Hydrogen-deuterium exchange in alcohols

The phrase conducted tour mechanism was coined by Cram to describe the removal of a proton by a base and its subsequent return to a different face of the same molecule from which it was removed [Ij. Originally, the conducted tour mechanism was postulated to explain the observation that rates of racemization of deuterated carbon acids were faster than hydrogen-deuterium exchange in solutions of potassium r rr-butoxide/rerf-butyl alcohol. Thus, the basic catalyst takes hydrogen or deuterium on a conducted tour of the substrate from one face of the molecule to the [other] (ref. [1], p. 101). This process was envisioned as a rotation of the carbanion within the solvent cage. We now recognize that the secondary amine forms a mixed aggregate with the enolate, such that the reprotonation (and perhaps conformational motion) is intrasupramolecular. ... 

The reaction takes place extremely rapidly and if D2O is present in excess all the alcohol is con verted to ROD This hydrogen-deuterium exchange can be catalyzed by either acids or bases If D30 is the catalyst in acid solution and DO the catalyst in base wnte reasonable reaction mech anisms for the conversion of ROH to ROD under conditions of (a) acid catalysis and (b) base catalysis... 

The treatment of cyclopenta[c]thiapyrane (31a) with potassium t-butoxide in 95° O-deuterated /-butyl alcohol at room temperature resulted in more than 90"o deuteration at the 1- and 3-positions and less than 5" , at the 4-, 5-, 6-, and 7-positions.86 Thus, the hydrogen-deuterium exchange takes place at the positions calculated by quantum chemistry (see Section IV,B). Completely analogous is the same reaction in the indeno[2,l-fi]-1-benzopyrane system (44).126... 

During the course of base-catalyzed exchange in O-deuterated alcohols, the vinylic hydrogen in the a position to the ketone is replaced by deuterium, in addition to the hydrogens activated by enolization. Thus, under these conditions the exchange of androst-l-en-3-one (16, R = H) gives a trideuterio derivative (18) instead of the expected 4,4-d2 analog (16, R = D). " (For other examples see compounds 13, 19, 21, 23, 26 and 27.) Incorporation of this deuterium is due to rapidly reversible alcohol addition (16 -+17) and elimination (17 18) which competes with the enolization step. " ... 

In the presence of strong alkali, the rhodium analog of 62, or RhCl(C8H,2)PPh3, hydrogenates aliphatic ketones at 1 atm and 20°C, and after treatment with borohydride the systems similarly reduce aromatic ketones to the alcohols (526). Deuterium exchange data for acetone reduction were interpreted in terms of hydrogen transfer within a mononuclear hydroxy complex containing substrate bound in the enol form (63). 

Studies in deuterated water have shown that the hydroxyl proton does not end up in the ethanal formed. The decomposition of the 2-hydroxyethyl is not a simple P-elimination to palladium hydride and vinyl alcohol, which then isomerises to ethanal. Instead, the four protons stemming from ethene are all present in the initial ethanal product [6] (measured at 5 °C in order to suppress deuterium/hydrogen exchange in the product) and most authors have therefore accepted an intramolecular hydride shift as the key-step of the mechanism (see Figure 15.2). There remains some doubt as to how the hydride shift takes place. 

Another important piece of mechanistic evidence comes from the fact that acetaldehyde obtained from ethylene and deuterium-labeled water does not have any deuterium incorporation. In other words, all four hydrogen atoms of the ethylene molecule are retained. This means vinyl alcohol, which would certainly exchange the hydroxyl proton with deuterium, cannot be a free shortlived intermediate that tautomerizes to acetaldehyde. However, in spite of many elegant synthetic accomplishments in organometallic chemistry, realistic model complexes of palladium for 8.4 to 8.7 remain unknown. 

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