DSE Chemistry Diagnostic: Carbon Chemistry

Unit Test 1: Alkanes vs Alkenes Reactivity

Question

(a) State the type of reaction that occurs when bromine water reacts with (i) ethane and (ii) ethene. Write equations for both. [4 marks]

(b) Explain why ethene reacts with bromine water but ethane does not, under normal conditions. [2 marks]

(c) A hydrocarbon $C$ decolourises bromine water and has the molecular formula $C_{4}H_{8}$. Draw and name all possible structural isomers of $C$. [4 marks]

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Worked Solution

(a) (i) Ethane + bromine: substitution reaction (requires UV light or heat).

$C_{2}H_{6} + Br_{2} \xrightarrow{UV} C_{2}H_{5}Br + HBr$

(ii) Ethene + bromine water: addition reaction (occurs at room temperature).

$C_{2}H_{4} + Br_{2} \rightarrow C_{2}H_{4}Br_{2}$

(1,2-dibromoethane)

(b) Ethane is an alkane containing only $C-C$ single bonds (sigma bonds). These are strong and do not easily break. Ethene is an alkene containing a $C=C$ double bond (one sigma + one pi bond). The pi bond is relatively weak and electron-rich, making it susceptible to attack by electrophiles like $Br_{2}$. The pi electrons are attracted to the electrophilic bromine molecule, initiating the addition reaction.

(c) $C_{4}H_{8}$ with a $C=C$ double bond (alkene isomers):

1. But-1-ene: $CH_{2}=CH-CH_{2}-CH_{3}$

2. But-2-ene (cis/trans isomers possible):

   • $CH_{3}-CH=CH-CH_{3}$ (cis-but-2-ene)
   • $CH_{3}-CH=CH-CH_{3}$ (trans-but-2-ene)

3. 2-Methylprop-1-ene: $CH_{2}=C(CH_{3})_{2}$

4. Cyclobutane (cyclic, not an alkene but satisfies $C_{4}H_{8}$ and does NOT decolourise bromine water -- exclude if bromine water test is specified)

5. Methylcyclopropane (cyclic, also does not decolourise bromine water -- exclude)

Since the question states $C$ decolourises bromine water, only the alkene isomers count:

1. But-1-ene
2. cis-But-2-ene
3. trans-But-2-ene
4. 2-Methylprop-1-ene

That is 4 structural isomers (3 positional + 1 branched chain; noting that cis/trans are stereoisomers of the same structural isomer, so structural isomers = 3: but-1-ene, but-2-ene, 2-methylprop-1-ene).

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Unit Test 2: Polymerisation of Alkenes

Question

(a) Draw the repeating unit of the polymer formed from propene ($CH_{2}=CHCH_{3}$) and name the polymer. [2 marks]

(b) Poly(chloroethene) (PVC) is made by polymerising chloroethene ($CH_{2}=CHCl$). Draw the repeating unit of PVC. [1 mark]

(c) A student claims that the polymerisation of propene produces a polymer with alternating single and double bonds. Explain why this is incorrect. [2 marks]

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Worked Solution

(a) Propene polymerises by addition polymerisation:

$n\ CH_{2}=CH-CH_{3} \rightarrow \left(-CH_{2}-CH(CH_{3})-\right)_{n}$

Repeating unit: $-CH_{2}-CH(CH_{3})-$

Polymer name: polypropene (or polypropylene).

(b) Chloroethene polymerises:

$n\ CH_{2}=CHCl \rightarrow \left(-CH_{2}-CHCl-\right)_{n}$

Repeating unit: $-CH_{2}-CHCl-$

(c) The student's claim is incorrect. In addition polymerisation, the pi bond of the $C=C$ double bond opens and forms new sigma bonds with adjacent monomer units. The double bond is consumed in the process. The resulting polymer chain contains only single bonds between carbon atoms. There are no remaining double bonds in the main polymer chain (unlike condensation polymers which may retain certain functional groups).

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Unit Test 3: Systematic Naming of Branched Compounds

Question

Name the following compounds using IUPAC systematic nomenclature:

(a) $CH_{3}-CH(CH_{3})-CH_{2}-CH_{3}$ [1 mark]

(b) $CH_{3}-CH_{2}-C(CH_{3})_{2}-CH_{2}-CH_{3}$ [2 marks]

(c) $CH_{3}-CHCl-CH_{2}-CH_{3}$ [1 mark]

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Worked Solution

(a) Longest carbon chain = 4 carbons (butane). Methyl group on carbon 2.

2-Methylbutane

(b) The structure is: C1($CH_{3}$)-C2($CH_{2}$)-C3($C(CH_{3})_{2}$)-C4($CH_{2}$)-C5($CH_{3}$).

The longest carbon chain has 5 carbons (pentane), with two methyl groups on carbon 3.

3,3-Dimethylpentane

(c) $CH_{3}-CHCl-CH_{2}-CH_{3}$: 4-carbon chain, chlorine on carbon 2.

2-Chlorobutane

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Integration Test 1: Multi-Step Synthesis Planning

Question

Starting from ethene ($CH_{2}=CH_{2}$), outline a reaction pathway to produce ethyl ethanoate ($CH_{3}COOC_{2}H_{5}$). For each step, give the reagent(s), conditions, and type of reaction. [6 marks]

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Worked Solution

Step 1: Ethene to ethanol

$CH_{2}=CH_{2} + H_{2}O \xrightarrow{H_{3}PO_{4}, 300\degree C, 60 \text{ atm}} CH_{3}CH_{2}OH$

Reagent: Steam ($H_{2}O$) Conditions: Phosphoric acid catalyst, 300$^{\circ}$C, 60 atm Type: Hydration (addition)

Step 2: Ethanol to ethanoic acid (oxidation)

$CH_{3}CH_{2}OH \xrightarrow{K_{2}Cr_{2}O_{7}/H^{+}} CH_{3}CHO \xrightarrow{K_{2}Cr_{2}O_{7}/H^{+}} CH_{3}COOH$

Reagent: Acidified potassium dichromate(VI) ($K_{2}Cr_{2}O_{7}/H_{2}SO_{4}$) Conditions: Reflux (for complete oxidation to carboxylic acid) Type: Oxidation

Step 3: Esterification

$CH_{3}COOH + CH_{3}CH_{2}OH \rightleftharpoons CH_{3}COOC_{2}H_{5} + H_{2}O$

Reagent: Ethanoic acid + ethanol Conditions: Concentrated sulphuric acid catalyst, gentle heating (reflux) Type: Esterification (condensation)

Note: Steps 1 and 2 produce both ethanol (from step 1) and ethanoic acid (from step 2). These are then combined in step 3 to form the ester. Alternatively, step 1 ethanol can be used directly, and a separate batch is fully oxidised to ethanoic acid for the esterification.

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Integration Test 2: Polymer Properties + Monomer Identification

Question

Polymer $P$ has the following repeating unit: $(-CH_{2}-CHCl-)_n$.

(a) Identify the monomer of polymer $P$ and draw its structure. [2 marks]

(b) Polymer $P$ is rigid and used for water pipes, while poly(ethene) is flexible and used for plastic bags. Explain the difference in properties in terms of intermolecular forces. [3 marks]

(c) Polymer $P$ releases toxic hydrogen chloride gas when burned in a limited supply of air. Write a balanced equation for this reaction and state why this is a concern for fire safety. [3 marks]

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Worked Solution

(a) Monomer: chloroethene (vinyl chloride), $CH_{2}=CHCl$.

(b) PVC (polymer $P$) has chlorine atoms attached to the polymer chain. The $C-Cl$ bond is polar, creating permanent dipole-dipole interactions between polymer chains. These stronger intermolecular forces make PVC rigid and tough.

Poly(ethene) has only non-polar $C-H$ and $C-C$ bonds. The only intermolecular forces are London dispersion forces, which are relatively weak. This makes poly(ethene) flexible and soft.

(c) When PVC burns in limited air:

$(-CH_{2}-CHCl-)_n + \frac{n}{2}O_{2} \rightarrow nC + nHCl + nH_{2}O$

Or more precisely:

$(-CH_{2}-CHCl-)_n + \frac{5n}{2}O_{2} \rightarrow 2nCO_{2} + nH_{2}O + nHCl$

(in sufficient oxygen, but in limited oxygen, incomplete combustion produces $C$ and $CO$ as well).

The release of toxic, corrosive $HCl$ gas is a serious fire safety concern because:

1. $HCl$ is highly toxic if inhaled, causing respiratory damage.
2. $HCl$ is corrosive and can damage equipment and structures.
3. It poses a danger to firefighters and building occupants.

This is why PVC is not recommended for use in environments where fire safety is critical without appropriate flame retardants.

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Integration Test 3: Naming + Reaction Prediction

Question

Consider the following alcohols:

• $A$: $CH_{3}CH_{2}CH_{2}CH_{2}OH$ (butan-1-ol)
• $B$: $CH_{3}CH_{2}CH(OH)CH_{3}$ (butan-2-ol)
• $C$: $(CH_{3})_{3}COH$ (2-methylpropan-2-ol)

(a) Classify each alcohol as primary, secondary, or tertiary. [2 marks]

(b) Arrange the three alcohols in order of increasing reactivity with acidified potassium dichromate(VI). Explain. [3 marks]

(c) Predict the organic product when $C$ is heated with acidified $K_{2}Cr_{2}O_{7}$. Explain. [2 marks]

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Worked Solution

(a) $A$ (butan-1-ol): The $-OH$ group is attached to a carbon bonded to one other carbon (and two hydrogens). Primary alcohol.

$B$ (butan-2-ol): The $-OH$ group is attached to a carbon bonded to two other carbons (and one hydrogen). Secondary alcohol.

$C$ (2-methylpropan-2-ol): The $-OH$ group is attached to a carbon bonded to three other carbons (no hydrogens). Tertiary alcohol.

(b) Increasing reactivity with acidified $K_{2}Cr_{2}O_{7}$:

$C \lt B \lt A$

The oxidation of alcohols involves removing a hydrogen atom from the carbon bearing the $-OH$ group. Primary alcohols are most easily oxidised because the $C-H$ bond on the alpha carbon is accessible. Tertiary alcohols are not oxidised at all by $K_{2}Cr_{2}O_{7}$ because the carbon bearing the $-OH$ group has no hydrogen atoms to remove. Secondary alcohols are oxidised to ketones.

(c) $C$ (tertiary alcohol) is not oxidised by acidified $K_{2}Cr_{2}O_{7}$. No reaction occurs (or only the orange-to-green colour change does not happen). The carbon bearing the $-OH$ group in a tertiary alcohol has no alpha hydrogen, so oxidation cannot proceed. The only way to break down a tertiary alcohol is through more drastic conditions such as dehydration to form an alkene, or complete combustion.