Hydrogen iodide, elimination

For 2-amino-4- m-nitrophenyl) seienazole, the yield is particularly high. This has been explained by the oxidizing effect of the nitro group, which liberates iodine from the hydrogen iodide eliminated in the condensation reaction. 

A considerable extension of the synthetic utility of the hypoiodite reaction is achieved if the steroid hypoiodite (2) is generated from the alcohol and acetyl hypoiodite and then decomposed in a nonpolar solvent. In this case ionic hydrogen iodide elimination in the 1,5-iodohydrin intermediate (3) is slow, thereby allowing (3) to be converted into an iodo hypoiodite (5). 

A variation of the general method for the synthesis of 2-amino-selenazoles is to avoid the use of the free a-halogenocarbonyl compound and in its place react the corresponding ketone and iodine with selenourea.This procedure is also taken from thiazole chemistry. By contrast with thiourea, the reaction with selenourea needs a longer reaction time and the work up of the reaction mixture is somewhat more difficult. Usually an excess of the ketone is used. In the preparation of 2-amino-4-( n-nitrophenyl)selenazole, a very high yield, calculated on the amount of iodine used, was obtained. To explain this peculiar result, the oxidative action of the nitro group was invoked. This liberates free iodine from some of the hydrogen iodide eliminated in the condensation reaction, and the free iodine then re-enters into the reaction. 

Bicyclic lactones8,60 are readily prepared from the corresponding cycloalkenyl acetic acids via iodolactonization and subsequent hydrogen iodide elimination. They are interesting starting materials for the stereoselective synthesis of civ-substituted cycloalkenes 25 and 26. The regiose-lectivity is determined by charge repulsion between the soft carbanion and the carboxylate anion. 

Hydrogen iodide is easily eliminated by strong bases from perfluoroalky lethy 1 iodides to give terminal alkenes With perfluoroalkylpropyl iodides, however, replacement of iodine by nucleophiles predominates over the elimination reacUon [f] (equation 1)... 

Historically, the rhodium catalyzed carbonylation of methanol to acetic acid required large quantities of methyl iodide co-catalyst (1) and the related hydrocarboxylation of olefins required the presence of an alkyl iodide or hydrogen iodide (2). Unfortunately, the alkyl halides pose several significant difficulties since they are highly toxic, lead to iodine contamination of the final product, are highly corrosive, and are expensive to purchase and handle. Attempts to eliminate alkyl halides or their precursors have proven futile to date (1). 

The splitting of (XII) took place smoothly with constantboiling hydriodic acid to give the pure iodo acid. The fluorina-tion of (XIII) was more facile than that of (XIV). In fact, with the latter acid, hydrogen iodide was eliminated to some extent with the production of ethyl hept-6-enecarboxylate, which was effectively removed by conversion into the dibromide with bromine, followed by distillation. 

Diethyl malonate reacts with iodine under basic soliddiquid conditions (procedure 6.4.20 omitting the alkene) to produce tetraethyl ethane-1,1,2,2-tetracarboxylate (Scheme 6.28) [110] the ethenetetracarboxylate is also formed, presumably from the reaction of the initially formed iodomalonate with its carbanion and subsequent elimination of hydrogen iodide. 

Reaction (9) generates methyl iodide for the oxidative addition, and reaction (10) converts the reductive elimination product acetyl iodide into the product and it regenerates hydrogen iodide. There are, however, a few distinct differences [2,9] between the two processes. The thermodynamics of the acetic anhydride formation are less favourable and the process is operated much closer to equilibrium. (Thus, before studying the catalysis of carbonylations and carboxylations it is always worthwhile to look up the thermodynamic data ) Under standard conditions the AG values are approximately ... 

Reaction of quinoxaline with the fluorine-iodine-triethylamine system gives 2-fluoroquinoxaline 40 and 2,3-difluoroquinoxaline 41, whose yields depend on the fluorine usage (Equation 6) <1999J(P1)803>. Using 6-chloro-quinoxaline or 6,7-dichloroquinoxaline as the substrate, the monofluoro product is predominantly formed regardless of the amount of fluorine used. It is suggested that the reaction proceeds via attack of the fluoride ion on the a-carbon of an intermediate A -iodo quinoxalinium species followed by elimination of hydrogen iodide with triethylamine. 

Another useful example is the elimination of hydrogen iodide with KOH from the compounds RC(I)=CHN02, accessible by addition of NOjI (formed in situ from N2O4 and I2) to acetylenes [141] ... 

Similarly, alkylation of ent-l from ( + )-(/ )-malic acid with homogeranyl iodide under analogous conditions furnished 3 in 35% yield. Above — 45 °C elimination of hydrogen iodide from homogeranyl iodide to give the conjugated diene occurred more rapidly than alkylation of the dianion of en -l41. 

The mechanism for the reaction is believed to be as shown in Eq. 15.170 (start with CH3OH, lower right, and end with CHjCOOH, lower left).180 The reaction can be initiated with any rhodium salt, e.g., RhCl3, and a source of iodine, the two combining with CO to produce the active catalyst, IRItfCO y. The methyl iodide arises from the reaction of methanol and hydrogen iodide. Note that the catalytic loop involves oxidative addition, insertion, and reductive elimination, with a net production of acetic acid from the insertion of carbon monoxide into methanol. The rhodium shuttles between the +1 and +3 oxidation states. The cataylst is so efficient that the reaction will proceed at atmospheric pressure, although in practice the system is... 

Elimination of hydrogen iodide from polyfluoroalkyl iodides is facile and gives fluoroal-kenes. Hydrogen iodide is easily eliminated by strong bases (e. g., sodium hydroxide) from polyfluoroalkyl iodides to give the fluoroalkenes.8... 

Treatment of owK-l-iodo-2-(pcrfluoropropyl (cyclohexane (trans-H 1) with sodium methoxide gives 3-(perfluoropropyl)cyclohexene (33) in a yield of 64% via anti elimination and l-(per-fluoropropyl)cyclohexene (32) in a yield of 13% via syn elimination of hydrogen iodide.77... 

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