Chemistry 9701 · AS & A Level · States of matter

States of matter — practice question

Boron makes a large number of useful compounds.
(a)[1]

The compound diborane, $\text{B}_2\text{H}_6$, may be used as a rocket fuel. It is made by reacting boron trifluoride, $\text{BF}_3$, with sodium borohydride, $\text{NaBH}_4$. Balance this equation: $\dots\text{BF}_3 + \dots\text{NaBH}_4 \rightarrow \dots\text{B}_2\text{H}_6 + \dots\text{NaBF}_4$.

(b)[3]

Primary and secondary alcohols may be formed when carbonyl compounds react with $\text{NaBH}_4$, which supplies hydride ions, $\text{H}^-$. Complete the mechanism for the reaction of butanone with hydride ions, $\text{H}^-$, and draw the intermediate in the box. Include every required curly arrow and the relevant dipoles.

(c(i))[1]

Borane, $\text{BH}_3$, is used to make alcohols from alkenes. The reaction takes place in two steps. Suggest the type of reaction in step $1$.

(c)

Borane, $\text{BH}_3$, is used to make alcohols from alkenes. The reaction takes place in two steps. The $\text{BH}_2$ group from $\text{BH}_3$ bonds to the least substituted carbon atom of the double bond, and the remaining H from $\text{BH}_3$ bonds to the other carbon.

(c(ii))[1]

Diol Y can be made by the same method. Draw the structure of the diene that could be used to make diol Y.

(d(i))[3]

Benzene, $\text{C}_6\text{H}_6$, and borazine, $\text{B}_3\text{N}_3\text{H}_6$, both have planar cyclic structures. Describe the structure of and bonding in benzene, $\text{C}_6\text{H}_6$.

(d(ii))[1]

In borazine, $\text{B}_3\text{N}_3\text{H}_6$, the boron and nitrogen atoms alternate around the ring. Every ring atom has one hydrogen atom attached to it. All boron-nitrogen bonds in borazine are 0.144 nm long, whereas in simple compounds B-N and B=N bond lengths are 0.154 nm and 0.136 nm respectively. Suggest and draw the structure of borazine.

Worked solution & mark scheme

This 10-mark question has a full step-by-step worked solution and mark scheme. One marking point: Balanced equation: $4\text{BF}_3 + 3\text{NaBH}_4 \rightarrow 2\text{B}_2\text{H}_6 + 3\text{NaBF}_4$

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