(a) Diamond and graphite are both crystalline solids made from carbon atoms. Graphite conducts electricity. Diamond does not conduct electricity. (i) Name the type of lattice structure present in the crystalline solids diamond and graphite. (ii) Explain how graphite conducts electricity. (iii) Explain why diamond does not conduct electricity. (b) Separate samples of phosphorus(V) chloride and silicon(IV) chloride are each added to an excess of cold water. (i) Write an…

(a)(i) Giant molecular (giant covalent) lattice · (a)(ii) Delocalised electrons move through the structure / between the layers · (a)(iii) No delocalised electrons; all four outer electrons used in bonding · (b)(i) PCl5 + 4H2O -> H3PO4 + 5HCl; SiCl4 + 2H2O -> SiO2 + 4HCl · (b)(ii) phosphorus(V) chloride: white/yellow solid; silicon(IV) chloride: colourless liquid · (b)(iii) PCl5 gives a colourless solution; SiCl4 gives a white suspension/precipitate in a colourless liquid ·…

A sample of iron contains three different isotopes and has a relative atomic mass, $A_r$, of 55.8. (i) Define relative atomic mass. (ii) Table 2.1 shows the abundances of two of the isotopes present in the sample of iron. Table 2.1 isotope relative isotopic mass abundance/% --- --- --- $^{54}\text{Fe}$ 53.9 6.0 $^{56}\text{Fe}$ 55.9 91.9 Use Table 2.1 to calculate the relative isotopic mass of the third isotope of iron in the sample. Show your working. (b) Deduce the number…

(a)(i) The average (weighted) mass of an atom of an element compared to 1/12 of the mass of a carbon-12 atom. · (a)(ii) 56.9 · (b) 1 · (c) A circle representing a spherical 2s orbital. · (d) electrons: 23, neutrons · (e) Smaller / decreases; loss of outer shell and increased nuclear attraction.

9701/22 May/June 2025 — Chemistry Solved Past Paper

Q1MediumStates of MatterChemical Periodicity

(a)

Diamond and graphite are both crystalline solids made from carbon atoms. Graphite conducts electricity. Diamond does not conduct electricity.

3M

(i)

Name the type of lattice structure present in the crystalline solids diamond and graphite.

1M Similar

(ii)

Explain how graphite conducts electricity.

1M Similar

DifficultyEasy

Worked solution

Answer

Graphite conducts electricity because it has delocalised electrons that are free to move through the structure / between the layers.

Final answer

Delocalised electrons move through the structure / between the layers

Detailed explanation

Background Concept

Graphite has a layered structure. Each carbon atom is bonded to three others by strong covalent bonds, forming hexagonal sheets. The fourth outer electron of each carbon is not localised in a bond but is delocalised over the layer. These delocalised electrons are mobile and can act as charge carriers.

Understanding the Question

The question asks why graphite conducts electricity. It is an explanation question: you need to name the mobile charge carriers and say they can move through the structure.

Approach

Identify what carries charge in graphite. Since there are no ions, the carriers must be electrons. Show that some electrons are delocalised and free to move.

Step-by-Step Reasoning

In graphite, each carbon uses three of its four outer electrons to form sigma bonds with three neighbouring carbons. The fourth electron is delocalised across the layer. When a potential difference is applied, these delocalised electrons can move through the structure, or between the layers, so graphite conducts electricity.

Key Takeaways

Electrical conductivity requires mobile charged particles. In graphite, the mobile particles are delocalised electrons.

Common Mistakes

Saying graphite conducts because the layers slide over each other confuses electrical conductivity with lubricant behaviour. Saying "free electrons" without "delocalised" may be accepted, but "delocalised" is the precise term.

Things to Be Careful About

The mark scheme accepts "delocalised electrons move through the structure" or "delocalised electrons move through/between the layers". Avoid saying all electrons are free; only one electron per carbon is delocalised.

Techniques used

relate delocalised electrons to electrical conductivitydescribe graphite structure

(iii)

Explain why diamond does not conduct electricity.

1M Similar

(b)

Separate samples of phosphorus(V) chloride and silicon(IV) chloride are each added to an excess of cold water.

6M

(i)

Write an equation for each reaction.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

\text{PCl}_5 + 4\text{H}_2\text{O} \rightarrow \text{H}_3\text{PO}_4 + 5\text{HCl}

\text{SiCl}_4 + 2\text{H}_2\text{O} \rightarrow \text{SiO}_2 + 4\text{HCl}

Final answer

PCl5 + 4H2O -> H3PO4 + 5HCl; SiCl4 + 2H2O -> SiO2 + 4HCl

Detailed explanation

Background Concept

Covalent chlorides of non-metals often hydrolyse in water. Phosphorus(V) chloride and silicon(IV) chloride are covalent chlorides. With an excess of cold water, PCl5 is completely hydrolysed to phosphoric acid and hydrogen chloride: PCl5 + 4H2O -> H3PO4 + 5HCl. Silicon(IV) chloride is hydrolysed to silicon dioxide and hydrogen chloride: SiCl4 + 2H2O -> SiO2 + 4HCl.

Understanding the Question

Write balanced equations for each reaction. The stem says each chloride is added to an excess of cold water, so complete hydrolysis occurs.

Approach

Identify the products from the known hydrolysis reactions, then balance atoms on both sides.

Step-by-Step Reasoning

For PCl5: one phosphorus atom needs to end up in H3PO4; the five chlorine atoms need five HCl molecules. The five HCl plus the three H in H3PO4 require eight hydrogen atoms, so four water molecules are needed. Check oxygen: four water molecules supply four O atoms, matching H3PO4. For SiCl4: one silicon atom forms SiO2; the four chlorine atoms form four HCl. Two water molecules supply the two O atoms for SiO2 and the four H atoms for four HCl.

Key Takeaways

Hydrolysis equations must be balanced by atoms. Excess water ensures complete hydrolysis to the oxoacid/oxide and HCl.

Common Mistakes

Writing PCl5 + H2O -> POCl3 + 2HCl uses limited water, not excess. Forgetting to balance hydrogen or oxygen. Writing SiCl4 + 2H2O -> Si(OH)4 + 4HCl is not the expected product; the mark scheme wants SiO2.

Things to Be Careful About

Use the correct coefficients: 4 for water with PCl5, 2 for water with SiCl4. State symbols are not required by the mark scheme, but if used they must be correct.

Techniques used

write balanced hydrolysis equationsbalance atoms in chemical equations

(ii)

Describe the appearance of phosphorus(V) chloride and silicon(IV) chloride at room temperature.

phosphorus(V) chloride:

silicon(IV) chloride:

2M Similar

(iii)

Compare the appearance of the mixtures produced when each reaction is complete.

2M Similar

(c)

Some oxides are amphoteric.

2M

(i)

Describe what is meant by amphoteric.

1M Similar

(ii)

Identify the formula of a Period 3 oxide that is amphoteric.

1M Similar

(d)

The melting points of different oxides are shown in Table 1.1.

Table 1.1

oxide	melting point/ °C	force of attraction broken during melting
$\text{SO}_2$	–73
$\text{H}_2\text{O}$	0
$\text{SiO}_2$	1610
$\text{Na}_2\text{O}$	1132
$\text{MgO}$	2852

5M

(i)

Complete Table 1.1 by identifying the strongest force of attraction in each oxide that is broken during melting. Use the abbreviations below.

i.d. = instantaneous dipole–induced dipole
p.d. = permanent dipole–permanent dipole
H = hydrogen bond
C = covalent bond
I = ionic bond

3M Similar

(ii)

A student suggests the following hypothesis.

The stronger the covalent bond between atoms in non-metal oxides, the higher the melting point.

Use Table 1.1 to deduce if this hypothesis is true or false or if there is not enough information to make a conclusion. Explain your answer.

2M Similar

DifficultyMedium

Worked solution

Answer

Not enough information. $\text{SO}_2$ and $\text{H}_2\text{O}$ are molecular: melting breaks intermolecular forces (p.d. and hydrogen bonds), not covalent bonds. Only $\text{SiO}_2$ breaks covalent bonds on melting, so there is no comparison of covalent bond strengths across non-metal oxides.

Final answer

Not enough information – SO2 and H2O break intermolecular forces, not covalent bonds; only SiO2 breaks covalent bonds

Detailed explanation

Background Concept

Melting point is determined by the force that must be broken during melting. For molecular oxides, that force is an intermolecular force, not a covalent bond. For giant covalent oxides, covalent bonds are broken. Therefore a hypothesis about covalent bond strength can only be tested using substances that break covalent bonds on melting.

Understanding the Question

Use Table 1.1 to decide whether the hypothesis is true, false, or cannot be concluded, and explain your answer.

Approach

Look at what force is broken for each oxide in the table. If most oxides do not break covalent bonds, the data cannot test the hypothesis.

Step-by-Step Reasoning

SO2 and H2O are molecular; melting breaks p.d. forces and hydrogen bonds, not covalent bonds. Their melting points therefore tell us nothing about covalent bond strength. SiO2 is the only oxide in the table that breaks covalent bonds on melting. Na2O and MgO are ionic, so they are not non-metal oxides and break ionic bonds. Thus there is no comparison of covalent bond strengths among non-metal oxides, so the hypothesis cannot be confirmed or refuted. The conclusion is "not enough information".

Key Takeaways

To evaluate a hypothesis, the data must actually measure the quantity in question. Melting point only reflects covalent bond strength when covalent bonds are broken.

Common Mistakes

Concluding "false" because SO2 has a low melting point despite strong covalent bonds. The correct answer is "not enough information", because SO2 and H2O do not break covalent bonds on melting.

Things to Be Careful About

The mark scheme gives M1 for "not enough information" and M2 for "SO2 and/or H2O break intermolecular forces" or "only SiO2 breaks covalent bonds". Mention at least one of these.

Techniques used

evaluate hypothesis using evidencedistinguish intermolecular forces from covalent bonds

Q2Medium-EasyAtoms, Molecules and StoichiometryAtomic Structure

A sample of iron contains three different isotopes and has a relative atomic mass, $A_r$ , of 55.8.

(a)

4M

(i)

Define relative atomic mass.

2M Similar

DifficultyEasy

Worked solution

Answer

The average (weighted) mass of an atom of an element compared to one-twelfth of the mass of a carbon-12 atom.

Final answer

The average (weighted) mass of an atom of an element compared to 1/12 of the mass of a carbon-12 atom.

Detailed explanation

Background Concept

Relative atomic mass ( $A_r$ ) is a dimensionless quantity used to express the mass of atoms on a scale that is convenient for chemical calculations. Because individual atoms have extremely small masses (e.g., an iron atom is roughly $9.3 \times 10^{-23}$ g), chemists use a relative scale. The standard reference is the carbon-12 isotope, which is assigned an exact mass of 12 atomic mass units (u).

Understanding the Question

The question asks for the formal definition of relative atomic mass. This is a standard recall question testing precise terminology. No calculation is required, but the definition must be exact to earn both marks.

Approach

Recall the IUPAC definition of relative atomic mass, ensuring both key components are included: the 'average (weighted) mass of an atom' and the reference standard '1/12 of the mass of a carbon-12 atom'.

Step-by-Step Reasoning

M1: State that it is the average (or weighted) mass of an atom of the element. It must be an average because most elements exist as a mixture of isotopes with different masses.
M2: State what it is compared to: one-twelfth ( $\frac{1}{12}$ ) of the mass of a carbon-12 atom. Alternatively, it can be phrased as 'on a scale in which the mass of a carbon-12 atom is exactly 12'.

Key Takeaways

Definitions in chemistry must be precise. 'Average mass' is required, not just 'mass', to account for isotopic abundance. The reference is always $\frac{1}{12}$ of C-12, not 1/12 of any other atom or 1 atomic mass unit directly (though they are equivalent by definition).

Common Mistakes

Forgetting the word 'average' or 'weighted'.
Saying 'compared to a carbon-12 atom' without specifying 'one-twelfth of its mass'.
Using 'atomic mass unit' (amu) in the definition, which is circular reasoning.

Things to Be Careful About

Ensure state symbols or units are not added, as $A_r$ is dimensionless. The mark scheme accepts 'compared to $\frac{1}{12}$ mass of one atom of C-12' or 'on a scale in which a carbon-12 atom has a mass of exactly 12'.

Techniques used

define relative atomic mass

(ii)

Table 2.1 shows the abundances of two of the isotopes present in the sample of iron.

Table 2.1

isotope	relative isotopic mass	abundance/%
$^{54}\text{Fe}$	53.9	6.0
$^{56}\text{Fe}$	55.9	91.9

Use Table 2.1 to calculate the relative isotopic mass of the third isotope of iron in the sample. Show your working.

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Abundance of third isotope = $100 - 91.9 - 6.0 = 2.1\%$

$A_r = \frac{\sum (\text{relative isotopic mass} \times \text{abundance})}{100}$

$55.8 = \frac{53.9 \times 6.0 + 55.9 \times 91.9 + x \times 2.1}{100}$

$5580 = 323.4 + 5137.21 + 2.1x$

$5580 = 5460.61 + 2.1x$

$2.1x = 119.39$

$x = 56.9$

Answer

56.9

Final answer

56.9

Detailed explanation

Background Concept

The relative atomic mass ( $A_r$ ) of an element is the weighted average of the relative isotopic masses of all its naturally occurring isotopes, weighted by their relative abundances. The formula is:
$A_r = \frac{\sum (\text{relative isotopic mass} \times \text{relative abundance})}{100}$

Understanding the Question

You are given the $A_r$ of iron (55.8) and the masses and abundances of two of its three isotopes. You need to find the relative isotopic mass of the third isotope. First, you must determine the abundance of the third isotope, then use the $A_r$ formula to solve for the unknown mass.

Approach

Calculate the abundance of the third isotope by subtracting the given abundances from 100%.
Substitute all known values into the $A_r$ equation.
Solve the resulting algebraic equation for the unknown relative isotopic mass ( $x$ ).

Step-by-Step Reasoning

Step 1: The total abundance must equal 100%. The abundance of the third isotope is $100 - 91.9 - 6.0 = 2.1\%$ . (M1)
Step 2: Set up the equation for $A_r$ :
$55.8 = \frac{(53.9 \times 6.0) + (55.9 \times 91.9) + (x \times 2.1)}{100}$
Step 3: Multiply both sides by 100:
$5580 = 323.4 + 5137.21 + 2.1x$
Step 4: Combine the known terms:
$5580 = 5460.61 + 2.1x$
Step 5: Rearrange to solve for $x$ :
$2.1x = 5580 - 5460.61 = 119.39$
$x = \frac{119.39}{2.1} = 56.852...$
Step 6: Round to an appropriate number of significant figures (one decimal place, matching the given data):
$x = 56.9$ (M2)

Key Takeaways

When using the $A_r$ formula, always ensure abundances sum to 100%. If one is missing, calculate it first. The algebraic manipulation is straightforward, but careful arithmetic is required to avoid errors.

Common Mistakes

Forgetting to calculate the missing abundance first.
Dividing by 100 incorrectly or forgetting to multiply $A_r$ by 100.
Rounding intermediate values too early, leading to a final answer that is slightly off.

Things to Be Careful About

Check your arithmetic. The mark scheme awards M1 for the correct abundance (2.1%) and M2 for the correct final answer (56.9) using error carried forward (ecf) if the abundance was wrong but the equation was set up correctly.

Techniques used

calculate relative isotopic mass from abundance and Ar

(b)

Deduce the number of pairs of electrons in the 3d sub-shell in an iron(II) ion.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

1

Final answer

1

Detailed explanation

Background Concept

The electron configuration of an atom describes the distribution of electrons in its atomic orbitals. For transition metals like iron (Fe, atomic number 26), the configuration is $[\text{Ar}] 3\text{d}^6 4\text{s}^2$ . When transition metals form positive ions (cations), electrons are removed from the outermost s-orbital first (the 4s orbital) before the d-orbitals (the 3d orbital). This is because the 4s electrons are higher in energy than the 3d electrons once the atom is ionized.

Understanding the Question

You need to deduce the number of pairs of electrons in the 3d sub-shell of an iron(II) ion ( $\text{Fe}^{2+}$ ). This requires determining the electron configuration of $\text{Fe}^{2+}$ and then applying Hund's rule to count the electron pairs in the 3d sub-shell.

Approach

Write the electron configuration for a neutral iron atom.
Remove two electrons to form the $\text{Fe}^{2+}$ ion (removing from 4s first).
Analyze the 3d sub-shell configuration to count the number of electron pairs.

Step-by-Step Reasoning

Step 1: Neutral iron (Fe, Z=26) has the configuration $1s^2 2s^2 2p^6 3s^2 3p^6 3\text{d}^6 4\text{s}^2$ or $[\text{Ar}] 3\text{d}^6 4\text{s}^2$ .
Step 2: To form $\text{Fe}^{2+}$ , remove two electrons. The first two electrons removed come from the 4s orbital. The configuration of $\text{Fe}^{2+}$ is $[\text{Ar}] 3\text{d}^6$ .
Step 3: The 3d sub-shell has 5 orbitals. According to Hund's rule, electrons fill empty orbitals singly before pairing up. For 6 electrons in 5 orbitals:
- 4 orbitals get 1 electron each (4 electrons).
- 1 orbital gets a second electron (2 electrons).
- Configuration: $\uparrow\downarrow, \uparrow, \uparrow, \uparrow, \uparrow$ .
Step 4: There is exactly 1 pair of electrons in the 3d sub-shell. (B1)

Key Takeaways

Always remove s-electrons before d-electrons when writing configurations for transition metal ions. Use Hund's rule to determine how electrons are distributed within a sub-shell.

Common Mistakes

Removing electrons from the 3d sub-shell first instead of the 4s sub-shell.
Forgetting Hund's rule and assuming electrons pair up immediately.
Counting total electrons in the 3d sub-shell (6) instead of pairs (1).

Things to Be Careful About

The question asks for 'pairs of electrons', not the number of unpaired electrons or total electrons. Read carefully.

Techniques used

deduce electron configuration of ioncount electron pairs

(c)

Sketch the shape of the lowest energy orbital in the shell with principal quantum number $n = 2$ .

1M Similar

(d)

Complete Table 2.2 to show information about particles in one ion of $^{56}\text{Fe}^{3+}$ .

Table 2.2

particle	number of particles present in one ion of $^{56}\text{Fe}^{3+}$
electrons
	30

2M Similar

DifficultyEasy

Worked solution

Answer

particle	number of particles present in one ion of $^{56}\text{Fe}^{3+}$
electrons	23
neutrons	30

Final answer

electrons: 23, neutrons

Detailed explanation

Background Concept

An atom or ion is composed of three subatomic particles: protons, neutrons, and electrons.

Protons determine the atomic number (Z) and the element's identity. For iron, Z = 26.
Neutrons contribute to the mass number (A). The number of neutrons is $A - Z$ .
Electrons determine the charge of the ion. For a neutral atom, electrons = protons. For a cation (positive ion), electrons = protons - charge. For an anion (negative ion), electrons = protons + charge.

Understanding the Question

You are given the ion $^{56}\text{Fe}^{3+}$ . The mass number (A) is 56, the atomic number (Z) for iron is 26, and the charge is +3. You need to fill in the number of electrons and identify the particle that has 30 present.

Approach

Calculate the number of protons (Z = 26 for Fe).
Calculate the number of neutrons (A - Z = 56 - 26 = 30).
Calculate the number of electrons (protons - charge = 26 - 3 = 23).
Match the calculated values to the table.

Step-by-Step Reasoning

Step 1: Iron (Fe) has an atomic number of 26, so there are 26 protons.
Step 2: The mass number is 56. Number of neutrons = $56 - 26 = 30$ . The table already has '30' in the second row, so the particle in the second row is neutrons. (M2)
Step 3: The ion has a +3 charge, meaning it has lost 3 electrons. Number of electrons = $26 - 3 = 23$ . (M1)
Step 4: Fill in the table:
- electrons: 23
- neutrons: 30

Key Takeaways

Always start with the atomic number (protons) and mass number. Neutrons = A - Z. Electrons = Z - charge (for cations).

Common Mistakes

Calculating electrons as protons + charge (which is for anions).
Forgetting that the charge is +3, not -3.
Misidentifying the particle with 30 as 'protons' (iron has 26 protons).

Things to Be Careful About

Ensure the particle names are spelled correctly ('neutrons', not 'neutron' if referring to the count, though 'neutron' is often accepted; the mark scheme says 'neutrons'). The number of electrons must be an integer.

Techniques used

calculate number of subatomic particles in an ion

(e)

The atomic radius of iron is $1.26 \times 10^{-10}\text{ m}$ .

Suggest the change to the radius, if any, after an iron atom reacts to produce an $\text{Fe}^{3+}$ ion. Explain your answer.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Smaller / decreases.
Loss of outer / valence shell (of electrons).
Loss of electrons so increase in (force of) attraction from nucleus.

Answer

The radius decreases. Iron loses its outer (4s) electrons to form $\text{Fe}^{3+}$ , resulting in the loss of the outermost electron shell. With fewer electrons and the same nuclear charge, the remaining electrons experience a greater effective nuclear attraction, pulling them closer to the nucleus.

Final answer

Smaller / decreases; loss of outer shell and increased nuclear attraction.

Detailed explanation

Background Concept

When a neutral atom loses electrons to form a positive ion (cation), its radius decreases. This happens for two main reasons:

Loss of an electron shell: If the electrons are removed from the outermost shell, that entire shell is lost, significantly reducing the size of the electron cloud.
Increased effective nuclear charge: With fewer electrons but the same number of protons in the nucleus, the remaining electrons experience a stronger pull from the nucleus (less shielding). This increased attraction pulls the electron cloud closer to the nucleus.

Understanding the Question

You are given the atomic radius of iron ( $1.26 \times 10^{-10}$ m) and asked to suggest the change in radius when it forms an $\text{Fe}^{3+}$ ion, and to explain why.

Approach

State whether the radius increases or decreases.
Explain the reason using the concepts of shell loss and nuclear attraction.

Step-by-Step Reasoning

Step 1: Iron (Fe) has the electron configuration $[\text{Ar}] 3\text{d}^6 4\text{s}^2$ . To form $\text{Fe}^{3+}$ , it loses the two 4s electrons and one 3d electron. The loss of the 4s electrons means the entire 4th shell (outermost shell) is lost. Therefore, the radius decreases / becomes smaller. (M1)
Step 2: Explain why. The ion has lost electrons (specifically from the outer shell). The nuclear charge (number of protons, 26) remains the same. With fewer electrons and the same nuclear charge, the force of attraction from the nucleus on the remaining electrons increases. (M2)
Alternatively, you can state that the loss of the outer shell means the remaining electrons are in a lower principal quantum number shell, which is inherently closer to the nucleus.

Key Takeaways

Cations are always smaller than their parent atoms. Anions are always larger. The explanation should reference either the loss of a shell or the increased effective nuclear charge (or both).

Common Mistakes

Saying the radius increases (confusing cations with anions).
Giving a vague explanation like 'because it has a positive charge'.
Forgetting to mention the loss of the outer shell, which is a major factor for transition metals losing s-electrons.

Things to Be Careful About

The mark scheme accepts 'loss of outer / valence shell' OR 'loss of electrons so increase in attraction from nucleus'. Providing both is the safest approach. Ensure you use the word 'decreases' or 'smaller', not 'shrinks' or 'gets smaller' (though 'gets smaller' might be accepted, 'decreases' is more scientific).

Techniques used

compare atomic and ionic radiiexplain using effective nuclear charge and shell loss

Q3Medium-HardHydrocarbonsAtoms, Molecules and StoichiometryReaction KineticsChemical Energetics

(a)

$\text{C}_3\text{H}_6$ reacts with $\text{HBr(g)}$ in an addition reaction.

7M

(i)

Define addition reaction.

1M Similar

DifficultyEasy

Worked solution

Answer

An addition reaction is one in which two (or more) reactants combine to form a single product.

Final answer

Two (or more) reactants combine to form only one product.

Detailed explanation

Background Concept

In organic chemistry, reactions are classified by what happens to the bonding framework. Addition reactions occur across a multiple bond (typically C=C or C≡C): the π-bond breaks and two new σ-bonds form, so that two molecules combine to give one larger molecule. This contrasts with substitution (one atom/group replaces another) and elimination (a small molecule is removed to form a multiple bond).

Understanding the Question

The command word is 'define', which asks for the precise meaning of the term 'addition reaction' in the context of organic chemistry. One mark is available, so a concise correct statement is sufficient.

Approach

State the defining feature: multiple reactants → single product. This is the hallmark that distinguishes addition from other reaction types.

Step-by-Step Reasoning

The key idea is that two molecules join together to give only one product. In the case of an alkene + HBr, the C=C double bond opens up and both the H and Br add across it, producing a single halogenoalkane molecule. The definition does not need to mention the mechanism or the type of bond — just the stoichiometric outcome.

Key Takeaways

Addition = two (or more) reactants → one product
This is characteristic of reactions involving multiple bonds (C=C, C≡C, C=O)
The definition is about the number of products, not the mechanism

Common Mistakes

Saying 'a reaction where atoms are added to a molecule' — this is vague and doesn't capture the 'one product' requirement
Confusing addition with substitution or elimination definitions
Not mentioning that only ONE product is formed

Things to Be Careful About

The mark scheme specifically requires the idea that two or more reactants make only one product. Simply saying 'addition across a double bond' without the 'one product' element may not score.

Techniques used

define a reaction typeidentify addition as combination of reactants into single product

(ii)

Complete Fig. 3.1 to show the mechanism for the addition reaction between $\text{C}_3\text{H}_6$ and $\text{HBr}$ to produce 2-bromopropane. Include charges, dipoles, lone pairs of electrons and curly arrows, as appropriate.

Fig. 3.1

4M Similar

DifficultyMedium

Worked solution

Answer

Step 1: Curly arrow from the C=C double bond to the H atom of H–Br.

Step 2: Curly arrow from the H–Br bond to the Br atom (H has $\delta+$ , Br has $\delta-$ ).

Intermediate: Secondary carbocation $(\text{CH}_3)_2\text{C}^+\text{H}$ and $\text{Br}^-$ .

Step 3: Curly arrow from a lone pair on $\text{Br}^-$ to the $\text{C}^+$ of the carbocation.

Final answer

Electrophilic addition mechanism: curly arrow from C=C to H of HBr, curly arrow from H-Br bond to Br (with dipole shown), secondary carbocation intermediate, curly arrow from Br⁻ lone pair to C⁺.

Detailed explanation

Background Concept

Electrophilic addition is the characteristic reaction of alkenes. The π-bond of C=C is a region of high electron density and acts as a nucleophile (electron-pair donor). The H–Br molecule is polarised ( $\text{H}^{\delta+}\text{–Br}^{\delta-}$ ) because bromine is more electronegative than hydrogen, so the H acts as the electrophile. The mechanism proceeds in two stages: (1) the π-electrons attack the electrophilic H, breaking the H–Br bond heterolytically to form a carbocation and Br⁻; (2) the Br⁻ (nucleophile) donates a lone pair to the electron-deficient carbocation carbon, forming the C–Br bond.

Understanding the Question

The question asks you to complete a given framework (Fig. 3.1) showing the mechanism for forming specifically 2-bromopropane. You must include charges, dipoles, lone pairs, and curly arrows. Four marks are available for four distinct features.

Approach

Identify the three stages: (1) attack of π-bond on H, (2) heterolytic cleavage of H–Br, (3) attack of Br⁻ on carbocation. Draw each with correct curly arrows (always from electron source to electron-poor site).

Step-by-Step Reasoning

M1 — Curly arrow from C=C to H: The double bond (π-electrons) is the electron source. The H of HBr is electron-poor ( $\delta+$ ) due to the polarisation of the H–Br bond. So the arrow starts at the C=C bond and points to H.

M2 — Curly arrow from H–Br bond to Br: When H accepts the electron pair from the π-bond, the H–Br bond must break. Both electrons in that bond go to Br (heterolytic fission), so the arrow starts on the H–Br bond and points to Br. The dipole ( $\delta+$ on H, $\delta-$ on Br) must be shown to justify why the bond breaks this way.

M3 — Correct intermediate: The carbocation must be the secondary one: $\text{CH}_3\text{–C}^+\text{H–CH}_3$ (2-bromopropane precursor). The positive charge is on the central carbon. Br⁻ is shown separately with its lone pairs and negative charge.

M4 — Curly arrow from Br⁻ lone pair to C⁺: The bromide ion is the nucleophile; it donates a lone pair to the electron-deficient carbocation carbon.

Key Takeaways

Curly arrows always show movement of electron pairs (from source to destination)
The dipole on HBr must be shown to justify the direction of bond breaking
The secondary carbocation is the intermediate for 2-bromopropane
Lone pairs on Br⁻ are the electron source for the second step

Common Mistakes

Drawing the curly arrow from H to the C=C (wrong direction — arrows go from electrons to electron-poor site)
Omitting the dipole on HBr (required for M2)
Drawing a primary carbocation instead of secondary
Not showing lone pairs on Br⁻
Drawing the arrow from the C⁺ to Br⁻ (backwards)

Things to Be Careful About

The dipole is essential for M2 — without it, that mark is lost
The intermediate must show the + charge clearly on the correct carbon
Lone pairs must be visible on Br⁻ for M4 to be awarded

Techniques used

draw curly arrows showing electron movementidentify the electrophile and nucleophileshow the carbocation intermediateinclude partial charges and dipoles

(iii)

Explain why the major product of this reaction is 2-bromopropane rather than 1-bromopropane.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

The secondary carbocation intermediate (leading to 2-bromopropane) is more stable than the primary carbocation (leading to 1-bromopropane).
This is because the central C⁺ has two methyl groups attached, which donate electron density by the positive inductive effect, stabilising the positive charge more effectively than the single ethyl group in the primary carbocation.

Final answer

The secondary carbocation is more stable due to the positive inductive effect of two methyl groups (rather than one ethyl group), so it forms preferentially, leading to 2-bromopropane as the major product.

Detailed explanation

Background Concept

When an unsymmetrical alkene reacts with a polar molecule like HBr, two possible carbocation intermediates can form. Markovnikov's rule states that the major product is the one formed via the more stable carbocation. Carbocation stability increases with the number of alkyl groups attached to the positively charged carbon (tertiary > secondary > primary) because alkyl groups have a positive inductive effect (+I): they push electron density towards the electron-deficient carbon, dispersing the positive charge.

Understanding the Question

The question asks WHY 2-bromopropane is the major product rather than 1-bromopropane. Two marks are available: one for identifying the more stable intermediate, and one for explaining WHY it is more stable (the inductive effect of alkyl groups).

Approach

Compare the two possible carbocations: secondary (from 2-bromopropane pathway) vs primary (from 1-bromopropane pathway). Explain the stability difference using the inductive effect.

Step-by-Step Reasoning

M1 — More stable intermediate: The pathway to 2-bromopropane involves a secondary carbocation $\text{CH}_3\text{–C}^+\text{H–CH}_3$ , while the pathway to 1-bromopropane involves a primary carbocation $\text{CH}_3\text{CH}_2\text{–C}^+\text{H}_2$ . The secondary carbocation is more stable, so it forms faster (lower activation energy for its formation), and is therefore the major intermediate.

M2 — Inductive effect explanation: The central carbon in the secondary carbocation has TWO methyl groups attached. Each methyl group exerts a positive inductive effect, pushing electron density towards the C⁺ and stabilising the charge. In the primary carbocation, only ONE ethyl group is attached to the C⁺, providing less stabilisation. Two methyl groups give greater total +I effect than one ethyl group.

Key Takeaways

Markovnikov's rule is explained by carbocation stability
More alkyl groups on C⁺ = more stable = major product
The positive inductive effect is the physical basis for this stability ordering
It is the NUMBER of alkyl groups directly attached to C⁺ that matters

Common Mistakes

Saying 'the secondary carbocation is more stable because it has more carbon atoms' — this is vague; you must mention the inductive effect
Saying 'tertiary is more stable than secondary' — irrelevant here since neither pathway gives a tertiary carbocation
Confusing the inductive effect with hyperconjugation (while hyperconjugation is the real physical explanation, at AS level the inductive effect is what is expected)
Saying '2 methyl groups are better than 1 ethyl group' without mentioning the inductive effect

Things to Be Careful About

You must explicitly mention the inductive effect (or electron-donating effect of alkyl groups) for M2
The comparison should be between the two relevant carbocations, not a general statement about stability

Techniques used

compare carbocation stabilityapply positive inductive effect of alkyl groupslink intermediate stability to major product

(b)

$\text{H}_2\text{(g)}$ and nickel are added to alkene X. Fig. 3.2 shows how the concentration of X changes with time.

Fig. 3.2

2M

(i)

Use Fig. 3.2 to identify the limiting reagent in this reaction. Explain your answer.

1M Similar

DifficultyEasy

Worked solution

Answer

$\text{H}_2$ is the limiting reagent. The concentration of X does not fall to zero (it levels off at approximately 0.12 mol dm $^{-3}$ ), so some X remains unreacted at the end, meaning $\text{H}_2$ must have been used up first.

Final answer

H₂ is the limiting reagent because some X (alkene) remains at the end (concentration does not reach zero).

Detailed explanation

Background Concept

In a reaction between two reagents, the limiting reagent is the one that is completely consumed first, stopping the reaction. On a concentration-time graph, the limiting reagent's concentration falls to zero, while the excess reagent's concentration levels off at a non-zero value.

Understanding the Question

Fig. 3.2 shows the concentration of alkene X decreasing with time. The curve levels off at approximately 0.12 mol dm⁻³ and does NOT reach zero. The question asks you to identify the limiting reagent and explain your reasoning.

Approach

If X's concentration doesn't reach zero, X is in excess. Therefore H₂ must be the limiting reagent (it ran out first, preventing further reaction of X).

Step-by-Step Reasoning

The graph shows [X] starting at 1.0 mol dm⁻³ and decreasing to about 0.12 mol dm⁻³, then remaining constant. Since [X] ≠ 0 at the end, X was not fully consumed. The reaction stopped because the other reactant (H₂) was used up. Therefore H₂ is the limiting reagent.

Key Takeaways

The reagent whose concentration does NOT reach zero is in excess
The reaction stops when the limiting reagent is exhausted
A plateau on a concentration-time graph indicates reaction completion

Common Mistakes

Saying X is the limiting reagent because its concentration decreased (it decreased but didn't reach zero)
Not providing the explanation (the mark requires both identification AND reasoning)

Things to Be Careful About

The mark scheme requires BOTH parts: naming H₂ as limiting AND stating that X remains. Either alone is insufficient.

Techniques used

interpret concentration-time graphidentify limiting reagent from non-zero final concentration

(ii)

Use Fig. 3.2 to describe how the gradient changes as the reaction proceeds. State what this shows about the rate during this reaction.

1M Similar

DifficultyEasy

Worked solution

Answer

The gradient (slope) of the curve decreases with time (becomes less steep). Since the gradient represents the rate of reaction, this shows that the rate decreases as the reaction proceeds, eventually reaching zero when the gradient becomes zero (horizontal line).

Final answer

The gradient decreases over time, showing that the rate of reaction decreases as the reaction proceeds.

Detailed explanation

Background Concept

On a concentration-time graph, the gradient at any point equals the instantaneous rate of reaction. A steep gradient means a fast reaction; a shallow gradient means a slow reaction. As reactants are consumed, their concentration decreases, leading to fewer successful collisions per unit time, so the rate falls.

Understanding the Question

The question asks you to describe how the gradient changes and state what this tells you about the rate. One mark is available for linking the gradient change to the rate change.

Approach

Observe that the curve starts steep and becomes progressively flatter. State that gradient = rate, so decreasing gradient means decreasing rate.

Step-by-Step Reasoning

At t = 0, the curve is steep (large negative gradient) → high initial rate. As time progresses, the curve becomes less steep → rate decreases. Eventually the curve becomes horizontal (gradient = 0) → rate = 0 (reaction has stopped, limiting reagent exhausted). The mark scheme accepts either 'as gradient decreases, rate decreases' or 'rate decreases until the reaction stops when gradient = 0'.

Key Takeaways

Gradient of concentration-time graph = rate at that moment
Decreasing gradient = decreasing rate
Zero gradient = reaction has stopped
The rate decreases because reactant concentration decreases (fewer collisions)

Common Mistakes

Saying 'the concentration decreases' without linking it to gradient and rate
Confusing the shape of the curve with the rate (the curve is concave up, but the key feature is that it gets less steep)

Things to Be Careful About

You must explicitly link the gradient to the rate — describing the curve shape alone without mentioning what it means for the rate will not earn the mark.

Techniques used

interpret gradient of concentration-time graphlink gradient to rate of reaction

(c)

Hydrocarbon Y contains two $\text{C}=\text{C}$ groups. There are no other functional groups present.

Y reacts with an excess of $\text{H}_2\text{(g)}$ to produce hexane, $\text{C}_6\text{H}_{14}$ .

Table 3.1

bond	energy/ $\text{kJ mol}^{-1}$
C−C	350
C=C	610
C≡C	840
H−H	436
C−H	410

Use Table 3.1 to calculate the enthalpy change per mole of $\text{C}_6\text{H}_{14}$ produced in this reaction.

2M Similar

DifficultyMedium

Worked solution

Working

Y is a diene with two C=C bonds that hydrogenates to hexane (C₆H₁₄), so Y = C₆H₁₀.

$\text{C}_6\text{H}_{10} + 2\text{H}_2 \rightarrow \text{C}_6\text{H}_{14}$

Only bonds that change need be considered:

Bonds broken:
$2 \times (\text{C=C}) + 2 \times (\text{H–H}) = 2(610) + 2(436) = 1220 + 872 = 2092 \text{ kJ mol}^{-1}$

Bonds made:
$4 \times (\text{C–H}) + 2 \times (\text{C–C}) = 4(410) + 2(350) = 1640 + 700 = 2340 \text{ kJ mol}^{-1}$

$\Delta H = 2092 - 2340 = -248 \text{ kJ mol}^{-1}$

Answer

$\Delta H = -248$ kJ mol $^{-1}$

Final answer

-248 kJ mol⁻¹

Detailed explanation

Background Concept

Enthalpy change can be calculated from bond energies using: ΔH = Σ(bonds broken) − Σ(bonds made). Bond breaking is endothermic (energy absorbed) and bond making is exothermic (energy released). Only bonds that actually change between reactants and products need be considered — bonds that are present in both and unchanged cancel out.

Understanding the Question

Hydrocarbon Y has two C=C groups and no other functional groups. It reacts with excess H₂ to produce hexane (C₆H₁₄). We must first deduce Y's molecular formula, then calculate ΔH per mole of hexane produced using the given bond energies.

Approach

Deduce Y's formula: hexane is C₆H₁₄. Adding 2H₂ (4 H atoms) to Y gives C₆H₁₄, so Y = C₆H₁₀ (a diene).
Identify bonds that change: two C=C become two C-C, and two H-H become four C-H.
Apply ΔH = bonds broken − bonds made.

Step-by-Step Reasoning

Deducing Y: C₆H₁₄ − 2H₂ = C₆H₁₀. This is a diene (two C=C bonds, consistent with the question statement).

Bonds broken (reactants side, bonds that disappear):

2 × C=C: 2 × 610 = 1220 kJ mol⁻¹
2 × H–H: 2 × 436 = 872 kJ mol⁻¹
Total broken = 2092 kJ mol⁻¹

Bonds made (products side, new bonds formed):

Each C=C → C-C conversion creates 1 additional C–C bond (the π-bond becomes a σ-bond to H). Two C=C → two C-C: 2 × 350 = 700 kJ mol⁻¹
Each H adds to form a C–H bond. Four H atoms added (2 from each H₂): 4 × 410 = 1640 kJ mol⁻¹
Total made = 2340 kJ mol⁻¹

ΔH = 2092 − 2340 = −248 kJ mol⁻¹

The negative sign confirms the reaction is exothermic, as expected for hydrogenation.

Key Takeaways

ΔH = Σ(bonds broken) − Σ(bonds made)
Only bonds that change need be considered (the 'shortcut' method)
Hydrogenation of alkenes/dienes is always exothermic
Deduce the unknown formula from the product and the known stoichiometry

Common Mistakes

Forgetting that each C=C → C-C change also creates 2 new C-H bonds (total 4 C-H for two C=C)
Using the full bond count for both molecules (Method 1 in the mark scheme) and making arithmetic errors
Getting the sign wrong (bonds made − bonds broken instead of the correct order)
Not realising Y is C₆H₁₀ (a diene with 2 C=C bonds)

Things to Be Careful About

The question asks for ΔH per mole of C₆H₁₄ produced, so the stoichiometry must be 1:1 with hexane
The mark scheme accepts either the full method or the shortcut; both give −248 kJ mol⁻¹
Units must be kJ mol⁻¹

Techniques used

determine molecular formula of diene from productidentify bonds broken and bonds madeapply ΔH = Σ(bonds broken) − Σ(bonds made)

(d)

Hydrocarbon Y reacts with $\text{H}_2\text{(g)}$ .

Fig. 3.3 shows the distribution of energies of $\text{H}_2\text{(g)}$ at temperature $T$ .

Area A represents the number of molecules with energy greater than or equal to the activation energy, $E_A$ , at temperature $T$ .

Fig. 3.3

3M

(i)

Annotate Fig. 3.3 to show the effect of adding nickel to Y and $\text{H}_2\text{(g)}$ at temperature $T$ .

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

A new, lower $E_A$ (labelled $E_{A(\text{catalysed})}$ ) is marked to the LEFT of the original $E_A$ on the energy (x) axis. The curve shape remains unchanged (same temperature).

Final answer

A lower E_A is marked to the left of the original E_A on the x-axis of the Maxwell-Boltzmann distribution.

Detailed explanation

Background Concept

A catalyst provides an alternative reaction pathway with a lower activation energy ( $E_A$ ). On a Maxwell-Boltzmann distribution curve, this is shown by moving the $E_A$ marker to the left on the energy axis. The curve itself does NOT change shape (because the temperature is unchanged — the distribution of molecular energies depends only on temperature). What changes is the proportion of molecules with energy ≥ the new, lower $E_A$ .

Understanding the Question

The question asks you to annotate the given Maxwell-Boltzmann curve (Fig. 3.3) to show the effect of adding nickel (a catalyst for hydrogenation) at the same temperature T. One mark is available.

Approach

Draw a new vertical dashed line to the LEFT of the original $E_A$ line on the x-axis, and label it as the catalysed $E_A$ .

Step-by-Step Reasoning

The original $E_A$ line is already shown on Fig. 3.3.
Adding a catalyst lowers $E_A$ but does NOT change the energy distribution (temperature is constant).
So we keep the same curve and add a new $E_A$ marker to the left of the original.
The area under the curve to the right of the new $E_A$ will be larger than area A.

Key Takeaways

Catalyst lowers $E_A$ → shown by moving the $E_A$ marker leftward
The MB curve shape is unchanged (same temperature)
A larger proportion of molecules now exceed the (lower) $E_A$

Common Mistakes

Changing the shape of the curve (this would represent a temperature change, not catalysis)
Drawing the new $E_A$ to the RIGHT (that would mean higher activation energy)
Not labelling the new $E_A$ clearly

Things to Be Careful About

The curve must remain the same shape — only the $E_A$ position changes. The mark is for showing the new $E_A$ to the left of the original on the x-axis.

Techniques used

annotate Maxwell-Boltzmann distribution with catalysed activation energyshow reduced E_A on energy axis

(ii)

Area B (not labelled on Fig. 3.3) represents the number of molecules with energy greater than or equal to the activation energy when nickel is added at temperature $T$ .

State the difference, if any, between areas A and B. Explain the significance of your answer on the rate of hydrogenation of Y. Give your answer in terms of collisions.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Area B is larger than area A, so the rate of hydrogenation increases.
This is because a greater proportion of molecules now have energy ≥ the (lower) activation energy, leading to an increase in the frequency of effective (successful) collisions.

Final answer

B is larger than A; rate increases due to increased frequency of effective collisions.

Detailed explanation

Background Concept

On a Maxwell-Boltzmann distribution, the area under the curve to the right of $E_A$ represents the fraction (or number) of molecules with sufficient energy to react. When a catalyst lowers $E_A$ , this area increases — meaning more molecules can react per unit time. In collision theory terms, the rate depends on the frequency of effective collisions (collisions with energy ≥ $E_A$ and correct orientation). More molecules exceeding the lower $E_A$ means more effective collisions per second, hence a faster rate.

Understanding the Question

Area A = molecules with energy ≥ original $E_A$ . Area B = molecules with energy ≥ catalysed (lower) $E_A$ . The question asks: (1) the difference between A and B, and (2) the significance for rate, expressed in terms of collisions. Two marks.

Approach

State that B > A (the lower $E_A$ means more molecules qualify). Then link this to collision theory: more molecules with sufficient energy → more effective collisions → faster rate.

Step-by-Step Reasoning

M1 — B larger, rate increases: Since the catalysed $E_A$ is lower (to the left), the area to its right (B) is necessarily larger than the area to the right of the original $E_A$ (A). A larger area means more molecules can react, so the rate increases.

M2 — Frequency of effective collisions: The increased area B means a greater proportion of molecules have energy ≥ $E_A$ . In collision theory, only collisions with energy ≥ $E_A$ are 'effective' (can lead to reaction). With more such molecules present, the frequency of effective collisions increases, explaining the higher rate.

Key Takeaways

Lower $E_A$ → larger area under MB curve to the right → more molecules can react
Rate increase is explained by increased frequency of effective collisions
The total number of collisions doesn't change (same temperature, same concentration) — what changes is the proportion that are effective

Common Mistakes

Saying 'more collisions occur' (wrong — total collision frequency is unchanged; it's the proportion that are effective that increases)
Saying 'the catalyst provides more energy to molecules' (wrong — catalysts lower the energy requirement, they don't increase molecular energies)
Not linking the area difference to collisions (the question specifically asks for an answer 'in terms of collisions')

Things to Be Careful About

You must use the word 'effective' or 'successful' when describing collisions
The explanation must be in terms of collisions as the question specifies
Don't confuse this with the temperature effect (where the curve shape changes)

Techniques used

compare shaded areas on MB distributionlink area to frequency of effective collisionsconnect catalyst to increased rate via collision theory

(e)

Alkene Z contains two $\text{C}=\text{C}$ bonds. Z reacts with an excess of hot concentrated acidified $\text{KMnO}_4$ to produce only $\text{CH}_3\text{COCH}_3$ , $\text{HOOCCH}_2\text{COOH}$ , $\text{CO}_2$ and $\text{H}_2\text{O}$ .

Suggest the structure of Z.

2M Similar

DifficultyMedium-Hard

Worked solution

Working

Hot concentrated acidified $\text{KMnO}_4$ cleaves each C=C bond:

$\text{R}_2\text{C}=$ → ketone
$\text{RCH}=$ → carboxylic acid
$\text{CH}_2=$ → $\text{CO}_2$

From the products:

$\text{CH}_3\text{COCH}_3$ (propanone) → one end was $(\text{CH}_3)_2\text{C}=$
$\text{HOOCCH}_2\text{COOH}$ (propanedioic acid) → middle fragment was $=\text{CHCH}_2\text{CH}=$
$\text{CO}_2$ → other end was $=\text{CH}_2$

Assembling the two C=C bonds:

$(\text{CH}_3)_2\text{C}=\text{CHCH}_2\text{CH}=\text{CH}_2$

Answer

$\text{CH}_2\text{=CHCH}_2\text{CH=C(CH}_3)_2$ (5-methylhexa-1,4-diene)

Final answer

CH₂=CHCH₂CH=C(CH₃)₂ (5-methylhexa-1,4-diene)

Detailed explanation

Background Concept

Hot concentrated acidified potassium manganate(VII) is a powerful oxidising agent that cleaves C=C double bonds completely. Each carbon of the double bond is oxidised based on its substituents:

If the carbon has two alkyl groups (R₂C=), it becomes a ketone (R₂C=O)
If the carbon has one alkyl group and one H (RCH=), it becomes a carboxylic acid (RCOOH)
If the carbon has two H atoms (=CH₂), it becomes CO₂ (+ H₂O)

For a diene with two C=C bonds, both are cleaved, giving up to four fragments (though some may be on the same carbon chain if the two C=C bonds are separated by a saturated chain).

Understanding the Question

Z has exactly two C=C bonds and no other functional groups. Oxidative cleavage gives ONLY propanone, propanedioic acid (HOOCCH₂COOH), CO₂, and H₂O. We must deduce Z's structure by working backwards from these fragments.

Approach

Identify what each product tells us about the C=C carbon that produced it.
Assign fragments to the two ends of each C=C bond.
Assemble the complete molecule.

Step-by-Step Reasoning

Propanone (CH₃COCH₃): This is a ketone, so one carbon of a C=C had two methyl groups: $(\text{CH}_3)_2\text{C}=$

CO₂: This comes from a $=\text{CH}_2$ terminus (carbon with two H's).

HOOCCH₂COOH (propanedioic acid): This is a dicarboxylic acid. Each COOH group came from a =CH– carbon (one H, one alkyl chain). The CH₂ in the middle was already saturated. So this fragment must have been between the two C=C bonds: $=\text{CHCH}_2\text{CH}=$

Assembling: We need to connect $(\text{CH}_3)_2\text{C}=$ to $=\text{CHCH}_2\text{CH}=$ and $=\text{CHCH}_2\text{CH}=$ to $=\text{CH}_2$ :

$(\text{CH}_3)_2\text{C}=\text{CHCH}_2\text{CH}=\text{CH}_2$

Written from the other end: $\text{CH}_2=\text{CHCH}_2\text{CH}=\text{C(CH}_3)_2$

Verification:

First C=C: $(\text{CH}_3)_2\text{C}=\text{CH}...$ → propanone + $...\text{COOH}$ ✓
Second C=C: $...\text{CH}=\text{CH}_2$ → $...\text{COOH}$ + CO₂ ✓
Middle: HOOCCH₂COOH ✓

Key Takeaways

Oxidative cleavage of alkenes is a powerful structural determination tool
Work backwards: each product tells you the substitution pattern of the C=C carbon
A dicarboxylic acid product means the =CH– fragment was between two C=C bonds
The number of products tells you how many C=C bonds were cleaved

Common Mistakes

Putting the two C=C bonds adjacent (conjugated diene) — this would not give the observed products
Forgetting that CO₂ comes from a terminal =CH₂ group
Not realising HOOCCH₂COOH must be the middle fragment (connected to both C=C bonds)
Drawing a structure with the wrong number of carbons (must total 6)

Things to Be Careful About

The question says 'only' these products, so no other fragments are possible
Z must have exactly two C=C bonds and nothing else
The molecular formula must be C₆H₁₀ (check: 6 carbons, 10 hydrogens for a diene)

Techniques used

work backwards from oxidative cleavage productsdeduce alkene fragments from carbonyl/carboxyl/CO₂ productsassemble diene structure from fragments

Q4MediumHydroxy CompoundsAtoms, Molecules and StoichiometryHalogen CompoundsNitrogen CompoundsCarboxylic Acids and DerivativesIntroduction to Organic ChemistryOrganic Synthesis

(a)

Propanoic acid, $\text{CH}_3\text{CH}_2\text{COOH}$ , reacts with reducing agent Q to produce propan-1-ol, $\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}$ .

2M

(i)

Suggest the formula of reducing agent Q.

1M Similar

(ii)

Complete the equation to show the reduction of propanoic acid to propan-1-ol. Use [H] to represent one atom of hydrogen from Q.

......\text{CH}_3\text{CH}_2\text{COOH} + ......[\text{H}] \rightarrow ......

1M Similar

(b)

Propan-1-ol is converted to compound T in a three-step synthesis.

In step 1, $\text{SOCl}_2$ is added to propan-1-ol to produce compound E.

In step 2, E reacts with a suitable reagent to produce butanenitrile.

In step 3, butanenitrile is heated with $\text{NaOH(aq)}$ .

5M

(i)

Name the type of reaction that occurs in step 1.

1M Similar

(ii)

Identify the reagent and conditions required in step 2.

1M Similar

(iii)

Construct an equation to describe the reaction of butanenitrile with $\text{NaOH(aq)}$ in step 3.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

\text{CH}_3\text{CH}_2\text{CH}_2\text{CN} + \text{NaOH} + \text{H}_2\text{O} \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{CO}_2\text{Na} + \text{NH}_3

Final answer

CH3CH2CH2CN + NaOH + H2O -> CH3CH2CH2CO2Na + NH3

Detailed explanation

Background Concept

Nitriles are hydrolysed by aqueous acid or alkali. With hot aqueous NaOH, the C $\equiv$ N is hydrolysed to a carboxylate salt plus ammonia: $\text{RCN} + \text{NaOH} + \text{H}_2\text{O} \rightarrow \text{RCO}_2\text{Na} + \text{NH}_3$ . The nitrogen leaves as NH $_3$ under alkaline conditions (under acidic conditions it would be the ammonium ion).

Understanding the Question

Construct the balanced equation for butanenitrile ( $\text{CH}_3\text{CH}_2\text{CH}_2\text{CN}$ ) heated with NaOH(aq).

Approach

Write the carbon skeleton R = $\text{CH}_3\text{CH}_2\text{CH}_2$ –, convert CN to CO $_2$ Na, put nitrogen out as NH $_3$ , then balance with NaOH and H $_2$ O.

Step-by-Step Reasoning

Organic product: sodium butanoate, $\text{CH}_3\text{CH}_2\text{CH}_2\text{CO}_2\text{Na}$ (M1).
Balance: left has N (in CN), right needs NH $_3$ — 3 H required. NaOH supplies Na and one O–H; H $_2$ O supplies the remaining H and O. Count: C 4=4, H 5+1+2=8 vs 5+3=8, N 1=1, O 1+1=2 vs 2, Na 1=1. Balanced (M2).
Alternative accepted form: $\text{RCN} + \text{NaOH} + 2\text{H}_2\text{O} \rightarrow \text{RCO}_2\text{Na} + \text{NH}_4\text{OH}$ .

Key Takeaways

Alkaline hydrolysis of a nitrile gives the carboxylate salt + NH $_3$ ; acidic hydrolysis gives the free acid + NH $_4^+$ .

Common Mistakes

Writing the free acid $\text{CH}_3\text{CH}_2\text{CH}_2\text{COOH}$ instead of the sodium salt in alkaline conditions.
Writing NH $_4^+$ or NH $_4$ Cl instead of NH $_3$ under alkaline conditions.
Forgetting H $_2$ O as a reactant, leaving the equation unbalanced.

Things to Be Careful About

Butanenitrile is $\text{CH}_3\text{CH}_2\text{CH}_2\text{CN}$ (4 carbons total) — do not write propanenitrile. Both marks depend on the correct organic product first, then the rest of the equation.

Techniques used

construct the balanced alkaline hydrolysis equation of a nitrile

(iv)

Name the type of reaction that occurs in step 3.

1M Similar

(c)

Compounds A, B and C belong to the alcohol homologous series. Each molecule of A, B and C contains four saturated carbon atoms.

3M

(i)

Identify the type of hybridisation shown in the saturated carbon atoms of all alcohols.

1M Similar

(ii)

Identify the gas produced when $\text{Na(s)}$ is added to separate samples of each alcohol.

1M Similar

(iii)

Describe the role of $\text{Na(s)}$ when it reacts with alcohols.

1M Similar

(d)

Table 4.1 shows the results of two tests on separate samples of A, B and C.

Table 4.1

compound	heat under reflux with acidified $\text{K}_2\text{Cr}_2\text{O}_7$	warm with alkaline $\text{I}_2\text{(aq)}$
A	remains orange	no visible change
B	orange to green	no visible change
C	orange to green	pale yellow precipitate

3M

(i)

Deduce the structure of A.

1M Similar

DifficultyMedium

Worked solution

Answer

A is not oxidised by acidified $\text{K}_2\text{Cr}_2\text{O}_7$ (stays orange) and gives no iodoform reaction, so A is a tertiary alcohol: (CH $_3$ ) $_3$ COH, 2-methylpropan-2-ol (tert-butanol).

Final answer

(CH3)3COH (2-methylpropan-2-ol)

Detailed explanation

Background Concept

Acidified dichromate(VI) (orange) oxidises primary alcohols to aldehydes/acids and secondary alcohols to ketones, being reduced to green Cr $^{3+}$ . Tertiary alcohols have no C–H on the carbinol carbon and resist oxidation, so the reagent stays orange. The tri-iodomethane (iodoform) test with I $_2$ /NaOH gives a pale yellow CHI $_3$ precipitate only for alcohols containing the $\text{CH}_3\text{CH(OH)}$ – group (and methyl ketones).

Understanding the Question

Use Table 4.1: A stays orange (no oxidation) and shows no visible change with I $_2$ /NaOH (no iodoform). Deduce A's structure, given A is a saturated C4 alcohol.

Approach

No oxidation $\Rightarrow$ tertiary alcohol. Among C4 alcohols, the only tertiary one is 2-methylpropan-2-ol. Cross-check: it lacks the $\text{CH}_3\text{CH(OH)}$ – group, consistent with the negative iodoform test.

Step-by-Step Reasoning

Orange $\rightarrow$ orange means dichromate is not reduced $\Rightarrow$ A is not oxidised $\Rightarrow$ tertiary alcohol.
C4 tertiary alcohol: central C bonded to OH and three CH $_3$ groups = (CH $_3$ ) $_3$ COH.
Negative iodoform test confirms no $\text{CH}_3\text{CH(OH)}$ – group. ✓

Key Takeaways

Dichromate distinguishes 1°/2° (green) from 3° (stays orange); iodoform identifies $\text{CH}_3\text{CH(OH)}$ –.

Common Mistakes

Suggesting butan-1-ol or butan-2-ol — both would turn dichromate green.
Confusing 'no visible change' in the iodoform test with a positive result.

Things to Be Careful About

Write the structure clearly: (CH $_3$ ) $_3$ COH; the name 2-methylpropan-2-ol is also acceptable.

Techniques used

interpret acidified dichromate test resultsinterpret the tri-iodomethane testdeduce alcohol structure from test data

(ii)

Deduce a possible name for B.

1M Similar

(iii)

Deduce the identity of C.

1M Similar

Q5MediumAtoms, Molecules and StoichiometryIntroduction to Organic ChemistryAnalytical Techniques

The structure of vitamin C is shown in Fig. 5.1.

Fig. 5.1

(a)

Deduce the empirical formula of vitamin C.

1M Similar

DifficultyEasy

Worked solution

Answer

C $_3$ H $_4$ O $_3$

Final answer

C3H4O3

Detailed explanation

Background Concept

A molecular formula shows the actual number of atoms of each element in one molecule of a compound. An empirical formula shows the simplest whole-number ratio of these atoms. To convert a molecular formula to an empirical formula, you divide all the subscripts by their greatest common divisor.

Understanding the Question

The question provides the displayed (structural) formula of vitamin C (ascorbic acid) and asks for its empirical formula. You must count every atom in the structure—including those on hydroxyl (–OH) groups and implicit hydrogens on ring carbons—and then reduce the ratio to its simplest form.

Approach

Count the total number of carbon (C), hydrogen (H), and oxygen (O) atoms in the displayed structure.
Write the molecular formula.
Divide all subscripts by their highest common factor to obtain the empirical formula.

Step-by-Step Reasoning

Count carbons: The five-membered ring contains 4 carbon atoms. The side chain (–CH(OH)–CH $_2$ OH) contains 2 carbon atoms. Total C = 6.
Count oxygens: The ring has 1 ether oxygen and 1 carbonyl oxygen (C=O). The ring has 2 enolic hydroxyl groups (–OH). The side chain has 2 hydroxyl groups. Total O = 1 + 1 + 2 + 2 = 6.
Count hydrogens: The ring CH has 1 H. The two enolic –OH groups have 2 H. The side chain –CH(OH)– has 2 H (1 on C, 1 on O). The terminal –CH $_2$ OH has 3 H (2 on C, 1 on O). Total H = 1 + 2 + 2 + 3 = 8.
Molecular formula: C $_6$ H $_8$ O $_6$ .
Empirical formula: Divide all subscripts by 2 to get C $_3$ H $_4$ O $_3$ .

Key Takeaways

Always remember to count implicit hydrogens on saturated carbons and explicit hydrogens on functional groups like –OH when deducing a formula from a displayed structure.

Common Mistakes

Forgetting the hydrogen on the ring carbon (C4) that is not part of a double bond or hydroxyl group.
Miscounting the hydrogens on the side chain.
Stopping at the molecular formula (C $_6$ H $_8$ O $_6$ ) instead of simplifying to the empirical formula.

Things to Be Careful About

Ensure you simplify the ratio fully. C $_6$ H $_8$ O $_6$ is not the empirical formula; C $_3$ H $_4$ O $_3$ is.

Techniques used

count atoms from displayed formulasimplify to empirical formula

(b)

The mass of vitamin C present in 150.0 g of lemon is found in an experiment.

stage 1 All the vitamin C in 150.0 g of lemon is extracted and dissolved in water to make $100.0\text{ cm}^3$ of solution L.

stage 2 A $25.0\text{ cm}^3$ sample of solution L is titrated with $5.00 \times 10^{-3}\text{ mol dm}^{-3}\text{ I}_2\text{(aq)}$ .

Fig. 5.2

Exactly $36.65\text{ cm}^3$ of $\text{I}_2\text{(aq)}$ reacts with the $25.0\text{ cm}^3$ sample of solution L.

[ $M_r$ : vitamin C, 176]

4M

(i)

Calculate the amount, in mol, of $\text{I}_2\text{(aq)}$ added in the titration.

1M Similar

DifficultyEasy

Worked solution

Working

$n = c \times V = 5.00 \times 10^{-3} \times \frac{36.65}{1000}$
$n = 1.833 \times 10^{-4} \text{ mol}$

Answer

$1.83 \times 10^{-4}$ mol

Final answer

1.83e-4 mol

Detailed explanation

Background Concept

In titrations, the amount (in moles) of a solution can be calculated using the equation $n = c \times V$ , where $n$ is the amount in moles, $c$ is the concentration in mol dm $^{-3}$ , and $V$ is the volume in dm $^3$ . If the volume is given in cm $^3$ , it must be divided by 1000 to convert to dm $^3$ .

Understanding the Question

You are given the concentration of the iodine solution ( $5.00 \times 10^{-3}$ mol dm $^{-3}$ ) and the volume used in the titration (36.65 cm $^3$ ). You need to calculate the number of moles of I $_2$ that reacted.

Approach

Convert the volume from cm $^3$ to dm $^3$ by dividing by 1000.
Multiply the concentration by the volume in dm $^3$ to find the moles of I $_2$ .

Step-by-Step Reasoning

Volume of I $_2$ = 36.65 cm $^3$ = $36.65 / 1000$ dm $^3$ = $0.03665$ dm $^3$ .
Concentration of I $_2$ = $5.00 \times 10^{-3}$ mol dm $^{-3}$ .
Moles of I $_2$ = $5.00 \times 10^{-3} \times 0.03665 = 1.8325 \times 10^{-4}$ mol.
Rounding to 3 significant figures (consistent with the concentration), we get $1.83 \times 10^{-4}$ mol. The mark scheme accepts $1.833 \times 10^{-4}$ or $1.8325 \times 10^{-4}$ .

Key Takeaways

Always check your units. Concentration is in mol dm $^{-3}$ , so volume must be in dm $^3$ . A common error is forgetting to divide cm $^3$ by 1000.

Common Mistakes

Forgetting to convert cm $^3$ to dm $^3$ , resulting in an answer that is 1000 times too large.
Using the wrong number of significant figures.

Things to Be Careful About

Keep full precision in intermediate calculations (e.g., use $1.8325 \times 10^{-4}$ ) to avoid rounding errors in subsequent parts of the question.

Techniques used

calculate moles from concentration and volume

(ii)

Use your answer to (b)(i) to calculate the percentage by mass of vitamin C present in 150.0 g of lemon.

(If you were unable to calculate a value for the amount of $\text{I}_2\text{(aq)}$ in (b)(i), use the value $7.65 \times 10^{-4}\text{ mol}$ . This is not the correct value.)

2M Similar

DifficultyMedium

Worked solution

Working

From the equation in Fig 5.2, the molar ratio of vitamin C to I $_2$ is 1 : 1.

Moles of vitamin C in 25.0 cm $^3$ sample = $1.8325 \times 10^{-4}$ mol.

Moles of vitamin C in 100.0 cm $^3$ of solution L:
$1.8325 \times 10^{-4} \times \frac{100.0}{25.0} = 7.33 \times 10^{-4} \text{ mol}$

Mass of vitamin C in 150.0 g of lemon:
$\text{Mass} = \text{moles} \times M_r = 7.33 \times 10^{-4} \times 176 = 0.1290 \text{ g}$

Percentage by mass of vitamin C:
$\% \text{ by mass} = \frac{0.1290}{150.0} \times 100 = 0.0860\%$

Answer

0.0860 %

Final answer

0.0860 %

Detailed explanation

Background Concept

To find the percentage by mass of a substance in a mixture, you need the mass of the substance and the total mass of the mixture. The mass of the substance can be found from its amount in moles and its relative formula mass ( $M_r$ ). In a titration, the stoichiometry of the reaction allows you to relate the moles of the titrant (I $_2$ ) to the moles of the analyte (vitamin C).

Understanding the Question

You have titrated a 25.0 cm $^3$ aliquot of a 100.0 cm $^3$ solution containing all the vitamin C extracted from 150.0 g of lemon. You need to find the total mass of vitamin C in the 150.0 g of lemon and then express it as a percentage.

Approach

Use the 1:1 molar ratio from the reaction equation to find moles of vitamin C in the 25.0 cm $^3$ sample.
Scale up the moles to find the total moles of vitamin C in the 100.0 cm $^3$ solution.
Convert moles to mass using the given $M_r$ of vitamin C (176).
Calculate the percentage by mass: $(\text{mass of vitamin C} / 150.0) \times 100$ .

Step-by-Step Reasoning

Moles of I $_2$ : From part (b)(i), moles of I $_2$ = $1.8325 \times 10^{-4}$ mol.
Moles of vitamin C in 25.0 cm $^3$ : The reaction is vitamin C + I $_2$ → M + 2I $^-$ + 2H $^+$ . The ratio is 1:1, so moles of vitamin C = $1.8325 \times 10^{-4}$ mol.
Total moles of vitamin C in 100.0 cm $^3$ : The 25.0 cm $^3$ sample is $1/4$ of the total volume. Total moles = $1.8325 \times 10^{-4} \times 4 = 7.33 \times 10^{-4}$ mol.
Mass of vitamin C: Mass = moles $\times M_r = 7.33 \times 10^{-4} \times 176 = 0.129008$ g.
Percentage by mass: $(0.129008 / 150.0) \times 100 = 0.086005\%$ . Rounding to 3 significant figures gives 0.0860%.

Key Takeaways

Always scale up the moles from the aliquot to the total volume before calculating the mass. Do not forget to divide by the total mass of the sample (150.0 g) to get the percentage.

Common Mistakes

Forgetting to multiply by 4 to account for the full 100.0 cm $^3$ volume.
Dividing by the wrong mass (e.g., using 25.0 g instead of 150.0 g).
Rounding too early in the calculation, leading to a loss of accuracy.

Things to Be Careful About

The mark scheme allows error carried forward (ecf) from part (b)(i). If you used the suggested incorrect value of $7.65 \times 10^{-4}$ mol, your mass would be $0.5386$ g and the percentage would be $0.359\%$ . Ensure your final answer has the correct number of significant figures (3 sf is appropriate here).

Techniques used

use stoichiometric ratio from balanced equationcalculate mass from moles and molar masscalculate percentage by mass

(iii)

The progress of the reaction of vitamin C with $\text{I}_2\text{(aq)}$ to produce M is monitored using infrared spectroscopy.

Table 5.1 indicates the presence of some absorptions in the infrared spectrum of vitamin C. Complete Table 5.1 to predict which of these absorptions, if any, are present in the infrared spectrum of M.

Table 5.1

absorption/ $\text{cm}^{-1}$	present in spectrum of vitamin C	present in spectrum of M
1500–1680	✓
2850–2950	✓
3200–3650	✓

Table 5.2

bond	functional groups containing the bond	characteristic infrared absorption range (in wavenumbers)/ $\text{cm}^{-1}$
C–O	hydroxy, ester	1040–1300
C=C	aromatic compound, alkene	1500–1680
C=O	amide carbonyl, carboxyl ester	1640–1690 1670–1740 1710–1750
C≡N	nitrile	2200–2250
C–H	alkane	2850–2950
N–H	amine, amide	3300–3500
O–H	carboxyl hydroxy	2500–3000 3200–3650

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

absorption / cm $^{-1}$	present in spectrum of vitamin C	present in spectrum of M
1500–1680	✓	✗
2850–2950	✓	✓
3200–3650	✓	✓

Reasoning:

1500–1680 cm $^{-1}$ : This range corresponds to C=C stretching. Vitamin C has a C=C bond, but compound M has only C=O and C–O bonds in that region (the C=C is oxidised to C=O). Thus, this absorption is not present in M.
2850–2950 cm $^{-1}$ : This range corresponds to C–H stretching (alkane). Both vitamin C and M contain C–H bonds (alkane framework). Thus, this absorption is present in M.
3200–3650 cm $^{-1}$ : This range corresponds to O–H stretching (hydroxy). Vitamin C has hydroxyl groups. Compound M still has hydroxyl groups on the side chain (–CH(OH)–CH $_2$ OH). Thus, this absorption is present in M.

Final answer

1500–1680: ✗; 2850–2950: ✓; 3200–3650: ✓

Detailed explanation

Background Concept

Infrared (IR) spectroscopy identifies functional groups based on the absorption of infrared radiation, which causes bonds to vibrate. Different bond types absorb at characteristic wavenumber ranges:

C=C (alkene): 1500–1680 cm $^{-1}$
C–H (alkane): 2850–2950 cm $^{-1}$
O–H (hydroxy): 3200–3650 cm $^{-1}$
C=O (carbonyl): 1640–1750 cm $^{-1}$
C–O (hydroxy, ester): 1040–1300 cm $^{-1}$

Understanding the Question

You are given the structures of vitamin C and compound M (the product of oxidation by I $_2$ ). You must complete a table indicating which of three IR absorption ranges are present in the spectrum of M. You can use the provided Table 5.2 to match ranges to bonds and functional groups.

Approach

Identify the bonds present in compound M by comparing its structure to vitamin C.
Match the bonds in M to the absorption ranges in Table 5.2.
Determine which absorptions from Table 5.1 will be present or absent in M.

Step-by-Step Reasoning

1500–1680 cm $^{-1}$ (C=C): Vitamin C has a C=C double bond in the ring. In compound M, this C=C bond has been oxidised to two C=O (carbonyl) groups. The C=C bond is gone. Therefore, the absorption at 1500–1680 cm $^{-1}$ is absent (✗) in M.
2850–2950 cm $^{-1}$ (C–H alkane): Both vitamin C and compound M contain an alkane carbon framework (–CH $_2$ – and –CH– groups). The C–H bonds are unchanged. Therefore, this absorption is present (✓) in M.
3200–3650 cm $^{-1}$ (O–H hydroxy): Vitamin C has four hydroxyl (–OH) groups. In compound M, the two enolic –OH groups on the ring are oxidised to C=O groups, but the two hydroxyl groups on the side chain (–CH(OH)–CH $_2$ OH) remain intact. Therefore, an O–H absorption is still present (✓) in M.

Key Takeaways

When comparing IR spectra of reactant and product, focus on which bonds are broken and which are formed. Bonds that remain unchanged will still show their characteristic absorptions.

Common Mistakes

Assuming that because vitamin C is oxidised, all O–H groups disappear. Remember to check the side chain.
Confusing the C=C absorption range (1500–1680) with the C=O absorption range (1640–1750). The C=O absorption is not listed in Table 5.1, so you only need to worry about the C=C disappearing.

Things to Be Careful About

Ensure you use the exact symbols (✓ or ✗) requested by the table. A blank or dash may be marked incorrect depending on the examiner's strictness, though the mark scheme accepts ✗, –, or blank for 'not present'.

Techniques used

identify functional groups in reactant and productpredict IR absorption changes

Chemistry 9701/22 — May/June 2025

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

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Things to Be Careful About

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Understanding the Question

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Understanding the Question

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Understanding the Question

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Background Concept

Understanding the Question

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Background Concept

Understanding the Question

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Step-by-Step Reasoning

Key Takeaways

Common Mistakes