A sample of iron is analysed using a mass spectrometer. The mass spectrum shows three isotopes of iron are present in the sample. (a) Define isotopes. (b) Fig. 2.1 shows the mass spectrum of the sample of iron. (i) Use Fig. 2.1 to calculate the relative atomic mass, $A_r$, of iron to one decimal place. Show your working. (ii) Complete Table 2.1 to show the number of protons and nucleons in one atom of $^{56}\text{Fe}$. Table 2.1 particle number of particles in one atom of…

(a) Atoms of the same element with the same number of protons but different number of neutrons. · (b)(i) 55.9 · (b)(ii) protons: 26, nucleons: 56 · (c) 7 · (d) Fe(g) -> Fe+(g) + e- · (e) Same

9701/21 May/June 2025 — Chemistry Solved Past Paper

Q1MediumChemical BondingGroup 2Atoms, Molecules and StoichiometryChemical PeriodicityElectrochemistry

(a)

Solid sodium conducts electricity. Sodium oxide conducts electricity when molten but not when solid.

3M

(i)

Name the type of bonding present in sodium and in sodium oxide.

bonding in sodium .............................................................................................................

bonding in sodium oxide ...................................................................................................

1M Similar

(ii)

Explain how solid sodium conducts electricity.

1M Similar

(iii)

Explain why sodium oxide conducts electricity when molten.

1M Similar

(b)

Separate samples of sodium and sodium oxide are each added to an excess of cold water.

4M

(i)

Write an equation for the reaction of sodium with cold water.

1M Similar

(ii)

Write an equation for the reaction of sodium oxide with cold water.

1M Similar

(iii)

Complete Table 1.1.

Do not refer to temperature changes when considering observations for these reactions.

Table 1.1

	sodium	sodium oxide
one similarity in observation on addition to cold water
one difference in observation on addition to cold water

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

	sodium	sodium oxide
similarity	both solids disappear, giving colourless solutions
difference	effervescence (hydrogen gas bubbles)	no effervescence

Final answer

Similarity: both dissolve/disappear giving colourless solution; difference: sodium effervesces, sodium oxide does not.

Detailed explanation

Background Concept

Sodium reacts vigorously with water producing hydrogen gas (effervescence); sodium oxide simply dissolves/reacts to form sodium hydroxide solution with no gas. Both leave NaOH in solution.

Understanding the Question

Complete the comparison table with one observation common to both and one that differs. The instruction not to mention temperature change rules out 'heat released' as an answer.

Approach

Think about what you would see: disappearance of solid, colour of solution, bubbling.

Step-by-Step Reasoning

Similarity: both solids disappear (sodium reacts away, Na2O dissolves/reacts) and both give a colourless solution of NaOH. Difference: sodium produces hydrogen gas, seen as effervescence; sodium oxide produces no gas, so no effervescence.

Key Takeaways

Both give alkaline NaOH solution, but only the metal gives a gas.

Common Mistakes

Citing temperature rise (explicitly excluded by the question), or describing sodium hydroxide solution as not colourless.

Things to Be Careful About

Do not refer to temperature changes — the question forbids it.

Techniques used

compare qualitative observations of two reactions

(c)

State the oxidation number of the Period 3 elements bonded to $\text{Cl}$ in $\text{NaCl}$ and $\text{PCl}_5$ .

Explain the difference in the oxidation number.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Oxidation number of Na in NaCl: $+1$ ; oxidation number of P in $\text{PCl}_5$ : $+5$ .
The difference arises because Na has only one valence (outer-shell) electron, so it can lose only one electron to chlorine, while P has five valence electrons and can share all five with chlorine atoms (expanding to five covalent bonds).

Final answer

Na: +1; P: +5; Na has 1 valence electron, P has 5.

Detailed explanation

Background Concept

Oxidation number reflects the number of electrons an atom loses or shares. Chlorine is more electronegative than both Na and P, so in each chloride the bonding electrons are assigned to Cl, giving Cl an oxidation number of −1. The Period 3 element's oxidation number then equals the number of electrons it has lost or is sharing.

Understanding the Question

Two tasks: state the oxidation numbers, and explain why they differ — the explanation must be in terms of the number of valence electrons.

Approach

Apply the rule that Cl is −1 in these chlorides, so the element's oxidation number balances the chlorines. Then explain via valence electron count.

Step-by-Step Reasoning

In NaCl, one Cl is −1, so Na is +1: sodium, in Group 1, has a single 3s valence electron and loses it to form Na+. In PCl5, five Cl atoms at −1 each total −5, so P is +5: phosphorus has five valence electrons (3s2 3p3) and forms five covalent bonds, sharing all five with the more electronegative chlorine atoms. The maximum oxidation number an element shows in its chlorides equals its group number / number of valence electrons.

Key Takeaways

Oxidation number in a chloride = number of valence electrons involved in bonding; Cl is −1 in chlorides.

Common Mistakes

Giving P as +3 (confusing with PCl3), or explaining the difference in terms of reactivity rather than valence electron count.

Things to Be Careful About

Both oxidation numbers AND the valence-electron explanation are needed for the two marks — the explanation must reference the number of outer-shell electrons.

Techniques used

deduce oxidation numbersrelate oxidation number to number of valence electrons

(d)

Table 1.2 shows melting points of some oxides.

Table 1.2

oxide	melting point/ °C
$\text{SO}_2$	–73
$\text{H}_2\text{O}$	0
$\text{SO}_3$	17
$\text{SiO}_2$	1610
$\text{MgO}$	2852
$\text{Al}_2\text{O}_3$	2072

A student suggests the following hypothesis.

The higher the oxidation number of the element combined with oxygen, the higher the melting point of the oxide.

Use Table 1.2 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

The hypothesis is false.

Comparing oxides of the same element: sulfur has oxidation number +4 in $\text{SO}_2$ (m.p. −73 °C) and +6 in $\text{SO}_3$ (m.p. +17 °C) — here the higher oxidation number does give the higher melting point. However, comparing across elements, Si has oxidation number +4 in $\text{SiO}_2$ (m.p. 1610 °C) while S has only +4 in $\text{SO}_2$ (m.p. −73 °C) and S in $\text{SO}_3$ has +6 (m.p. 17 °C) — the same or higher oxidation number gives a far lower melting point. Since $\text{SO}_3$ (+6) has a much lower melting point than $\text{SiO}_2$ (+4), a higher oxidation number does not consistently give a higher melting point, so the hypothesis is false.

Final answer

False — e.g. SO3 (oxidation number +6) has a much lower melting point (17 °C) than SiO2 (+4, 1610 °C).

Detailed explanation

Background Concept

Melting point depends on structure and bonding (giant ionic/covalent vs simple molecular), not directly on oxidation number. SiO2 is giant covalent (very high m.p.), MgO and Al2O3 are ionic lattices (high m.p.), while SO2, SO3 and H2O are simple molecular with weak intermolecular forces (low m.p.).

Understanding the Question

The hypothesis links oxidation number to melting point. You must use the table to decide: true, false, or insufficient data — and justify with comparisons.

Approach

Assign oxidation numbers to each element in its oxide, then look for pairs that contradict the hypothesis. A single counter-example makes it false.

Step-by-Step Reasoning

Oxidation numbers: S is +4 in SO2 and +6 in SO3; Si is +4 in SiO2; Mg is +2 in MgO; Al is +3 in Al2O3. Within sulfur, SO3 (+6, 17 °C) > SO2 (+4, −73 °C), which superficially supports the hypothesis. But SO3 (+6) melts at 17 °C while SiO2 (+4) melts at 1610 °C and MgO (+2) at 2852 °C — higher oxidation numbers with far lower melting points. These counter-examples disprove the hypothesis, so it is false. The real determinant is structure/bonding: giant structures melt high, simple molecular oxides melt low.

Key Takeaways

Melting point reflects structure and bonding type, not oxidation number. When testing a hypothesis against data, look specifically for counter-examples.

Common Mistakes

Concluding 'true' from the SO2/SO3 pair alone; concluding 'not enough information' without attempting comparisons; comparing oxides of different elements without assigning oxidation numbers.

Things to Be Careful About

All three elements are needed for the two marks: the verdict (false), a comparison/statement of oxidation numbers, and the correct link to melting points. The comparison must pair oxides whose oxidation numbers and melting points contradict the trend.

Techniques used

evaluate a hypothesis against datacompare oxidation numbers with melting points

Q2MediumAtomic StructureAtoms, Molecules and Stoichiometry

A sample of iron is analysed using a mass spectrometer. The mass spectrum shows three isotopes of iron are present in the sample.

(a)

Define isotopes.

1M Similar

(b)

Fig. 2.1 shows the mass spectrum of the sample of iron.

Fig. 2.1

4M

(i)

Use Fig. 2.1 to calculate the relative atomic mass, $A_r$ , of iron to one decimal place.

Show your working.

2M Similar

DifficultyMedium-Easy

Worked solution

Working

A_r = \frac{(5.9 \times 54) + (91.9 \times 56) + (2.2 \times 57)}{100}

A_r = \frac{318.6 + 5146.4 + 125.4}{100}

A_r = 55.9

Answer

55.9

Final answer

55.9

Detailed explanation

Background Concept

The relative atomic mass ( $A_r$ ) of an element is the weighted average mass of its atoms compared to 1/12th the mass of a carbon-12 atom. When a sample contains multiple isotopes, the $A_r$ is calculated by multiplying the mass of each isotope by its relative abundance (as a fraction or percentage), summing these products, and dividing by the total abundance (usually 100% if percentages are used).

Understanding the Question

The question provides a mass spectrum (Fig. 2.1) showing three isotopes of iron with their m/e values (which correspond to their mass numbers for singly charged ions) and their percentage abundances. The task is to calculate the relative atomic mass to one decimal place.

Approach

Use the formula for relative atomic mass:
$A_r = \frac{\sum (\text{isotope mass} \times \% \text{ abundance})}{100}$
Substitute the values from the spectrum and compute the result.

Step-by-Step Reasoning

Identify the isotopes and their data from Fig. 2.1:
- $^{54}\text{Fe}$ : mass = 54, abundance = 5.9%
- $^{56}\text{Fe}$ : mass = 56, abundance = 91.9%
- $^{57}\text{Fe}$ : mass = 57, abundance = 2.2%
Set up the calculation:
$A_r = \frac{(5.9 \times 54) + (91.9 \times 56) + (2.2 \times 57)}{100}$
Calculate the numerator:
- $5.9 \times 54 = 318.6$
- $91.9 \times 56 = 5146.4$
- $2.2 \times 57 = 125.4$
- Sum = $318.6 + 5146.4 + 125.4 = 5590.4$
Divide by 100:
$A_r = \frac{5590.4}{100} = 55.904$
Round to one decimal place as required: 55.9.

Key Takeaways

Always ensure abundances are converted to fractions (divide by 100) or the sum is divided by 100 at the end. The m/e value for a singly charged ion ( $z=1$ ) in a mass spectrometer is numerically equal to the mass number of the isotope.

Common Mistakes

Forgetting to divide the sum by 100 (using percentages directly as if they were fractions).
Using the wrong mass values (e.g., using exact isotopic masses instead of mass numbers, though for this level, mass numbers from the m/e axis are expected).
Rounding errors during intermediate steps.

Things to Be Careful About

Check the significant figures and decimal places required. Here, one decimal place is specified. The sum of percentages is $5.9 + 91.9 + 2.2 = 100.0\%$ , confirming the data is complete.

Techniques used

calculate relative atomic mass from mass spectrum

(ii)

Complete Table 2.1 to show the number of protons and nucleons in one atom of $^{56}\text{Fe}$ .

Table 2.1

particle	number of particles in one atom of $^{56}\text{Fe}$
protons
nucleons

2M Similar

DifficultyEasy

Worked solution

Answer

particle	number of particles in one atom of $^{56}\text{Fe}$
protons	26
nucleons	56

Final answer

protons: 26, nucleons: 56

Detailed explanation

Background Concept

The nuclide symbol $^{A}_{Z}\text{X}$ provides key information about an atom:

$Z$ (atomic number or proton number) = number of protons.
$A$ (mass number or nucleon number) = number of protons + number of neutrons.
For a neutral atom, number of electrons = number of protons.

Iron (Fe) has an atomic number of 26, meaning it always has 26 protons.

Understanding the Question

The question asks for the number of protons and nucleons in one atom of $^{56}\text{Fe}$ . The superscript 56 is the mass number (nucleon number).

Approach

Identify the atomic number of iron (from periodic table or prior knowledge) for protons. The mass number is given directly in the symbol for nucleons.

Step-by-Step Reasoning

Iron (Fe) is element 26, so it has 26 protons. This is constant for all iron isotopes.
The symbol $^{56}\text{Fe}$ indicates a mass number ( $A$ ) of 56. Nucleons are protons + neutrons, so the number of nucleons is 56.
Fill in the table: protons = 26, nucleons = 56.

Key Takeaways

The number of protons defines the element and does not change between isotopes. The mass number (nucleons) changes due to different neutron numbers.

Common Mistakes

Confusing nucleons with neutrons. Nucleons = protons + neutrons. If asked for neutrons in $^{56}\text{Fe}$ , it would be $56 - 26 = 30$ .
Looking up the wrong atomic number for iron.

Things to Be Careful About

Ensure the correct terminology: 'protons' (subatomic particle) vs 'nucleons' (collective term for protons and neutrons in the nucleus).

Techniques used

identify protons and nucleons from nuclide symbol

(c)

Deduce the number of pairs of electrons in the shell with principal quantum number $n = 3$ in an Fe atom.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

7

Working

Electron configuration of Fe ( $Z=26$ ): $1s^2 2s^2 2p^6 3s^2 3p^6 3d^6 4s^2$
Shell $n=3$ contains: $3s^2 3p^6 3d^6$
Total electrons in $n=3$ shell = $2 + 6 + 6 = 14$
Number of pairs = $14 / 2 = 7$

Final answer

7

Detailed explanation

Background Concept

The principal quantum number $n$ designates the electron shell. For a given $n$ , the subshells are $s$ ( $n=1$ ), $p$ ( $n=2$ ), $d$ ( $n=3$ ), etc. The $n=3$ shell can hold up to 18 electrons (one $3s$ , three $3p$ , five $3d$ orbitals).

Understanding the Question

The question asks for the number of pairs of electrons in the shell with principal quantum number $n=3$ in a neutral iron atom (Fe, $Z=26$ ).

Approach

Write the full electron configuration of Fe.
Identify all subshells with $n=3$ ( $3s$ , $3p$ , $3d$ ).
Sum the electrons in these subshells.
Divide by 2 to get the number of pairs.

Step-by-Step Reasoning

Iron has 26 electrons. Following the Aufbau principle:
$1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^6$ (or $3d^6 4s^2$ )
The shell with $n=3$ includes the $3s$ , $3p$ , and $3d$ subshells.
Electrons in $n=3$ : $3s^2$ (2 electrons) + $3p^6$ (6 electrons) + $3d^6$ (6 electrons) = 14 electrons.
Each pair consists of 2 electrons. Number of pairs = $14 / 2 = 7$ .

Key Takeaways

When counting electrons in a shell, include all subshells with that principal quantum number, even if they are filled after a higher shell (like $4s$ filling before $3d$ ).

Common Mistakes

Forgetting to include the $3d$ electrons when counting electrons in the $n=3$ shell. Many students stop at $3p^6$ and miss the $3d^6$ electrons.
Counting individual electrons instead of pairs.
Using the wrong electron configuration for transition metals (remember $4s$ fills before $3d$ , but when writing configuration by shell, $3d$ belongs to $n=3$ ).

Things to Be Careful About

The question asks for pairs of electrons, not total electrons. Double-check the final division by 2.

Techniques used

write electron configurationcount electron pairs

(d)

Write an equation to represent the first ionisation energy of iron.

1M Similar

(e)

Suggest how the value for the first ionisation energy of $^{54}\text{Fe}$ compares to the first ionisation energy of $^{56}\text{Fe}$ . Explain your answer in terms of the factors that affect ionisation energy.

4M Similar

DifficultyMedium

Worked solution

Answer

The first ionisation energy of $^{54}\text{Fe}$ is the same as that of $^{56}\text{Fe}$ .

Explanation:

Both isotopes have the same number of protons (same nuclear charge).
Both have the same electronic configuration (same shielding).
Therefore, the nuclear attraction to the outermost electron is the same.
Consequently, the energy required to remove the outermost electron is unaffected.

Final answer

Same

Detailed explanation

Background Concept

Ionisation energy depends on three main factors:

Nuclear charge (number of protons): Greater nuclear charge increases attraction to electrons, increasing IE.
Shielding (inner electrons): Inner electrons shield outer electrons from the full nuclear charge, decreasing IE.
Distance from the nucleus: Outer electrons further from the nucleus experience less attraction, decreasing IE.

Isotopes of an element have the same number of protons and electrons but different numbers of neutrons. Neutrons do not affect electrostatic attraction significantly.

Understanding the Question

Compare the first ionisation energy of $^{54}\text{Fe}$ and $^{56}\text{Fe}$ and explain using factors affecting ionisation energy.

Approach

State whether they are the same or different.
Explain using the factors: nuclear charge, shielding/electronic configuration, and resulting attraction.

Step-by-Step Reasoning

Comparison: The values are the same. Isotopes have identical chemical properties and ionisation energies.
Nuclear charge: Both $^{54}\text{Fe}$ and $^{56}\text{Fe}$ have 26 protons. Therefore, the nuclear charge is identical.
Electronic configuration / Shielding: Both have 26 electrons arranged identically ( $1s^2 2s^2 2p^6 3s^2 3p^6 3d^6 4s^2$ ). The number of inner (shielding) electrons is the same.
Attraction: Since nuclear charge and shielding are the same, the effective nuclear charge felt by the outermost electron is the same. The slight mass difference does not significantly affect electron-nucleus distance or attraction.
Conclusion: The energy required to remove the outermost electron (first IE) is the same for both isotopes.

Key Takeaways

Isotopes have the same ionisation energy because ionisation energy depends on electrostatic forces, which are determined by protons and electrons, not neutrons.

Common Mistakes

Saying the IE is different because of different mass or different number of neutrons.
Failing to mention 'same nuclear charge' or 'same number of protons'.
Not linking the factors back to 'nuclear attraction' or 'energy required'.
Saying 'same mass' affects it (mass has negligible effect on electronic structure).

Things to Be Careful About

The mark scheme specifically looks for: same electronic configuration/same shielding, same number of protons/same nuclear charge, same nuclear attraction, and no change. Ensure all these logical steps are covered in the explanation.

Techniques used

compare ionisation energies of isotopes

Q3MediumReaction KineticsEquilibria

(a)

Hexene reacts with hydrogen gas to produce hexane.

\text{reaction 1} \quad \text{C}_6\text{H}_{12}(\text{l}) + \text{H}_2(\text{g}) \rightarrow \text{C}_6\text{H}_{14}(\text{l})

Fig. 3.1 shows the distribution of energies of $\text{H}_2(\text{g})$ molecules at temperature $T$ .

Fig. 3.1

6M

(i)

Sketch on Fig. 3.1 the shape of the curve for the same sample of $\text{H}_2(\text{g})$ molecules when the temperature is increased.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

The new curve (dashed) should:

Start at the origin.
Have a lower peak than the original curve.
Have the peak shifted to the right (higher energy).
Cross the original curve once, to the right of the original peak.

Final answer

See sketch description above

Detailed explanation

Background Concept

A Maxwell-Boltzmann distribution curve shows the distribution of molecular energies in a gas at a specific temperature. The area under the curve represents the total number of molecules (which is constant if the amount of gas doesn't change). The peak of the curve represents the most probable energy. The curve starts at the origin (no molecules have zero energy in a continuous distribution, though practically it starts very low) and has a long tail to the right.

Understanding the Question

The question asks to sketch the curve for the same sample of $\text{H}_2(\text{g})$ at a higher temperature. We are given the original curve at temperature $T$ with activation energy $E_A$ marked.

Approach

When temperature increases, the average kinetic energy of the molecules increases. This means:

The total area under the curve must remain the same (same number of molecules).
To keep the area constant while shifting to higher energies (right), the peak must lower.
The curve must spread out, so it extends further to the right.
The curve must start at the origin (energy = 0, number of molecules = 0).
Because the new curve is lower at low energies but higher at high energies, it must cross the original curve at some point.

Step-by-Step Reasoning

Start at origin: At 0 energy, there are effectively 0 molecules. Both curves start here.
Peak lower: Since the average energy increases, the distribution spreads out. To maintain the same area (total number of molecules), the height of the peak must decrease.
Peak to the right: The most probable energy (peak) increases with temperature. So the new peak is to the right of the old peak.
Crossing: At low energies, the higher temperature curve is lower (fewer molecules have low energy). At high energies (the tail), the higher temperature curve is higher (more molecules have high energy). Therefore, they must cross once. The crossing point is to the right of the original peak.

Key Takeaways

For a Maxwell-Boltzmann curve at a higher temperature: lower peak, shifted right, starts at origin, crosses original curve once to the right of the original peak, longer tail.

Common Mistakes

Drawing the peak higher (wrong, area must be constant).
Shifting the peak left (wrong, average energy increases).
Not starting at the origin.
Drawing the new curve entirely above the old one (violates conservation of molecules/area).

Things to Be Careful About

The curve must cross the original curve. If it doesn't, it implies more molecules total or a different distribution shape.
The crossing point must be to the right of the original peak.

Techniques used

sketch Maxwell-Boltzmann distribution at higher temperatureinterpret energy distribution curves

(ii)

Explain why increasing the temperature increases the rate of reaction 1.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Greater proportion of $\text{H}_2$ molecules have energy greater than or equal to the activation energy ( $\ge E_A$ ).
This leads to an increase in the frequency of successful collisions between $\text{H}_2$ and $\text{C}_6\text{H}_{12}$ .

Final answer

See answer above

Detailed explanation

Background Concept

The rate of reaction depends on the frequency of successful collisions. For a collision to be successful, the colliding particles must have energy equal to or greater than the activation energy ( $E_A$ ). The Maxwell-Boltzmann distribution shows how many molecules have a given energy.

Understanding the Question

Explain why increasing temperature increases the rate of reaction 1 (hydrogenation of hexene). This is a standard kinetics explanation linking temperature to rate.

Approach

Look at the distribution curve at higher temperature (from part i). The area to the right of $E_A$ is larger.
This means more molecules have enough energy to react.
More energetic molecules collide more frequently and successfully.

Step-by-Step Reasoning

Mark 1 (M1): At a higher temperature, the curve shifts right and flattens. The area under the curve to the right of $E_A$ (the shaded region representing molecules with $E \ge E_A$ ) is larger. So, a greater proportion (or percentage) of molecules have energy $\ge E_A$ . Note: It's not just 'more molecules', it's a 'greater proportion' because the total number of molecules is constant.
Mark 2 (M2): Because more molecules have the required activation energy, a higher fraction of collisions are successful. This increases the frequency of successful collisions (between $\text{H}_2$ and hexene), thus increasing the rate.

Key Takeaways

Rate increases with temperature because a greater proportion of molecules have energy $\ge E_A$ , leading to more frequent successful collisions. It is NOT because molecules move faster (though they do, the primary reason for the exponential rate increase is the energy distribution change).

Common Mistakes

Saying 'molecules move faster' (this is true but doesn't fully explain the rate increase in terms of activation energy; the mark scheme specifically looks for the proportion with energy $\ge E_A$ ).
Saying 'more molecules have energy $\ge E_A$ ' without mentioning 'proportion' or 'fraction' (technically the number increases if volume changes, but usually we assume constant amount, so proportion is the key word. However, 'greater proportion' is the standard mark scheme wording).
Forgetting to mention 'successful' collisions.

Things to Be Careful About

Must mention activation energy or $E_A$ .
Must mention 'proportion' or 'fraction' or 'percentage'.

Techniques used

relate temperature to kinetic energy distributionexplain rate increase using collision theory

(iii)

State the role of nickel when it is added to reaction 1.

1M Similar

(iv)

Annotate Fig. 3.2 to show the effect of adding nickel to reaction 1 at temperature $T$ .

Fig. 3.2

1M Similar

(b)

Define Le Chatelier’s principle.

2M Similar

(c)

Reaction 2 shows the equilibrium reaction between X(g) and Y(g) to produce Z(g) in a sealed container.

\text{reaction 2} \quad a\text{X}(\text{g}) + b\text{Y}(\text{g}) \rightleftharpoons c\text{Z}(\text{g})

Fig. 3.3 shows the effect of changing pressure on the percentage yield of Z(g) at two different temperatures, 300 K and 350 K.

Fig. 3.3

Deduce two conclusions about reaction 2 using Fig 3.3.

2M Similar

DifficultyMedium

Worked solution

Answer

The forward reaction is exothermic.
(Reasoning: At a given pressure, the yield is lower at 350 K than at 300 K. Increasing temperature decreases yield, so equilibrium shifts left (reverse direction). Reverse is endothermic, so forward is exothermic.)
$(a + b) > c$ .
(Reasoning: Increasing pressure increases the yield of Z. Equilibrium shifts to the side with fewer moles of gas to reduce pressure. Since yield of Z increases, the right side (products) has fewer moles. So moles of reactants $(a+b)$ > moles of products $(c)$ .)

Final answer

Forward reaction is exothermic; $(a + b) > c$

Detailed explanation

Background Concept

Le Chatelier's principle applies to equilibria:

Temperature: Increasing temperature favors the endothermic direction. If forward is exothermic, increasing T shifts equilibrium left (reactants), decreasing yield of products.
Pressure: Increasing pressure favors the side with fewer moles of gas. If reactants have more moles than products, increasing pressure shifts equilibrium right (products), increasing yield.

Understanding the Question

Reaction 2: $a\text{X}(\text{g}) + b\text{Y}(\text{g}) \rightleftharpoons c\text{Z}(\text{g})$ . Graph shows yield of Z vs pressure at 300 K and 350 K. Deduce two conclusions.

Approach

Analyze the effect of temperature (compare the two curves) and the effect of pressure (slope of curves).

Step-by-Step Reasoning

Conclusion 1 (Temperature effect): Look at the curves. At any fixed pressure (e.g., 200 atm), the yield at 300 K is higher than at 350 K. This means increasing temperature (from 300 to 350) decreases the yield of Z. According to Le Chatelier, increasing T shifts equilibrium in the endothermic direction. Since yield decreases, shift is to the left (reverse). So reverse is endothermic, meaning the forward reaction is exothermic.
Conclusion 2 (Pressure effect): Look at the curves. As pressure increases (x-axis), yield of Z increases (y-axis). Increasing pressure shifts equilibrium to the side with fewer gas moles. Since yield of Z (product) increases, the equilibrium shifts to the right. Therefore, the product side has fewer moles than the reactant side. Reactants have $(a+b)$ moles, products have $c$ moles. So $(a + b) > c$ .

Key Takeaways

Higher T -> lower yield => Exothermic forward reaction.
Higher P -> higher yield => Fewer moles on product side.

Common Mistakes

Getting the temperature effect backwards (saying endothermic).
Getting the pressure effect backwards (saying $a+b < c$ ).
Not stating the conclusion clearly (e.g., just writing 'exothermic' without context, though usually 'forward reaction is exothermic' is the mark).

Things to Be Careful About

The graph has 'yield / %'. Higher up means more product.
300 K is the upper curve (higher yield). 350 K is the lower curve (lower yield).

Techniques used

deduce exothermic/endothermic from temperature-yield graphdeduce mole ratio from pressure-yield graph

Q4MediumAtoms, Molecules and StoichiometryElectrochemistryHydroxy CompoundsAnalytical Techniques

The structure of vitamin C is shown in Fig. 4.1.

Fig. 4.1

(a)

Deduce the empirical formula of vitamin C.

1M Similar

DifficultyEasy

Worked solution

Answer

$\text{C}_3\text{H}_4\text{O}_3$

Final answer

C3H4O3

Detailed explanation

Background Concept

An empirical formula represents the simplest whole-number ratio of atoms of each element in a compound. To find it from a structural formula, you first determine the molecular formula by counting all atoms, then divide by their greatest common divisor.

Understanding the Question

The question asks for the empirical formula of vitamin C based on its skeletal structure shown in Fig. 4.1. You must count the number of carbon, hydrogen, and oxygen atoms in the molecule.

Approach

Count the total number of C, H, and O atoms in the molecular structure to get the molecular formula.
Divide the subscripts by their highest common factor to obtain the simplest whole-number ratio.

Step-by-Step Reasoning

Looking at Fig. 4.1:

Carbon atoms: There are 4 carbons in the five-membered ring (including the carbonyl carbon and the two alkene carbons, plus the ring junction carbon) and 2 carbons in the side chain ( $-\text{CH}(\text{OH})-\text{CH}_2\text{OH}$ ). Total C = 6.
Oxygen atoms: There is 1 ring oxygen, 1 carbonyl oxygen ( $=\text{O}$ ), and 4 hydroxyl oxygens ( $-\text{OH}$ ). Total O = 6.
Hydrogen atoms: The side chain $-\text{CH}_2\text{OH}$ has 3 H, the $-\text{CH}(\text{OH})-$ has 2 H, the ring junction $-\text{CH}-$ has 1 H, and the two ring $-\text{OH}$ groups have 2 H. Total H = 3 + 2 + 1 + 2 = 8.

Molecular formula: $\text{C}_6\text{H}_8\text{O}_6$ .
Dividing by 2 gives the empirical formula: $\text{C}_3\text{H}_4\text{O}_3$ .

Key Takeaways

Always count carefully from skeletal structures: vertices and ends of lines are carbons, and hydrogens attached to carbons are implied to satisfy four bonds.

Common Mistakes

Forgetting to count hydrogens on carbons (they are not drawn in skeletal structures).
Miscounting the ring atoms or the side chain.

Things to Be Careful About

Ensure you are giving the empirical formula (simplest ratio), not the molecular formula. The question specifically asks to "deduce the empirical formula".

Techniques used

count atoms from skeletal structuresimplify to empirical formula

(b)

The concentration of vitamin C is found by titration with $\text{I}_2(\text{aq})$ .

Fig. 4.2

A vitamin C tablet is dissolved in water to produce $200.0\text{ cm}^3$ of vitamin C solution.
$5.00\text{ cm}^3$ of this vitamin C solution is added to a flask with approximately $150\text{ cm}^3$ of water and an indicator.
Exactly $28.40\text{ cm}^3$ of $5.00 \times 10^{-4}\text{ mol dm}^{-3}$ $\text{I}_2(\text{aq})$ reacts with the sample of vitamin C solution in the flask.

[ $M_r$ : vitamin C, 176]

6M

(i)

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

1M Similar

(ii)

Use your answer to (b)(i) to calculate the mass, in g, of vitamin C in the tablet. Show your working.

(If you were unable to calculate a value for the amount of $\text{I}_2(\text{aq})$ in (b)(i), use the value $2.64 \times 10^{-4}\text{ mol}$ . This is not the correct value.)

2M Similar

DifficultyMedium

Worked solution

Working

From Fig. 4.2, the reaction ratio is 1 mol vitamin C : 1 mol $\text{I}_2$ .

Amount of vitamin C in $5.00 \text{ cm}^3$ sample = amount of $\text{I}_2$ = $1.42 \times 10^{-5} \text{ mol}$ .

Mass of vitamin C in $5.00 \text{ cm}^3$ sample:

\text{mass} = n \times M_r = 1.42 \times 10^{-5} \text{ mol} \times 176 \text{ g mol}^{-1} = 2.4992 \times 10^{-3} \text{ g}

The original solution volume is $200.0 \text{ cm}^3$ , which is $200.0 / 5.00 = 40$ times the sample volume.

Total mass of vitamin C in the tablet:

\text{total mass} = 2.4992 \times 10^{-3} \text{ g} \times \frac{200.0}{5.00} = 0.099968 \text{ g}

Answer

$0.099968 \text{ g}$ (or $0.100 \text{ g}$ to 3 s.f.)

Final answer

0.099968 g

Detailed explanation

Background Concept

In a titration, the moles of titrant used can be used to find the moles of analyte via the stoichiometric ratio from the balanced equation. If the analyte solution was diluted or if a sample was taken from a larger volume, a scaling factor must be applied to find the total amount in the original sample.

Understanding the Question

You have calculated the moles of $\text{I}_2$ in $5.00 \text{ cm}^3$ of the vitamin C solution. You need to find the total mass of vitamin C in the entire $200.0 \text{ cm}^3$ solution (which came from one tablet). The $M_r$ of vitamin C is 176.

Approach

Use the 1:1 stoichiometry from Fig. 4.2 to find moles of vitamin C in the $5.00 \text{ cm}^3$ sample.
Convert moles to mass using $M_r = 176$ .
Scale up the mass by the ratio of total volume to sample volume ( $200.0 / 5.00$ ).

Step-by-Step Reasoning

Moles of vitamin C in $5.00 \text{ cm}^3$ = $1.42 \times 10^{-5} \text{ mol}$ (from 1:1 ratio).
Mass in $5.00 \text{ cm}^3$ = $1.42 \times 10^{-5} \times 176 = 2.4992 \times 10^{-3} \text{ g}$ .
Scaling factor = $200.0 / 5.00 = 40$ .
Total mass = $2.4992 \times 10^{-3} \times 40 = 0.099968 \text{ g}$ .

Alternatively, calculate total moles first: $1.42 \times 10^{-5} \times (200.0 / 5.00) = 5.68 \times 10^{-4} \text{ mol}$ , then mass = $5.68 \times 10^{-4} \times 176 = 0.099968 \text{ g}$ .

Key Takeaways

Always track the volume scaling factor in titrations where a sample is taken from a larger prepared solution.

Common Mistakes

Forgetting to multiply by the dilution factor ( $200/5$ ).
Using the wrong $M_r$ or not using $M_r$ at all.
Incorrect stoichiometric ratio (though here it is 1:1, students might assume 2:1 from the $2\text{I}^-$ produced).

Things to Be Careful About

The mark scheme accepts $0.099968$ . Ensure significant figures are reasonable (3 s.f. gives $0.100 \text{ g}$ , which is also acceptable, but carrying extra digits is safer during calculation).

Techniques used

use stoichiometric ratio from balanced equationcalculate mass from moles and molar massscale up for dilution factor

(iii)

Deduce the role of $\text{I}_2(\text{aq})$ in the reaction in Fig. 4.2.

1M Similar

(iv)

Suggest two reasons why hot concentrated acidified potassium manganate(VII) is not a suitable reagent for producing Q from vitamin C.

2M Similar

DifficultyMedium

Worked solution

Answer

It will also oxidise the alcohol (hydroxy) groups (primary and secondary) in vitamin C.
It will break the carbon-carbon double bond (C=C) in the ring.

Final answer

See answer

Detailed explanation

Background Concept

Potassium manganate(VII) ( $\text{KMnO}_4$ ) in hot concentrated acidified conditions is a very strong oxidising agent. It can oxidise:

Primary alcohols to carboxylic acids.
Secondary alcohols to ketones.
Carbon-carbon double bonds (C=C), often cleaving them to form carbonyl compounds or carboxylic acids.

Iodine ( $\text{I}_2$ ) is a milder oxidising agent that selectively oxidises the enediol group in vitamin C to a diketone without affecting other functional groups like isolated alcohols or breaking C=C bonds under these conditions.

Understanding the Question

You need to suggest two reasons why hot concentrated acidified $\text{KMnO}_4$ is not suitable for producing compound Q from vitamin C. This means $\text{KMnO}_4$ will cause unwanted side reactions.

Approach

Look at the functional groups in vitamin C (Fig. 4.1) that are not the target of the oxidation to Q. Identify which of these will react with a strong oxidising agent like $\text{KMnO}_4$ .

Step-by-Step Reasoning

Vitamin C contains:

Two secondary alcohol groups on the side chain ( $-\text{CH}(\text{OH})-$ and $-\text{CH}_2\text{OH}$ ).
A carbon-carbon double bond (C=C) in the ring.
The enediol group (which is the target).

$\text{KMnO}_4$ is too strong and will:

Oxidise the primary alcohol ( $-\text{CH}_2\text{OH}$ ) to a carboxylic acid ( $-\text{COOH}$ ) and the secondary alcohol ( $-\text{CH}(\text{OH})-$ ) to a ketone ( $-\text{C}(=\text{O})-$ ).
Oxidatively cleave the C=C double bond in the ring, breaking the ring structure.

Therefore, it does not give the desired product Q selectively.

Key Takeaways

Strong oxidising agents like $\text{KMnO}_4$ are not selective and will react with multiple functional groups. Milder agents like $\text{I}_2$ are needed for selective oxidations.

Common Mistakes

Only mentioning one reason (e.g., only alcohols or only C=C).
Saying "it will oxidise vitamin C" (that's the point, the question asks why it's not suitable for producing Q specifically).

Things to Be Careful About

Ensure you mention both the alcohol groups and the C=C bond. The mark scheme specifically looks for these two points.

Techniques used

evaluate reagent selectivitypredict side reactions with functional groups

(c)

Predict two absorptions that will be seen in the infrared spectra of both vitamin C and Q. Describe the relevant bond and the specific functional group that is responsible for each absorption identified.

Fig. 4.3

Table 4.1

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

Any two of the following:

wavenumber / $\text{cm}^{-1}$	bond	functional group
1040 – 1300	C–O	ester / hydroxy
3200 – 3650	O–H	hydroxy / alcohols
2850 – 2950	C–H	alkane
1710 – 1750	C=O	ester

Final answer

See answer

Detailed explanation

Background Concept

Infrared (IR) spectroscopy measures the absorption of infrared radiation by molecules, causing bond vibrations. Different bonds absorb at characteristic wavenumbers ( $\text{cm}^{-1}$ ). The provided table lists absorption ranges for various bonds and their associated functional groups.

Understanding the Question

You need to predict two absorptions that will be seen in the IR spectra of both vitamin C and compound Q. This means you must identify bonds/functional groups that are present in both structures. Looking at Fig. 4.3:

Vitamin C has: C=O (ester), C-O (ester, hydroxy), O-H (hydroxy), C-H (alkane), C=C (alkene).
Compound Q has: C=O (ester, ketone/diketone), C-O (ester), C-H (alkane). It has lost the C=C and the two enediol O-H groups, but retains the side-chain O-H and the ester C=O and C-O.

Common bonds in both: C-O, O-H (from side chain), C-H, C=O (ester).

Approach

Identify the functional groups present in both molecules.
Match these to the bonds and wavenumber ranges in Table 4.1.
Ensure the functional group name matches the bond type in the table.

Step-by-Step Reasoning

C-O bond: Both have ester groups and vitamin C (and Q's side chain) has hydroxy groups. Range: 1040–1300 $\text{cm}^{-1}$ .
O-H bond: Both have hydroxy (alcohol) groups on the side chain. Range: 3200–3650 $\text{cm}^{-1}$ .
C-H bond: Both have alkane parts (side chain). Range: 2850–2950 $\text{cm}^{-1}$ .
C=O bond: Both have ester groups. Range: 1710–1750 $\text{cm}^{-1}$ .

Note: C=C is only in vitamin C, not Q. The C=O from the new diketone in Q is in the carbonyl range (1670-1740), but the ester C=O is definitely in both (1710-1750).

Key Takeaways

When asked for absorptions common to two molecules, focus on functional groups that remain unchanged or are present in both. Always pair the bond, the functional group, and the wavenumber range correctly from the table.

Common Mistakes

Including C=C (only in vitamin C).
Giving the wavenumber range for C=O of a ketone/carboxyl (1640-1690 or 1670-1740) instead of the ester (1710-1750) if not careful, though the ester is definitely in both.
Forgetting to specify the functional group or giving an incorrect match (e.g., C-O with carboxyl).

Things to Be Careful About

The question asks for "two complete rows". You must provide the wavenumber range, the bond, and the functional group. Any two valid rows from the table that apply to both molecules will score.

Techniques used

interpret IR absorption tableidentify common functional groups in structures

Q5MediumCarboxylic Acids and DerivativesIntroduction to Organic ChemistryEquilibriaHydroxy CompoundsHalogen CompoundsCarbonyl Compounds

Compound A contains the elements carbon, hydrogen and oxygen only.
When A is heated with $\text{H}_2\text{SO}_4(\text{aq})$ , compounds B and C are produced, as shown in Fig. 5.1.

Fig. 5.1

(a)

Draw the structure of A.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

The structure is diethyl propanedioate (diethyl malonate).

Final answer

Diethyl propanedioate structure (see diagram)

Detailed explanation

Background Concept

Esters are derivatives of carboxylic acids where the acidic hydrogen is replaced by an alkyl group. They have the general formula RCOOR'. Esters can be hydrolysed back to the parent carboxylic acid and alcohol. Acid-catalysed hydrolysis is the reverse of esterification:

$\text{RCOOR}' + \text{H}_2\text{O} \xrightleftharpoons{\text{H}^+} \text{RCOOH} + \text{R}'\text{OH}$

If a compound produces a dicarboxylic acid and two molecules of an alcohol upon hydrolysis with water, it must be a diester (an ester formed from a dicarboxylic acid and a monoalcohol, or a dicarboxylic acid and a diol, but here the stoichiometry 1:2 suggests a diester of a monoalcohol).

Understanding the Question

The reaction scheme shows Compound A reacting with 2 molecules of water in the presence of aqueous sulfuric acid and heat to produce Compound B (HOOCCH₂COOH, propanedioic acid/malonic acid) and 2 molecules of Compound C (CH₃CH₂OH, ethanol). This is a hydrolysis reaction. We need to draw the structure of the starting ester, Compound A.

Approach

Identify the reaction type: Acid-catalysed hydrolysis of an ester.
Identify the products: Propanedioic acid (a dicarboxylic acid with 3 carbons) and ethanol (a 2-carbon alcohol).
Reconstruct the ester: The ester links form between the carboxylic acid groups of B and the hydroxyl groups of C. Since there are 2 moles of ethanol and 1 mole of acid, A is the diethyl ester of propanedioic acid.
Draw the skeletal structure.

Step-by-Step Reasoning

Identify the reaction: The reagents H₂SO₄(aq) and heat with water indicate acid hydrolysis of an ester.
Identify products: B is HOOCCH₂COOH (propanedioic acid). C is CH₃CH₂OH (ethanol).
Reconstruct A: The reaction is $\text{A} + 2\text{H}_2\text{O} \rightarrow \text{HOOCCH}_2\text{COOH} + 2\text{CH}_3\text{CH}_2\text{OH}$ . This means A is formed by condensing propanedioic acid with 2 ethanol molecules, losing 2 water molecules.
Structure: The central part is the -CH₂- from the acid. The two -COOH groups become -COOCH₂CH₃ groups.
Drawing: The skeletal structure shows a central CH₂ connected to two carbonyl carbons (C=O), each connected to an oxygen, which is connected to an ethyl group (CH₂CH₃).

Key Takeaways

Hydrolysis of esters breaks the ester link (C-O single bond next to carbonyl) to give the acid and alcohol.
A diester hydrolyses to give one dicarboxylic acid molecule and two monoalcohol molecules.
Skeletal structures omit C and H atoms on the carbon chain, showing vertices and ends of lines as carbons.

Common Mistakes

Drawing the full structural formula instead of the skeletal formula when not specified (though skeletal is standard for organic chem at this level, full structural is often accepted if correct, but skeletal is preferred for diesters to avoid clutter).
Incorrectly placing the ester link (e.g., drawing an ether linkage).
Forgetting the carbonyl group (C=O) in the ester.

Things to Be Careful About

Ensure the ester links are correct: R-C(=O)-O-R'. Do not draw R-O-C(=O)-R' (which is the same, but ensure the carbonyl is on the acid side).
The central carbon chain is 3 carbons long (C-C-C), with the ester groups on C1 and C3.

Techniques used

deduce ester structure from hydrolysis productsdraw skeletal formula

(b)

Name the type of reaction that occurs when A is heated with $\text{H}_2\text{SO}_4(\text{aq})$ .

1M Similar

(c)

When aqueous $\text{Na}_2\text{CO}_3$ is added to separate samples of B and C, effervescence is observed with B only.

2M

(i)

Complete the equation to describe the reaction between B and an excess of aqueous $\text{Na}_2\text{CO}_3$ .

...... $\text{HOOCCH}_2\text{COOH}$ + ...... $\text{Na}_2\text{CO}_3 \rightarrow$ ......

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

$\text{HOOCCH}_2\text{COOH} + \text{Na}_2\text{CO}_3 \rightarrow \text{NaOOCCH}_2\text{COONa} + \text{CO}_2 + \text{H}_2\text{O}$

Final answer

HOOCCH2COOH + Na2CO3 -> NaOOCCH2COONa + CO2 + H2O

Detailed explanation

Background Concept

Carboxylic acids are weak acids, but they are stronger than carbonic acid (H₂CO₃). Therefore, carboxylic acids react with carbonates (like Na₂CO₃) and hydrogen carbonates (like NaHCO₃) to produce a salt, water, and carbon dioxide gas. This is a standard test for carboxylic acids (effervescence/bubbling due to CO₂).

For a dicarboxylic acid like propanedioic acid (HOOCCH₂COOH), there are two acidic protons. If excess carbonate is present, both protons are replaced by sodium ions to form the disodium salt.

General reaction: $2\text{RCOOH} + \text{Na}_2\text{CO}_3 \rightarrow 2\text{RCOONa} + \text{CO}_2 + \text{H}_2\text{O}$
Or for a dicarboxylic acid (H₂A): $\text{H}_2\text{A} + \text{Na}_2\text{CO}_3 \rightarrow \text{Na}_2\text{A} + \text{CO}_2 + \text{H}_2\text{O}$

Understanding the Question

Compound B is HOOCCH₂COOH (propanedioic acid). We need to complete the equation for its reaction with aqueous Na₂CO₃. The question states 'excess' Na₂CO₃, ensuring full neutralization to the disodium salt.

Approach

Identify the acid: HOOCCH₂COOH (diprotic).
Identify the base: Na₂CO₃.
Products: Disodium salt (NaOOCCH₂COONa), CO₂, H₂O.
Balance: 1 mole of acid reacts with 1 mole of carbonate to give 1 mole of salt, 1 mole CO₂, 1 mole H₂O.

Step-by-Step Reasoning

Reactants: HOOCCH₂COOH + Na₂CO₃.
The two -COOH groups react with the CO₃²⁻ ion.
The products are the salt where H⁺ is replaced by Na⁺: NaOOCCH₂COONa.
The carbonate ion (CO₃²⁻) reacts with 2H⁺ to form H₂O and CO₂.
Equation: $\text{HOOCCH}_2\text{COOH} + \text{Na}_2\text{CO}_3 \rightarrow \text{NaOOCCH}_2\text{COONa} + \text{H}_2\text{O} + \text{CO}_2$ .
Check balancing: C: 3+2=5 left, 4+1=5 right. Na: 2 left, 2 right. H: 4 left, 2+2=4 right. O: 4+3=7 left, 4+1+2=7 right. Balanced.

Key Takeaways

Carboxylic acids react with carbonates to release CO₂.
Dicarboxylic acids form disodium salts with carbonates.
The stoichiometry is 1:1 for H₂A + Na₂CO₃.

Common Mistakes

Writing the acid salt (NaHOOCCH₂COONa) instead of the normal salt. With excess carbonate, the normal salt forms.
Forgetting to balance the equation (e.g., writing 2 acid molecules with 1 carbonate, which is wrong for a diprotic acid reacting with a divalent base; actually 2 RCOOH + Na2CO3 -> 2 RCOONa + H2O + CO2. For HOOCCH2COOH, it's 1:1).
Writing H₂CO₃ as a product instead of CO₂ + H₂O.

Things to Be Careful About

Ensure the salt formula is correct: NaOOCCH₂COONa. The carboxylate groups are -COO⁻Na⁺.
State symbols are not explicitly required in the mark scheme for this part, but if included, (aq) for reactants and (aq) for salt, (g) for CO₂, (l) for H₂O.

Techniques used

balance acid-carbonate equation

(ii)

Suggest why C does not react with aqueous $\text{Na}_2\text{CO}_3$ in a similar type of reaction to B.

1M Similar

(d)

A student suggests a two-step synthesis to produce B, as shown in Fig. 5.2.

Fig. 5.2

3M

(i)

Identify the reagents and conditions required in steps 1 and 2.

step 1 ................................................................................................................

step 2 ................................................................................................................

2M Similar

DifficultyMedium

Worked solution

Answer

Step 1: NaOH(aq) (or KOH(aq)) and heat (or warm).

Step 2: Acidified potassium dichromate(VI) (or K₂Cr₂O₇ / H⁺) and heat under reflux.

Final answer

Step 1: NaOH(aq), heat. Step 2: Acidified K2Cr2O7, heat under reflux.

Detailed explanation

Background Concept

Step 1: Haloalkane to Alcohol
Halogenoalkanes can be converted to alcohols by nucleophilic substitution. The reagent is aqueous alkali (NaOH or KOH). The hydroxide ion (OH⁻) acts as a nucleophile, attacking the carbon bonded to the halogen. Conditions require heating to ensure the reaction goes to completion. If alcoholic KOH is used, elimination (to form an alkene) would occur instead.

Step 2: Primary Alcohol to Carboxylic Acid
Primary alcohols can be oxidized first to aldehydes and then to carboxylic acids. To get the carboxylic acid directly, an oxidizing agent like acidified potassium dichromate(VI) (K₂Cr₂O₇ / H₂SO₄) is used. The reaction must be carried out under reflux to prevent the escape of the intermediate aldehyde or volatile alcohol, allowing full oxidation to the carboxylic acid. Distillation would stop at the aldehyde.

Understanding the Question

We need to convert 1,3-dichloropropane (ClCH₂CH₂CH₂Cl) to propane-1,3-diol (HOCH₂CH₂CH₂OH) in Step 1, and then to propanedioic acid (HOOCCH₂COOH, Compound B) in Step 2.

Approach

Step 1: Replace -Cl with -OH. This is nucleophilic substitution. Reagent: NaOH(aq). Condition: Heat.
Step 2: Oxidize primary alcohol groups (-CH₂OH) to carboxylic acid groups (-COOH). Reagent: Acidified K₂Cr₂O₇. Condition: Heat under reflux.

Step-by-Step Reasoning

Step 1: ClCH₂CH₂CH₂Cl → HOCH₂CH₂CH₂OH. Two Cl atoms replaced by two OH groups. Reagent: NaOH(aq) (aqueous is crucial to avoid elimination). Condition: Heat (reflux is good but 'heat' is often accepted for substitution, though reflux is safer). Mark scheme says 'NaOH(aq) AND heat'.
Step 2: HOCH₂CH₂CH₂OH → HOOCCH₂COOH. Primary alcohols oxidized to carboxylic acids. Reagent: Acidified potassium dichromate(VI) (K₂Cr₂O₇ with dilute H₂SO₄). Condition: Heat under reflux (to ensure full oxidation).

Key Takeaways

Haloalkane + NaOH(aq) + heat → Alcohol (substitution).
Haloalkane + NaOH(alc) + heat → Alkene (elimination).
Primary alcohol + acidified K₂Cr₂O₇ + heat under reflux → Carboxylic acid.
Primary alcohol + acidified K₂Cr₂O₇ + heat under distillation → Aldehyde.

Common Mistakes

Using alcoholic NaOH for Step 1 (would give alkene).
Using KMnO₄ for Step 2 (acceptable in some contexts, but dichromate is standard for A-Level organic synthesis unless specified).
Forgetting 'acidified' for the dichromate.
Forgetting 'heat under reflux' for Step 2 (would stop at aldehyde or not react fully).

Things to Be Careful About

Specify 'aqueous' for NaOH in Step 1.
Specify 'acidified' and 'heat under reflux' for Step 2.

Techniques used

identify reagents for nucleophilic substitutionidentify reagents for oxidation of alcohol to carboxylic acid

(ii)

Identify the type of reaction that $\text{HOCH}_2\text{CH}_2\text{CH}_2\text{OH}$ undergoes in step 2.

1M Similar

(e)

B reacts with an excess of reducing agent R to produce D.

An excess of $\text{PCl}_5$ is added to D.

A vigorous reaction occurs.
Misty fumes are seen.
Organic compound E is produced.

4M

(i)

Identify the formula of the reducing agent R.

1M Similar

(ii)

Complete the equation to describe the reaction of B with an excess of R.
Use [H] to represent one atom of hydrogen from R.

...... $\text{HOOCCH}_2\text{COOH}$ + ......[H] $\rightarrow$ ......

2M Similar

DifficultyMedium

Worked solution

Working

Reduction of carboxylic acid to alcohol:
$\text{R-COOH} + 4[\text{H}] \rightarrow \text{R-CH}_2\text{OH} + \text{H}_2\text{O}$

Compound B is HOOCCH₂COOH (2 carboxyl groups).
Total [H] needed = 2 × 4 = 8 [H].
Product D is HOCH₂CH₂CH₂OH.

$\text{HOOCCH}_2\text{COOH} + 8[\text{H}] \rightarrow \text{HOCH}_2\text{CH}_2\text{CH}_2\text{OH} + 2\text{H}_2\text{O}$

Answer

$\text{HOOCCH}_2\text{COOH} + 8[\text{H}] \rightarrow \text{HOCH}_2\text{CH}_2\text{CH}_2\text{OH} + 2\text{H}_2\text{O}$

Final answer

HOOCCH2COOH + 8[H] -> HOCH2CH2CH2OH + 2H2O

Detailed explanation

Background Concept

The reduction of a carboxylic acid to a primary alcohol using [H] (representing a reducing agent like LiAlH₄ followed by H⁺/H₂O) can be represented as:
$\text{RCOOH} + 4[\text{H}] \rightarrow \text{RCH}_2\text{OH} + \text{H}_2\text{O}$

The carboxyl group (-COOH) gains 4 hydrogen atoms: 2 to reduce the C=O to CH-OH (actually C=O to CH₂-OH involves adding 2H to C and 2H to O, net 4H, and losing the original O as water).
Let's trace atoms: -COOH (C, 2O, 1H) -> -CH₂OH (C, 1O, 3H). Change: +2H to C, +2H to O (replacing =O with -OH and adding H to C). Wait.
-COOH + 4[H] -> -CH₂OH + H₂O.
Left: C, 2O, 1H + 4H = C, 2O, 5H.
Right: C, 1O, 3H + H₂O (2H, 1O) = C, 2O, 5H. Balanced.

Understanding the Question

Compound B is propanedioic acid: HOOC-CH₂-COOH. It has two carboxylic acid groups. We are using excess reducing agent, so both groups are reduced to primary alcohols.
Product D is propane-1,3-diol: HOCH₂-CH₂-CH₂OH.
We need to write the balanced equation using [H].

Approach

Write the reduction half-equation for one -COOH group: -COOH + 4[H] -> -CH₂OH + H₂O.
Since there are two -COOH groups, multiply by 2: 2(-COOH) + 8[H] -> 2(-CH₂OH) + 2H₂O.
Add the central CH₂ group back.
Full equation: HOOCCH₂COOH + 8[H] -> HOCH₂CH₂CH₂OH + 2H₂O.

Step-by-Step Reasoning

Reactant: HOOCCH₂COOH.
Reagent: 8[H] (4 per carboxyl group × 2 groups).
Organic Product: HOCH₂CH₂CH₂OH (propane-1,3-diol).
By-product: 2H₂O (one per carboxyl group reduced).
Check balance:
- Left: C₃H₄O₄ + 8H = C₃H₁₂O₄.
- Right: C₃H₈O₂ + 2H₂O = C₃H₁₂O₄.
- Balanced.

Key Takeaways

Reduction of -COOH to -CH₂OH requires 4[H].
Dicarboxylic acids require 8[H] to become diols.
Water is a by-product of this reduction.

Common Mistakes

Using 4[H] total instead of 8[H] (forgetting there are two acid groups).
Forgetting the water by-product.
Writing the wrong organic product (e.g., aldehyde).

Things to Be Careful About

The question asks to use [H]. Do not write LiAlH₄ in the equation.
Ensure the organic product is correct: HOCH₂CH₂CH₂OH.

Techniques used

balance reduction equation with [H]

(iii)

Name organic compound E.

1M Similar

Q6MediumCarbonyl CompoundsHydroxy Compounds

Three bottles of colourless liquids labelled F, G and H contain separate pure samples of the compounds ethanal, propanal or propanone but not necessarily in that order.

(a)

State the functional group in ethanal, propanal and propanone.

1M Similar

(b)

State a reagent and the relevant observation that confirm that F, G and H have the same functional group.

reagent .....................................................................................................................................

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

2M Similar

(c)

Separate samples of F, G and H are each tested with Tollens’ reagent and with alkaline $\text{I}_2(\text{aq})$ . The observations are shown in Table 6.1.

Table 6.1

	Tollens’ reagent	alkaline $\text{I}_2(\text{aq})$
F	no observable change	pale yellow precipitate
G	silver mirror	pale yellow precipitate
H	silver mirror	no precipitate

4M

(i)

Use Table 6.1 to name the organic compounds in bottles F, G and H.

F = .............................................

G = .............................................

H = .............................................

2M Similar

DifficultyMedium

Worked solution

Answer

F = propanone
G = ethanal
H = propanal

Final answer

F = propanone, G = ethanal, H = propanal

Detailed explanation

Background Concept

Tollens' reagent (ammoniacal silver nitrate) oxidises aldehydes but not ketones; the Ag+ is reduced to metallic silver, seen as a silver mirror. Alkaline iodine gives the tri-iodomethane (iodoform) test: a pale yellow precipitate of $\text{CHI}_3$ forms with compounds containing the $\text{CH}_3\text{CO}-$ group (methyl ketones and ethanal) or the $\text{CH}_3\text{CH(OH)}-$ group (oxidised in situ to $\text{CH}_3\text{CO}-$ ).

The three candidates:

ethanal, $\text{CH}_3\text{CHO}$ : aldehyde AND has $\text{CH}_3\text{CO}-$ skeleton → Tollens' positive, iodoform positive
propanal, $\text{CH}_3\text{CH}_2\text{CHO}$ : aldehyde but no $\text{CH}_3\text{CO}-$ group → Tollens' positive, iodoform negative
propanone, $\text{CH}_3\text{COCH}_3$ : ketone with $\text{CH}_3\text{CO}-$ group → Tollens' negative, iodoform positive

Understanding the Question

Table 6.1 gives each bottle's response to the two tests. 'No observable change' with Tollens' means no aldehyde; 'silver mirror' means aldehyde; iodoform precipitate means a methyl carbonyl group.

Approach

Work through each bottle: use Tollens' first to split aldehydes (G, H) from the ketone (F), then use the iodoform result to separate ethanal (positive) from propanal (negative).

Step-by-Step Reasoning

F: no change with Tollens' → not an aldehyde → propanone (the only ketone). Its pale yellow precipitate with iodine is consistent, since propanone has the $\text{CH}_3\text{CO}-$ group.
G: silver mirror → aldehyde; yellow precipitate with iodine → contains $\text{CH}_3\text{CO}-$ → ethanal (the only aldehyde with a methyl carbonyl unit).
H: silver mirror → aldehyde; no precipitate with iodine → no $\text{CH}_3\text{CO}-$ group → propanal.

Key Takeaways

Tollens' distinguishes aldehyde from ketone; the iodoform test distinguishes ethanal from other aldehydes and identifies methyl ketones. Using two tests in combination pins down each structure.

Common Mistakes

Swapping G and H by forgetting that ethanal, not propanal, gives the iodoform test (propanal's carbonyl carbon is bonded to an ethyl group, not a methyl group). Assuming all aldehydes give the iodoform test.

Things to Be Careful About

The iodoform test responds to the $\text{CH}_3\text{CO}-$ group specifically — the methyl must be directly attached to the carbonyl carbon. Propanal is $\text{CH}_3\text{CH}_2\text{CHO}$ , so it fails this test.

Techniques used

interpret Tollens' test resultsinterpret the tri-iodomethane (iodoform) testdeduce compound identities from test observations

(ii)

Identify the yellow precipitate produced when alkaline $\text{I}_2(\text{aq})$ is added to separate samples of F and G.

1M Similar

(iii)

Compound J does not contain the same functional group as F, G and H. Compound J also reacts with alkaline $\text{I}_2(\text{aq})$ to produce a pale yellow precipitate.

Suggest the structure of compound J.

1M Similar

DifficultyMedium

Worked solution

Answer

Any secondary alcohol containing the $\text{CH}_3\text{CH(OH)}-$ group, e.g. propan-2-ol, $\text{CH}_3\text{CH(OH)CH}_3$ (or ethanol, $\text{CH}_3\text{CH}_2\text{OH}$ )

Final answer

CH3CH(OH)R, e.g. propan-2-ol

Detailed explanation

Background Concept

The iodoform test is positive for two structural features: the $\text{CH}_3\text{CO}-$ group (methyl ketones and ethanal) and the $\text{CH}_3\text{CH(OH)}-$ group. Alcohols of the latter type are oxidised by the alkaline iodine (an oxidising mixture) in situ to the corresponding methyl carbonyl compound, which then gives the yellow $\text{CHI}_3$ precipitate. Ethanol ( $\text{CH}_3\text{CH}_2\text{OH}$ ) is the only primary alcohol that qualifies, as it oxidises first to ethanal.

Understanding the Question

J does NOT contain the carbonyl group, yet still gives a pale yellow precipitate with alkaline iodine. We must suggest a structure with a different functional group that still passes the iodoform test.

Approach

Ask: which non-carbonyl structures give a positive iodoform test? Answer: alcohols containing $\text{CH}_3\text{CH(OH)}-$ , i.e. secondary alcohols with a methyl on the carbinol carbon, plus ethanol.

Step-by-Step Reasoning

Alkaline iodine oxidises the $\text{CH}_3\text{CH(OH)}-$ unit to $\text{CH}_3\text{CO}-$ , which then undergoes the iodoform reaction to give the pale yellow $\text{CHI}_3$ precipitate. So J can be any $\text{CH}_3\text{CH(OH)R}$ where R is H or an alkyl group — for example propan-2-ol ( $\text{CH}_3\text{CH(OH)CH}_3$ ) or ethanol ( $\text{CH}_3\text{CH}_2\text{OH}$ , where R = H). These are alcohols, not carbonyl compounds, satisfying the condition on J's functional group.

Key Takeaways

The iodoform test detects $\text{CH}_3\text{CO}-$ OR $\text{CH}_3\text{CH(OH)}-$ ; the alcohol route works because iodine/alkali oxidises the alcohol first.

Common Mistakes

Suggesting a methyl ketone or ethanal — these contain the carbonyl group, which J must not. Suggesting butan-2-ol-type alcohols is fine, but suggesting a primary alcohol other than ethanol (e.g. propan-1-ol) is wrong — it lacks the $\text{CH}_3\text{CH(OH)}-$ unit.

Things to Be Careful About

The general structure is $\text{CH}_3\text{CH(OH)R}$ with R = H or alkyl; the OH must be on the carbon bearing the $\text{CH}_3$ group.

Techniques used

apply the iodoform test to secondary alcohols and ethanolsuggest a structure consistent with test results

Chemistry 9701/21 — May/June 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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Working

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Understanding the Question

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