Take the 2-minute tour ×
Chemistry Stack Exchange is a question and answer site for scientists, academics, teachers and students. It's 100% free, no registration required.

Is it possible to synthesise ethers from carboxylic acids, and if so, how? By ether I mean an ether in general ($\ce{R-O-R'}$), not just specific examples.

I theorised one could perform Fischer esterification, $\ce{ROH +R'COOH -> RCOOR'}$, and then perform Wolff-Kishner reduction on the carbonyl group. The problem with this theory is that it assumes that the single-bonded oxygen between the $\ce{R}$ and $\ce{R'}$ doesn't influence the reduction process.

share|improve this question
    
Wolff Kishner wouldn't work on an ester.. –  sencer Dec 12 '12 at 0:25

3 Answers 3

up vote 7 down vote accepted

This is an interesting question. I can tell you that Wolff-Kishner won't work - both the hydrazine and the hydroxide required will attack the ester in a substitution fashion (with a loss of -OR'). Other reductions might work.

For example a Google search for "reduce ester to ether" brought up the following:

enter image description here

The reaction appears to be complete within 3 hours with moderate to good (but not stellar) yields. The abstract and key figures are available for free at www.organic-chemistry.org. The paper is published in the Journal of Organic Chemistry.

It appears that $\ce{Et3SiH}$ is the terminal reductant, but most of the work is being done by indium. The combination of $\ce{Et3SiH}$ and $\ce{InBr3}$ produces the active reductant $\ce{HInBr3}$, which provides the equivalent of $\ce{H.}$. The second $\ce{H.}$ comes from $\ce{Et3SiH}$. The proposed mechanism from the paper is as follows:

enter image description here

Other mild sequential one-electron reductions with a source of $\ce{H.}$ would also likely work. The two-electron reductions, like Wolff-Kishner ($\ce{NH2NH2 +NaOH}$) and hydride-transfer (e.g. $\ce{LiAlH4}$), are all heavy-handed and nucleophilic with the tendency to take apart or over-reduce the ester.

share|improve this answer
    
Could you try to simplify this answer? I'm a grade 12 graduate that's done an organic chemistry unit and I was hoping for something I could understand. Like, for instance, your first equation I couldn't understand for the life of me. –  Brenton Horne Dec 6 '12 at 14:36
    
Oh and will these reactions be valid for the formation of 2-arachidonyl glyceryl ether (has the carbonyl group removed from 2-arachidonoylglycerol) from 2-arachidonoylglycerol. (An ester formed from the Fischer esterification of arachidonic acid with glycerol) –  Brenton Horne Dec 6 '12 at 14:43
2  
@BrentonHorne - The information in both of your comments needs to be in your question. Your question contained enough knowledge and proper use of vocabulary that I suspected you to have at least completed a two semester sequence in organic chemistry at the university level. The notation in my answer would be easy to understand for someone at that level. –  Ben Norris Dec 6 '12 at 21:52
1  
@BrentonHorne - and secondly, A glance at the paper suggests that it might not be. No unsaturated esters were used, and esters of secondary alcohols were over-reduced. There are no reactions in the sophomore-level Organic Chemistry book that will do what you want. Most chemists learn about these kind of reactions in graduate school. –  Ben Norris Dec 6 '12 at 21:54
    
Oh thanks, I'll take that as a compliment (that you thought I sounded knowledgeable enough to be a graduate student -- I am studying math at that level) and thanks for answering my question -- I have to learn more before I can comprehend what you're saying the solution of my problem is. –  Brenton Horne Dec 6 '12 at 22:13

If you want symmetrical ethers you could do the following

Reduce carboxylic acid to alcohol using Lithium aluminium hydride and then perform bimolecular dehydration on it using concentrated sulphuric acid.

$$\begin{matrix} \ce{R-COOH& ->[\large\ce{LiAlH4}]& R-OH\\ Carboxylic\ acid& & Alcohol\\ 2\ R-OH& ->[\large\ce{conc. H2SO4}]& R-O-R\\ Alcohol& & Symmetrical\\ & & Ether} \end{matrix}$$


For asymmetrical ethers yo could d o the following

Reduce carboxylic acid of both the consituent alkyl groups of ether using above method.
React one of the alcohols with Sodium metal to form Sodium alkoxide, and then make the alkoxide react with the other alcohol to form ether.

$$\begin{matrix} \ce{R-COOH& ->[\large\ce{LiAlH4}]& R-OH& (reduction)\\ R'-COOH& ->[\large\ce{LiAlH4}]& R'-OH& (reduction)\\ \hline \\ R-OH& ->[\ce{Na}]& R-O- Na+& (generation\ of\ nucleophile)\\ Alcohol& & Alkoxide&\\ \hline \\ R'-OH\ +\ R-O- Na+& ->& R-O-R'& (attack\ of\ nucleophile)\\ Alcohol\qquad Alkoxide& & Asymmetric\ Ether& } \end{matrix}$$ NOTE: This is only from a theoretical point of view

share|improve this answer
    
I do ,not think this would work: nucleophilic substitution on a non-activated alcohol group is not very efficient (as OH- is not a good leaving group). Ether synthesis in basic conditions usually requires an alkoxide and halide alkyl (Williamson ether synthesis). –  PLD Jan 20 '13 at 10:18
    
@PLD didn't think about that, but is the symmetric ether synthesis correct –  Aditya Sriram Jan 20 '13 at 11:06
    
@ Aditya Sriram : yes, acidic dehydrating conditions are standard for ether synthesis, and proceeds through a nucleophilic reaction on the protonated alcohol, which leaves H2O as a good leaving group. –  PLD Jan 20 '13 at 13:26
    
@PLD what if I converted one reduced alcohol to alkyl halide using thionyl chloride(which gives a good yield) and then reacted the alky halide and alcohol as you said - Williamson's synthesis. –  Aditya Sriram Jan 22 '13 at 16:23

If you're willing to start with an ester, the following might serve as a side note to the excellent answer given by Ben Norris:

Matthias Beller and coworkers reported on the conversion of esters to ethers using silanes $\ce{R3SiH}$ in the presence of $\ce{Fe3(CO)12}$ as catalyst.

Reductions of esters to ethers using $\ce{LiAlH4}$ or $\ce{NaBH4}$ can be carried out in the presence of $\ce{BF3\cdot Et2O}$.

If the ester can be cyclic, i.e. a lactone, trichlorsilane in the presence of a radical initiator or under uv light will yield cyclic ethers.

If you're willing to add a carbon atom to your ester or lactone and convert it to an enol ether, the Tebbe olefination is your friend.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.