Abstract

The reaction of the three isomeric tributylboranes (tri- n-butyl, triisobutyl and tri- sec-butyl) with bromine in the dark gives rise to both butyl bromide and hydrogen bromide when carbon tetrachloride is used as a solvent. The rate of disappearance of the borane and bromine are essentially equal and decreases in the order sec-butyl > n-butyl > isobutyl. However, the corresponding butyl bromide appears at a much slower rate and the formation of hydrogen bromide is quite rapid during the initial stages of the reaction. The amount of hydrogen bromide produced in the reaction reaches a peak in 1 h and then decreases with time. Similar results are obtained in cyclohexane. In methylene chloride, the rate of initial disappearance of bromine and tributylborane compares closely to the results obtained in carbon tetrachloride and cyclohexane. However, butyl bromide is formed with essentially the same rate as the rate of disappearance of the borane. Moreover, hydrogen bromide is formed in only minor amounts and the yields of alkyl bromides are high. In tetrahydrofuran, tri-n-butylborane and tri-sec-butylborane react at a rate similar to the rate similar to the rate of formation of the corresponding bromobutanes. This reaction is proposed to involve a slow, direct electrophilic attack of bromine, or its complex with THF, on tributylborane. Whereas in carbon tetrachloride, cyclohexane and methylene chloride, a fast, initially free-radical bromination, followed by a slow cleavage of the resulting a-bromoorganoborane with hydrogen bromide, takes place. Evidence supporting this mechanism is given. Competitive bromination studies reveal that the α-hydrogen in trialkylboranes is highly reactive toward free-radical bromination in the dark reaction. As an important synthetic application of this new reaction, the preparation of alkyl bromides is presented.

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