Chemistry 9701 · AS & A Level · Hydroxy compounds

Hydroxy compounds — practice question

Boron makes a wide range 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: $\ldots\text{BF}_3 + \ldots\text{NaBH}_4 \rightarrow \ldots\text{B}_2\text{H}_6 + \ldots\text{NaBF}_4$.

(b)[3]

Primary and secondary alcohols can be produced by reacting carbonyl compounds with $\text{NaBH}_4$, which acts as a source of 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 all required curly arrows and the relevant dipoles.

(c(i))[1]

Borane, $\text{BH}_3$, is used to make alcohols from alkenes. This reaction happens 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 occurs 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 prepare 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. Each atom in the ring has one hydrogen atom attached. Every boron-nitrogen bond in borazine is $0.144\,\text{nm}$ long, whereas in simple compounds B-N and B=N bond lengths are $0.154\,\text{nm}$ and $0.136\,\text{nm}$ 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: Correctly 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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