LiOH hydrolysis of methyl 2,2-dimethoxyacetate not giving product?

Question

I am forming 2,2-dimethoxyacetic acid from its methyl ester, and below is the experimental protocol I am following (from Tetrahedron 2016, 72 (3), 420–430):

2,2-Dimethoxy Acetic Acid Formation

1. Add lithium hydroxide monohydrate ($\pu{89.5 mmol}$, $\pu{3.75 g}$, $\pu{1.2 equiv}$) to a solution of methyl dimethoxyacetate ($\pu{74.6 mmol}$, $\pu{10 g}$, $\pu{1 equiv}$) in dioxane ($\pu{45 mL}$) and water ($\pu{45 mL}$) at $\pu{0 ^{\circ}C}$.
2. Stir the mixture for $\pu{1 h}$ at $\pu{0 ^{\circ}C}$ and one additional hour at room temperature.
3. Add sodium hydroxide ($\pu{1M}$ aq soln, $\pu{75 mL}$) to the mixture.
4. Extract the mixture with diethyl ether ($\pu{375 mL}$).
5. Acidify the recovered aqueous layer with hydrogen chloride ($\pu{6N}$ aq soln) until a pH of 1-2.
6. Extract the organics with diethyl ether ($\pu{375 mL}$).
7. Dry the combined organic layers with magnesium sulfate and filter.
8. Evaporate the solvent slowly under reduced pressure to obtain 2,2-dimethoxyacetic acid.

I repeated this reaction many times but every time I check the $\mathrm{^{13}C}$-NMR, I only see a single peak at $\delta \: \pu{67.12 ppm}$. I believe this corresponds to C-OMe. By theory (and by literature) I should see 3 peaks. $\mathrm{^{1}H}$-NMR does show 6:1 ratio of singlets at $\delta \: \pu{3.704 ppm}$ (6H) and $\delta \: \pu{3.458 ppm}$ (1H) but it is not well aligned with literature; By theory, a single hydrogen atom adjacent to $\ce{C=O}$ group should be more deshielded than $\delta \: \pu{3.458 ppm}$ (i.e. be placed at higher ppm). I thought through what I have done and I still can't figure out which step has gone wrong. Any thoughts and ideas? Would appreciate any advice.

Answer 1

If your reaction has worked and you get the expected product, you should see following NMR data as the sought literature reported:

$\mathrm{^{1}H}$-NMR: $\delta$ 3.44 (s, $\ce{6H}$), 4.85 (s, $\ce{1H}$), and 10.32 (broad s, $\ce{1H}$).

$\mathrm{^{13}C}$-NMR: $\delta$ 54.1 ($\ce{2CH3}$), 98.6 ($\ce{CH}$), and 170.7 ($\ce{C_q}$).

Instead, you got two peaks, $\delta$ 3.704 (s, $\ce{6H}$) and $\delta$ 3.458 (s, $\ce{1H}$), in your $\mathrm{^{1}H}$-NMR. And, you got only one peak at $\delta$ 67.12 in your $\mathrm{^{13}C}$-NMR. All this data telling me is your reaction has not worked as expected. What you have is 1,4-dioxane during the ether extract, $\mathrm{^{1}H}$-NMR of which gives only a singlet at $\delta$ 3.71 when $\ce{CDCl3}$ is the solvent. It also gives one peak at $\delta$ 67.14 in $\mathrm{^{13}C}$-NMR under same solvent (Ref.1).

I believe your compound may still be in the aqueous phase. You may need to saturate the aqueous phase with salt ($\ce{NaCl}$) before extract the product to organic phase. Note that acetic acid is very soluble in water. Therefore, I expect 2,2-dimethoxyacetic acid to be highly soluble in water as well (a $\ce{C4(H2O)4}$ compound like a monosaccharide).

References:

1. H. E. Gottlieb, V. Kotlyar, A. Nudelman, “NMR Chemical Shifts of Common Laboratory Solvents as Trace Impurities,” J. Org. Chem. 1997, 62(21), 7512–7515 (DOI: 10.1021/jo971176v).

Answer 2

The answer from @Mathew Mahindaratne in my opinion correctly identifies what the OP has isolated. It remains to suggest how to get the required product. It may be a matter of salting out the aqueous phase, but there are other possibilities. Some observers may feel that given a literature prep from a respected group there is no need to alter it, however the OP has not been able to replicate it. Accordingly I offer this procedure.

Dissolve the methyl dimethoxy acetate in a mixture of 9:1 THF/water at room temperature. Add $\pu{1.1eq}$ of $\ce{LiOH}$ monohydrate. Leave to stir at room temperature overnight. Concentrate under reduced pressure. Dissolve the residue in a minimum amount of water and cool in ice bath. Wash with $\ce{Et2O}$. Acidify using a slight excess of $\pu{2N} \; \ce{HCl}$ ($\mathrm{pH}$ really doesn't need to go below 3.0 as $\mathrm{p}K_\mathrm{a}$ of product is probably around 3.7). Extract $\times 3$ with $\ce{Et2O}$. Dry the combined organic extract, filter and concentrate. This should minimise any change of hydrolysing the acetal.