Problem 1

Boron hydrides

\ce{B_{x}H_{y}}

were developed by Alfred Stock; William Lipscomb won the 1976 Nobel Prize for elucidating their bonding. (a) Using mass percent of boron and molar mass, derive the molecular formulae of two boron hydrides: A is a liquid at 25 °C, 83.1 % B,

M = 65.1

g mol

^{-1}

; B is a solid, 88.5 % B,

M = 122.2

g mol

^{-1}

. (b) The styx code counts B–H–B bridges (s), three-centre BBB bonds (t), two-centre B–B bonds (y) and

\ce{BH2}

groups (x);

\ce{B2H6}

is 2002. Propose a structure for tetraborane

\ce{B4H10}

with styx = 4012. (c) A compound

\ce{B4CCl6O}

has two types of B atoms (tetrahedral : trigonal planar = 1:3) and a

\ce{C#O}

triple bond; suggest a structure. (d) From

D(\ce{B-Cl}) = 443

D(\ce{Cl-Cl}) = 242

kJ mol

^{-1}

\Delta_fH^{\circ}(\ce{BCl3(g)}) = -403

and

\Delta_fH^{\circ}(\ce{B2Cl4(g)}) = -489

kJ mol

^{-1}

, estimate the B–B bond dissociation enthalpy in

\ce{B2Cl4}

. (e) In a reaction scheme starting from

\ce{B2H6}

, give the boron-containing products: 1 =

\ce{B2H6}

\ce{NaOH}

/then

\ce{CH3OH}

\ce{H+}

; 2 = product of

\ce{B2H6}

\ce{C6H6}

\ce{O2}

chemistry identified by cryoscopy (

\Delta T_f = 0.205

°C for 0.312 g in 25.0 g benzene,

K_f = 5.12

kg K mol

^{-1}

); 3 =

\ce{B2H6}

\ce{Cl2}

; 4 =

\ce{B2H6}

\ce{NH3}

(1:1 adduct); 5 = thermal product of 4, b.p. 55 °C.

Step 4 of 4: Diborane reaction products

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Intuition

Each transformation replaces B–H–B bridges by ordinary two-centre bonds — methoxy, chloro, or the $\ce{B-N}$ $\pi$ system of borazine.

M_2 = \dfrac{K_f\, m}{\Delta T_f\, m_{solv}} = \dfrac{5.12 \times 0.312}{0.205 \times 0.0250} = 312\ \text{g mol}^{-1} \Rightarrow 2:\ \ce{B3O3(C6H5)3};\quad 1:\ \ce{B(OCH3)3},\ 3:\ \ce{BCl3},\ 4:\ \ce{H3B-NH3}\ (\ce{BNH6}),\ 5:\ \ce{B3N3H6}

Analysis

Cryoscopy gives $M = 312$ g mol $^{-1}$ , matching the trimeric boroxine $\ce{B3O3(C6H5)3}$ (phenylboronic anhydride). Methanolysis yields trimethyl borate $\ce{B(OCH3)3}$ , chlorination gives $\ce{BCl3}$ , ammonia gives the adduct $\ce{H3B\bond{<-}NH3}$ , and its thermolysis gives borazine $\ce{B3N3H6}$ (b.p. 55 °C, inorganic benzene).