Problem

The reaction of InBr with an excess of HCBr3 in 1,4- dioxane (C4H8O2) leads to compound A...

The reaction of InBr with an excess of HCBr3 in 1,4- dioxane (C4H8O2) leads to compound A which is an adduct of 1,4-dioxane and contains 21.4% In. During the reaction, the indium is oxidized. The 1H NMR spectrum of A shows signals at δ 36 ppm (singlet) and δ 3.6ppm (multiplet) in a ratio 1:8. Treatment of A with two molar equivalents of InBr followed by addition of IPh4P]Br yields the salt B which contains 16.4% In and 34.2% Br. The 1H NMR spectrum of B exhibits signals in the range δ 8.01- 7.71 ppm and a singlet at δ 0.20 ppm with relative integrals of 60:1. Suggest identities for A and B.

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Solutions For Problems in Chapter 19