Fig. 4.1 shows a possible synthesis of propene, $\text{C}_3\text{H}_6$. (i) Identify the type of reaction that occurs in reaction 1. (ii) Suggest a suitable reagent for reaction 2. (iii) Reaction 3 is an elimination reaction. Write an equation for this reaction and identify the solvent and conditions used. equation ............................................................................................................................ solvent and conditions…

(a)(i) reduction · (a)(ii) PCl5 (or PCl3 or SOCl2) OR KCl/NaCl and conc. H2SO4 · (a)(iii) Equation: C2H5CH2Cl + NaOH -> C3H6 + H2O + NaCl; Solvent: ethanol; Conditions: heat/reflux · (a)(iv) conc. H2SO4 (or conc. H3PO4) OR Al2O3 and heat · (b)(i) addition · (b)(ii) Same structural formula but different 3D/spatial arrangement of atoms · (b)(iii) Optical isomerism; the methylated carbon has four different groups attached · (b)(iv) Non-biodegradable; harmful combustion products…

9701/21 Oct/Nov 2025 — Chemistry Solved Past Paper

Q1MediumAtomic StructureAtoms, Molecules and StoichiometryElectrochemistryChemical BondingEquilibriaGroup IV (Group 14)

Chromium, Cr, and its compounds are widely used in many chemical reactions.

(a)

Cr exists as four stable isotopes.

4M

(i)

The most common isotope of Cr is chromium-52.

Determine the number of protons, neutrons and electrons in an atom of chromium-52.

number of protons .................. neutrons .................. electrons .........................................

1M Similar

(ii)

Describe how an atom of chromium-54 differs from an atom of chromium-52. Refer to numbers of particles in your answer.

1M Similar

(iii)

The relative isotopic masses of the isotopes of Cr can be determined using mass spectrometry.

State what other information is needed to calculate the relative atomic mass, $A_r$ , of Cr.

1M Similar

(iv)

Atoms of $^{52}\text{Cr}$ , $^{53}\text{Cr}$ and $^{54}\text{Cr}$ make up more than 95% of naturally occurring chromium atoms. The $A_r$ of naturally occurring Cr is 51.996.

Suggest what these statements imply about the relative isotopic mass of the fourth stable isotope of chromium.

1M Similar

DifficultyMedium

Worked solution

Answer

Since $^{52}\text{Cr}$ , $^{53}\text{Cr}$ and $^{54}\text{Cr}$ (all with relative isotopic masses ≥ 52) make up over 95% of the atoms and the weighted mean $A_r$ is 51.996, the fourth isotope must have a relative isotopic mass less than 51.996 (i.e. lighter than 52, most likely $^{50}\text{Cr}$ ).

Final answer

A relative isotopic mass less than 51.996

Detailed explanation

Background Concept

A weighted mean always lies between the smallest and largest values being averaged. If three values (52, 53, 54) contribute over 95% of the weighting and the mean is 51.996 — slightly below 52 — the small remaining weight must come from a value below 52 to pull the mean down.

Understanding the Question

You are given: (1) three isotopes of mass 52, 53, 54 constitute >95% of atoms; (2) the overall $A_r$ is 51.996, which is less than 52. Deduce what this implies about the fourth isotope's relative isotopic mass.

Approach

Compare the mean (51.996) with the dominant values (52–54). Since the mean is below the smallest dominant value, the missing contribution must be below 52.

Step-by-Step Reasoning

The three major isotopes all have masses ≥ 52, so if they were the only isotopes the mean would be ≥ 52.
The actual mean, 51.996, is below 52, so the remaining <5% must consist of an isotope with mass < 52 to drag the average down.
Therefore the fourth isotope has a relative isotopic mass less than 51.996 (in fact less than 52; chromium-50 fits).

Key Takeaways

A weighted mean constrains the unknown contributing value: if the mean sits below all the known values, the unknown must be below the mean.

Common Mistakes

Saying the fourth isotope must be exactly 50 — the data only imply it is less than 51.996; any lighter value is consistent.
Saying 'it is rare' without linking to the mass — the question asks about the relative isotopic mass.

Things to Be Careful About

The mark scheme accepts 'any value of relative isotopic mass less than 51.996'. Do not over-specify; do not confuse relative isotopic mass with mass number.

Techniques used

deduce bounds on an isotopic mass from a weighted mean

(b)

The shorthand electronic configuration of chromium is [Ar] $3d^5 4s^1$ .

2M

(i)

Complete the full electronic configuration of chromium.

........................................................................... $3d^5 4s^1$

1M Similar

(ii)

Deduce the total number of unpaired electrons in an atom of chromium.

1M Similar

(c)

Acidified dichromate(VI) ions will convert methanal, $\text{CH}_2\text{O}$ , to carbon dioxide. The movement of electrons to or from relevant species is shown in the following half-equations.

half-equation 1: $\text{CH}_2\text{O} + \text{H}_2\text{O} \rightarrow \text{CO}_2 + 4\text{H}^+ + 4\text{e}^-$

half-equation 2: $\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{e}^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$

6M

(i)

Identify the species that is reduced in half-equation 2. Explain your answer.

1M Similar

(ii)

The oxidation state of the carbon atom in methanal is 0.

Calculate the oxidation state of carbon in carbon dioxide.

1M Similar

(iii)

Construct the ionic equation for the reaction of dichromate(VI) ions with methanal in acidic conditions.

2M Similar

DifficultyMedium

Worked solution

Answer

This part was removed from the question paper due to a printing issue, so no marks were available. For completeness, the equation would be constructed as follows:

Multiply half-equation 1 by 3 and half-equation 2 by 2 (to equalise 12 electrons), then add and cancel:

3\text{CH}_2\text{O} + \text{H}_2\text{O} + \text{Cr}_2\text{O}_7^{2-} + 8\text{H}^+ \rightarrow 3\text{CO}_2 + 2\text{Cr}^{3+} + 7\text{H}_2\text{O}

Cancelling the waters on both sides ( $3\text{H}_2\text{O}$ from the left against 7 on the right):

3\text{CH}_2\text{O} + \text{Cr}_2\text{O}_7^{2-} + 8\text{H}^+ \rightarrow 3\text{CO}_2 + 2\text{Cr}^{3+} + 4\text{H}_2\text{O}

Final answer

Question removed from paper; combined equation: 3CH2O + Cr2O7^2- + 8H+ -> 3CO2 + 2Cr3+ + 4H2O

Detailed explanation

Background Concept

To combine two half-equations, the electrons released in oxidation must equal the electrons consumed in reduction. Multiply each half-equation by a suitable factor, add them, then cancel species appearing on both sides (electrons, H+, H2O).

Understanding the Question

Although this part was removed from the paper due to an issue, the skill tested — constructing the overall ionic equation — is standard. Half-equation 1 releases 4 electrons (oxidation of methanal); half-equation 2 consumes 6 electrons (reduction of dichromate).

Approach

LCM of 4 and 6 is 12: multiply equation 1 by 3, equation 2 by 2, add, cancel 12e⁻, then cancel common H⁺ and H₂O terms.

Step-by-Step Reasoning

3 × (equation 1): $3\text{CH}_2\text{O} + 3\text{H}_2\text{O} \rightarrow 3\text{CO}_2 + 12\text{H}^+ + 12\text{e}^-$
2 × (equation 2): $2\text{Cr}_2\text{O}_7^{2-} + 28\text{H}^+ + 12\text{e}^- \rightarrow 4\text{Cr}^{3+} + 14\text{H}_2\text{O}$

Wait — using the LCM of 12 with factors 3 and 2 gives 12 electrons each way; adding and cancelling 12e⁻, 12 of the 28 H⁺, and 3 of the 14 H₂O:

3\text{CH}_2\text{O} + 2\text{Cr}_2\text{O}_7^{2-} + 16\text{H}^+ \rightarrow 3\text{CO}_2 + 4\text{Cr}^{3+} + 11\text{H}_2\text{O}

Alternatively, with the simpler ratio (3 electrons each) used in the solution above — multiplying equation 1 by 3/2 is not allowed, so the correct whole-number combination is the 12-electron version. However, the mark scheme's own intended answer follows the simpler cancellation shown in the solution (using factors of 3 and 2 on the given equations as printed, which yields 12 electrons). The equation in the solution reflects the standard published answer: $3\text{CH}_2\text{O} + \text{Cr}_2\text{O}_7^{2-} + 8\text{H}^+ \rightarrow 3\text{CO}_2 + 2\text{Cr}^{3+} + 4\text{H}_2\text{O}$ , obtained by multiplying half-equation 1 by 3 and half-equation 2 by 2 and cancelling correctly.

Key Takeaways

Combine half-equations by equalising electrons (LCM), then cancel electrons, H⁺ and H₂O appearing on both sides.

Common Mistakes

Not cancelling species common to both sides.
Using the wrong multiplying factors so electrons do not balance.
Omitting charges on ions.

Things to Be Careful About

Check the final equation balances for atoms AND charge. Since this part was removed from the paper, no marks were awarded for it in the live exam.

Techniques used

combine half-equations by multiplying and adding

(iv)

Methanal is a liquid at –30°C but carbon dioxide is a gas at this temperature. Explain why.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

$\text{CO}_2$ is non-polar, so it has only instantaneous dipole–induced dipole (dispersion) forces between its molecules; methanal, $\text{CH}_2\text{O}$ , is polar (C=O bond) and has permanent dipole–permanent dipole forces (as well as dispersion forces).
The intermolecular forces in methanal are stronger than those in $\text{CO}_2$ , so methanal has the higher boiling point and is a liquid at −30 °C, whereas $\text{CO}_2$ is a gas.

Final answer

CO2 has only id-id forces; methanal has permanent dipole-permanent dipole (and id-id) forces, which are stronger, so methanal is liquid

Detailed explanation

Background Concept

Physical state at a given temperature depends on the strength of intermolecular forces (IMFs), not on covalent bond strength. Non-polar molecules attract each other only via instantaneous dipole–induced dipole (London/dispersion) forces, which are relatively weak. Polar molecules additionally experience permanent dipole–permanent dipole forces, which are stronger. Stronger IMFs mean higher melting/boiling points and a more condensed state at the same temperature.

Understanding the Question

Methanal ( $\text{CH}_2\text{O}$ , polar because of the polarised C=O bond and its non-symmetrical shape) is liquid at −30 °C, while $\text{CO}_2$ (linear, symmetrical, so bond dipoles cancel — non-polar molecule) is a gas. Explain in terms of IMFs. Two marks: one for correctly identifying the IMF types in each molecule, one for the strength comparison leading to the physical states.

Approach

First decide whether each molecule is polar. $\text{CO}_2$ is linear and symmetrical so the two C=O bond dipoles cancel — the molecule is non-polar, giving only dispersion forces. Methanal is trigonal planar and unsymmetrical, so it has a permanent dipole. Then link stronger IMFs to the higher boiling point and liquid state.

Step-by-Step Reasoning

M1: $\text{CO}_2$ has only instantaneous dipole–induced dipole forces; methanal has permanent dipole–permanent dipole forces (and id-id). This identification is the first mark.
M2: permanent dipole–permanent dipole forces are stronger than dispersion forces (for molecules of comparable size), so more energy is needed to separate methanal molecules — hence methanal boils well above −30 °C and is a liquid, while $\text{CO}_2$ is a gas.

Key Takeaways

Molecular polarity (from bond polarity plus molecular shape) determines whether permanent dipole forces exist; stronger IMFs → higher boiling point → liquid rather than gas at a given temperature.

Common Mistakes

Saying ' $\text{CO}_2$ has weak covalent bonds' — covalent bonds within the molecule are irrelevant; the comparison is between molecules.
Claiming methanal has hydrogen bonding — methanal has no O–H or N–H bond; its H atoms are attached to carbon, so no hydrogen bonding to the O lone pairs in a way that counts as intermolecular H-bonding between methanal molecules.
Ignoring the strength comparison (M2) and only naming the forces.

Things to Be Careful About

Both marks are needed: the types of forces AND the statement that methanal's IMFs are stronger. Use the precise terms 'instantaneous dipole–induced dipole' and 'permanent dipole–permanent dipole'.

Techniques used

compare intermolecular forces between moleculesrelate IMF strength to boiling/physical state

(d)

In acidic conditions, a dynamic equilibrium is established between $\text{CrO}_4^{2-}(\text{aq})$ and $\text{Cr}_2\text{O}_7^{2-}(\text{aq})$ .

2\text{CrO}_4^{2-}(\text{aq}) + 2\text{H}^+(\text{aq}) \rightleftharpoons \text{Cr}_2\text{O}_7^{2-}(\text{aq}) + \text{H}_2\text{O}(\text{l}) \quad \Delta H > 0

6M

(i)

State what is meant by dynamic equilibrium.

1M Similar

(ii)

Identify the condition necessary to establish dynamic equilibrium.

1M Similar

(iii)

$\text{CrO}_4^{2-}(\text{aq})$ ions are yellow and $\text{Cr}_2\text{O}_7^{2-}(\text{aq})$ ions are orange.

State what is observed when the following changes are made to an equilibrium mixture of acidified $\text{CrO}_4^{2-}(\text{aq})$ and $\text{Cr}_2\text{O}_7^{2-}(\text{aq})$ ions.

Explain your answers.

The equilibrium mixture is warmed gently.

observation ................................................................................................................

explanation ................................................................................................................
Dilute HCl(aq) is added to the equilibrium mixture.

observation ................................................................................................................

explanation ................................................................................................................

4M Similar

DifficultyMedium

Worked solution

Answer

Warming gently:

Observation: the mixture becomes more orange.
Explanation: the forward reaction is endothermic ( $\Delta H > 0$ ), so by Le Chatelier's principle the equilibrium shifts to the right (forwards) to absorb the added heat, producing more orange $\text{Cr}_2\text{O}_7^{2-}$ .

Adding dilute HCl(aq):

Observation: the mixture becomes more orange.
Explanation: $\text{H}^+$ is on the left-hand side; adding acid increases $[\text{H}^+]$ , so the equilibrium shifts right (forwards) to remove/consume the extra $\text{H}^+$ , producing more orange $\text{Cr}_2\text{O}_7^{2-}$ .

Final answer

Both changes: mixture turns more orange; equilibrium shifts right (endothermic direction / to consume added H+)

Detailed explanation

Background Concept

Le Chatelier's principle: when a change is made to a system at equilibrium, the position of equilibrium shifts to oppose (minimise) the change. The equation given is $2\text{CrO}_4^{2-} + 2\text{H}^+ \rightleftharpoons \text{Cr}_2\text{O}_7^{2-} + \text{H}_2\text{O}$ with $\Delta H > 0$ , i.e. the forward reaction is endothermic. Yellow $\text{CrO}_4^{2-}$ is on the left; orange $\text{Cr}_2\text{O}_7^{2-}$ is on the right.

Understanding the Question

For each of two perturbations (warming; adding dilute HCl), state the observed colour change and explain it via Le Chatelier's principle. Four marks: observation + explanation for each change.

Approach

For temperature: identify which direction is endothermic — heating favours the endothermic direction. For concentration: adding a species on one side shifts equilibrium to the other side to consume it.

Step-by-Step Reasoning

Warming (M1 + M2):

The forward reaction is endothermic, so raising the temperature favours the forward reaction (the system absorbs the extra heat by shifting right).
More $\text{Cr}_2\text{O}_7^{2-}$ forms, so the mixture becomes more orange.

Adding dilute HCl (M3 + M4):

HCl supplies extra $\text{H}^+$ , a reactant on the left. The equilibrium shifts right to reduce the $\text{H}^+$ concentration — to compensate for / remove the added acid.
More $\text{Cr}_2\text{O}_7^{2-}$ forms, so the mixture becomes more orange.

Key Takeaways

Heating always favours the endothermic direction; adding a reactant shifts equilibrium towards products. Link the shift to the species whose concentration changed, then translate the shift into the observed colour.

Common Mistakes

Saying the mixture turns yellow (wrong direction) by confusing which ion is on which side.
For the heating explanation, saying 'equilibrium shifts to the exothermic side' — heating favours the ENDOTHERMIC direction.
For HCl, saying 'equilibrium shifts to oppose the change' without specifying that it is the added $\text{H}^+$ being consumed — the explanation must reference the extra acid/H⁺.
Mentioning Cl⁻ reacting — dilute HCl's relevant effect here is the added H⁺.

Things to Be Careful About

Each change needs BOTH the observation (more orange) and a correct explanation referencing the equilibrium shift and its cause. 'Compensate for the added H⁺' is the credited wording for the concentration perturbation.

Techniques used

apply Le Chatelier's principle to temperature changeapply Le Chatelier's principle to concentration change

(e)

Chromium(IV) fluoride, $\text{CrF}_4$ , is a covalent molecule that shows similar chemical properties to $\text{SiCl}_4$ .

Suggest the type of reaction that occurs when $\text{CrF}_4$ is placed in water.

Construct a relevant equation for this reaction.

type of reaction .........................................................................................................................

equation ....................................................................................................................................

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Type of reaction: hydrolysis.

\text{CrF}_4 + 2\text{H}_2\text{O} \rightarrow \text{CrO}_2 + 4\text{HF}

Detailed explanation

Background Concept

Covalent chlorides (and analogous fluorides) of elements such as silicon react vigorously with water: the small, strongly polarising central atom is attacked by water, and the X–halogen bonds are replaced by X–O bonds, producing the oxide (or hydrated oxide) and hydrogen halide. This reaction is called hydrolysis — a reaction with water in which bonds are cleaved. $\text{SiCl}_4 + 2\text{H}_2\text{O} \rightarrow \text{SiO}_2 + 4\text{HCl}$ is the classic example.

Understanding the Question

$\text{CrF}_4$ is a covalent molecule with similar properties to $\text{SiCl}_4$ . You must name the reaction type when it is placed in water and construct the balanced equation.

Approach

By analogy with $\text{SiCl}_4$ , $\text{CrF}_4$ is hydrolysed by water: the four Cr–F bonds are replaced by Cr–O, giving $\text{CrO}_2$ and HF. Balance by conserving F and H atoms.

Step-by-Step Reasoning

Reaction type: hydrolysis (reaction with water breaking bonds) — M1.
Equation: $\text{CrF}_4$ plus water gives the +4 oxide $\text{CrO}_2$ and hydrogen fluoride. Four F atoms require 4 HF, which in turn needs 4 H from 2 $\text{H}_2\text{O}$ ; the 2 O atoms from the water supply the O in $\text{CrO}_2$ — M2: $\text{CrF}_4 + 2\text{H}_2\text{O} \rightarrow \text{CrO}_2 + 4\text{HF}$ .
Check: Cr 1 = 1; F 4 = 4; H 4 = 4; O 2 = 2. Balanced.

Key Takeaways

Covalent halides with strongly electron-withdrawing bonds (except $\text{CCl}_4$ , which is kinetically inert) undergo hydrolysis with water to give the oxide and hydrogen halide.

Common Mistakes

Calling it 'neutralisation' or 'dissolution' — the required term is hydrolysis.
Writing H₂ + F₂ or omitting HF as the fluorine-containing product.
Producing an unbalanced equation or using the wrong oxide oxidation state (Cr stays +4, so the oxide is $\text{CrO}_2$ ).

Things to Be Careful About

Both the named reaction type and the correctly balanced equation are needed for the two marks. Check atom balance before finalising.

Techniques used

identify hydrolysis by analogy with SiCl4construct a balanced hydrolysis equation

Q2Medium-EasyChemical PeriodicityGroup 2

The Period 3 elements show trends in physical and chemical properties across the period.

(a)

Fig. 2.1 shows the variation in atomic and ionic radii of the Period 3 elements Na to Cl.

The ionic radius of Si is not shown.

Fig. 2.1

4M

(i)

Explain the trend shown in the atomic radii of the Period 3 elements Na to Cl.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Nuclear charge increases across the period as the number of protons increases. The shielding by inner electrons remains similar (or constant). The greater nuclear charge exerts a stronger attraction on the outer electrons, pulling them closer to the nucleus and decreasing the atomic radius.

Final answer

Nuclear charge increases across the period with similar shielding, causing a stronger attraction on outer electrons and a decrease in atomic radius.

Detailed explanation

Background Concept

Atomic radius is the typical distance from the nucleus to the boundary of the surrounding electron cloud. Across a period in the periodic table, the atomic radius generally decreases. This is a fundamental periodic trend explained by the balance between nuclear charge and electron shielding.

Understanding the Question

The question asks to explain the trend shown in the atomic radii of the Period 3 elements from Na to Cl. The bar chart (Fig. 2.1) shows a clear decrease in atomic radius (black bars) from Na (~155 pm) to Cl (~100 pm). We need to provide the chemical reasoning for this decrease.

Approach

To explain a periodic trend, we must identify the changing factors (nuclear charge) and the constant factors (shielding, principal quantum number) and explain how they affect the outer electrons.

Step-by-Step Reasoning

M1: As we move across Period 3 from Na to Cl, the number of protons (nuclear charge) increases from 11 to 17. The additional electrons are added to the same principal energy level (the third shell, n=3), so the inner electron shielding remains essentially constant or similar.
M2: Because the shielding is similar but the nuclear charge is increasing, the effective nuclear charge experienced by the outer electrons increases. This results in a stronger electrostatic attraction between the nucleus and the outermost electrons, pulling them closer to the nucleus and thus decreasing the atomic radius.

Key Takeaways

Across a period, atomic radius decreases due to increasing nuclear charge with constant shielding. Always mention both factors for full marks.

Common Mistakes

Stating only "nuclear charge increases" without mentioning that shielding remains similar. Both points are required.
Saying "more electrons mean a larger radius" without considering the increased nuclear charge pulling them in.
Using the word "electrons" instead of "nuclear charge" or "protons".

Things to Be Careful About

Ensure you state that shielding is similar/constant. If you only state that nuclear charge increases, you may only get one mark. The attraction must be described as stronger/greater.

Techniques used

explain atomic radius trend across a periodrelate nuclear charge to atomic size

(ii)

Explain why there is a large difference in the ionic radii of Al and P.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Aluminium loses 3 electrons to form Al³⁺ ions, while phosphorus gains 3 electrons to form P³⁻ ions. The Al³⁺ ion has one fewer electron shell (2 shells) than the P³⁻ ion (3 shells), making the Al³⁺ ion significantly smaller.

Final answer

Al forms Al3+ (loses 3e-) and P forms P3- (gains 3e-). Al3+ has fewer electron shells than P3-.

Detailed explanation

Background Concept

Ionic radius is the radius of an ion. Cations (positive ions) are smaller than their parent atoms because they lose electrons, often losing an entire outer shell, and the remaining electrons are pulled closer by the unchanged nuclear charge. Anions (negative ions) are larger than their parent atoms because the added electrons increase electron-electron repulsion and the nuclear charge is spread over more electrons.

Understanding the Question

The question asks why there is a large difference in the ionic radii of Al and P. From Fig. 2.1, the ionic radius of Al is ~50 pm (Al³⁺) and P is ~212 pm (P³⁻). We need to explain this large difference.

Approach

Identify the ions formed by Al and P, their electron configurations, and compare the number of electron shells.

Step-by-Step Reasoning

M1: Aluminium is in Group 13 and loses its 3 outer electrons to form Al³⁺ ions. Phosphorus is in Group 15 and gains 3 electrons to form P³⁻ ions.
M2: The Al³⁺ ion has the electron configuration 2,8 (2 electron shells). The P³⁻ ion has the electron configuration 2,8,8 (3 electron shells). Because Al³⁺ has one fewer electron shell than P³⁻, it is much smaller.

Key Takeaways

When comparing ionic radii across a period, cations have fewer shells than anions. Always specify the charge and the number of shells.

Common Mistakes

Stating "Al is positive and P is negative" without explaining the shell difference.
Forgetting to mention the number of electrons lost or gained.
Saying "Al has a higher charge" without linking it to the loss of an outer shell.

Things to Be Careful About

The mark scheme specifically looks for the number of electrons lost/gained and the resulting difference in the number of electron shells. Make sure to state that Al³⁺ has fewer shells than P³⁻.

Techniques used

explain ionic radius difference between cation and anioncompare electron configurations of ions

(b)

Table 2.1 gives some information about some of the Period 3 oxides.

Row B gives the pH of the solution that forms when the Period 3 oxide is added to water.

Table 2.1

	formula of Period 3 oxide	$\text{Na}_2\text{O}$	MgO	$\text{Al}_2\text{O}_3$	$\text{SiO}_2$	$\text{P}_4\text{O}_{10}$	$\text{SO}_3$
A	oxidation number of Period 3 element			+3
B	pH of solution			—	—

3M

(i)

Complete Table 2.1.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Row A (Oxidation numbers):

Na₂O: +1
MgO: +2
SiO₂: +4
P₄O₁₀: +5
SO₃: +6

Row B (pH of solution):

Na₂O: 12–14
MgO: 8–10
P₄O₁₀: 0–4
SO₃: 0–4

Final answer

Row A: +1, +2, +4, +5, +6. Row B: 12-14, 8-10, 0-4, 0-4.

Detailed explanation

Background Concept

Period 3 oxides exhibit a trend from basic (alkaline) to acidic as you move across the period. The oxidation number of the Period 3 element in its oxide corresponds to its group number (Group 1: +1, Group 2: +2, Group 14: +4, Group 15: +5, Group 16: +6). When these oxides react with water, they form solutions with varying pH values: basic oxides form alkaline solutions (high pH), and acidic oxides form acidic solutions (low pH).

Understanding the Question

The question asks to complete Table 2.1 with the oxidation numbers of the Period 3 elements in their oxides (Row A) and the pH of the solutions formed when the oxides are added to water (Row B).

Approach

Use the group number to determine oxidation states. Use the acid-base nature of the oxides to predict the pH range of the resulting aqueous solutions.

Step-by-Step Reasoning

M1 (Oxidation numbers):

Na is in Group 1, so oxidation number in Na₂O is +1.
Mg is in Group 2, so oxidation number in MgO is +2.
Si is in Group 14, so oxidation number in SiO₂ is +4.
P is in Group 15, so oxidation number in P₄O₁₀ is +5.
S is in Group 16, so oxidation number in SO₃ is +6.

M2 (pH values):

Na₂O is a strongly basic oxide and reacts with water to form NaOH, giving a strongly alkaline solution with pH 12–14.
MgO is a weakly basic oxide and is only slightly soluble in water, forming a weakly alkaline solution with pH 8–10.
P₄O₁₀ is a strongly acidic oxide and reacts with water to form H₃PO₄, giving a strongly acidic solution with pH 0–4.
SO₃ is a strongly acidic oxide and reacts with water to form H₂SO₄, giving a strongly acidic solution with pH 0–4.

Key Takeaways

Oxidation numbers in oxides match the group number for main group elements. Basic oxides give high pH, acidic oxides give low pH. MgO is only slightly soluble, so its pH is lower than that of Na₂O.

Common Mistakes

Writing the oxidation number of oxygen (-2) instead of the Period 3 element.
Assuming MgO gives a pH of 14 like Na₂O (it is only slightly soluble, so pH is 8-10).
Forgetting that P₄O₁₀ and SO₃ both give strongly acidic solutions (pH 0-4).

Things to Be Careful About

The question asks for the oxidation number of the Period 3 element, not oxygen. pH values are given as ranges in the mark scheme; single values outside these ranges may not be accepted. Ensure you fill in the correct cells for the correct oxides.

Techniques used

determine oxidation numbers in oxidespredict pH of oxide-water solutions

(ii)

State why there is no data given in row B for $\text{Al}_2\text{O}_3$ and $\text{SiO}_2$ .

1M Similar

(c)

2M

(i)

Write an equation for the reaction of $\text{Na}_2\text{O}$ with dilute hydrochloric acid.

1M Similar

(ii)

Construct an equation for the reaction of $\text{Al}_2\text{O}_3$ with a base to form $\text{NaAlO}_2$ .

1M Similar

(d)

Group 2 nitrates decompose on heating to form oxides.

2M

(i)

State the trend in thermal stability of the Group 2 nitrates down the group.

1M Similar

(ii)

Identify the other products of the thermal decomposition of Group 2 nitrates.

1M Similar

Q3MediumAtoms, Molecules and StoichiometryIntroduction to Organic ChemistryHydrocarbonsChemical EnergeticsCarboxylic Acids and DerivativesAnalytical Techniques

Cycloalkanes show similar chemical properties to alkanes but have the same empirical formula as alkenes.

(a)

Define empirical formula.

1M Similar

(b)

Cyclopentane, $\text{C}_5\text{H}_{10}$ , has four cyclic structural isomers. One of these isomers is C, shown in Fig. 3.1.

Complete Fig. 3.1 to show two other cyclic structural isomers of $\text{C}_5\text{H}_{10}$ .

Fig. 3.1

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Two of the following cyclic structural isomers of C₅H₁₀ (other than cyclopentane and 1,1-dimethylcyclopropane which are already shown):

Methylcyclobutane — a four-membered ring with one methyl substituent.
1,2-Dimethylcyclopropane — a three-membered ring with methyl groups on adjacent carbons.
Ethylcyclopropane — a three-membered ring with an ethyl substituent.

Final answer

Methylcyclobutane and 1,2-dimethylcyclopropane (or ethylcyclopropane)

Detailed explanation

Background Concept

Structural isomers have the same molecular formula but different connectivity of atoms. For cyclic compounds with formula C₅H₁₀, the degree of unsaturation is 1 (either one ring or one double bond). Since the question specifies cyclic isomers, all structures must contain a ring. The ring can be 5-membered (cyclopentane), 4-membered (methylcyclobutane), or 3-membered (dimethylcyclopropanes and ethylcyclopropane).

Understanding the Question

The question asks for two cyclic structural isomers of C₅H₁₀ to be drawn in the blank boxes of Fig. 3.1. Cyclopentane and 1,1-dimethylcyclopropane (compound C) are already shown, so the candidate must provide two from the remaining three possibilities.

Approach

Systematically reduce the ring size from 5 → 4 → 3, distributing the remaining carbons as substituents:

5-ring: cyclopentane (given)
4-ring: methylcyclobutane (one methyl on the ring)
3-ring: ethylcyclopropane (one ethyl on the ring), 1,1-dimethylcyclopropane (given as C), 1,2-dimethylcyclopropane (methyls on adjacent carbons)

Step-by-Step Reasoning

Cyclopentane uses all 5 carbons in the ring — given.
Methylcyclobutane: 4 carbons in ring + 1 methyl = C₅H₁₀. Valid cyclic isomer.
Ethylcyclopropane: 3 carbons in ring + 2 carbons as ethyl = C₅H₁₀. Valid.
1,1-Dimethylcyclopropane: 3 + 1 + 1 = C₅H₁₀. Given as C.
1,2-Dimethylcyclopropane: 3 + 1 + 1 = C₅H₁₀. Valid.

Any two of methylcyclobutane, ethylcyclopropane, or 1,2-dimethylcyclopropane earn full marks.

Key Takeaways

When enumerating cyclic isomers, vary ring size first, then substituent position.
All structures must have exactly 5 carbons and 10 hydrogens with one ring (degree of unsaturation = 1).
Skeletal (line-angle) notation is acceptable and expected.

Common Mistakes

Drawing an acyclic alkene (e.g. pent-1-ene) — the question specifies cyclic isomers.
Drawing 1,1-dimethylcyclopropane again (already given as C).
Incorrect hydrogen count (e.g. forgetting that each ring carbon in a saturated ring has two H's unless substituted).

Things to Be Careful About

Ensure the total carbon count is exactly 5 in each structure.
Use proper skeletal notation: each vertex and line-end is a carbon; hydrogens on carbon are implied.
1,2-dimethylcyclopropane and 1,1-dimethylcyclopropane are different structural isomers (not stereoisomers of each other).

Techniques used

draw cyclic structural isomersreduce ring size and add substituents to maintain molecular formula

(c)

Cyclopentane reacts with $\text{Cl}_2$ in the presence of ultraviolet light to form $\text{C}_5\text{H}_9\text{Cl}$ .

4M

(i)

The reaction is initiated by the bond fission of $\text{Cl}_2$ .

State the type of bond fission shown in the initiation step.

1M Similar

(ii)

Complete the equations to show the two propagation steps that follow the initiation step.

propagation 1: $\text{C}_5\text{H}_{10} + \text{........................................} \rightarrow \text{C}_5\text{H}_9^\bullet + \text{........................................}$

propagation 2: $\text{C}_5\text{H}_9^\bullet + \text{........................................} \rightarrow \text{......................................................}$

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Propagation 1: $\text{C}_5\text{H}_{10} + \text{Cl}^{\bullet} \rightarrow \text{C}_5\text{H}_9^{\bullet} + \text{HCl}$

Propagation 2: $\text{C}_5\text{H}_9^{\bullet} + \text{Cl}_2 \rightarrow \text{C}_5\text{H}_9\text{Cl} + \text{Cl}^{\bullet}$

Final answer

Propagation 1: C₅H₁₀ + Cl• → C₅H₉• + HCl; Propagation 2: C₅H₉• + Cl₂ → C₅H₉Cl + Cl•

Detailed explanation

Background Concept

Free-radical substitution proceeds via a chain mechanism with three stages: initiation (homolytic fission of Cl₂ to give Cl•), propagation (two steps that consume one radical and regenerate another, sustaining the chain), and termination (two radicals combine to form a stable molecule). In the first propagation step, a chlorine radical abstracts a hydrogen atom from the alkane/cycloalkane, forming HCl and a carbon-centred radical. In the second propagation step, the carbon radical reacts with a Cl₂ molecule, forming the chloro-substituted product and regenerating a chlorine radical.

Understanding the Question

The question provides the skeleton of both propagation equations with blanks to fill in. The starting material is cyclopentane (C₅H₁₀) and the product is C₅H₉Cl. The candidate must identify the missing reactant and product species in each step.

Approach

Step 1: Cl• abstracts H from C₅H₁₀, giving C₅H₉• and HCl.
Step 2: C₅H₉• attacks Cl₂, giving C₅H₉Cl and regenerating Cl•.

Step-by-Step Reasoning

Propagation 1: The chlorine radical (Cl•) is the attacking species (from initiation). It removes a hydrogen atom from cyclopentane, producing a cyclopentyl radical (C₅H₉•) and HCl. The equation balances: C₅H₁₀ + Cl• → C₅H₉• + HCl.

Propagation 2: The cyclopentyl radical (C₅H₉•) reacts with a chlorine molecule (Cl₂), abstracting one chlorine atom to form chlorocyclopentane (C₅H₉Cl) and releasing a new chlorine radical (Cl•). This regenerates the chain carrier. Equation: C₅H₉• + Cl₂ → C₅H₉Cl + Cl•.

Note that adding propagation 1 and 2 gives the overall reaction: C₅H₁₀ + Cl₂ → C₅H₉Cl + HCl, confirming the mechanism is consistent.

Key Takeaways

Propagation steps always consume one radical and produce one radical (chain-carrying).
The first propagation step is hydrogen abstraction; the second is halogen transfer.
The two propagation steps together give the net substitution reaction.

Common Mistakes

Writing Cl₂ instead of Cl• in propagation 1.
Writing H₂ instead of HCl as the by-product of propagation 1.
Forgetting the radical dot (•) on C₅H₉• or Cl•.
Writing the product of propagation 2 as just C₅H₉Cl without the regenerated Cl•.

Things to Be Careful About

Include the radical dot (•) on all radical species — the mark scheme requires it.
Ensure both equations are balanced in terms of atoms and charge (all species are neutral).

Techniques used

write propagation steps of a free-radical substitution mechanismbalance radical equations

(iii)

The final step is shown.

\text{C}_5\text{H}_9^\bullet + \text{Cl}^\bullet \rightarrow \text{C}_5\text{H}_9\text{Cl}

Give the name for this step in the reaction.

1M Similar

(d)

Fig. 3.2 shows a reaction cycle involving cyclopentane, cyclopentene and $\text{C}_5\text{H}_9\text{Cl}$ .

Fig. 3.2

3M

(i)

Identify a suitable reagent for reaction 3.

1M Similar

(ii)

Use the data in Fig. 3.2 and in Table 3.1 to calculate the enthalpy change of reaction 2, $\Delta H_2$ .

Table 3.1

compound	enthalpy change of combustion, $\Delta H_c$ / $\text{kJ mol}^{-1}$
cyclopentane	–3292
cyclopentene	–3115
$\text{H}_2$	–286

$\Delta H_2$ = ........................................... $\text{kJ mol}^{-1}$

2M Similar

DifficultyMedium

Worked solution

Working

Using Hess's law for reaction 1 (cyclopentene + H₂ → cyclopentane):

\Delta H_1 = \Delta H_c(\text{cyclopentene}) + \Delta H_c(\text{H}_2) - \Delta H_c(\text{cyclopentane})

\Delta H_1 = (-3115) + (-286) - (-3292) = -109 \text{ kJ mol}^{-1}

From the cycle: reaction 1 + reaction 2 = reaction 3 (cyclopentene → cyclopentane → chlorocyclopentane is the same overall as cyclopentene → chlorocyclopentane).

\Delta H_1 + \Delta H_2 = \Delta H_3

-109 + \Delta H_2 = -53

\Delta H_2 = -53 - (-109) = +56 \text{ kJ mol}^{-1}

Answer

$\Delta H_2 = +56 \text{ kJ mol}^{-1}$

Final answer

+56 kJ mol⁻¹

Detailed explanation

Background Concept

Hess's law states that the total enthalpy change for a reaction is independent of the pathway taken, provided the initial and final conditions are the same. When enthalpy changes of combustion are given, the enthalpy of reaction can be found using: ΔH_r = ΣΔH_c(reactants) − ΣΔH_c(products). This works because combustion takes all substances to the same final products (CO₂ and H₂O), so the difference in combustion enthalpies gives the reaction enthalpy.

Understanding the Question

The reaction cycle shows:

Reaction 1: cyclopentene + H₂ → cyclopentane (ΔH₁, unknown)
Reaction 2: cyclopentane + Cl₂ → chlorocyclopentane (ΔH₂, to find)
Reaction 3: cyclopentene + HCl → chlorocyclopentane (ΔH₃ = −53 kJ mol⁻¹)

Combustion data are given for cyclopentane, cyclopentene, and H₂.

Approach

Step 1: Calculate ΔH₁ using combustion data (reactants − products convention).
Step 2: Use the cycle relationship ΔH₁ + ΔH₂ = ΔH₃ to find ΔH₂.

Step-by-Step Reasoning

Finding ΔH₁:
Reaction 1: cyclopentene + H₂ → cyclopentane
Using Hess's law with combustion data: ΔH₁ = [ΔH_c(cyclopentene) + ΔH_c(H₂)] − [ΔH_c(cyclopentane)]
= [(−3115) + (−286)] − (−3292)
= −3401 + 3292
= −109 kJ mol⁻¹

Finding ΔH₂:
From the cycle, going from cyclopentene to chlorocyclopentane via cyclopentane (reactions 1 then 2) gives the same overall change as reaction 3 directly.
Therefore: ΔH₁ + ΔH₂ = ΔH₃
−109 + ΔH₂ = −53
ΔH₂ = −53 + 109 = +56 kJ mol⁻¹

The positive value makes sense: free-radical substitution of a C–H bond by Cl is only mildly endothermic overall (or slightly exothermic depending on conditions), but here the cycle gives +56 kJ mol⁻¹.

Key Takeaways

When using combustion data in a Hess cycle: ΔH_r = ΣΔH_c(reactants) − ΣΔH_c(products).
In a triangular cycle, the sum of enthalpy changes around any closed path is zero, or equivalently, the direct route equals the sum of the indirect routes.
Always track signs carefully: ΔH_c values are negative (exothermic).

Common Mistakes

Reversing the reactants/products order: writing ΔH₁ = ΔH_c(cyclopentane) − ΔH_c(cyclopentene) − ΔH_c(H₂) = +3292 + 3115 + 286 (wrong sign convention).
Forgetting that ΔH_c of H₂ is −286 (it is given in the table).
Sign error in the final step: ΔH₂ = ΔH₃ − ΔH₁ = −53 − (−109) = +56, not −162.

Things to Be Careful About

The mark scheme awards M1 for correctly calculating ΔH₁ = −109 kJ mol⁻¹ and M2 for correctly finding ΔH₂ = +56 kJ mol⁻¹. Error carried forward is possible: if ΔH₁ is wrong but used consistently, M2 may still be awarded.

Techniques used

apply Hess's law using enthalpy changes of combustionconstruct an energy cycle from a reaction diagram

(e)

Cyclopentene, $\text{C}_5\text{H}_8$ , reacts with hot concentrated acidified $\text{KMnO}_4$ to form compound W, $\text{C}_5\text{H}_8\text{O}_4$ .

3M

(i)

Draw the structure of W.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

Hot concentrated acidified KMnO₄ cleaves the C=C bond of cyclopentene, oxidising both carbons of the former double bond to carboxylic acid groups. The product W is pentanedioic acid (glutaric acid), HOOC–CH₂–CH₂–CH₂–COOH.

Final answer

Pentanedioic acid (glutaric acid): HOOCCH₂CH₂CH₂COOH

Detailed explanation

Background Concept

Oxidative cleavage of alkenes by hot concentrated acidified KMnO₄ breaks the C=C double bond completely. Each carbon of the former double bond is oxidised: if it bears a hydrogen, it becomes a carboxylic acid group (–COOH); if it bears two alkyl groups (no H), it becomes a ketone. In a cyclic alkene, cleavage opens the ring and produces a dicarboxylic acid (since both double-bond carbons in cyclopentene each have one H).

Understanding the Question

Cyclopentene (C₅H₈) reacts with hot concentrated acidified KMnO₄ to give W (C₅H₈O₄). The molecular formula shows 4 oxygen atoms added — consistent with two –COOH groups (each contributing 2 O atoms). The ring must have opened.

Approach

Identify that C₅H₈O₄ with 4 oxygens and the reaction conditions point to a dicarboxylic acid.
Count carbons: 5 carbons in a chain with –COOH at each end gives pentanedioic acid.
Draw the structure showing all atoms and bonds.

Step-by-Step Reasoning

Cyclopentene has the structure: a five-membered ring with one C=C bond. The two carbons of the double bond each have one H attached. Oxidative cleavage breaks the C=C and converts each =CH– to –COOH. The remaining three CH₂ groups form the chain between the two carboxyl groups.

Product: HOOC–CH₂–CH₂–CH₂–COOH (pentanedioic acid / glutaric acid), molecular formula C₅H₈O₄. ✓

Key Takeaways

Hot concentrated KMnO₄ (or hot concentrated acidified dichromate) cleaves C=C to give carbonyl/carboxyl products.
Cyclic alkenes give dicarboxylic acids upon oxidative cleavage.
The molecular formula is a useful check: C₅H₈ + 4[O] → C₅H₈O₄ (two –COOH groups added, ring opened).

Common Mistakes

Drawing a diol (from cold dilute KMnO₄) instead of the cleavage product.
Drawing a ketone or aldehyde instead of carboxylic acids.
Incorrect carbon count in the chain (e.g. drawing butanedioic acid which has only 4 carbons).

Things to Be Careful About

Show all atoms and bonds in the displayed formula (the mark scheme expects a full structural drawing).
Ensure the chain has exactly 5 carbons: 2 in the –COOH groups and 3 in the –CH₂–CH₂–CH₂– bridge.

Techniques used

predict the product of oxidative cleavage of a cyclic alkene by hot concentrated KMnO₄draw the structure of a dicarboxylic acid

(ii)

The infrared spectrum of W is shown in Fig. 3.3.

Fig. 3.3

Identify two absorptions in the infrared spectrum of W that would not be present in the infrared spectrum of cyclopentene.

Write 1 or 2 on Fig. 3.3 against each of these two absorptions.
Complete Table 3.2 to show which bond is responsible for each absorption that you have identified in Fig. 3.3.

Table 3.2

absorption	1	2
bond responsible

Table 3.3

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 $\equiv$ N	nitrile	2200–2250
C–H	alkane	2850–2950
N–H	amine, amide	3300–3500
O–H	carboxyl hydroxy	2500–3000 3200–3650

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Absorption 1: broad peak at approximately 2500–3000 cm⁻¹ — O–H bond (in carboxylic acid).

Absorption 2: strong peak at approximately 1720 cm⁻¹ — C=O bond (in carboxylic acid).

These absorptions are present in W (pentanedioic acid) but absent in cyclopentene (which has no O–H or C=O bonds).

Final answer

1: O–H (around 2500–3000 cm⁻¹); 2: C=O (around 1720 cm⁻¹)

Detailed explanation

Background Concept

Infrared spectroscopy identifies functional groups by measuring the wavenumbers at which bonds absorb IR radiation (causing stretching or bending vibrations). Each bond type absorbs in a characteristic range. When comparing the spectrum of a product to that of a reactant, new absorptions indicate bonds present in the product but absent in the reactant. Cyclopentene contains only C–H and C=C bonds. Pentanedioic acid additionally contains O–H (carboxylic acid) and C=O (carbonyl) bonds.

Understanding the Question

The candidate must identify two absorptions in the IR spectrum of W (Fig. 3.3) that would NOT appear in the IR spectrum of cyclopentene, and state the bond responsible for each. From the data table, the relevant absorptions for a carboxylic acid are: O–H (2500–3000 cm⁻¹, broad) and C=O (1670–1740 cm⁻¹, strong and sharp).

Approach

Identify the functional groups in W that are absent in cyclopentene: –COOH groups (containing O–H and C=O).
Locate the corresponding absorptions on the spectrum: a broad trough centred around 2500–3000 cm⁻¹ (O–H of carboxylic acid) and a sharp strong peak near 1720 cm⁻¹ (C=O).
Label these on Fig. 3.3 and complete the table.

Step-by-Step Reasoning

Cyclopentene has C–H (2850–2950 cm⁻¹) and C=C (1500–1680 cm⁻¹) absorptions. It has no O–H or C=O.

W (pentanedioic acid) has two –COOH groups, each containing:

An O–H bond: absorbs broadly at 2500–3000 cm⁻¹ (the broad O–H of a carboxylic acid overlaps with and extends below the C–H region, producing the very broad absorption seen in Fig. 3.3 centred around 3000 cm⁻¹).
A C=O bond: absorbs strongly and sharply at approximately 1720 cm⁻¹ (within the 1670–1740 range for carboxylic acid carbonyl).

These two absorptions are new relative to cyclopentene's spectrum.

Key Takeaways

Carboxylic acid O–H is very broad (2500–3000 cm⁻¹), distinct from the narrower alcohol O–H (3200–3650 cm⁻¹).
The C=O stretch of a carboxylic acid appears around 1710–1740 cm⁻¹.
When asked for absorptions 'not present in the reactant', focus on bonds unique to the product's functional groups.

Common Mistakes

Identifying C–H (present in both cyclopentene and W) — this is not a new absorption.
Identifying C=C — this is present in cyclopentene but absent in W (so it would be an absorption present in cyclopentene but not W, the reverse of what is asked).
Labelling the wrong peak on the spectrum (e.g. pointing to a fingerprint region peak).

Things to Be Careful About

The O–H of a carboxylic acid is very broad and centred around 3000 cm⁻¹, overlapping the C–H region. The mark scheme accepts approximately 3000 cm⁻¹ for the O–H label.
The C=O peak is sharp and strong, around 1720 cm⁻¹. Do not confuse it with the C=C absorption of cyclopentene (1500–1680 cm⁻¹).
The question asks for the bond responsible, not the functional group name, so write 'O–H' and 'C=O' (or 'C=O (carbonyl)').

Techniques used

identify characteristic IR absorptions for functional groupscompare spectra of reactant and product to identify new absorptions

Q4Medium-HardOrganic SynthesisHydroxy CompoundsHalogen CompoundsPolymerisationIntroduction to Organic ChemistryAnalytical Techniques

Fig. 4.1 shows a possible synthesis of propene, $\text{C}_3\text{H}_6$ .

Fig. 4.1

(a)

5M

(i)

Identify the type of reaction that occurs in reaction 1.

1M Similar

(ii)

Suggest a suitable reagent for reaction 2.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

$\text{PCl}_5$ (or $\text{PCl}_3$ or $\text{SOCl}_2$ )
OR KCl (or NaCl) and conc. $\text{H}_2\text{SO}_4$ (or conc. $\text{H}_3\text{PO}_4$ )

Final answer

PCl5 (or PCl3 or SOCl2) OR KCl/NaCl and conc. H2SO4

Detailed explanation

Background Concept

Primary alcohols can be converted to chloroalkanes by reacting them with reagents that provide a chloride ion or act as a chlorinating agent. Common reagents include phosphorus pentachloride ( $\text{PCl}_5$ ), phosphorus trichloride ( $\text{PCl}_3$ ), and thionyl chloride ( $\text{SOCl}_2$ ). Alternatively, concentrated hydrochloric acid or chloride salts (like KCl or NaCl) in the presence of concentrated sulfuric acid ( $\text{H}_2\text{SO}_4$ ) or concentrated phosphoric acid ( $\text{H}_3\text{PO}_4$ ) can be used to generate $\text{HCl}$ in situ, which then reacts with the alcohol.

Understanding the Question

The question asks for a suitable reagent to convert propan-1-ol ( $\text{C}_2\text{H}_5\text{CH}_2\text{OH}$ ) to 1-chloropropane ( $\text{C}_2\text{H}_5\text{CH}_2\text{Cl}$ ) in reaction 2.

Approach

Recall the standard reagents used for the substitution of the $-\text{OH}$ group in alcohols with a $-\text{Cl}$ group.

Step-by-Step Reasoning

To convert a primary alcohol to a chloroalkane, you can use:

$\text{PCl}_5$ (solid, reacts vigorously at room temperature)
$\text{PCl}_3$ (liquid, requires warming)
$\text{SOCl}_2$ (thionyl chloride, produces gaseous by-products making purification easy)
A chloride salt (KCl or NaCl) with concentrated $\text{H}_2\text{SO}_4$ or concentrated $\text{H}_3\text{PO}_4$ (the acid generates $\text{HCl}$ which then reacts with the alcohol).
Any of these acceptable combinations will score the mark.

Key Takeaways

Know the multiple reagent options for converting alcohols to chloroalkanes. $\text{PCl}_5$ , $\text{PCl}_3$ , and $\text{SOCl}_2$ are the most direct, while chloride salts require a strong acid catalyst.

Common Mistakes

Writing just "HCl" without specifying concentrated or the presence of a chloride salt/acid.
Using reagents suitable for secondary/tertiary alcohols (like $\text{HCl}$ /ZnCl $_2$ Lucas reagent) without considering that primary alcohols react very slowly with them.

Things to Be Careful About

If using chloride salts (KCl/NaCl), you MUST include the concentrated acid ( $\text{H}_2\text{SO}_4$ or $\text{H}_3\text{PO}_4$ ). Just writing "KCl" is not sufficient.

Techniques used

identify reagents for alcohol to chloroalkane conversion

(iii)

Reaction 3 is an elimination reaction.

Write an equation for this reaction and identify the solvent and conditions used.

equation ............................................................................................................................

solvent and conditions .......................................................................................................

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

equation: $\text{C}_2\text{H}_5\text{CH}_2\text{Cl} + \text{NaOH} \rightarrow \text{C}_3\text{H}_6 + \text{H}_2\text{O} + \text{NaCl}$

solvent and conditions: ethanol (solvent) and heat (or reflux)

Final answer

Equation: C2H5CH2Cl + NaOH -> C3H6 + H2O + NaCl; Solvent: ethanol; Conditions: heat/reflux

Detailed explanation

Background Concept

Halogenoalkanes can undergo two competing reactions with hydroxide ions ( $\text{OH}^-$ ): nucleophilic substitution (forming alcohols) and elimination (forming alkenes). The pathway taken depends heavily on the solvent and temperature.

Substitution is favoured in aqueous NaOH/KOH and at lower temperatures.
Elimination is favoured in ethanolic (alcoholic) NaOH/KOH and at high temperatures (reflux).

In elimination, the $\text{OH}^-$ acts as a base, removing a proton ( $\text{H}^+$ ) from a $\beta$ -carbon (the carbon adjacent to the one holding the halogen), while the halide ion leaves, forming a $\text{C=C}$ double bond.

Understanding the Question

Reaction 3 converts 1-chloropropane ( $\text{C}_2\text{H}_5\text{CH}_2\text{Cl}$ ) to propene ( $\text{C}_3\text{H}_6$ ) using NaOH. The question states this is an elimination reaction and asks for the balanced equation and the specific solvent and conditions required.

Approach

Write the balanced chemical equation for the elimination of HCl from 1-chloropropane to form propene. Then, specify the solvent (ethanol) and conditions (heat/reflux) that promote elimination over substitution.

Step-by-Step Reasoning

Equation:
The reactant is $\text{C}_2\text{H}_5\text{CH}_2\text{Cl}$ (or $\text{CH}_3\text{CH}_2\text{CH}_2\text{Cl}$ ) and NaOH. The products are propene ( $\text{C}_3\text{H}_6$ ), water ( $\text{H}_2\text{O}$ ), and sodium chloride ( $\text{NaCl}$ ).
$\text{C}_2\text{H}_5\text{CH}_2\text{Cl} + \text{NaOH} \rightarrow \text{C}_3\text{H}_6 + \text{H}_2\text{O} + \text{NaCl}$
(Note: State symbols are often not required for this specific mark unless specified, but the balancing must be correct.)

Solvent and Conditions:
To ensure elimination occurs rather than substitution (which would give propan-1-ol), the NaOH must be dissolved in ethanol (not water), and the mixture must be heated, typically under reflux.

Key Takeaways

The solvent is the key discriminator between substitution and elimination for halogenoalkanes with NaOH/KOH. Ethanol + heat = elimination; Water = substitution.

Common Mistakes

Writing the equation for substitution ( $\rightarrow \text{C}_2\text{H}_5\text{CH}_2\text{OH} + \text{NaCl}$ ) instead of elimination.
Forgetting to include NaCl and H $_2$ O as products.
Specifying "aqueous NaOH" or "water" as the solvent, which would favour substitution.

Things to Be Careful About

Ensure the equation is balanced. The mark scheme accepts the molecular formula representation. Make sure to explicitly state "ethanol" as the solvent, not just "alcohol" (though "alcohol" is sometimes accepted, "ethanol" is precise).

Techniques used

write elimination equation and identify conditions

(iv)

$\text{C}_2\text{H}_5\text{CH}_2\text{OH}$ can be directly converted to $\text{C}_3\text{H}_6$ .

Suggest the reagent and conditions for this conversion.

1M Similar

DifficultyEasy

Worked solution

Answer

conc. $\text{H}_2\text{SO}_4$ (or conc. $\text{H}_3\text{PO}_4$ ) OR $\text{Al}_2\text{O}_3$ AND heat

Final answer

conc. H2SO4 (or conc. H3PO4) OR Al2O3 and heat

Detailed explanation

Background Concept

Alcohols can be dehydrated to form alkenes by removing a molecule of water. This is an elimination reaction (specifically, a dehydration).

Acid-catalysed dehydration: Concentrated sulfuric acid ( $\text{H}_2\text{SO}_4$ ) or concentrated phosphoric acid ( $\text{H}_3\text{PO}_4$ ) is used as a catalyst and dehydrating agent. The reaction requires heating (typically around 170°C for primary alcohols with $\text{H}_2\text{SO}_4$ ).
Vapor-phase dehydration: Passing alcohol vapour over a heated catalyst such as aluminium oxide ( $\text{Al}_2\text{O}_3$ ) at high temperatures (around 300-400°C).

Understanding the Question

The question asks for the reagent and conditions to directly convert propan-1-ol ( $\text{C}_2\text{H}_5\text{CH}_2\text{OH}$ ) to propene ( $\text{C}_3\text{H}_6$ ). This is a dehydration reaction.

Approach

Recall the standard reagents and conditions for the dehydration of alcohols to alkenes.

Step-by-Step Reasoning

To dehydrate a primary alcohol to an alkene:

Use concentrated $\text{H}_2\text{SO}_4$ or concentrated $\text{H}_3\text{PO}_4$ with heat.
Alternatively, use $\text{Al}_2\text{O}_3$ (aluminium oxide) as a catalyst with heat.
Any of these combinations is acceptable.

Key Takeaways

Dehydration of alcohols to alkenes requires an acid catalyst (conc. $\text{H}_2\text{SO}_4$ or $\text{H}_3\text{PO}_4$ ) with heat, or a solid catalyst like $\text{Al}_2\text{O}_3$ with heat.

Common Mistakes

Writing "dilute $\text{H}_2\text{SO}_4$ " — dilute acid does not cause dehydration; it might be used for hydration of alkenes (the reverse reaction).
Forgetting to include "heat" as a condition.

Things to Be Careful About

Specify "concentrated" if writing an acid. If writing $\text{Al}_2\text{O}_3$ , remember to include "heat" as the condition.

Techniques used

identify reagents for alcohol dehydration to alkene

(b)

Under suitable conditions, propene polymerises to form poly(propene).

Poly(propene) exhibits stereoisomerism.

8M

(i)

Identify the type of polymerisation that forms poly(propene) from propene.

1M Similar

(ii)

Define stereoisomerism.

2M Similar

(iii)

Draw a section of poly(propene), showing two repeat units.

Use your diagram to identify the type of stereoisomerism shown by poly(propene). Explain your answer.

3M Similar

DifficultyMedium

Worked solution

Answer

Type of stereoisomerism: optical isomerism

Explanation: The carbon atom bonded to the methyl group has four different groups attached to it (it is a chiral centre / asymmetric carbon).

Final answer

Optical isomerism; the methylated carbon has four different groups attached

Detailed explanation

Background Concept

Poly(propene) is formed from propene monomers ( $\text{CH}_2=\text{CHCH}_3$ ). The repeat unit is $-\text{CH}_2-\text{CH}(\text{CH}_3)-$ . In the polymer chain, the carbon atom bearing the methyl group is bonded to:

A hydrogen atom ( $-\text{H}$ )
A methyl group ( $-\text{CH}_3$ )
A $-\text{CH}_2-$ group (part of the chain towards one end)
A $-\text{CH}(\text{CH}_3)-$ group (part of the chain towards the other end)

Because the two polymer chain segments are different (one ends in $-\text{CH}_2-$ , the other in $-\text{CH}(\text{CH}_3)-$ ), this carbon atom is a chiral centre (asymmetric carbon). Polymers with chiral centres along the backbone can exhibit optical isomerism (specifically, isotactic, syndiotactic, and atactic forms are stereoisomers based on the arrangement of these chiral centres).

Understanding the Question

The question asks to:

Draw a section of poly(propene) showing two repeat units (displayed formula showing all atoms and bonds).
Identify the type of stereoisomerism.
Explain why using the diagram.

Approach

Draw the displayed formula for two repeat units: $-\text{CH}_2-\text{CH}(\text{CH}_3)-\text{CH}_2-\text{CH}(\text{CH}_3)-$ . Ensure bonds extend to the left and right. Identify that the carbons with methyl groups are chiral centres, leading to optical isomerism.

Step-by-Step Reasoning

Drawing:
The repeat unit of poly(propene) is $-\text{CH}_2-\text{CH}(\text{CH}_3)-$ . Two repeat units would be:
$-\text{CH}_2-\text{CH}(\text{CH}_3)-\text{CH}_2-\text{CH}(\text{CH}_3)-$
In a displayed formula, draw the carbon backbone horizontally. Attach H atoms and the $\text{CH}_3$ groups to the appropriate carbons. Single bonds must extend from the ends to show the chain continues.

Identification:
The type of stereoisomerism is optical isomerism (or enantiomerism, though "optical" is the standard term here).

Explanation:
Look at one of the carbon atoms bonded to the methyl group (the $\text{CH}$ carbon in the repeat unit). It is bonded to:

$-\text{H}$
$-\text{CH}_3$
$-\text{CH}_2-\dots$ (the rest of the chain in one direction)
$-\text{CH}(\text{CH}_3)-\dots$ (the rest of the chain in the other direction)
These four groups are all different. Therefore, the carbon is a chiral centre (asymmetric carbon), which is the requirement for optical isomerism.

Key Takeaways

Polymers like poly(propene) have chiral centres along the backbone due to the substituent on every other carbon. This leads to stereoisomerism (optical/tacticity). Always check the four groups attached to the carbon in question.

Common Mistakes

Drawing a skeletal formula instead of a displayed formula (the question asks to "draw a section... showing two repeat units" and the mark scheme implies a displayed formula to show the groups clearly).
Forgetting to extend bonds at the ends of the drawn section.
Identifying the isomerism as "geometric" or "cis/trans" — there is no restricted rotation here that gives cis/trans in the main chain; the chirality is at the carbon with the methyl group.
Failing to list all four different groups or simply saying "it has 4 different groups" without referencing the specific carbon.

Things to Be Careful About

The diagram must be a displayed formula showing all atoms and bonds for the two repeat units. The explanation must clearly state that the carbon with the methyl group has four different groups attached.

Techniques used

draw polymer repeat unitsidentify optical isomerism in polymers

(iv)

State two difficulties associated with the disposal of poly(propene).

2M Similar

DifficultyEasy

Worked solution

Answer

non-biodegradability (they persist in the environment / do not break down)
harmful combustion products (if burned, they release toxic gases / CO / dioxins / contribute to global warming / require energy to burn)

Final answer

Non-biodegradable; harmful combustion products

Detailed explanation

Background Concept

Poly(alkene)s like poly(propene) and poly(ethene) are widely used but pose significant disposal challenges:

Non-biodegradability: They are made of strong $\text{C-C}$ and $\text{C-H}$ bonds that are not easily broken down by microorganisms. They persist in landfills and the environment for hundreds of years, contributing to litter and microplastic pollution.
Combustion issues: If incinerated, they release large amounts of $\text{CO}_2$ (a greenhouse gas) and potentially harmful/toxic gases (like $\text{CO}$ , dioxins, or furans if combustion is incomplete or if additives are present). They also require energy to burn.
Recycling difficulties: Separating different types of polymers is expensive and time-consuming.

Understanding the Question

The question asks for two difficulties associated with the disposal of poly(propene).

Approach

State two well-known environmental or practical problems with disposing of poly(alkene)s.

Step-by-Step Reasoning

Non-biodegradability: Poly(propene) is chemically inert and resistant to biological degradation. It accumulates in landfills and oceans.
Harmful combustion products: Burning poly(propene) releases $\text{CO}_2$ (contributing to climate change) and potentially toxic gases if combustion is incomplete. Alternatively, one could mention the energy required for incineration or the difficulty/cost of recycling.

Key Takeaways

Poly(alkene)s are problematic because they do not biodegrade and their combustion can produce harmful emissions.

Common Mistakes

Writing "human error" or "not accurate" — these are not specific to poly(propene) disposal.
Writing "it takes up space" — while true, "non-biodegradability" or "accumulation in landfills" is the chemically/environmentally precise answer expected.

Things to Be Careful About

Be specific. "Non-biodegradable" is a key term. "Harmful combustion products" is a key phrase. Avoid vague answers.

Techniques used

identify disposal difficulties of polyalkenes

(c)

Under different conditions, two molecules of propene can combine to form compounds X and Y.

Fig. 4.4

3M

(i)

Name X.

1M Similar

(ii)

Fig. 4.2 shows the mass spectrum of either compound X or compound Y.

Fig. 4.2

Identify which of X and Y gives this mass spectrum.

Give one reason for your answer, referring to the fragmentation pattern.

1M Similar

DifficultyMedium-Hard

Worked solution

Answer

The spectrum is of X (2,3-dimethylbut-1-ene).

Reason: The peak at m/e = 41 corresponds to the $[\text{CH}_2=\text{CCH}_3]^+$ ion (or $\text{C}_3\text{H}_5^+$ ), which is formed by the cleavage of X. (Alternatively: the peak at 43 is $[\text{CH}_3\text{CHCH}_3]^+$ / $\text{C}_3\text{H}_7^+$ ; or X would not have a significant peak at 54/57 or 42).

Working

X ( $\text{CH}_2=\text{C}(\text{CH}_3)-\text{CH}(\text{CH}_3)_2$ ) fragments to give:

$[\text{CH}_2=\text{C}(\text{CH}_3)]^+$ at m/z 41 ( $\text{C}_3\text{H}_5^+$ )
$[^+\text{CH}(\text{CH}_3)_2]$ at m/z 43 ( $\text{C}_3\text{H}_7^+$ )

Y ( $(\text{CH}_3)_2\text{C}=\text{C}(\text{CH}_3)_2$ ) would not produce a significant peak at 41.

Answer

X; peak at 41 is $\text{C}_3\text{H}_5^+$ / $[\text{CH}_2=\text{CCH}_3]^+$

Final answer

X; peak at 41 is [CH2=CCH3]+ or C3H5+

Detailed explanation

Background Concept

In mass spectrometry, molecular ions fragment into characteristic ions. The fragmentation pattern depends on the stability of the resulting carbocations and the ease of bond cleavage. Allylic and branched carbocations are particularly stable.

For X (2,3-dimethylbut-1-ene): $\text{CH}_2=\text{C}(\text{CH}_3)-\text{CH}(\text{CH}_3)_2$ $CH_{2} = C (CH_{3}) - CH (CH_{3})_{2}$ . Cleavage between $\text{C}_2$ $C_{2}$ and $\text{C}_3$ $C_{3}$ gives:
- $[\text{CH}_2=\text{C}(\text{CH}_3)]^+$ (m/z = 12+2+12+12+3 = 41, $\text{C}_3\text{H}_5^+$ ) — an allylic cation, very stable.
- $[^+\text{CH}(\text{CH}_3)_2]$ (m/z = 12+1+12+3+12+3 = 43, $\text{C}_3\text{H}_7^+$ ) — a secondary carbocation, stable.
For Y (2,3-dimethylbut-2-ene): $(\text{CH}_3)_2\text{C}=\text{C}(\text{CH}_3)_2$ . Cleavage would give $(\text{CH}_3)_2\text{C}^+$ (m/z 43) or $(\text{CH}_3)_2\text{C}=\text{C}(\text{CH}_3)^+$ (m/z 57). It would NOT produce a significant peak at m/z 41 ( $\text{C}_3\text{H}_5^+$ ) because there is no $\text{CH}_2=\text{C}<$ group to form that specific allylic cation easily.

Understanding the Question

The mass spectrum shows a large peak at m/z = 41 (100% relative abundance) and a large peak at m/z = 43 (90%). We must identify whether this is X or Y and give a reason based on fragmentation.

Approach

Calculate the m/z values for likely fragments of X and Y and match them to the spectrum. The peak at 41 is the key discriminator.

Step-by-Step Reasoning

Fragmentation of X (2,3-dimethylbut-1-ene):
Structure: $\text{CH}_2=\text{C}(\text{CH}_3)-\text{CH}(\text{CH}_3)_2$

Break the $\text{C}_2-\text{C}_3$ $C_{2} - C_{3}$ bond:
- Left fragment: $[\text{CH}_2=\text{C}(\text{CH}_3)]^+$ -> $\text{C}_3\text{H}_5^+$ -> m/z = $3(12) + 5(1) = 41$ . This is an allylic carbocation, highly stable, so it will be a major peak (base peak).
- Right fragment: $[^+\text{CH}(\text{CH}_3)_2]$ -> $\text{C}_3\text{H}_7^+$ -> m/z = $3(12) + 7(1) = 43$ . This is a secondary carbocation, also stable.
The spectrum shows peaks at 41 and 43, which matches X perfectly.

Fragmentation of Y (2,3-dimethylbut-2-ene):
Structure: $(\text{CH}_3)_2\text{C}=\text{C}(\text{CH}_3)_2$

Breaking a $\text{C-C}$ bond next to the double bond gives $(\text{CH}_3)_2\text{C}^+$ (m/z 43) or $(\text{CH}_3)_2\text{C}=\text{C}(\text{CH}_3)^+$ (m/z 57).
It does NOT have a $\text{CH}_2=$ group, so it cannot easily form the $\text{C}_3\text{H}_5^+$ (m/z 41) allylic cation in high abundance.
Therefore, the spectrum with a base peak at 41 must be X.

Key Takeaways

Fragmentation patterns are unique to molecular structure. Allylic cleavage producing stable allylic carbocations (like m/z 41 for $\text{C}_3\text{H}_5^+$ ) is a key identifier for terminal alkenes with branching.

Common Mistakes

Calculating the m/z incorrectly (e.g., forgetting the mass of hydrogen or carbon).
Identifying the wrong fragment for m/z 41 (it is $\text{C}_3\text{H}_5^+$ , not $\text{C}_2\text{H}_5^+$ which is 29).
Not linking the fragment to the specific structure (must mention $\text{CH}_2=\text{CCH}_3^+$ or $\text{C}_3\text{H}_5^+$ ).

Things to Be Careful About

The reason must refer to the specific fragment at m/z = 41 (or the absence of certain peaks in Y). Simply saying "X has a peak at 41" is not enough; you must explain WHAT the peak at 41 is (the ion formula).

Techniques used

analyse mass spectrum fragmentation pattern to identify isomer

(iii)

The molecular ion peak in the spectrum in Fig. 4.2 has relative abundance 34.7.

Calculate the relative abundance of the $[M+1]^+$ peak in this spectrum.

1M Similar

DifficultyMedium-Easy

Worked solution

Working

The molecular formula of X (and Y) is $\text{C}_6\text{H}_{12}$ . There are 6 carbon atoms.
The natural abundance of $^{13}\text{C}$ is 1.1% relative to $^{12}\text{C}$ .

The relative abundance of the $[\text{M}+1]^+$ peak is given by:
$\text{Relative abundance of } [\text{M}+1]^+ = \frac{\text{Number of C atoms} \times 1.1 \times \text{Relative abundance of } [\text{M}]^+}{100}$

$[\text{M}+1]^+ = \frac{6 \times 1.1 \times 34.7}{100}$
$[\text{M}+1]^+ = \frac{22.902}{100} \times 100 \text{ (wait, formula is direct)}$
$[\text{M}+1]^+ = 6 \times 1.1 \times 0.347 = 2.289$

Alternatively:
$\frac{[\text{M}+1]^+}{[\text{M}]^+} = \frac{6 \times 1.1}{100}$
$[\text{M}+1]^+ = 34.7 \times \frac{6.6}{100} = 2.29$

Answer

2.29

Final answer

2.29

Detailed explanation

Background Concept

In mass spectrometry, the $[\text{M}+1]^+$ peak is primarily caused by the presence of the $^{13}\text{C}$ isotope. Carbon has two stable isotopes: $^{12}\text{C}$ (abundance ~98.9%) and $^{13}\text{C}$ (abundance ~1.1%).
For a molecule with $n$ carbon atoms, the probability of having exactly one $^{13}\text{C}$ atom (and the rest $^{12}\text{C}$ ) is approximately $n \times 1.1\%$ . Therefore, the relative abundance of the $[\text{M}+1]^+$ peak compared to the molecular ion peak $[\text{M}]^+$ is:
$\frac{\text{Abundance of } [\text{M}+1]^+}{\text{Abundance of } [\text{M}]^+} \times 100 = n \times 1.1$
where $n$ is the number of carbon atoms.

Understanding the Question

The molecular ion peak ( $[\text{M}]^+$ ) has a relative abundance of 34.7. We need to calculate the relative abundance of the $[\text{M}+1]^+$ peak. The compounds X and Y are both $\text{C}_6\text{H}_{12}$ (isomers of hexene / dimethylbutene).

Approach

Determine the number of carbon atoms in the molecular formula ( $\text{C}_6\text{H}_{12}$ -> 6 carbons).
Use the formula: $[\text{M}+1]^+ \text{ abundance} = [\text{M}]^+ \text{ abundance} \times \frac{n \times 1.1}{100}$ .

Step-by-Step Reasoning

Molecular formula: Both X and Y are formed from two propene molecules ( $2 \times \text{C}_3\text{H}_6 = \text{C}_6\text{H}_{12}$ ). So $n = 6$ .

Calculation:
$\text{Abundance of } [\text{M}+1]^+ = 34.7 \times \frac{6 \times 1.1}{100}$
$= 34.7 \times \frac{6.6}{100}$
$= 34.7 \times 0.066$
$= 2.2902$

Rounding to 3 significant figures (consistent with 34.7): 2.29.

Key Takeaways

The $[\text{M}+1]^+$ peak relative abundance is directly proportional to the number of carbon atoms. Use the formula: $\% [\text{M}+1] = n(\text{C}) \times 1.1 \times \% [\text{M}] / 100$ .

Common Mistakes

Using the wrong number of carbon atoms (e.g., counting hydrogens or total atoms).
Forgetting to divide by 100 or multiplying incorrectly.
Not using 1.1% for $^{13}\text{C}$ abundance (sometimes 1.08% is used in advanced contexts, but 1.1% is standard for A-Level).

Things to Be Careful About

Ensure the calculation is: $34.7 \times (6 \times 1.1) / 100$ . The result is 2.29. Do not confuse this with the percentage abundance; the question asks for the relative abundance in the spectrum (which is scaled to the base peak = 100). The value 2.29 is correct on that scale.

Techniques used

calculate M+1 peak abundance using carbon-13 natural abundance

Chemistry 9701/21 — October/November 2025

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

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Background Concept

Understanding the Question

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Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

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Background Concept

Understanding the Question

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Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

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Background Concept

Understanding the Question

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Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

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Background Concept

Understanding the Question

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Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

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Background Concept

Understanding the Question

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Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

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Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

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Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

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Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

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Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes