Sciencemadness Discussion Board - Dodecanol from sodium lauryl sulfate. Fatty alcohol esters.

Dodecanol from sodium lauryl sulfate. Fatty alcohol esters.

semiconductive - 29-11-2018 at 14:04

I'm experimenting with making inexpensive wax like esters for making hard wax models. The models have to be insoluble in water but easily melted and cleaned off of surfaces using alcohol. eg: warm isoproply or methanol.

The closest natural ester I tried is beeswax, but it softens in my hand at room temperature and sags; So I've been experimenting with synthesizing esters using various fatty acids and alcohols including palm oil, and palm alcohol from industrial sources to make a much stiffer/harder ester.

I found that sodium lauryl sulfate cooked with oxalic or tartaric acid, makes a very hard and stiff waxy ester when cooled to room temperature. It also melts easily at around 25 to 35C and miscs with warm alcohol. It's almost perfect ... However, it has sandy/grainy lumps and doesn't melt uniformly. I think that's because I can't remove the sodium sulfate from the ester and it doesn't sink to the bottom of a melted pot of ester.

I've tried several times to remove the sodium sulfate from Lauryl alcohol before cooking it with a carboxylic acid, but surprisingly the purification methods I've tried don't work.

For example:
1) Mix 50g of NaSO4-C12-H26 with 6.4g of CaOH2, and 12.8g of CaCl2.2H2O in water.

I expected the calcium to replace Na, because CaSO4 is insoluble in water and the "plaster" should sink to the bottom of the flask. Thats what happens when mixing sodium sulfate with calcium chloride. Sodium Chloride should not be very soluble in lauryl alcohol, and so I thought I'd perform the reaction in an excess of water. That *should* have (theoretically) resulted in lauryl alcohol (density 0.8x) floating on top of salty water with plaster on the bottom of the flask.

Therefore, I carried out the reaction in 600 to 800mL of water to insure there was plenty of extra water. ( The solid sodium lauryl sulfate powder fills about 60mL of space. )

The SLS very easily dissolves into that much water when heated.

After mixing in the strong base a white calcium scum sank to the bottom of the flask, mixed with a small amount of lauryl fatty alcohol. The rest of the lauryl alcohol, however, stayed mixed with the water and does not float.
In fact, when I chill the water ... the entire flask hardens into pasty lump of whitish material.

I've tried several variations on this theme ... using no CaCl, but only CaOH ... essentially the same result. Adding NaOH to the solidified whitish material after chilling: result, paste becomes liquidy again ... but the lauryl alcohol still doesn't float or separate from water (even after boiling for 8 hours.)

Adding excess sodium carbonate (Na2CO3), causes the lauryl alcohol to sink to the bottom of the flask!!! and mostly separate from the water. However, it does not solidify ... but rather acts like a very dense colloidal suspension. Since it sank, I think it must an adduct or still chemically bonded to a sulfate, somehow.

This is extremely curious. Why is Lauryl alcohol so difficult to separate from sodium or calcium sulfate?

I bought a pound of SLS off ebay, from "pro supply outlet", and since it does make a waxy ester melting from 25 to 30C, it really does seem to be lauryl alcohol in the detergent. But I'm totally at a loss as to how I can purify the lauryl alcohol (even crudely!).

Sodium Lauryl Sulfate (SLS) Usp/Kosher 1 Lb. Pack 9735
https://www.ebay.com/itm/202510889906?ViewItem=&item=202...

UC235 - 29-11-2018 at 16:13

Why make a solid with calcium hydroxide? Shouldn't refluxing NaOH solution with SLS give sodium sulfate solution and an oily upper layer of dodecanol? small amounts of unreacted SLS will emulsify the water and dodecanol together.

CRN114-Discriminator - 29-11-2018 at 20:30

Its amazing how long it takes some things to crystallize out. I wonder if it were kept in a liquid state if, given enough time, or enough liquification/solidification cycles, if the salts would settle out more readily. It is just a thought. I am sure that there is someone here, far more clever than myself, who will come up with something less time intensive and more straight forward. Good luck with your project.

DavidJR - 30-11-2018 at 12:27

Why not just distill the dodecanol? Boiling point is 259C so the use of reduced-pressure would probably be advisable.

semiconductive - 30-11-2018 at 14:21

Quote: Originally posted by UC235

Why make a solid with calcium hydroxide? Shouldn't refluxing NaOH solution with SLS give sodium sulfate solution and an oily upper layer of dodecanol? small amounts of unreacted SLS will emulsify the water and dodecanol together.

There is no particular reason; I'm familiar that when something becomes a solid, even if the reaction is somewhat unfavorable ... it still will precipitate out, and the reaction go to completion. Whereas, with refluxing I'm less certain how long it will take and how to tell when it's finished.

NaOH is more expensive than CaOH2, where I live. I have to order the lye, but pickeling lime can be bought at a local store, as well as calcium chloride. I tried refluxing with straight NaOH, and it made a sticky mess with no clear separation of dodecyl chloride from water; so I figured, adding calcium would improve the separation ...

However, with Calcium, I totally didn't expect the reacted SLS to *sink* to the bottom. It's just theoretical, like your explanation ... So I'm curious as to what really happened.

Today, I tried refluxing NaCl 10grams, with 10grams of SLS in 50g water. In theory, sulfuric acid is stronger than HCl, and so ... the sodium should mostly transfer to Na2SO4, and the excess HCl would become Dodecyl chloride ...
Dodecyl chloride disassociates in water; but is supposedly very insoluble.

Pure SLS in water, has solubility of about 0.1gram/gram.,
So, I expected less than 5g to dissolve in 50g of water; but since Dodecyl Chloride has a mp. of -6C, I figured all of it would go liquid, at least, with time / refluxing as the reaction proceeded at ~100C.

When I heated it strongly, 100C, and put a 24/40 plug with keck clip on top, (keck clip will leak steam, but not let much foam out), it all dissolved.
There was a bi-layer separation after about 4 hours of heating, with the dodecanol on top. I used a 50mL platt bottom round boiler flask, and over half of the solution was the top layer.
Therefore, I think much of the water had to be inside the top layer. When I let it cool, the top layer solidified and shrank a lot. I then drained the water layer from underneath, and it weighed about 35 grams. So, over half the water came out.

But, clearly most of the SLS did not convert to dodecanol, or else the HCl and Na2SO4, stayed miscable with the alcohol and kept water with it. The NaCl salt did increase the density of the water, and the fatty alcohol did float this time. So, at least that helped the separation.

I added 35g of fresh water and 7g of NaCl, to mimmic the previous solution concentration. I hoped that would remove as much Na2SO4 as possible. I then added 4.3g of Na2CO3 (washing soda), to see if I could drive the reaction to completion by refluxing.

Then I added a 300mm vigeruex column, on top of a 50mL platt boiling flash (about 50% airspace, as 50mL is half full.) and tried boiling it again.
It massively foamed, right up to the top of the column. I lost a little bit of solution, (messy!)

So I added another refluxing column, allihn, (600mm), and started boiling it again.
Very slowly the foam converted to liquidy droplets that float on the water.
After about 6 hours of strong boiling, about 10% of the vigereux column still had foam in it, but the bubbles were getting noiticably bigger and more clear (less white).

I think the foam dissolves a little bit in the oily droplets, and the reaction takes place in the boiling water. I think this because oily droplets condense on the walls of the boiling flask, but not so much on the vigereux column.

So, then I tried adding about 2 tea-spoons of 2-propanol (eg:ISOProply), to see if I could dissolve the foam. (Bad idea, the water was still boiling hot). It shot out most of the foam over 600mm of refluxing columns and onto the stove.

The foam, however, completely disappeared. After letting it cool to room-temp, though, there was no bi-layer. It looks like either all the dodecanol ejected, or else adding iso-propyl made dodecanol miscable.

[Edited on 30-11-2018 by semiconductive]

[Edited on 30-11-2018 by semiconductive]

semiconductive - 30-11-2018 at 14:33

Quote: Originally posted by DavidJR

Why not just distill the dodecanol? Boiling point is 259C so the use of reduced-pressure would probably be advisable.

I only have a fish-tank pump, modified to produce vacuum ... so I can only get a mild vacuum similar to an aspirator.

The SLS browns/carbonizes when heated to near the boiling temperature.
I had to clean out two boilng flasks the first time I tried, and even pirannah solution using 90% H2SO4 hardware store drain cleaner + HCL (30%) + potassium nitrate + 30% H2O2, didn't attack it well.

My Prianah solution normally will slowly eat graphite out of glass frit ... but for some reason, SLS/dodcanol is very resistant to it.
So, manually cleaning with a brush was mandatory ... and a pain. Only palmolive detergent seemed to help.

I've ordered a gas wash bottle/condensation flask and a drink chiller off ebay, to see about recovering low boiling point alcohols from a mixed solution set-up.

That way, I can do a boiler flask, a condeser flask, a vacuum line, and then an alcohol recovery trap that is chilled. With IPA/Ethanol/or Methanol in the mix, I' hoping the boiling point should be greatly reduced and some dodecanol will come over in the vapors.

If I can recover the alcohol, maybe I can recycle it (like a soxhlet extractor would do.).

[Edited on 30-11-2018 by semiconductive]

egret - 1-12-2018 at 10:20

Did you tried an acid catalysed hydrolysis? Heat the SLS in dilute sulphuric acid close to boiling with a good stirring.

http://booksc.org/book/1613229/22e4f0

happyfooddance - 1-12-2018 at 20:11

Maybe you could transesterify by refluxing with excess methanol and a pinch of H2SO4 as a catalyst. Distill off the methanol and neutralize the small amount of H2SO4. Decant and done.

Edit: Also, many sources sell SLES (sodium lauryl ether sulfate) labeled as SLS and vice-versa so you might want to test your product first if you haven't already

[Edited on 12-2-2018 by happyfooddance]

clearly_not_atara - 1-12-2018 at 21:44

Calcium lauryl sulfate is soap scum, that insoluble residue that forms when you use soap in hard water. Most calcium salts of lipophilic acids are poorly soluble. And of course it's not easy to hydrolyse something that doesn't dissolve in water!

Furthermore calcium hydroxide will only basify to pH 11 or so because of its limited solubility.

However, the reaction of calcium hydroxide and sodium carbonate produces sodium hydroxide:

Ca(OH)2 (aq) + Na2CO3 (aq) >> NaOH (aq) + CaCO3 (s)

So if you can buy calcium hydroxide, you can certainly either make or buy sodium carbonate (boil a solution of baking soda) and therefore make NaOH.

Then you should have no trouble hydrolysing your soap.

semiconductive - 7-12-2018 at 13:15

Quote: Originally posted by clearly_not_atara

Calcium lauryl sulfate is soap scum, that insoluble residue that forms when you use soap in hard water. Most calcium salts of lipophilic acids are poorly soluble. And of course it's not easy to hydrolyse something that doesn't dissolve in water!

Furthermore calcium hydroxide will only basify to pH 11 or so because of its limited solubility.

However, the reaction of calcium hydroxide and sodium carbonate produces sodium hydroxide:

Ca(OH)2 (aq) + Na2CO3 (aq) >> NaOH (aq) + CaCO3 (s)

So if you can buy calcium hydroxide, you can certainly either make or buy sodium carbonate (boil a solution of baking soda) and therefore make NaOH.

Then you should have no trouble hydrolysing your soap.

Wow. I will have to try that! It's another example of the precipitation of an insoluble, driving a reaction to completion. Yes, if that works ... it would be a very inexepensive way to make NaOH aqueous solution from CaOH.

semiconductive - 7-12-2018 at 13:55

Quote: Originally posted by egret

Did you tried an acid catalysed hydrolysis? Heat the SLS in dilute sulphuric acid close to boiling with a good stirring.

http://booksc.org/book/1613229/22e4f0

Thanks for the link.
The quantitative findings will be helpful, when I try again next week I can use that to estimate the progress of the reaction (time wise), if it's slow.

FYI:
Yes; I tried acid catalyzing, because Fisher estrification is the best known process , and it was the first thing I thought of. I didn't use Sulfuric, though, I used HCl. I was hoping that the remaining chloride could be removed by long boiling after the reaction was finished. Adding a strong acid didn't seem to change things, much, though because the reaction still stalled.

I'm curious, what is the point of trying to hydrolyze using extra sulfuric acid as a catlyst ... when there is already sulfuric acid in the sodium sulfate (SL-S) ?

I mean, the existing sulfuric acid itself already forms an ester with lauryl alcohol.
That's what SLS is.... So, as the sodium sulfate hydrolyzes, the result is an acid... (NaHSO4, is a strong acid.); and that acid should have the same kind of effect as excess sufuric acid would have ?

My present thinking it this:

The hydrolysis of SLS does take place with just NaCl mixed with SLS. Hydrolysis is slow, but at least it happens. I think the will go to completion if boiled long enough with plenty of excess salt water solution since NaCl is pretty much insoluble in alcohol; the 1Dodecanol separates out easily as the NaCl water is dense, and 1-dodecanol floats on it very well.

So; I plan to:
1) hydrolyse SLS to 1-do decanol using NaCl & maybe adding excess sulfuric or HCl, 2) Remove salts/acids from the hydrolysis by separatory funnel to purify dodecanol and remove all sodium. 3) Do Fischer estrification using strong acid as a catalyst; Probably via HCl, and maybe adding methanol. Any Methyl-Chloride formed can be distilled off.

Unfortunately; I broke my boiling flask during an accident, so I'm waiting for more glassware before trying again.

The main problem with SLS is that it is *SO* foaming, that it tends to insulate the liquid at the bottom during boiling and during hydrolysis. The result is periodic steam explosions that cause the flask to "jump"/"bump" violently. I didn't have the erlenmeyer flask securely clamped to a lab stand and it got off the stove during a paritcularly violent bump... *!@$#.

That foam also appears to majorly slow down the reaction. eg: it ties up chemical and prevents mixing in the hot liquid below.

EDIT: I got my pyrex 1000mL beaker, today, so I tried an experiment.
10g of SLS, 20g NaCl (sea salt), 150g of water, 1 tsp of 30% w HCl.
I Dissolved SLS, then added salt. Brought to a boil, and waited about 15 minutes. Surprisingly, it didn't foam up massively this time!!!! The minor suds disappeared within about five minutes, and a film of alcohol could be seen forming on the surface.

I put it in the cooler, to solidify 1-decanol.
Then I poked two holes in the solid surface, to allow the salt water to drain out.
I rinsed once with distilled water, and then removed all the alcohol to a tare weighted 50mL platt flask and re-melted. Another layer of water formed underneath the 1-decanol, about 35 to 40% the volume of the alcohol .... it was a LOT of water. So I put it back in the cooler a second time ... poked two holes, and drained water once more.

Remelting a second time, no more water showed up ....
I weighed it, and got 5.975 grams of material.
So, then I brought it up to a simmer at 110C, for about a half hour to drive off any water dissolved in the alcohol. Steam bubbles popped for a while, and gradually decreased. A small film of scum formed on the very bottom, and I decided to discard that and keep only the liquid alcohol. With the scum included, it weighed 5.4grams; With the scum discarded, I ended up with 5.35 grams. There was a minor loss, max of 50mg, when transferring the 1decanol from the original flask to the re-boiler; I could have gotten as high as 5.4 grams (max) if I had been more careful in the transfer. Small amounts of decanol boil out with the steam, and condense on the glass and are hard to recover.

Theory vs. Practice:
10g SLS @ 288.372g/mol gives 34.68m mol of SLS (and decanol.)
34.68m mol of decanol @ 158.28g/mol = 5.489grams.

My yield, then, was 5.35 / 5.489 = 97%
If I hadn't lost any when transferring, it might have been as high as 99%.

So, I think just using table salt and a few drops of HCl is probably a good way to hydrolyze SLS. It's pretty efficient.

I have both citric acid ( tri-protic carboxylic acid w/ 1-OH alcohol like bond available.), and tartaric acid. I'll try making a waxy ester with each of them, and see how hard they are and what the melting points are like.

[Edited on 8-12-2018 by semiconductive]

semiconductive - 7-12-2018 at 19:29

Quote: Originally posted by happyfooddance

Maybe you could transesterify by refluxing with excess methanol and a pinch of H2SO4 as a catalyst. Distill off the methanol and neutralize the small amount of H2SO4. Decant and done.

Edit: Also, many sources sell SLES (sodium lauryl ether sulfate) labeled as SLS and vice-versa so you might want to test your product first if you haven't already

[Edited on 12-2-2018 by happyfooddance]

From the previous experiment, the molecular weights all agree very closely with theory for SLS. So, I think the chances are very good that I have true SLS, and not SLeS.

semiconductive - 7-12-2018 at 19:43

Quote:

Quote: Originally posted by clearly_not_atara

I know that would be true if the lauryl was a fatty acid ... I mean, soap scum is usually calcium bonded to a fatty acid.

From experience, I know Sodium Stearate will loose sodium ... and become calcium - di-sterate when CaOH is added.

I've used that reaction to recover steric acid from bar soap, and separate it from the glycerine.
eg: I wash the calcium di-sterate with water to get rid of glycerine, and then dissolve calcium di-sterate in HCl.
Steric acid will float on top...

But, in the case of SLS, the Sulfate only has the ability to form 2 bonds. So, I don't quite see how a single sulfate can be chemically bonded to both a calcium and also to the lauryl alocohol ...

Wouldn't that require 3 bonds ?

eg: The possible ions are ... 2[H]+ [SO4]-- , [Ca]++ , [Na]+ and [1decanol]+

So ... I suppose calcium scum could be doing something like this, with two sulfates ??
[1decanol+] : [ SO4 -- ] : [ Ca++ ] : [ SO4 - ] : [1decanol+]

So, why would it do that ... rather than just form [Ca++] [SO4--] and 1decanol?

eg: gypsum + free alcohol.

Does the sulfate molecule have some geometric shape and location of ionizations so that multiple different calciums atoms will bond in preference to a single calcium?

In that case,
Is Gypsum, CaSO4, really a crystal with bonds like NaCl, such that no Na is really bonded to any particular Cl, but is resonating with all local neighbors?
eg: Are Calcium bonds in gypsum resonating with multiple different sulfates?

[Edited on 8-12-2018 by semiconductive]

[Edited on 8-12-2018 by semiconductive]

stamasd - 8-12-2018 at 04:41

Quote: Originally posted by semiconductive

Quote: Originally posted by clearly_not_atara

This was in fact the first known method of producing aqueous NaOH (lye solution) used in soapmaking since antiquity... mixing soda ash (sodium carbonate) with Ca(OH)2 (slaked lime) would give a lye solution much more effective for making soap than the original soda ash.

OTOH if you're fatty alcohols that you're after, there are at least 2 which are available OTC and not expensive: cetyl alcohol (C16) and stearyl alcohol (C18). You can buy them from soapmaking supply stores.

[Edited on 8-12-2018 by stamasd]

semiconductive - 8-12-2018 at 14:43

Quote: Originally posted by stamasd

OTOH if you're fatty alcohols that you're after, there are at least 2 which are available OTC and not expensive: cetyl alcohol (C16) and stearyl alcohol (C18). You can buy them from soapmaking supply stores.

[Edited on 8-12-2018 by stamasd]

Yup... definitely cheaper.
Though I don't see stearyl alcohol at the local stores or on e-bay, so I'll have to look more for that.

Cetyl alcohol for soap-making, $32/3LB ~= $10.67 / LB
SLS ( $15.75/Lb ---> 55% lauryl alcholol) ~= $ 28 / LB 1-dodecanol.

There's five desirable properties that I'm after:
Low cost, low toxicity, high alcohol solubility, little to no water solubility, higher melting point/stiffer ester at room temperature to hold it's shape.

To be honest, I don't really care if it's an ester or a different chemical, so long as it has the properites to be a useful "wax" like substance.

eg: My idea is to make wax originals, then rinse them out with alcohol rather than "burn" them out, as is traditional for lost wax molds. Therefore, I would prefer the final chemical to be soluble in room temperature ethanol, methanol, or isopropyl; but I don't mind heating if I can get almost all the ester/wax out of the mold without "burning" anything. It's the carbon monoxide which shortens the life of electric kiln elements.

With respect to accidental ingestion of chemicals....
Lauryl alcohol is supposedly half as toxic as ethanol; so that was another reason I thought it would be a good starting point. Low toxicity is also why I'm now trying to make esters with either tartaric acid, or citric acid, rather than oxalic acid.

Some fatty alcohols, supposedly, are even good for you and dissolve bad cholesterol from blood vein walls. I will definitely look into cetyl alcohol, it's a longer chain than lauryl ... ? so, it should have a higher melting point when made into an ester.

[Edited on 8-12-2018 by semiconductive]

semiconductive - 8-12-2018 at 15:21

Fisher esterification of 1-dodecanol with tartaric acid : half failure.

I tried to esterify 1-dodecanol with tartaric acid. I ended up with a bi-layer, having two substances with different refractive indexes and which both solidify upon cooling.

I started with 5.4g of well boiled and pefectly clear 1-dodecanol to make sure it was "dry" of all water.
Then I measured out enough tartaric acid (di-carboxylic acid) to esterify all the 1-dodecanol.

(5.4g / 186g/mol ) * 0.5 (for di-carboxylic) * 150.087g/mol = 2.17g of tartaric acid.

I placed them together in a 50mL platt flask, melted at 115C with powdery tartaric acid on bottom, and then added three drops of 30% wt HCl.
Immediately, two layers of liquid formed; it was as if the tartaric acid "melted" ... so I continued heating over-night. The only result was a slight darkening and a loss of mass ( I've noticed that dodecanol vaporizes slowly when > 100C ). I lost close to a gram in 16 hours of heating.

The bi-layer is still there, with less than 1/3 the liquid in the bottom layer. ( see photo. ) Note, the top layer is more discolored than the bottom. Dodecanol does not appear to yellow, by itself, when heated ... so the color change is likely due to the acid.

Upon cooling, both the top and bottom layer solidified into a waxy like substance.
Upon bringing to a "boil" the bottom layer boiled away and re-condensed to a solid on the flask walls. The top layer did not boil.
The melting point of the bottom layer appears to be around 45C, and it boils at around 140-160C.

I did not get a melting temperature of the top layer, but it seems to melt at a slightly higher temperature than the bottom. (Maybe I'm wrong). Pure dodecanol is supposed to melt at 25C.

I tried re-heating with a drop of 90% H2SO4 to see if reaction could be driven to completion ... but no change was observed.

I did a sanity check of the quality of the tartaric acid:
I tried heating pure crystalline tartaric acid at 110C; I wanted to make sure I didn't have a race-mix variety with water of crystallization eg: .2H2O. I measured less than 20mg lost out of 3grams in 12 hours of heating. So, I don't think there is any risk of the bi-layer being due to insufficient tartaric acid .... because weighing was not skewed by water of crystallization. I noticed that the acid browns, very slowly, at 120C. Therefore, I think the darkening of the top layer of dodecanol (over-night) is because the dissolved tartaric acid oxidizes mildly at 110-160C.

As far as I can tell, I used sufficient tartaric acid to esterify all the dodecanol; and I should have a uniform substance and not a bi-layer.

I tried adding a little more tartaric acid (200mg) to the mixture, to see if the bottom layer would grow in size. The acid appeared to dissolve slowly (took about an hour), but no noticeable change of lower bi-layer width to top was seen. So, I think the acid is just dissolving equally well in both layers.

I'm wondering what the liklihood is that only one of the two proton sites on the di-carboxylic acid are estrifying, giving me two different substances?

Any comments / suggestions?

[Edited on 8-12-2018 by semiconductive]

happyfooddance - 8-12-2018 at 17:26

There are a lot of misconceptions to unpack in all these posts, but I will just say that fischer/speier esterification is a reversible process that is an equilibrium reaction, and you almost always need to use one of the components in a large excess to get good yields. Also, a product of the reaction is water, which needs to be removed as it shifts the equilibrium towards formation of reactants.

semiconductive - 8-12-2018 at 18:54

Quote: Originally posted by happyfooddance

There are a lot of misconceptions to unpack in all these posts, but I will just say that fischer/speier esterification is a reversible process that is an equilibrium reaction, and you almost always need to use one of the components in a large excess to get good yields. Also, a product of the reaction is water, which needs to be removed as it shifts the equilibrium towards formation of reactants.

The temperature is >100C, therefore all water formed will boil off (slowly).

But what is the definition of "large excess"?
Would adding a large excess of tartaric acid be of use? or does the excess have to be on the alcohol side? (and why?)

DraconicAcid - 8-12-2018 at 19:03

You would generally use a four-fold excess of one or the other. In this case, you want an excess of the alcohol, or you'll only get the monoester instead of the diester you want.

happyfooddance - 8-12-2018 at 19:43

Quote: Originally posted by semiconductive

But what is the definition of "large excess"?
Would adding a large excess of tartaric acid be of use? or does the excess have to be on the alcohol side? (and why?)

Adding more acid or more alcohol and how much is a question that is answered by considering many factors, such as B.P. of mixture (products must form and not decompose, preferably water distills out), solubility of reactants, and very importantly, ease of isolation of the resulting ester (and or recovery of solvents if that's important to you). Also cost is often a factor.

Writing down a table of physical constants of the reactants and products for your specific reaction will help you decide. You can compare their boiling points and solubilities/co-solubilities and other info and make a strategic choice.

For example, some acids you can remove by neutralization and they will stay in the aqueous layer (and even help separate the ester), some you wouldn't want to or it would form a solid mass. Some are easy to distill (formic, acetic). Same thing with the alcohols and ester product, their properties during and after the reaction dictate their proportions.

semiconductive - 17-12-2018 at 14:15

Quote: Originally posted by happyfooddance

Writing down a table of physical constants of the reactants and products for your specific reaction will help you decide. You can compare their boiling points and solubilities/co-solubilities and other info and make a strategic choice.

For example, some acids you can remove by neutralization and they will stay in the aqueous layer (and even help separate the ester), some you wouldn't want to or it would form a solid mass. Some are easy to distill (formic, acetic). Same thing with the alcohols and ester product, their properties during and after the reaction dictate their proportions.

You're right ... isolation of pure substances is the real problem.

Unfortunately, I don't see any data for the di-ester's properties; it's apparently not something well studied. So, I don't see any way to analytically come to a conclusion.

I got new glassware today; including some stuff to aid in vacuum distillation. I have bottles meant for separating liquid from gas in a vacuum line. ( eg: Gas wash bottle like, but with a drip tube and not a perforated bottom ).

I looked up what properties I could...

Dodecol is strange stuff. The boiling point is 259C; To distil it (for example from sodium lauryl sulfate) requires the boiling points of reactants from result to be different enough to effectively separate;

To get an idea of an lauryl ester's B.P. vs. pure alcohol, I looked up SLS ester, vs. dodecanol.

Strangely .... SLS's boiling point is unexpectedly much *lower*, at 204-207C than the pure alcohol's B.P. (259C). Not only that, SLS B.P = SLS M.P. ; Therefore, SLS must sublime rather than just melt. SO, distillation of the pure alcohol is impractical....

http://www.thegoodscentscompany.com/data/rw1000091.html

That might not be the case with the di-ester, but it means common sense for most substances (mixed) vs. pure, where mixtures have lower B.P. than pure ... doesn't necessarily apply for fatty esters.

Dodecyl chloride, OTOH, has a boiling point of only 140C. That's a temperature much lower than the alcohol, and easy to reach on a stove. Dodecyl chloride is decomposed by water, so I can easily get the pure alcohol from Dodecy chloride.

Another issues I discovered, is that the dodecanol made previously by liquid phase separation is not very pure; SLS dissolves into floating dodecanol, and a few perfecnt of SLS remains in it. My yield (by weight) is about 97%, but there could easily be round 3-5% SLS still dissolved in the alcohol ... so the actual yield is lower.

The final issue I'm running into, is that SLS contamination automatically means water contamination. When steam forms in impure SLS-Dodecanol mixtures, it takes lauryl alcohol with it at much lower temperatures than lauryl alcohol's boiling point. I can see it reflux on the earlenmeyer flask wall with temperatures as low as 105C. However, I can't get the alcohol/SLS to separate from the steam even by using a vigeruex column. When the SLS condenses on the vigereux column wall, it absorbs the steam....

The only way I can think to purify the alcohol is to make a substance with much lower boiling point than water contaminated SLS.

I'm thinking, maybe I could make a closed loop system to run H-CL gas in a loop, bubbling it through a moist SLS and table salt mixture; so that as chloride ester is formed, I can distill it off at 150C and have it condense in a chilled container.

My fish tank vacuum pump is immune to H-CL, and has a vacuum inlet and pressure outlet; so I can take the output (pressure) side and recycle the gas.

Thoughts? (I'll set up the distilling apparatus, and post pictures in the next post.)

semiconductive - 17-12-2018 at 14:21

Quote: Originally posted by DraconicAcid

You would generally use a four-fold excess of one or the other. In this case, you want an excess of the alcohol, or you'll only get the monoester instead of the diester you want.

Thank you. That makes perfect sense.... !