(a) The halogens chlorine and bromine react readily with hydrogen. $$ X_2(g) + H_2(g) \rightarrow 2HX(g) \quad [X = Cl \text{ or } Br] $$ (i) Describe how you could carry out this reaction using chlorine. (ii) Describe two observations you would make if this reaction was carried out with bromine. (iii) Use bond energy data from the Data Booklet to calculate the $\Delta H^\ominus$ for this reaction when $X = Cl$, $$ \Delta H^\ominus =…

(a)(i) Burn hydrogen in chlorine or expose the mixture to UV light. · (a)(ii) Red-brown colour disappears; steamy white fumes produced; container gets warm. · (a)(iii) -182 kJ mol^-1; -103 kJ mol^-1 · (a)(iv) The H-Br bond is weaker than the H-Cl bond. · (b)(i) UV light / sunlight · (b)(ii) +22 kJ mol^-1 · (b)(iii) The reaction is endothermic. · (c)(i) Breaking of a bond where each atom gets one electron, forming radicals. · (c)(ii) •CH2Cl; C-Br bond is weaker. · (d) See…

Chemistry 9701/41 — October/November 2011

Cambridge A-Level · A Level Structured Questions · worked solutions for every part, with the mark scheme

questions

100

marks

120

minutes

Topics Nitrogen Compounds · Introduction to A Level Organic Chemistry · Hydroxy Compounds · Polymerisation · Halogen Compounds · Chemical Energetics · +7 more

Q1Halogen CompoundsChemical EnergeticsIntroduction to A Level Organic ChemistryFree sample

(a)

The halogens chlorine and bromine react readily with hydrogen.

X_2(g) + H_2(g) \rightarrow 2HX(g) \quad [X = Cl \text{ or } Br]

(i)

Describe how you could carry out this reaction using chlorine.

DifficultyEasy

Worked solution

Answer

Burn hydrogen gas in chlorine gas (or expose a mixture of hydrogen and chlorine gases to ultraviolet light/sunlight).

Final answer

Burn hydrogen in chlorine or expose the mixture to UV light.

Detailed explanation

Background Concept

The reaction between hydrogen and halogens ( $H_2 + X_2 \rightarrow 2HX$ ) is a redox reaction that proceeds via a free-radical chain mechanism. For chlorine, the bond dissociation energy is high enough that the reaction does not occur in the dark at room temperature, but the energy from ultraviolet light is sufficient to initiate homolytic fission of the $Cl-Cl$ bond, generating chlorine radicals ( $Cl\cdot$ ) which start the chain reaction.

Understanding the Question

The question asks for the experimental method to carry out the reaction between hydrogen and chlorine. It requires identifying the specific conditions needed to initiate the reaction.

Approach

Recall the standard laboratory demonstration or the industrial synthesis conditions. The key is that the reaction is photochemical (requires light) or combustion-based.

Step-by-Step Reasoning

Initiation: The reaction needs energy to start. Heat alone is often not the primary method described in the context of the explosive nature (though it can work, light is the classic trigger for the explosive reaction). The mark scheme specifies "burn or shine light/uv".
Method 1 (Combustion): Hydrogen is burned in a jet of chlorine. The flame indicates the reaction is occurring, producing hydrogen chloride gas.
Method 2 (Photochemical): A mixture of hydrogen and chlorine gases is exposed to ultraviolet light (or bright sunlight). This triggers a rapid, often explosive, reaction.

Key Takeaways

The reaction between hydrogen and chlorine is initiated by UV light or combustion.
It is a radical substitution/addition process.

Common Mistakes

Suggesting "heating" as the primary method without mentioning light or burning, as the specific initiation by light is the key feature of this reaction's mechanism (photolysis).
Confusing the conditions for Chlorine (light/burning) with Iodine (heating/catalyst).

Things to Be Careful About

Ensure the answer mentions light or burning. Just saying "mix the gases" is incorrect as they won't react in the dark.

Techniques used

describe the initiation conditions for a photochemical reactionidentify the role of light in radical formation

(ii)

Describe two observations you would make if this reaction was carried out with bromine.

DifficultyMedium-Easy

Worked solution

Answer

The red-brown colour of the bromine vapour disappears (decolourises).
Steamy white fumes of hydrogen bromide are produced.
The container becomes warm/hot.

Final answer

Red-brown colour disappears; steamy white fumes produced; container gets warm.

Detailed explanation

Background Concept

Bromine is a volatile liquid that produces red-brown vapour. The reaction with hydrogen produces hydrogen bromide ( $HBr$ ), which is a colourless gas. Like $HCl$ , $HBr$ fumes in moist air because it dissolves in atmospheric water vapour to form tiny droplets of hydrobromic acid. The reaction is exothermic, releasing heat.

Understanding the Question

The question asks for two observations when hydrogen reacts with bromine. Observations must be things seen, heard, or felt (temperature change), not chemical conclusions.

Approach

Identify the reactant's appearance, the product's appearance/behaviour, and the energy change.

Step-by-Step Reasoning

Reactant Colour: Bromine vapour is red-brown. As it reacts, it is consumed, so the colour fades or disappears.
Product Appearance: $HBr$ is a gas that forms acidic fumes in moist air. This appears as "steamy white fumes".
Energy Change: The reaction is exothermic (as calculated in part iii), so the surroundings (the container) will get warm or hot.

Key Takeaways

Halogen vapours have distinct colours (Cl: pale green, Br: red-brown, I: purple).
Hydrogen halides ( $HX$ ) fume in moist air.
Reactions of halogens with hydrogen are exothermic.

Common Mistakes

Saying "colourless gas produced" (you can't see a colourless gas, but you can see the fumes it forms).
Forgetting the temperature change.
Confusing the colour of bromine (red-brown) with iodine (purple/black).

Things to Be Careful About

Be specific about the colour change: "red-brown disappears".
Describe the fumes as "steamy" or "white".

Techniques used

predict observations based on chemical propertiesrelate reaction exothermicity to temperature change

(iii)

Use bond energy data from the Data Booklet to calculate the $\Delta H^\ominus$ for this reaction when

$X = Cl$ ,

\Delta H^\ominus = \text{........................................ } \text{kJ mol}^{-1}

$X = Br$ ,

\Delta H^\ominus = \text{........................................ } \text{kJ mol}^{-1}

DifficultyMedium-Easy

Worked solution

Working

For $X = Cl$ :
Bonds broken: $1 \times \text{H-H}$ and $1 \times \text{Cl-Cl}$
Bonds made: $2 \times \text{H-Cl}$

\Delta H^\ominus = \Sigma E_{\text{bonds broken}} - \Sigma E_{\text{bonds made}}

\Delta H^\ominus = (436 + 244) - (2 \times 431)

\Delta H^\ominus = 680 - 862 = -182 \text{ kJ mol}^{-1}

For $X = Br$ :
Bonds broken: $1 \times \text{H-H}$ and $1 \times \text{Br-Br}$
Bonds made: $2 \times \text{H-Br}$

\Delta H^\ominus = (436 + 193) - (2 \times 366)

\Delta H^\ominus = 629 - 732 = -103 \text{ kJ mol}^{-1}

Answer

$\Delta H^\ominus (Cl) = -182 \text{ kJ mol}^{-1}$
$\Delta H^\ominus (Br) = -103 \text{ kJ mol}^{-1}$

Final answer

-182 kJ mol^-1; -103 kJ mol^-1

Detailed explanation

Background Concept

The enthalpy change of a reaction can be estimated using mean bond enthalpies. The principle is that energy is absorbed to break bonds (endothermic) and energy is released when bonds are formed (exothermic).
$\Delta H = \text{Energy absorbed to break bonds} - \text{Energy released to form bonds}$

Understanding the Question

Calculate $\Delta H^\ominus$ for the reaction $H_2(g) + X_2(g) \rightarrow 2HX(g)$ for Chlorine and Bromine using Data Booklet values.

Approach

List the bonds broken in the reactants.
List the bonds made in the products.
Look up the bond enthalpy values.
Substitute into the equation: $\Delta H = \Sigma E_{\text{broken}} - \Sigma E_{\text{made}}$ .

Step-by-Step Reasoning

Chlorine Case:

Reactants: $H-H$ (436 kJ/mol), $Cl-Cl$ (244 kJ/mol). Total broken = $436 + 244 = 680$ .
Products: $2 \times H-Cl$ (431 kJ/mol). Total made = $2 \times 431 = 862$ .
Calculation: $680 - 862 = -182$ kJ/mol.

Bromine Case:

Reactants: $H-H$ (436 kJ/mol), $Br-Br$ (193 kJ/mol). Total broken = $436 + 193 = 629$ .
Products: $2 \times H-Br$ (366 kJ/mol). Total made = $2 \times 366 = 732$ .
Calculation: $629 - 732 = -103$ kJ/mol.

Key Takeaways

Bond enthalpy calculations give estimates, not exact values (as they use mean values).
The formula is Broken minus Made.

Common Mistakes

Reversing the formula (Made minus Broken), which gives the wrong sign.
Forgetting to multiply the product bond energy by the coefficient (2).
Using the wrong values from the Data Booklet (e.g., confusing bond enthalpy with enthalpy of atomisation).

Things to Be Careful About

Ensure the sign is negative for exothermic reactions.
Check the units are kJ mol $^{-1}$ .

Techniques used

apply Hess's law using bond enthalpiescalculate enthalpy change from bond breaking and making

(iv)

What is the major reason for the difference in these two $\Delta H^\ominus$ values?

DifficultyMedium-Easy

Worked solution

Answer

The H-Br bond is weaker (has a lower bond enthalpy) than the H-Cl bond, so less energy is released when the H-Br bonds form.

Final answer

The H-Br bond is weaker than the H-Cl bond.

Detailed explanation

Background Concept

The enthalpy change of reaction depends on the balance between energy required to break reactant bonds and energy released forming product bonds. If the product bonds are stronger (higher bond enthalpy), the reaction is more exothermic. If they are weaker, it is less exothermic.

Understanding the Question

Explain why $\Delta H$ for bromine (-103) is less negative than for chlorine (-182).

Approach

Compare the bond enthalpies of the bonds broken and made. The H-H bond is constant. The halogen-halogen bonds differ (Cl-Cl 244 vs Br-Br 193). The hydrogen-halogen bonds differ significantly (H-Cl 431 vs H-Br 366).

Step-by-Step Reasoning

Look at the products: $2 \times H-Cl$ releases $862$ kJ. $2 \times H-Br$ releases $732$ kJ.
The difference in energy released is $130$ kJ.
Look at reactants: Breaking $Cl-Cl$ costs $244$ . Breaking $Br-Br$ costs $193$ . Difference is $51$ kJ saved.
Net effect: The much larger drop in energy released by forming the weaker H-Br bond (compared to H-Cl) dominates the calculation, making the bromine reaction less exothermic.
Major reason: The H-Br bond is weaker than the H-Cl bond.

Key Takeaways

Bond strength decreases down the group for H-X bonds.
Weaker bonds formed means less energy released, so less exothermic reaction.

Common Mistakes

Blaming the halogen-halogen bond strength alone (while it contributes, the H-X bond difference is the major factor here).
Saying "Bromine is less reactive" (this is a consequence, not the bond energy explanation).

Things to Be Careful About

Focus on the bond enthalpy values.

Techniques used

compare bond strengthsexplain enthalpy trends

(b)

Some halogens also react readily with methane.

CH_4(g) + X_2(g) \rightarrow CH_3X(g) + HX(g)

(i)

What conditions are needed to carry out this reaction when $X$ is bromine, $Br$ ?

DifficultyEasy

Worked solution

Answer

Ultraviolet light (UV) or sunlight.

Final answer

UV light / sunlight

Detailed explanation

Background Concept

The reaction between methane and halogens is a free-radical substitution. It requires an initiation step where the halogen molecule undergoes homolytic fission to produce radicals. This requires energy, typically supplied by ultraviolet light.

Understanding the Question

Identify the conditions for $CH_4 + Br_2 \rightarrow CH_3Br + HBr$ .

Approach

Recall the mechanism requirements. Heat alone is usually insufficient or leads to other reactions; light is the specific initiator.

Step-by-Step Reasoning

The bond $Br-Br$ needs to break to start the chain reaction.
UV light provides the necessary energy for photolysis.

Key Takeaways

Free radical substitution of alkanes requires UV light.

Common Mistakes

Saying "heat" or "catalyst" (not applicable here).

Things to Be Careful About

Specify UV or sunlight, not just "light" (though "light" is often accepted, UV is more precise for the energy needed).

Techniques used

recall reaction conditions for free radical substitution

(ii)

Use bond energy data from the Data Booklet to calculate the $\Delta H^\ominus$ of this reaction for the situation where $X$ is iodine, $I$ .

\Delta H^\ominus = \text{........................................ } \text{kJ mol}^{-1}

DifficultyMedium-Easy

Worked solution

Working

Reaction: $CH_4 + I_2 \rightarrow CH_3I + HI$

Bonds broken:
$1 \times \text{C-H}$ (in methane) and $1 \times \text{I-I}$
$E_{\text{broken}} = 410 + 151 = 561 \text{ kJ mol}^{-1}$

Bonds made:
$1 \times \text{C-I}$ (in iodomethane) and $1 \times \text{H-I}$
$E_{\text{made}} = 240 + 299 = 539 \text{ kJ mol}^{-1}$

\Delta H^\ominus = E_{\text{broken}} - E_{\text{made}} = 561 - 539 = +22 \text{ kJ mol}^{-1}

Answer

$\Delta H^\ominus = +22 \text{ kJ mol}^{-1}$

Final answer

+22 kJ mol^-1

Detailed explanation

Background Concept

In a substitution reaction like $CH_4 + X_2 \rightarrow CH_3X + HX$ , one C-H bond in the alkane is broken, and the X-X bond is broken. A new C-X bond and a new H-X bond are formed.

Understanding the Question

Calculate $\Delta H$ for the reaction of methane with iodine.

Approach

Identify bonds broken: C-H and I-I.
Identify bonds made: C-I and H-I.
Retrieve values from Data Booklet: C-H (410), I-I (151), C-I (240), H-I (299).
Calculate: (Sum broken) - (Sum made).

Step-by-Step Reasoning

Broken: $410 + 151 = 561$ .
Made: $240 + 299 = 539$ .
$\Delta H = 561 - 539 = +22$ .

Key Takeaways

Substitution involves breaking one bond of each type in reactants and forming one bond of each type in products.
Positive $\Delta H$ indicates an endothermic reaction.

Common Mistakes

Forgetting to include the H-I bond formed.
Using the wrong C-H bond energy (methane has 4 identical C-H bonds, but only one is broken).
Arithmetic errors.

Things to Be Careful About

The sign is positive (+22), not negative.

Techniques used

apply Hess's law using bond enthalpiesidentify bonds broken and made in substitution

(iii)

Hence suggest why it is not possible to make iodomethane, $CH_3I$ , by this reaction.

DifficultyMedium-Easy

Worked solution

Answer

The reaction is endothermic ( $\Delta H$ is positive), meaning it is not energetically favourable / the activation energy is too high / the equilibrium lies far to the left.

Final answer

The reaction is endothermic.

Detailed explanation

Background Concept

For a reaction to occur readily, it generally needs to be exothermic or have a low activation energy. An endothermic reaction requires a continuous input of energy to proceed and is often thermodynamically unfavourable at standard conditions (depending on entropy).

Understanding the Question

Explain why iodomethane cannot be made by this direct reaction, based on the calculated $\Delta H$ .

Approach

The calculated $\Delta H$ is $+22$ kJ mol $^{-1}$ . This means the reaction absorbs energy. Without a continuous supply of energy, the reaction will not proceed. Also, the reverse reaction (exothermic) is favoured.

Step-by-Step Reasoning

The reaction is endothermic.
Endothermic reactions are less likely to occur spontaneously compared to exothermic ones.
Therefore, the reaction does not happen readily / is not feasible.

Key Takeaways

Iodine does not react with methane because the reaction is endothermic.
Reactivity of halogens with methane decreases down the group (F > Cl > Br > I).

Common Mistakes

Saying "iodine is not reactive" without linking it to the energy change calculated.
Mentioning entropy without data (stick to the enthalpy argument provided by the calculation).

Things to Be Careful About

Link the answer back to the "Hence" in the question (use the result from part ii).

Techniques used

interpret enthalpy change for feasibility

(c)

Halogenoalkanes can undergo homolytic fission in the upper atmosphere.

(i)

Explain the term homolytic fission.

DifficultyEasy

Worked solution

Answer

Homolytic fission is the breaking of a covalent bond where each atom retains one electron from the shared pair, forming radicals.

Final answer

Breaking of a bond where each atom gets one electron, forming radicals.

Detailed explanation

Background Concept

Covalent bonds involve shared pairs of electrons. When they break, the electrons can be distributed in two ways:

Heterolytic fission: One atom takes both electrons, forming ions (cation and anion).
Homolytic fission: Each atom takes one electron, forming neutral species with unpaired electrons called free radicals.

Understanding the Question

Define "homolytic fission".

Approach

State the mechanism of bond breaking (one electron each) and the product (radicals).

Step-by-Step Reasoning

Bond breaks.
Electrons split equally (1 each).
Radicals are produced.

Key Takeaways

Homolytic = equal split.
Product = Radical.
Fish-hook arrows are used to show movement of single electrons.

Common Mistakes

Confusing with heterolytic fission (ions formed).
Forgetting to mention that radicals are formed.

Things to Be Careful About

Use the word "radical" or "unpaired electron".

Techniques used

define bond fission types

(ii)

Suggest the most likely organic radical that would be formed by the homolytic fission of bromochloromethane, $CH_2BrCl$ . Explain your answer.

DifficultyMedium-Easy

Worked solution

Answer

The radical formed is $\cdot CH_2Cl$ (chloromethyl radical).
The C-Br bond is weaker than the C-Cl bond (has a lower bond enthalpy), so it breaks more easily.

Final answer

•CH2Cl; C-Br bond is weaker.

Detailed explanation

Background Concept

In a molecule with multiple different bonds (like C-Cl and C-Br), the bond with the lowest bond enthalpy (weakest bond) will break first during homolytic fission. Bond enthalpy generally decreases down the group for carbon-halogen bonds (C-F > C-Cl > C-Br > C-I) because the halogen atoms get larger and the overlap with carbon's orbitals becomes less effective.

Understanding the Question

Identify the radical from $CH_2BrCl$ and explain why.

Approach

Identify the bonds: C-H, C-Cl, C-Br.
Compare C-Cl and C-Br bond strengths. C-Br is weaker.
Break C-Br. The electrons go one to Br, one to the Carbon.
The organic part is $CH_2Cl$ with an unpaired electron.

Step-by-Step Reasoning

Bond enthalpy C-Cl is approx 338 kJ/mol. Bond enthalpy C-Br is approx 276 kJ/mol.
C-Br is weaker.
Fission of C-Br produces $Br\cdot$ and $\cdot CH_2Cl$ .
Therefore, the organic radical is $\cdot CH_2Cl$ .

Key Takeaways

Weakest bond breaks first.
Down the group, C-X bond strength decreases.

Common Mistakes

Choosing the $\cdot CH_2Br$ radical (incorrect because C-Cl is stronger).
Forgetting the dot for the radical.

Things to Be Careful About

Draw the structure with the dot on the carbon.

Techniques used

compare bond enthalpiespredict radical formation

(d)

The reaction between propane and chlorine produces a mixture of many compounds, four of which are structural isomers with the molecular formula $C_3H_6Cl_2$ .

Draw the structural or skeletal formulae of these isomers, and indicate any chiral atoms with an asterisk (*).

DifficultyMedium

Worked solution

Answer

The four structural isomers of $C_3H_6Cl_2$ are:

1,1-dichloropropane: $CH_3CH_2CHCl_2$
1,2-dichloropropane: $CH_3CHClCH_2Cl$ (Chiral carbon is C2)
1,3-dichloropropane: $ClCH_2CH_2CH_2Cl$
2,2-dichloropropane: $CH_3CCl_2CH_3$

Chiral atom identification:
In 1,2-dichloropropane, the central carbon (C2) is bonded to four different groups: $-H$ , $-CH_3$ , $-Cl$ , and $-CH_2Cl$ . Therefore, C2 is a chiral centre and should be marked with an asterisk (*).

Final answer

See diagram; 1,2-dichloropropane has a chiral centre.

Detailed explanation

Background Concept

Structural isomers have the same molecular formula but different connectivity. For $C_3H_6Cl_2$ , we are looking for dichloro-substituted propanes. Chiral centres (stereocentres) are carbon atoms bonded to four different groups.

Understanding the Question

Draw the 4 structural isomers of $C_3H_6Cl_2$ and mark the chiral atom(s).

Approach

Draw the carbon skeleton (propane: C-C-C).
Place two chlorine atoms in all unique positions.
- Both on C1: 1,1-dichloropropane.
- Both on C2: 2,2-dichloropropane.
- One on C1, one on C2: 1,2-dichloropropane.
- One on C1, one on C3: 1,3-dichloropropane.
- (Note: 2,3 is same as 1,2 due to symmetry; 3,3 is same as 1,1).
Check each for chirality.

Step-by-Step Reasoning

1,1-dichloropropane ( $CH_3-CH_2-CHCl_2$ ): C1 has two Cl (not chiral). C2 has two H (not chiral). C3 has three H (not chiral). No chiral centre.
2,2-dichloropropane ( $CH_3-CCl_2-CH_3$ ): C2 has two Cl and two Methyls (not chiral). No chiral centre.
1,3-dichloropropane ( $ClCH_2-CH_2-CH_2Cl$ ): Symmetrical. C1/C3 have two H. C2 has two H. No chiral centre.
1,2-dichloropropane ( $CH_3-CHCl-CH_2Cl$ ): Look at C2. It is bonded to:
1. $-H$
2. $-Cl$
3. $-CH_3$ (Methyl)
4. $-CH_2Cl$ (Chloromethyl)
  These four groups are all different. Therefore, C2 is chiral.

Key Takeaways

Systematic substitution helps avoid missing isomers.
Chirality requires 4 different groups attached to a carbon.
Symmetry often eliminates chirality in 1,1 and 2,2 isomers.

Common Mistakes

Drawing 1,2 and 2,3 as different isomers (they are the same molecule flipped).
Identifying the wrong carbon as chiral (e.g. C1 in 1,2-dichloropropane has two Hydrogens, so it is not chiral).
Missing one of the isomers.

Things to Be Careful About

Ensure the asterisk is on the correct carbon in 1,2-dichloropropane.
Draw clear skeletal or displayed formulae.

Techniques used

draw structural isomersidentify chiral centres

The rest of this paper

7 more questions

Q2Reaction Kinetics · Equilibria15M
Q3Transition Elements14M
Q4Nitrogen Compounds · Hydroxy Compounds14M
Q5Hydrocarbons · Hydroxy Compounds · Introduction to A Level Organic Chemistry · Organic Synthesis · Carboxylic Acids and Derivatives9M
Q6Polymerisation · Nitrogen Compounds10M
Q7Nitrogen Compounds · Analytical Techniques10M
Q8Polymerisation10M

Loading the full paper…