AS & A Level Chemistry 3.3 Metallic bonding: Exam Style Questions Paper 2

Question

The reducing agent $\mathrm{LiAlH}_4$ can be synthesised by reacting aluminium chloride with lithium hydride, $\mathrm{LiH}$.

(a) (i) At $200^{\circ} \mathrm{C}$, aluminium chloride exists as $\mathrm{Al}_2 \mathrm{Cl}_6(\mathrm{~g})$.
Draw the structure of $\mathrm{Al}_2 \mathrm{C} l_6(\mathrm{~g})$, showing fully any coordinate (dative covalent) bonds in the molecule.[2]

(ii) At $1000^{\circ} \mathrm{C}$, aluminium chloride exists as $\mathrm{AlCl} l_3(\mathrm{~g})$.
State the bond angle in $\mathrm{AlCl}_3(\mathrm{~g})$.
$\circ[1]$

(iii) Lithium hydride contains the ions $\mathrm{Li}^{+}$and $\mathrm{H}^{-}$.
State the electronic configuration of these two ions.
$\mathrm{Li}^{+}$
$\mathrm{H}^{-}$[1]

(iv) $\mathrm{LiAlH}_4$ decomposes slowly to form $\mathrm{LiAl}(\mathrm{s})$ and $\mathrm{H}_2(\mathrm{~g})$.
$
\mathrm{LiAlH}_4(\mathrm{~s}) \rightarrow \mathrm{LiAl}(\mathrm{s})+2 \mathrm{H}_2(\mathrm{~g})
$
LiAl(s) shows metallic bonding.Describe metallic bonding [1]

(b) $\mathrm{LiAlH}_4$ cannot be used in aqueous solution because it reacts with water to produce $\mathrm{LiOH}(\mathrm{aq})$, $\mathrm{H}_2(\mathrm{~g})$ and a white precipitate which is soluble in excess sodium hydroxide.
Identify the white precipitate. [1]

(c) Two students try to prepare 2-hydroxybutanoic acid in the laboratory.

Both students oxidise butane-1,2-diol to form $\mathbf{P}$ in reaction 1 . One student then reduces $\mathbf{P}$ using $\mathrm{LiAlH}_4$. $\mathbf{Q}$ is formed. The other student reduces $\mathbf{P}$ using $\mathrm{NaBH}_4$. $\mathbf{R}$ is formed.

(i) State the reagents and conditions required for reaction 1.[2]
(ii) Only one of the students successfully prepares 2‑hydroxybutanoic acid.
Identify which of Q or R is 2‑hydroxybutanoic acid and explain the difference between reactions 2 and 3.

A third student prepares 2‑hydroxybutanoic acid using propanal as the starting material. In
step 1 the student reacts propanal with a mixture of NaCN and HCN.

(iii) Draw the mechanism for the reaction of propanal with the mixture of NaCN and HCN to form S.
● Identify the ion that reacts with propanal.
● Draw the structure of the intermediate of the reaction.
● Include all charges, partial charges, lone pairs and curly arrows.

(iv) Complete the equation for the reaction in step 2, when S is heated under reflux with HCl(aq)

$
\mathrm{C}_2 \mathrm{H}_5 \mathrm{CH}(\mathrm{OH}) \mathrm{CN}+ \rightarrow \mathrm{C}_2 \mathrm{H}_5 \mathrm{CH}(\mathrm{OH}) \mathrm{COOH}+
$ [1]

(v) The infrared spectrum of an organic compound is shown. The organic compound is either S or 2‑hydroxybutanoic acid.

Deduce the identity of the compound. Give two reasons for your answer.
In your answer, identify any relevant absorptions above $1500 \mathrm{~cm}^{-1}$ in the spectrum and the bonds that correspond to these absorptions. [2] [Total: 17]

▶️Answer/Explanation

Ans:

(a)(i)  M1: correct representation of $\mathrm{Al}_2 \mathrm{Cl} l_6$, dot and cross or line diagram

M2: TWO correct co-ordinate bonds identified

(a)(ii) 120

(a)(iii)  $\mathrm{Li}^{+}$is $1 \mathrm{~s}^2 \quad \mathrm{H}^{-}$is $1 \mathrm{~s}^2$

(a)(iv) (Lattice of) cations / positive ions surrounded by delocalised electrons’

(b)  $\mathrm{Al}(\mathrm{OH})_3 /$ aluminium hydroxide

(c)(i) M1: potassium dichromate[(VI)]
M2: acid(ified) AND (heat under) reflux

3(c)(ii) (M1: correct identity of
R and statement re: reaction 3 ONLY ketone reduced)
R (is 2-hydroxybutanoic acid) AND as (only) C=O / ketone reduced

(M2: correct explanation re: strength of reducing agents) $\mathrm{NaBH}_4$ cannot reduce the $\mathrm{COOH} /$ carboxylic acid OR
$\mathrm{LiAlH}_4$ can reduce the $\mathrm{COOH} /$ carboxylic acid

3(c)(iii)

M1: Presence of :CN (if bonding shown, must be unambiguous triple bond)
M2: curly arrow from :CN lone pair to carbonyl carbon
M3: correct dipole AND curly arrow from double bond to oxygen
M4: correct intermediate drawn

(c)(iv) $\mathrm{C}_2 \mathrm{H}_5 \mathrm{CH}(\mathrm{OH}) \mathrm{CN}+\mathbf{H C l}+\mathbf{2} \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{C}_2 \mathrm{H}_5 \mathrm{CH}(\mathrm{OH}) \mathrm{COOH}+\mathbf{N H}_4 \mathbf{C l}$

(c)(v) Any two of three absorption references:

•  absorption 2200-2250 $\left(\mathrm{cm}^{-1}\right)$ shows presence of $\mathrm{C} \equiv \mathrm{N}$
•  lack of absorption at $1680-1730\left(\mathrm{~cm}^{-1}\right)$ shows lack of $C=O$
• lack of absorption at $2500-3000\left(\mathrm{~cm}^{-1}\right)$ shows lack of $\mathrm{RCO}_2-\mathrm{H} / \mathrm{O}-\mathrm{H}$ in $\mathrm{RCO}_2 \mathrm{H}$