouldn't that let the next hydroxyl leave easier?

Magpie- Thanks, I agree!

DJ- I plan on reporting mass in the paper, however for this specific case I need to use density calculations. My balance reads milligrams, and therefore has a capacity of 120g. I couldn't have possibly weighed the sample in a plastic cup! :D

I believe that order of addition is everything. In the lab I work in, I have gone from 5% yields to 90% yields just from the order of addition (well, in extreme cases like Swern oxidation). My thinking is that if the acid and bromide are mixed FIRST, then it will form HBr which will dissolve in the water I added. When the alcohol is added dropwise, the few molecules of alcohol are literally 'swimming' in an ocean of HBr and given no choice but to SN2 instead of turn to ether. When I added the acid to the NaBr, I did it very slowly and the acid was very cold. I didn't notice too many fumes, but the solution turned orange.

After I received the crude product, I washed, dried, distilled and dried again. I have not taken mass before and after drying, again, I simply don't have the capacity.




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[*] posted on 2-4-2016 at 19:51


I have a revised edition of the Robertson’s “Laboratory Practice of Organic Chemistry” (4th edition, 1962) which (more or less) corresponds to reference 1 in the Joseph, et al. J Chem Educ paper cited earlier. I had the actual edition cited but got rid of it during a recent clean out of my library. Following is a summary of the preparation of butyl bromide in the above mentioned text:

In a 500 ml flask fitted with a reflux condenser place 0.4 mole n-butyl alcohol and 0.5 mole of potassium bromide or sodium bromide. If sodium bromide is used, note any water of crystallization and account for this water in the amount of sodium bromide used as well as in the amount of water used in the following step.

Prepare a solution of sulfuric acid by adding 55 ml conc sulfuric acid to 30 ml cold water [This is where you account for water of crystallization in NaBr-AvB]. The acid solution is cooled and then added to the reaction flask with good mixing.

Reflux the reaction mixture for 30 minutes after which ABOUT 95% OF THE POSSIBLE PRODUCT WILL HAVE BEEN PRODUCED [my emphasis-AvB]. Further reaction for another 30 minutes only provides 1-2 grams of additional product.

Set the reaction for downward distillation and distill until no more oily product is observed. Separate the product from the aqueous layer. Wash the crude butyl bromide with 25 ml cold conc sulfuric acid, followed by 25 ml 1M NaOH then water. Dry the product over about 3 grams of CaCl2 or MgSO4 letting the product dry for 1 day or more. The dried product needs to be free of emulsified water. If during the following distillation any water is observed in the distillate, the product must be redried.

Finally, filter and distill collecting pure n-butyl bromide bp 99-103. Estimated yield, 45 grams or ca 88%.


Hope this helps - AvB
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[*] posted on 2-4-2016 at 21:02


The very first Orgsyn prep (Vol 1, 1921, pg 3 (http://www.orgsyn.org/Content/pdfs/procedures/CV1P0025.pdf)) is relevant here. A series of simple alkyl bromides are produced by various methods. BuBr is produced in 90-95% yield by three different methods. However, the scale is very large and mechanical/evaporative losses are thus minimized.

I have run several preps directly from or modified from this paper. 2 runs of ethyl bromide, 2 isopropyl bromide, 6-bromohexanoic acid and all have worked splendidly. I strongly suspect that chasing any string of references back ends with either this orgsyn paper or it's references.
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[*] posted on 3-4-2016 at 03:35


Quote: Originally posted by ScienceHideout  
DJ- I plan on reporting mass in the paper, however for this specific case I need to use density calculations. My balance reads milligrams, and therefore has a capacity of 120g. I couldn't have possibly weighed the sample in a plastic cup! :D

I see. I've had similar issues with exceeding the balance capacity before, myself. My advice is to invest in a larger capacity balance with less resolution. Personally, I have a pair of Mettler balances, 160x0.001 g and 1600x0.01 g. I'm sure even 0.1 g resolution would be fine as a large scale balance, but I got a good deal. It should not be too expensive to find something suitable. The other alternative is to work within the limits of your equipment, but that doesn't sound very fun, does it?

Quote: Originally posted by ScienceHideout  

I believe that order of addition is everything. In the lab I work in, I have gone from 5% yields to 90% yields just from the order of addition (well, in extreme cases like Swern oxidation). My thinking is that if the acid and bromide are mixed FIRST, then it will form HBr which will dissolve in the water I added. When the alcohol is added dropwise, the few molecules of alcohol are literally 'swimming' in an ocean of HBr and given no choice but to SN2 instead of turn to ether. When I added the acid to the NaBr, I did it very slowly and the acid was very cold. I didn't notice too many fumes, but the solution turned orange.

I agree that order of addition can significantly affect the outcome of a reaction. However, to quote myself (with added emphasis):
Quote: Originally posted by DJF90  
I'm not sure how much of an issue the order of addition is in this case. Alkene and ether formation are really only an issue at elevated temperature with primary alcohols, so as long as the components are brought together at eg. room temperature it shouldn't make much difference.

To elaborate using ethanol as an example (because I'm familiar with the data), the formation of ether is only appreciable at temperatures exceeding 100 *C, and becomes preparatively useful at ca 140 *C. Ethylene formation occurs at a higher temperature (ca 180 *C) I'd be willing to bet that butanol behaves similarly, and your worries about ether formation are unfounded. I guess we'll never know without access to a GC.

Quote: Originally posted by AvBaeyer  
I have a revised edition of the Robertson’s “Laboratory Practice of Organic Chemistry” (4th edition, 1962) which (more or less) corresponds to reference 1 in the Joseph, et al. J Chem Educ paper cited earlier.

Thanks again for that Av. It looks to me as if the text was revised to incorporate the findings of the J. Chem. Ed. paper, because the portionwise addition of sulfuric acid and longer reflux period are no longer present.

[Edited on 3-4-2016 by DJF90]
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[*] posted on 3-4-2016 at 09:02


So I was busy cleaning the lab this morning after all my hard work. I think that my BuBr yield is more than satisfactory, and I will go on to the next step. Maybe after I improve the grignard yield I will come back to the bromination.

But my next idea is to quantify the amount of water in my ether and THF. The idea is simple. First I will dry the solvent over sodium (scary, but whatever) and that will be my standard for a completely anhydrous solvent. Then, I will add a scoop of anhydrous CuSO4 and sonicate. I will pop it into my spectrometer and find the transmittance in the blue range. Then, I will take .95 mL of the anhydrous stuff and mix it with .05 mL of water. Again I will add a scoop of copper sulfate, but this time the transmittance will be higher. Lastly, I will take some from the bottle and then mix it with copper sulfate and sonicate. Then I can estimate the amount of water using Beers law.




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[*] posted on 3-4-2016 at 09:30


Quote: Originally posted by ScienceHideout  

But my next idea is to quantify the amount of water in my ether and THF. The idea is simple. First I will dry the solvent over sodium (scary, but whatever) and that will be my standard for a completely anhydrous solvent.


I have always dried my ether (recovered from starting fluid) over 3A mole sieves thinking that my sodium was too precious. However, use of sodium is a standard way for obtaining absolute ether. Brewster tells how to do it and so does Vogel. If the solvent only has a trace of water then use of sodium should not be scary. Others can advise you better than me.

[Edited on 3-4-2016 by Magpie]




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[*] posted on 5-4-2016 at 16:36


I thought the comparison of various procedures for making n-BuBr very interesting and the expected yields showed considerable variation. To provide further comparison I constructed an Excel spreadsheet comparing 3 student lab procedures from books that I have. Three observations jumped out at me:

1. The two older procedures have two distillations, one immediately following the reaction at reflux. The most recent one does not.

2. NaBr*2H20 may be a common form of NaBr, making me wonder about my pool chemical grade sodium bromide.

3. The amount of chemicals used has a half-life of 30 years.

My Excel spreadsheet is attached. Feel free to add to it the parameters of your procedure(s).

Attachment: Comparison of Some Methods for Making n-Butyl Bromide.xls (25kB)
This file has been downloaded 113 times




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[*] posted on 5-4-2016 at 17:40


Thanks, I will take a look at this more during my free time.

As far as bromide, I suppose one thing that I can do is run a test. 1g pool-grade NaBr in excess AgNO3- how much precipitate does it produce?




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[*] posted on 30-8-2016 at 14:19


Twelve principles of Green Chemistry:

1. It is better to prevent waste than to treat or clean up waste after it is formed.

2. Synthetic methods should be designed to maximize the incorporation of all materials used in the process into the final product.

3. Wherever practicable, synthetic methodologies should be designed to use and generate substances that possess little or no toxicity to human health and the environment.

4. Chemical products should be designed to preserve efficacy of function while reducing toxicity.

5. The use of auxiliary substances (e.g. solvents, separation agents, etc.) should be made unnecessary wherever possible and innocuous when used.

6. Energy requirements should be recognized for their environmental and economic impacts and should be minimized. Synthetic methods should be conducted at ambient temperature and pressure.

7. A raw material or feedstock should be renewable rather than depleting wherever technically and economically practicable.

8. Reduce derivatives – Unnecessary derivatization (blocking group, protection/deprotection, temporary modification) should be avoided whenever possible.

9. Catalytic reagents (as selective as possible) are superior to stoichiometric reagents.

10. Chemical products should be designed so that at the end of their function they do not persist in the environment and break down into innocuous degradation products.

11. Analytical methodologies need to be further developed to allow for real-time, in-process monitoring and control prior to the formation of hazardous substances.

12. Substances and the form of a substance used in a chemical process should be chosen to minimize potential for chemical accidents, including releases, explosions, and fires.

https://en.wikipedia.org/wiki/Green_chemistry#Principles


THF is a suspected cancerogen and forms peroxides. Greener alternatives are 2-MeTHF and cyclopentyl methyl ether with the latter having resistance to peroxide formation. The grignard reaction is not particularly a green reaction, it affords magnesium as stochiometric reagent instead of a catalyst for example. Of course a reaction can hardly suit all the criteria and there will be some trade-off between these criteria in most cases.

It's even more difficult to apply the green chemistry principles in an amateur scenario, so I would prefer the same reaction. Nevertheless, I would use diethyl ether, because it has a lower health impact.



[Edited on 30-8-2016 by Alice]
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[*] posted on 25-9-2016 at 22:03


I too was just about to reference the 12 principles of Green Chemistry, though I had sourced from the EPA website rather than Wiki...
https://www.epa.gov/greenchemistry/basics-green-chemistry#tw...

Please, for the love of all that is sacred, you've gotta stop calling this a "Green Synthesis."


[Edited on 9-26-2016 by Protium-1]
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[*] posted on 25-9-2016 at 23:48
Calling out premise, asking for reference


Chemists often praise the usefulness of their product by describing the context in which it solves or circumvents some other problem, or supersedes the prior art in some way. This you have done, it seems you describe tert-alcohols as a participant in a low-temp, non-hazardous-material-producing reaction to obtain Elemental Potassium and possibly Sodium in certain aliphatic solvents. Further, it seems that this remains your sole premise for calling this a "Green Reaction."

Aside from the argument that your actual reaction(s) involves very non-green solvents like THF, and aside from the fact that you've clearly put a lot of work into this, I find it very problematic your whole premise as stated in the Abstract and Intro and believe it possible you may have misunderstood something you had read in passing.

Please, correct me where I am wrong, and please DO provide at minimum one solid Reference.

I contend that there is no such reaction involving any tertiary-alcohol, nonetheless 3-methyl-heptan-3-ol, either as a reactant, co-solvent, catalyst, or otherwise participant in said reaction that produces elemental alkali metals such as Potassium or Sodium from their cationic form.

Assuming you can even get K+ ions to dissolve in aliphatic solution, by way of PTC or whatever, if a tert-alcohol itself were to be the reducing agent for producing elemental Potassium from K+ ions, the tert-alcohol would itself be oxidized. I pray you tell me what would the oxidized product of the tert-alcohol be?

If the tert-alcohol were acting merely as a cosolvent, catalyst, or otherwise spectator, any elemental alkali metal produced by unspecified reaction would simply deprotonate any remaining tert-alcohol, evolving hydrogen of course, and become ionized again.

I have tried to keep this short by not going into detail how tert-alcohols can for example be dehydrated and further oxidized into ketones and carboxylic acid by conc. aq. HNO3 because those products to would simply deprotonate in the presence of elemental alkali metals, once again to produce ions.

I hereby assert (at the risk of looking like a fool when the author of this paper provides any reputable reference otherwise) that tert-alcohols do not participate in any reaction that produces elemental alkali metals from their cations in any aliphatic (or any at all) solvent.

Unless shown otherwise, it should remain clear to see that the stated usefulness (and therefore the entire stated premise for this publication) of this 3-methyl-heptan-3-ol as a "Green" alternative to the electrolytic production of alkali metals is totally bogus.

[Edited on 9-26-2016 by Protium-1]
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[*] posted on 26-9-2016 at 00:54


Quote: Originally posted by Protium-1  
Chemists often praise the usefulness of their product by describing the context in which it solves or circumvents some other problem, or supersedes the prior art in some way. This you have done, it seems you describe tert-alcohols as a participant in a low-temp, non-hazardous-material-producing reaction to obtain Elemental Potassium and possibly Sodium in certain aliphatic solvents. Further, it seems that this remains your sole premise for calling this a "Green Reaction."

Aside from the argument that your actual reaction(s) involves very non-green solvents like THF, and aside from the fact that you've clearly put a lot of work into this, I find it very problematic your whole premise as stated in the Abstract and Intro and believe it possible you may have misunderstood something you had read in passing.

Please, correct me where I am wrong, and please DO provide at minimum one solid Reference.

I contend that there is no such reaction involving any tertiary-alcohol, nonetheless 3-methyl-heptan-3-ol, either as a reactant, co-solvent, catalyst, or otherwise participant in said reaction that produces elemental alkali metals such as Potassium or Sodium from their cationic form.

Assuming you can even get K+ ions to dissolve in aliphatic solution, by way of PTC or whatever, if a tert-alcohol itself were to be the reducing agent for producing elemental Potassium from K+ ions, the tert-alcohol would itself be oxidized. I pray you tell me what would the oxidized product of the tert-alcohol be?

If the tert-alcohol were acting merely as a cosolvent, catalyst, or otherwise spectator, any elemental alkali metal produced by unspecified reaction would simply deprotonate any remaining tert-alcohol, evolving hydrogen of course, and become ionized again.

I have tried to keep this short by not going into detail how tert-alcohols can for example be dehydrated and further oxidized into ketones and carboxylic acid by conc. aq. HNO3 because those products to would simply deprotonate in the presence of elemental alkali metals, once again to produce ions.

I hereby assert (at the risk of looking like a fool when the author of this paper provides any reputable reference otherwise) that tert-alcohols do not participate in any reaction that produces elemental alkali metals from their cations in any aliphatic (or any at all) solvent.

Unless shown otherwise, it should remain clear to see that the stated usefulness (and therefore the entire stated premise for this publication) of this 3-methyl-heptan-3-ol as a "Green" alternative to the electrolytic production of alkali metals is totally bogus.

[Edited on 9-26-2016 by Protium-1]

http://www.sciencemadness.org/talk/viewthread.php?tid=14970
Edit: just so its easier, here is the patent referring to the synthesis of potassium https://www.google.com/patents/US4725311

also, in the first page of the thread it was challenged whether this is a green synthesis.

[Edited on 26-9-2016 by HeYBrO]

[Edited on 26-9-2016 by HeYBrO]

[Edited on 26-9-2016 by HeYBrO]
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