Esterification & Reduction of 3,4,5-TMBA ?

[Mountain_Girl]

Question 1:
Why is the conversion of 3,4,5-TMBA to the benzyl alcohol usually done by making the ester first and then reducing with LAH ?
Why not reduce the acid directly with LAH ?
(This quest. has been asked before but afaik it wasn't answered)
Is it perhaps because the overall yields are higher ?

Question 2:
Can the esterification be done with MeOH/(H2SO4 or HCl) instead of (Me)2SO4 ?

Question 3:
Can the ester reduction be done with Na/EtOH instead of LAH ?

[uemura]

Why is the conversion of 3,4,5-TMBA to the benzyl alcohol usually done by making the ester first and then reducing with LAH ?
In the literature the 3,4,5 Trimethoxybenzoic-acid is made from gallic acid via methylation with DMS. DMS in excess also esterises the -cooh to its metyhlester. So you get the methylester in one step from gallic acid.
 
Why not reduce the acid directly with LAH ?
1) because of the above preparation. 2) the acid is likely not very soluable in ether but the ester is.

Can the esterification be done with MeOH/(H2SO4 or HCl) instead of (Me)2SO4 ?
Yes, of course.

Can the ester reduction be done with Na/EtOH instead of LAH ?
Think so, but Uemura isn't absolutely sure. Sometimes NaBH4 also reduces esters to alcohols.

There are also ways to arrive at the alcohol. But when you intend to go from the alcohol to the chloride, then to the nitrile and at the end to the amine, the alcohol to chloride step is a bit 'ugly'.
Carpe Diem

[Sunlight]

Check this :

https://www.thevespiary.org/rhodium/Rhodium/chemistry/phenethylamines.dibah.html

Preparation of 3,4-Disubstituted Benzyl Alcohols (II).

A. To a solution of 100 g of the aldehyde (I) in 1000 ml of methanol was added 40 g of NaBH4 in the course of 5 h. The mixture was left overnight. After the solvent had been distilled off, the residue was mixed with water and extracted with chloroform. The extract was washed twice with water and was dried with Na2SO4. The residue after the solvent had been driven off was the required alcohol. Yield 98%.

Then chlorinate the alcohol, make the nitrile and reduce it. It is also explained in the article.

[Mountain_Girl]

(Hey why didn't my text go blue?)

Uemura:
Thanks for the reply. Your answers make sense.
When you say the alc. to choride step is 'ugly', I assume you mean the yields are low ?
If I look at the synth on Rhodiums page (

https://www.thevespiary.org/rhodium/Rhodium/chemistry/mescaline.txt

), the yield is 13g chloride from 25g alc. = 48%, which ain't great.
Although there are better methods using SOCl2 that are available.
But that's all beside the point since SWIM actually would like to synth the aldehyde from the alc. using MnO2 if she could get a procedure for this.

Sunlight:
SWIMs target is actually the aldehyde .

[uemura]

Moutain_Girl,

yield is 13g chloride from 25g alc. = 48%, which ain't great.

In his youth Uemura did this once. Beside the low yield, attempts to xtalise the chloride have been difficult.

Another way to the aldehyde from the acid would be to prepare the acide chloride and reduce this one with NaBH4 to the aldehyde.

Carpe Diem

[Mountain_Girl]

Unfortunately swim has the TMBA, but no SOCl2,PCl5,PCl3 etc. to convert to the acid chloride, otherwise she would do it and try Rosenmund (got Pd).
That's why she's considering the alternative route:
1. TMBA + CH3OH/H2SO4  ^__> Methyl Ester
2. Methyl Ester + Na/EtOH ^__> TM Benzyl alc.
3. TM Benzyl alc. + MnO2 ^__> TM Benzaldehyde

Btw, swim has LiAlH4 but feels step 2 is safer (she may be wrong in this perception) with Na/EtOH, although yields may be lower.

[hest]

I think the routhe sound good, I'l give beilsten a look at it today (and charge your hive account for 10$  ). Iff you have a lot of LAH then use it to all the reductions, but i guess you want to use it later :-)

[hest]

Na and methanol sounds good (70%) NaBH4 works damm well, but that might not bee an option.
The best way to reduce with Na is to melt the Na in (dry)toluene, then stir and let the toluene cool down to 60°C now wee have a fine Na powder in toluene. To this you add your ester disolved in dry(use sodium) methanol and drip this into the toluen fast (keep the temp at 60-80°C) when all the ester is added you are home.
Use 6 mole of Na per mole of acid. (less might work)
This is from O.S. Col.vol.II page 372

[obia]

a little nomenclature note 3,4,5-TMBA of commerce is usually the benzaldehyde and I think it is used in making trimethoprim.. I forget the price exactly but it is less than $40 per kg in china for small qtys

[Mountain_Girl]

Thanks once again Hest, feel free to charge my Hive account as you see fit.
Sorry but more questions:
* Are you saying substitute methanol for ethanol ?
* Swim's Na is in kerosene. How to transfer it to toluene ?
* You say dry the methanol with sodium !?

[hest]

Is your Na a 'powder'(like Li) or in a stick(the usual way)?
Do you have acces to MeOH ?

I don't think the type of alcohol is importent, i look it up tomorow.
The alcohol has to be dry (as in 99% not the usual 96%) just add a bit of Na. (if you have the 96% ad 5mole% if you have 99% ad 1mole%)

[hest]

Alcohol type not importent. But drynes is werrye importent

[uemura]

TMBA + CH3OH/H2SO4  __> Methyl Ester

Uemura came across a note in one of his books yesterday, which said: methyl esters cannot be prepared by azeotrope destillations of MeOH, acid and H2SO4 . Just wanted to add this, so use an access of MeOH + acid + H2SO4, reflux and the isolate the ester.

Carpe Diem

[Mountain_Girl]

Hest:

* The Na is lumps in paraffin. I assume it can just be transferred lump by lump with forceps.
* MeOH is available but toluene is not. Maybe xylene (xylol actually) can be used instead.
* The type of alc. is probably not important, except that MeOH will be more reactive than EtOH from my understanding.
* I'm still confused about using Na to dry the alc. - will the Na not react with the alc. ? Isn't that the point of Na/alc. reduction ?

[I thought of something interesting: If MeOH is used in the reduction, it is both a reagent and a product. Should it then not be possible to add a small amount of MeOH & Na initially, and then only Na after that to react with the MeOH as it is formed. I was wondering if this might increase yields from an an equilibrium point of view, i.e. a product is being removed from the reaction as it is formed ?]

[Mountain_Girl]

Uemura:

methyl esters cannot be prepared by azeotrope destillations of MeOH, acid and H2SO4

My understanding is that when it is an ethyl ester that is being prepared, the water that is produced can be removed as it is formed as the EtOH/water azeotrope. This will drive the reaction forward (altough it should also remove EtOH but you'll be using a large excess of that anyway).

Since MeOH doesn't form an azeotrope with water, the same principle would not apply to the preparation of methyl esters. (All the MeOH would boil off first before any water).

Hmm..now you've got me thinking - would it be better to make the Me or Et ester... ?

[hest]

When yau add the Alc. solution to the Na two reaction are running. One, the red. of the acid and two the formation of NaOMe (or NaOEt).
You want the reaction to go fast, else you are just making NaOMe, thats why the OS. synt melt the sodium and let it solidfy into a slurry.
Wather in the alcohol lower the yeald alot, so you dont want anny wather in the alcohol. The point with Na is to make the wather to NaOH and H2 (Na+H2O->NaOH+½H2)

Xylene=xylol will work as well as the solvent.

So i sugest the following routhe
Put 5g Na into 250mL methanol, let it stir a day or two, we now have a mixture of NaOH NaOMe and MeOH, but no wather, (iff you want the pure dry alcohol just destil it off, but it is not nessecery)

Add 22g(1mol) Na to 250mL xylen. heat until the Na melts, then let it cool under stirring.
Disolve 35g(0.2mol) TMB-acid in 200mL dry methanol. Drip the solution to the Na/Xylen slurry in such a rate that it boils gently (5-10min?)
Let it cool down, add wather and seperate. Wash the xylen two times with wather, dry and ewaporate. You now have TMB-Alcohol (Expected yeald ~60%)
(this is all teory I have no personal experience)
 
 
 

[uemura]

Mountain_Girl,

My understanding is that when it is an ethyl ester that is being prepared, the water that is produced can be removed as it is formed as the EtOH/water azeotrope. This will drive the reaction forward (altough it should also remove EtOH but you'll be using a large excess of that anyway).

Aceotrope destillation is normally based on a non-polar solvent which forms an azeotrop with water (e.g. toluene). The touluene/water destills off, seperates when cooled and the touluene is moved back to the rxn (e.g with a DeanStark trap). This works for all alcohols but not MeOH (that's what the book says) because the MeOH destills also off with the water and the toluene so that no water seperates when cooled down.

Anyway, to make the methylester just take a big excess of MeOH (eg. 5 times molar amount), the acid and add H2SO4 to catalise the esterisation. The big amount of MeOH drives the equilibrioum towords the ester and after x hrs of relux, just dump the rxn into water and work up.

Carpe Diem

[Mountain_Girl]

Hest:

Thanks. I now conclude:

In terms of drying the MeOH, I should add just enough Na to react the water away. Any extra Na will just use up some MeOH. Okay that's cool.

In terms of the redn. of the ester it's become clear to me that I don't really understand the mechanism of the reaction. I think it's best that I go away and do some more reading before I ask any further questions.

[Mountain_Girl]

Uemura:

This works for all alcohols but not MeOH (that's what the book says) because the MeOH destills also off with the water and the toluene so that no water seperates when cooled down.

Okay that makes sense, but why wouldn't it happen with EtOH ?

after x hrs of relux, just dump the rxn into water and work up.

I assume the MeOH would be distilled off first ?

[Mountain_Girl]

Hest:

Last night I read this in Vogel:

*Na alone cannot be used for the complete removal of water in an alcohol owing to the equilibrium between the resultant NaOH and the alcohol,
NaOH + ROH = RONa + H2O
An ester must then be added to remove the NaOH by saponification.*
*

However I think this only becomes a consideration if perfectly anhydrous alcohol is required, i.e. Na will sufficiently dry the alcohol for swim's purposes.