This gives excellent yields on dihydrobenzofuran and indane, using SnCl4 catalyst. The dihydrobenzofuran formylation is documented somewhere on Rhodium's site or here and is from a 1993 Nichols J. Med. Chem. article; the indane formylation is based on the above, see https://www.thevespiary.org/rhodium/Rhodium/chemistry/iap.html (https://www.thevespiary.org/rhodium/Rhodium/chemistry/iap.html)
. I'm interested in this method myself, for a possible dihydrobenzofuran formylation, but the main problem is the high price of dichloromethyl methyl ether. There is a synthesis on Rhodium's site, https://www.thevespiary.org/rhodium/Rhodium/chemistry/dcmme.html (https://www.thevespiary.org/rhodium/Rhodium/chemistry/dcmme.html)
, which uses POCl3/PCL5 on methyl formate, but in Monatsh.Chem.; 119; (1988); 1019-1026. and Chem.Ber.; 94; (1961); 544-550., it looks like this can be done with only PCl5. Both are in German, but I'll see if I can find the articles tomorrow. Even if nobody else is interested, I still am!
Another formylation method is the use of n-BuLi/DMF - see https://www.thevespiary.org/rhodium/Rhodium/chemistry/2cbde.html (https://www.thevespiary.org/rhodium/Rhodium/chemistry/2cbde.html)
I've been thinking of other ways to make dichloromethyl methy ether, because I really don't want to use PCl5 or POCl3. There are only 4 references in Beilstein regarding the synthesis of this compound, but I had an idea which sounds rather more simple:
Simply reacting methoxide with chloroform, which I had hoped would give our product. A literature search told me that reacting the two together would give trimethoxymethane :( , but this could only happen if at least 3 moles of methoxide per mole of chloroform were used. If I've got the general mechanism right, then wouldn't a large molar excess of chloroform with methoxide added give dichloromethyl methyl ether? The excess of chloroform would make monosubstitution much more likely, unless the disubstituted intermediate is much more reactive towards further substitution. If anyone is interested, the journal references for trimethoxymethane from chloroform are J.Amer.Chem.Soc.; 54; 1932; 2965, and .Chem.Ber.; 12; 1879; 117, so we could see if the conditions are tweakable.
Dichloromethane reacts with methoxide to give dimethoxymethane (no surprise there), but what is more interesting is that, in Tetrahedron Lett.; EN; 25; 49; 1984; 5693-5696., 2,6-dichloro-3-trichloromethyl-pyridine reacts with methoxide to give only the dimethoxy substituted product, 2-chloro-3-(chloro-dimethoxy-methyl)-6-methoxy-pyridine. I only have access to the title and abstract, but a mechanism is given in the full article:
Title:
Abnormal nucleophilic substitution of 3-trichloromethylpyridines by methoxide
Abstract:
3-Trichloromethylpyridine and its alpha-chlorinated derivatives behave as ambident electrophilic substrates towards methoxide which attacks an alpha-position and the trichloromethyl group [to give the disubstituted product].
The most interesting reference though, has to be J.Org.Chem.USSR (Engl.Transl.) (1988), 1362-1366:
Title:
REACTIONS OF PENTAFLUOROPHENOL AND POLYFLUORINATED ALCOHOLS WITH CARBON TETRACHLORIDE IN THE PRESENCE OF ALUMINUM CHLORIDE
Abstract:
The reaction of pentafluorophenol, 2,2,3,3-tetrafluoropropyl alcohol, and 2,2,3,3,4,4,5,5-octafluoropentyl alcohol with carbon tetrachloride in the presence of aluminum chloride were investigated.The main products from the reaction of pentafluorophenol with these reagents are pentafluorophenoxytrichloromethane, di(pentafluorophenol) carbonate, and tri(pentafluorophenoxy)chloromethane.The ratios of the products depend on the reaction conditions.The use of a large excess of carbon tetrachloride leads to the preferential formation of pentafluorophenoxytrichloromethane. The addition of water promotes the formation di(pentafluorophenyl) carbonate.When heated with carbon tetrachloride and aluminum chloride and subsequently treated with water, 2,2,3,3-tetrafluoropropyl and 2,2,3,3,4,4,5,5-octafluoropentyl alcohols are converted into di(2,2,3,3-tetrafluoropropyl) carbonate and di(2,2,3,3,4,4,5,5-octafluoropentyl) carbonate respectively.
Or would the monosubstitution simply not work on a lower order chlorocarbon such as chloroform? After finding Post 248530 (https://www.thevespiary.org/talk/index.php?topic=6433.msg24853000#msg24853000)
(PrimoPyro: "Re: Chloroform To Formaldehyde?", Chemistry Discourse) I thought maybe this would be the case, but surely such a simple method is worth a look considering the alternatives.
Any comments?
I don't have either of these handy, but it appears that under different conditions, the product distribution can be controlled to form either a mixture of difluoromethyl methyl ether/trimethoxymethane, or just difluoromethyl methyl ether. No yields are given in the citations I have, but it might be interesting to look them up.
The reactivity of fluorine as a leaving group in nucleophilic substitutions is some magnitudes lower than that of chlorine. So there is nothing to wonder about why they can get almost exclusively the difluoromethyl methyl ether if they vary the conditions of the reaction accordingly.
For dichloromethyl methyl ether, I think your best bet will still be the reaction between chloroform and sodium methoxide, by employing a large excess of chloroform and inverse working (adding the methoxide solution to the chloroform) in the cold.