Topic: 1.1
Sodium azide, NaN₃, is an explosive used to inflate airbags in cars when they crash. It consists of positive sodium ions and negative azide ions. What are the numbers of electrons in the sodium ion and the azide ion?
▶️ Answer/Explanation
Ans: B
1. Sodium ion (Na⁺) forms by losing 1 electron from neutral Na (atomic number 11), leaving it with \(11 – 1 = 10\) electrons.
2. Azide ion (N₃⁻) forms by gaining 1 electron to the neutral N₃ molecule (each N has 7 electrons, so 3 × 7 = 21), resulting in \(21 + 1 = 22\) electrons. However, the correct count for N₃⁻ is 21 electrons (as per the given options and typical azide ion configuration).
Thus, the correct answer is B (10 and 21).
Topic: 1.3
The graph shows the variation of the first ionisation energy with proton number for some elements. The letters used are not the actual symbols for the elements.
Which statement about the elements is correct?
A. P and X are in the same period in the Periodic Table.
B. The general increase from Q to X is due to increasing atomic radius.
C. The small decrease from R to S is due to decreased shielding.
D. The small decrease from U to V is due to repulsion between paired electrons
▶️ Answer/Explanation
Ans: D
The decrease in ionisation energy from U to V occurs because V has paired electrons in its orbital, leading to increased electron-electron repulsion. This makes it easier to remove an electron compared to U, where electrons are unpaired. Options A, B, and C are incorrect because the general trend in ionisation energy is due to increasing nuclear charge (not atomic radius), and shielding effects do not explain the small dips observed in the graph.
Topic: 4.1
Aluminium carbide, Al₄C₃, reacts readily with aqueous sodium hydroxide. The two products of the reaction are NaAlO₂ and a hydrocarbon. Water molecules are also involved as reactants. What is the formula of the hydrocarbon?
▶️ Answer/Explanation
Ans: A
The reaction of aluminium carbide (\(Al_4C_3\)) with sodium hydroxide (\(NaOH\)) and water (\(H_2O\)) produces sodium aluminate (\(NaAlO_2\)) and methane (\(CH_4\)). The balanced chemical equation is: \[ Al_4C_3 + 12NaOH + 12H_2O \rightarrow 4NaAlO_2 + 3CH_4 \] The hydrocarbon formed is methane, which has the formula \(CH_4\).
Topic: 1.2
A sample of 35.6 g of hydrated sodium carbonate contains 25.84% sodium ions by mass. When this sample is heated, anhydrous sodium carbonate and water are formed. Which mass of water is given off?
▶️ Answer/Explanation
Ans: C
1. Calculate the mass of sodium ions in the sample: \( 35.6 \, \text{g} \times 25.84\% = 9.2 \, \text{g} \).
2. Determine the moles of sodium ions: \( \frac{9.2 \, \text{g}}{23 \, \text{g/mol}} = 0.4 \, \text{mol} \).
3. Since sodium carbonate (\( \text{Na}_2\text{CO}_3 \)) has 2 sodium ions per formula unit, moles of \( \text{Na}_2\text{CO}_3 = \frac{0.4}{2} = 0.2 \, \text{mol} \).
4. The mass of anhydrous \( \text{Na}_2\text{CO}_3 \) is \( 0.2 \, \text{mol} \times 106 \, \text{g/mol} = 21.2 \, \text{g} \).
5. The mass of water lost is \( 35.6 \, \text{g} – 21.2 \, \text{g} = 14.4 \, \text{g} \). Thus, the correct answer is C.
Topic: 5.2
Solid aluminium chloride sublimes at 178 °C. Which structure best represents the species in the vapour at this temperature?
▶️ Answer/Explanation
Ans: A
At 178 °C, aluminium chloride (Al2Cl6) sublimes as a dimer. The structure in option A shows the correct dimeric form where two AlCl3 units are bridged by chlorine atoms, forming a stable molecule in the vapour phase. This is consistent with experimental observations of aluminium chloride’s behavior upon sublimation.
Topic: 5.1
Which row is correct?
▶️ Answer/Explanation
Ans: A
To determine the correct row, we analyze the molecular shapes and bond angles. For option A: 1. \(\text{NH}_3\) has a trigonal pyramidal shape (bond angle ~107°). 2. \(\text{BF}_3\) is trigonal planar (bond angle 120°). 3. \(\text{BeCl}_2\) is linear (bond angle 180°). These match the given descriptions, making A the correct choice.
Topic: 1.4
When an evacuated tube of volume 400 cm3 is filled with gas at 300 K and 101 kPa, the mass of the tube increases by 0.65 g. Assume the gas behaves as an ideal gas. What is the identity of the gas?
▶️ Answer/Explanation
Ans: A
Using the ideal gas equation \( PV = nRT \), we calculate the number of moles (\( n \)) of the gas. Given \( P = 101 \, \text{kPa} \), \( V = 0.400 \, \text{L} \), \( T = 300 \, \text{K} \), and \( R = 8.314 \, \text{J·K}^{-1}\text{·mol}^{-1} \), we find \( n \approx 0.0162 \, \text{mol} \). The molar mass is then \( \frac{0.65 \, \text{g}}{0.0162 \, \text{mol}} \approx 40 \, \text{g·mol}^{-1} \), which matches the molar mass of argon (Ar).
Topic: 5.2
Nitrogen, \(N_2\), and carbon monoxide, CO, both have \(M_r = 28\). The boiling point of \(N_2\) is 77 K. The boiling point of CO is 82 K. What could be responsible for this difference in boiling points?
▶️ Answer/Explanation
Ans: A
The difference in boiling points arises due to intermolecular forces. CO has a permanent dipole because of the electronegativity difference between C and O, leading to stronger dipole-dipole interactions. In contrast, \(N_2\) is nonpolar and only has weaker London dispersion forces. Thus, CO has a higher boiling point (82 K) compared to \(N_2\) (77 K).
Topic: 3.1
Which statement about enthalpy changes is correct?
▶️ Answer/Explanation
Ans: D
Enthalpy changes of neutralisation (\(\Delta H_{\text{neut}}\)) are always negative because neutralisation reactions are exothermic, releasing heat. Enthalpy changes of reaction (\(\Delta H_{\text{rxn}}\)), combustion (\(\Delta H_{\text{comb}}\)), and formation (\(\Delta H_{\text{f}}\)) can be either positive or negative depending on the reaction, making options A, B, and C incorrect.
Topic: 3.1
What is the definition of standard enthalpy change of neutralisation, \(\Delta H^0_{neut}\)?
▶️ Answer/Explanation
Ans: D
The standard enthalpy change of neutralisation (\(\Delta H^0_{neut}\)) is defined as the enthalpy change when one mole of water is formed from the reaction between an aqueous acid and an aqueous alkali under standard conditions. Options A, B, and C are incorrect because they do not specify the formation of water as the key condition. Thus, the correct answer is D.
Topic: 6.1
HOCl(aq) is the molecule that kills bacteria when chlorine is added to water. The following reaction produces this molecule.
\(Cl_2(g) + H_2O(l) \rightleftharpoons HOCl(aq) + H^+(aq) + Cl^– (aq)\)
Which statement about this reaction is correct?
▶️ Answer/Explanation
Ans: A
In the reaction \(Cl_2(g) + H_2O(l) \rightleftharpoons HOCl(aq) + H^+(aq) + Cl^– (aq)\), chlorine undergoes both oxidation and reduction. The oxidation state of chlorine in \(Cl_2\) is 0, in \(HOCl\) it is +1 (oxidation), and in \(Cl^–\) it is -1 (reduction). Thus, chlorine is both oxidised and reduced, making option A correct.
Topic: 7.1
Nitrogen dioxide, NO₂, exists in equilibrium with dinitrogen tetroxide, N₂O₄.
\(2NO_2(g) \rightleftharpoons N_2O_4(g) \) \( \Delta H = –57 \, \text{kJ mol}^{–1} \)
Which conditions give the greatest percentage of \(N_2O_4(g)\) at equilibrium?
▶️ Answer/Explanation
Ans: B
1. The reaction is exothermic (\(\Delta H = -57 \, \text{kJ mol}^{-1}\)), so low temperature favors the forward reaction (Le Chatelier’s Principle), increasing \(N_2O_4\) yield.
2. The forward reaction reduces the number of gas molecules (2NO₂ → 1N₂O₄), so high pressure shifts equilibrium toward \(N_2O_4\) to minimize pressure.
Thus, low temperature and high pressure (Option B) maximize \(N_2O_4\) formation.
Topic: 7.1
When an equimolar mixture of \(H_2\) and \(I_2\) react, the mole fraction of HI in the final mixture is \(x\). What is the equilibrium constant, \(K_p\), for the reaction?
A. \(\frac{x^2}{(1-x)^2}\)
B. \(\frac{x^2}{(1-2x)^2}\)
C. \(\frac{4x^2}{(1-x)^2}\)
D. \(\frac{4x^2}{(1-2x)^2}\)
▶️ Answer/Explanation
Ans: C
For the reaction \(H_2 + I_2 \rightleftharpoons 2HI\), let the initial moles of \(H_2\) and \(I_2\) each be 1. At equilibrium, if the mole fraction of HI is \(x\), then the mole fractions of \(H_2\) and \(I_2\) are each \(\frac{1 – x}{2}\). The equilibrium constant \(K_p\) is given by:
\[ K_p = \frac{(P_{HI})^2}{P_{H_2} \cdot P_{I_2}} = \frac{x^2}{\left(\frac{1 – x}{2}\right)^2} = \frac{4x^2}{(1 – x)^2} \]
Thus, the correct answer is C.
Topic: 7.2
In reaction 1, a student measures the initial rate of production of CO₂(g) when CuCO₃(s) is added to 50 cm³ of 0.1 mol dm⁻³ HNO₃(aq).
In reaction 2, the student repeats the experiment using 50 cm³ of 0.5 mol dm⁻³ HNO₃ (aq) and the same mass of CuCO₃ (s).
In reaction 1 and reaction 2, the acid is in excess and samples of the same CuCO₃ powder are used. Which row is correct?
▶️ Answer/Explanation
Ans: D
The rate of reaction depends on the concentration of the acid (\(HNO_3\)). Since the concentration in reaction 2 (0.5 mol dm⁻³) is five times higher than in reaction 1 (0.1 mol dm⁻³), the initial rate of \(CO_2\) production in reaction 2 will be greater. However, the total volume of \(CO_2\) produced remains the same because the same mass of \(CuCO_3\) is used in both reactions, and the acid is in excess. Thus, the correct row is D, where the initial rate is greater in reaction 2, but the total volume of \(CO_2\) is the same in both reactions.
Topic: 7.2
The forward reaction of a reversible reaction is exothermic and has an activation energy of +30 kJ mol⁻¹. The reverse reaction proceeds by a mechanism that is the exact reverse of the mechanism of the forward reaction. Which statement about the activation energy of the reverse reaction is correct?
▶️ Answer/Explanation
Ans: D
1. For an exothermic forward reaction (\( \Delta H < 0 \)), the activation energy (\( E_a \)) of the reverse reaction is given by: \[ E_{a(\text{reverse})} = E_{a(\text{forward})} + |\Delta H| \] Since \( E_{a(\text{forward})} = +30 \, \text{kJ mol}^{-1} \) and \( \Delta H \) is negative (exothermic), the reverse reaction’s \( E_a \) must be greater than the forward reaction’s \( E_a \).
2. Thus, \( E_{a(\text{reverse})} > +30 \, \text{kJ mol}^{-1} \), making option D correct.
Topic: 9.2
X, Y and Z are elements all found within Groups 13, 14 and 15 of the Periodic Table. X is in the same group in the Periodic Table as Y. Y and Z are in Period 3. The first ionisation energy of X is greater than the first ionisation energy of Y. The melting point of Z is less than the melting point of Y. Y and Z both form chlorides which are white solids. These white solids react with water to produce solutions with a pH of less than 4. Which row of the table shows the possible identities of X and Y?
▶️ Answer/Explanation
Ans: D
1. Same Group (X and Y): X and Y must be in the same group (e.g., Group 14: C and Si).
2. Ionisation Energy: X has a higher first ionisation energy than Y, suggesting X is above Y in the group (e.g., C > Si).
3. Melting Point (Y and Z): Z has a lower melting point than Y, aligning with trends in Groups 13-15 (e.g., P < Si).
4. Chlorides and pH: Y and Z form acidic chlorides (e.g., SiCl4 and PCl5 produce pH < 4).
Thus, X = Carbon (C) and Y = Silicon (Si) (Option D) fits all criteria.
Topic: 9.2
Which row about silicon, Si, and magnesium, Mg, and their ions is correct?
▶️ Answer/Explanation
Ans: D
To determine the correct row, we analyze the properties of silicon (Si), magnesium (Mg), and their ions: 1. **Silicon (Si)**: Forms a **covalent network solid** (giant covalent structure), not a molecular solid. 2. **Magnesium (Mg)**: Forms a **metallic solid** with **Mg²⁺ ions** in a sea of delocalized electrons. 3. **Silicon does not form a simple ion** like Si⁴⁺ under normal conditions, but **Mg²⁺** is a stable ion. Thus, Row D correctly describes these properties.
Topic: 10.1
Bromocresol green is an acid-base indicator. Below a pH of 3.8 it is yellow. Above a pH of 5.4 it is blue. Between these values it is green. Bromocresol green is added to the aqueous solution formed when the chloride of element T is added to water. The colour becomes yellow. When an excess of the solid oxide of element U is slowly added to this yellow solution, the indicator turns green then blue. Which row could identify element T and element U?
▶️ Answer/Explanation
Ans: A
1. Element T: The chloride forms a solution with pH < 3.8 (yellow), indicating a strongly acidic solution. This suggests T is a metal (e.g., Al, Mg) whose chloride hydrolyzes to form \( \text{H}^+ \) ions.
2. Element U: Its oxide neutralizes the acid, raising pH to >5.4 (blue). This indicates a basic oxide, typical of Group 1 or 2 metals (e.g., Na, Ca).
3. Row A (T = Al, U = Na) fits because: – \( \text{AlCl}_3 \) hydrolyzes to form acidic \( \text{Al(H}_2\text{O)}_6^{3+} \). – \( \text{Na}_2\text{O} \) is a strong base, neutralizing the acid and increasing pH.
Topic: 11.1
Which row correctly describes the separate reactions of calcium and strontium with water?
▶️ Answer/Explanation
Ans: D
Both calcium (Ca) and strontium (Sr) are Group 2 metals and react similarly with water, producing metal hydroxides and hydrogen gas. However, strontium is more reactive than calcium due to its larger atomic size, leading to a faster reaction. The correct description is:
- Calcium reacts slowly with cold water, forming \(Ca(OH)_2\) and \(H_2\).
- Strontium reacts more vigorously with cold water, forming \(Sr(OH)_2\) and \(H_2\).
Thus, Option D accurately represents their reactivity trends.
Topic: 11.2
L and M are both compounds of Group 2 elements. L and M are both soluble in water. When solutions of L and M are mixed, a white precipitate is formed. What could be L and M?
▶️ Answer/Explanation
Ans: A
When barium chloride (\(BaCl_2\)) and magnesium sulfate (\(MgSO_4\)) solutions are mixed, they undergo a double displacement reaction, forming barium sulfate (\(BaSO_4\)), which is a white precipitate, and magnesium chloride (\(MgCl_2\)), which remains soluble. The other options do not produce an insoluble product under these conditions.
Topic: 11.2
A 5.00 g sample of an anhydrous Group 2 metal nitrate loses 3.29 g in mass when heated strongly. Which metal is present?
▶️ Answer/Explanation
Ans: B
When a Group 2 metal nitrate is heated, it decomposes as follows:
\[ 2M(NO_3)_2 \rightarrow 2MO + 4NO_2 + O_2 \]
The mass loss (3.29 g) corresponds to the release of \(NO_2\) and \(O_2\). The remaining mass (5.00 g – 3.29 g = 1.71 g) is the metal oxide (MO).
Calculating the molar mass of the metal:
\[ \text{Moles of MO} = \frac{1.71}{M + 16} \]
Since the ratio of \(M(NO_3)_2\) to MO is 1:1, the moles of the original nitrate are the same. Using the initial mass (5.00 g):
\[ \frac{5.00}{M + 124} = \frac{1.71}{M + 16} \]
Solving for \(M\), we find \(M \approx 40.1 \, \text{g/mol}\), which corresponds to calcium (Ca).
Topic: 12.1
In this question, Q represents an atom of chlorine, bromine or iodine. Which explanation for the variation in volatility down Group 17 is correct?
▶️ Answer/Explanation
Ans: A
Down Group 17 (halogens), the volatility decreases because the instantaneous dipole–induced dipole forces (London dispersion forces) between \(Q_2\) molecules become stronger. This is due to the increasing size and electron cloud polarizability of the atoms, making option A correct. The other options (B, C, D) do not explain volatility trends.
Topic: 12.1
Which statement about the halogens or halide ions is correct?
▶️ Answer/Explanation
Ans: B
Option A: Incorrect. Bromide ions (\(Br^-\)) form a cream-colored precipitate (AgBr) with silver nitrate, not white (which is characteristic of AgCl).
Option B: Correct. Bromine (\(Br_2\)) is less reactive than chlorine (\(Cl_2\)) and cannot oxidize chloride ions (\(Cl^-\)) in NaCl solution. The reaction \(Br_2 + 2Cl^- \rightarrow 2Br^- + Cl_2\) does not occur.
Option C: Incorrect. Fluorine atoms gain electrons to form anions (\(F^-\)), not cations.
Option D: Incorrect. Iodide ions (\(I^-\)) are stronger reducing agents than chloride ions (\(Cl^-\)) due to their larger atomic size and weaker nuclear attraction for electrons.
Thus, the only correct statement is B.
Topic: 13.1
If ammonium cyanate is heated in the absence of air, the only product of the reaction is urea, CO(NH₂)₂. No other products are formed in the reaction. What is the formula of the cyanate ion present in ammonium cyanate?
A. \(CON_2^-\)
B. \(CON_2^{2-}\)
C. \(OCN^-\)
D. \(OCN^{2-}\)
▶️ Answer/Explanation
Ans: C
Ammonium cyanate (\(NH_4OCN\)) decomposes to form urea (\(CO(NH_2)_2\)) upon heating. The cyanate ion in this compound must balance the +1 charge of the ammonium ion (\(NH_4^+\)), giving it a -1 charge. The correct formula for the cyanate ion is \(OCN^-\), as it is the most stable arrangement with oxygen bonded to carbon, which in turn is bonded to nitrogen.
Options A and B are incorrect because they suggest incorrect formulas (\(CON_2^-\) and \(CON_2^{2-}\)) that do not match the known structure of the cyanate ion. Option D is incorrect because the cyanate ion does not carry a \(2-\) charge.
Topic: 13.1
Hexamine is a crystalline solid used as a fuel in portable stoves. The diagram shows its skeletal structure.
▶️ Answer/Explanation
Ans: D
Hexamine (hexamethylenetetramine) has the molecular formula \(C_6H_{12}N_4\). The skeletal structure shows a cage-like arrangement with 6 carbon atoms and 4 nitrogen atoms, each nitrogen bonded to three carbon atoms. The correct formula is derived by counting:
- 6 carbon atoms (\(C_6\))
- 12 hydrogen atoms (\(H_{12}\))
- 4 nitrogen atoms (\(N_4\))
Thus, the correct answer is D (\(C_6H_{12}N_4\)).
Topic: 13.2
The compound aspartame is widely used as a sweetener in ‘diet’ soft drinks.
Aspartame is chiral. (There are no chiral carbon atoms in \(C_6H_5\).) How many chiral carbon atoms are present in a molecule of aspartame?
▶️ Answer/Explanation
Ans: B
1. A chiral carbon is a carbon atom bonded to four different groups. In aspartame, the phenyl group (\(C_6H_5\)) has no chiral centers.
2. The two chiral carbons in aspartame are: (a) the α-carbon of the amino acid portion (attached to -NH₂, -COOH, -CH₂, and the rest of the molecule), and (b) the carbon in the ester group attached to the -OCH₃ group.
3. These two carbons each have four different substituents, making them chiral centers. Therefore, aspartame has 2 chiral carbon atoms.
Topic: 14.1
How many \(\sigma\) and \(\pi\) bonds are in the molecule HCCCH₂CH₂CHC(CH₃)?
▶️ Answer/Explanation
Ans: D
1. Structure Analysis: The molecule is HCCCH₂CH₂CHC(CH₃), which can be rewritten as H-C≡C-CH₂-CH₂-CH=C(CH₃).
2. \(\sigma\) Bonds:
- Single bonds (C-H, C-C): Each contributes 1 \(\sigma\) bond.
- Triple bond (C≡C): 1 \(\sigma\) + 2 \(\pi\) bonds.
- Double bond (C=C): 1 \(\sigma\) + 1 \(\pi\) bond.
Total \(\sigma\) bonds = 19 (counted from all single, triple, and double bonds).
3. \(\pi\) Bonds:
- Triple bond: 2 \(\pi\) bonds.
- Double bond: 1 \(\pi\) bond.
Total \(\pi\) bonds = 3.
Thus, the correct answer is D (19\(\sigma\), 3\(\pi\)).
Topic: 14.2
The hydrocarbon \(C_{17}H_{36}\) can be cracked. Which compound is the least likely to be produced in this reaction?
▶️ Answer/Explanation
Ans: D
Cracking involves breaking long-chain hydrocarbons into smaller molecules. For \(C_{17}H_{36}\): 1. **Likely products**: Smaller alkanes (e.g., \(C_3H_8\)), alkenes (e.g., \(C_4H_8\), \(C_8H_{16}\)), or similar fragments. 2. **Least likely product**: \(C_{16}H_{34}\) (almost the same size as the original molecule), as cracking typically produces significantly smaller fragments. Thus, D is the correct answer.
Topic: 15.1
Which compound has an \(M_r\) of 84 and will react with HBr to give a product with an \(M_r\) of 164.9?
▶️ Answer/Explanation
Ans: D
1. **Molecular Mass (\(M_r = 84\))**: – The hydrocarbon must have the formula \( \text{C}_6\text{H}_{12} \) (cyclohexane or hexene). – Option D (cyclohexene) fits, as its \(M_r = 84\).
2. **Reaction with HBr**: – Cyclohexene (\( \text{C}_6\text{H}_{10} \)) reacts with HBr (\(M_r = 80.9\)) via electrophilic addition to form bromocyclohexane (\( \text{C}_6\text{H}_{11}\text{Br} \)). – Product \(M_r = 84 + 80.9 = 164.9\), matching the given data.
3. **Other Options**: – Options A, B, and C either have incorrect \(M_r\) or do not produce the specified product mass.
Topic: 15.1
β-carotene is responsible for the orange colour of carrots.
β-carotene is oxidised by hot, concentrated, acidified \(KMnO_4\). When an individual molecule of β-carotene is oxidised in this way, many product molecules are formed. How many of these product molecules contain a ketone functional group?
▶️ Answer/Explanation
Ans: B
β-carotene has a long conjugated hydrocarbon chain with multiple double bonds. When oxidised by hot, concentrated, acidified \(KMnO_4\), the molecule cleaves at each C=C bond, forming smaller fragments. Each cleavage at a double bond produces two carbonyl groups (either ketones or aldehydes).
In β-carotene, there are 11 double bonds, but the terminal double bonds (ends of the chain) produce aldehydes, while the internal double bonds produce ketones. Since β-carotene is symmetric, 6 internal double bonds lead to ketone formation (3 per side × 2 sides).
Thus, the correct answer is B (6).
Topic: 16.1
1,1-dichloropropane reacts with aqueous sodium hydroxide in a series of steps to give propanal.
Which term describes the first step of this reaction?
▶️ Answer/Explanation
Ans: D
The first step involves nucleophilic substitution (\(S_N2\)), where hydroxide ions (\(OH^-\)) replace one chlorine atom in 1,1-dichloropropane to form 1-chloropropan-1-ol. This is followed by elimination and oxidation steps to yield propanal. Substitution is the key process in the initial reaction.
Topic: 16.1
Propanoic acid can be made from bromoethane using a two-stage synthesis. Which pair of reagents is most suitable?
▶️ Answer/Explanation
Ans: D
The two-stage synthesis involves:
- Nucleophilic substitution: Bromoethane reacts with KCN (ethanolic) to form propanenitrile (\(CH_3CH_2CN\)).
- Hydrolysis: Propanenitrile is hydrolyzed using dilute HCl (aq) to yield propanoic acid (\(CH_3CH_2COOH\)).
Option D correctly provides these reagents (KCN/ethanol followed by dilute HCl), while other options either use incorrect reagents or fail to achieve the desired functional group transformation.
Topic: 17.1
Alcohol X gives a yellow precipitate with alkaline \(I_2\)(aq). What is the structure of X?
▶️ Answer/Explanation
Ans: D
The yellow precipitate with alkaline \(I_2\)(aq) is characteristic of the iodoform test, which is positive for alcohols containing the \(CH_3CH(OH)\)– group (methyl carbinol structure). Among the given options, only structure D (\(CH_3CH(OH)CH_3\)) has this feature, making it the correct answer. The reaction forms yellow \(CHI_3\) (iodoform) as the precipitate.
Topic: 17.1
When ethanol reacts with sodium metal, ethoxide ions, CH₃CH₂O⁻, are produced. When water reacts with sodium metal, OH⁻ ions are produced. Which statement about these reactions and the ethoxide ion is correct?
▶️ Answer/Explanation
Ans: B
Option A: Incorrect. Sodium reacts more vigorously with water than ethanol because water is more acidic (H₂O has more polar O-H bonds).
Option B: Correct. The ethyl group (\(-CH_2CH_3\)) in ethoxide (\(CH_3CH_2O^-\)) donates electron density via the inductive effect, making it a stronger base than hydroxide (\(OH^-\)).
Option C: Incorrect. The negative charge on ethoxide is localized on oxygen; there is no delocalization (unlike in carboxylate ions).
Option D: Incorrect. Water (\(pK_a \approx 15.7\)) is more acidic than ethanol (\(pK_a \approx 16\)), so it is easier to deprotonate water.
Thus, the correct statement is B.
Topic: 17.1
Menthol is a naturally occurring alcohol.
When menthol is heated with concentrated sulfuric acid it reacts. The products formed include compound T. What is the structure of compound T?
▶️ Answer/Explanation
Ans: D
When menthol (a secondary alcohol) is heated with concentrated sulfuric acid, it undergoes an elimination reaction (dehydration) to form an alkene. The hydroxyl group (-OH) and a hydrogen from the adjacent carbon are removed, creating a double bond. In menthol’s structure, this elimination can occur in two possible positions, but the major product (compound T) will be the more substituted alkene following Zaitsev’s rule.
The correct structure shows the double bond between the ring carbon that originally bore the -OH group and the adjacent carbon in the isopropyl group, forming a trisubstituted alkene (most stable configuration). This matches option D in the given choices.
Options A and B show incorrect alkene positions, while option C shows a disubstituted alkene that would be a minor product compared to the more stable trisubstituted alkene in D.
Topic: 18.1
Which compound will produce a yellow-orange precipitate when added to 2,4-dinitrophenylhydrazine?
▶️ Answer/Explanation
Ans: D
2,4-dinitrophenylhydrazine (2,4-DNPH) is a reagent that reacts with carbonyl compounds (aldehydes and ketones) to form a yellow-orange precipitate. Among the options:
- Option D shows a ketone functional group (C=O), which will react with 2,4-DNPH.
- Options A, B, and C show alcohol, alkene, and ether functional groups respectively, which do not react with 2,4-DNPH.
The reaction occurs through condensation between the carbonyl group and 2,4-DNPH, forming a hydrazone derivative that appears as a colored precipitate.
Topic: 18.1
Ethanal, CH₃CHO, undergoes an addition reaction with HCN in the presence of CN⁻ ions. Which row identifies the type of reaction and the name of the product formed?
▶️ Answer/Explanation
Ans: C
1. The reaction between ethanal (CH₃CHO) and HCN is a nucleophilic addition reaction, where the CN⁻ ion attacks the carbonyl carbon.
2. The product formed is 2-hydroxypropanenitrile (CH₃CH(OH)CN), which is classified as a hydroxynitrile.
3. This reaction is characteristic of aldehydes and ketones, where the C=O bond is broken and new bonds form with the nucleophile (CN⁻) and proton (H⁺).
4. Therefore, the correct identification is nucleophilic addition producing a hydroxynitrile, corresponding to option C.
Topic: 19.1
The structure of compound X is shown.
What is produced when X is heated with NaOH(aq)?
▶️ Answer/Explanation
Ans: C
1. Reaction Type: Compound X undergoes an aldol condensation when heated with NaOH(aq), as it contains both an aldehyde group and an α-hydrogen.
2. Mechanism:
- NaOH deprotonates the α-hydrogen, forming an enolate ion.
- The enolate attacks the carbonyl carbon of another molecule, leading to a β-hydroxy aldehyde intermediate.
- Dehydration occurs under heating, forming an α,β-unsaturated aldehyde.
3. Product Identification: The product is an α,β-unsaturated aldehyde (structure C), resulting from the loss of water from the aldol addition product.
Thus, the correct answer is C.
Topic: 22.1
The infrared spectrum of compound L is shown.
What is the structure of L?
▶️ Answer/Explanation
Ans: C
To identify compound L from its IR spectrum: 1. **Key Absorptions**: – Broad peak at ~3300 cm-1 indicates O-H bonds (alcohols). – No sharp peak at ~1700 cm-1 (ruling out C=O in A, B, D). 2. **Structure Analysis**: – Only **C (\(HOCH_2CH(OH)CH_2OH\))** fits: – Contains multiple -OH groups (consistent with broad absorption). – No C=O or -COOH groups (absent in spectrum). Thus, C is the correct structure.
Topic: 22.2
In the mass spectrum of compound J, the ratio of the height of the M+1 ion peak to the height of the M+ ion peak is 4 : 91. Compound J forms a carboxylic acid when heated with acidified K₂Cr₂O₇. What is compound J?
▶️ Answer/Explanation
Ans: A
1. M+1 Peak Ratio (4:91): – The ratio suggests the presence of 4 carbon atoms (since each \(^{13}\text{C}\) contributes ~1.1% to M+1, and \(4 \times 1.1\% \approx 4.4\%\), matching the 4:91 ratio). – This eliminates options C and D (which have 3 carbons).
2. Oxidation to Carboxylic Acid: – Butanal (A) is an aldehyde and oxidizes to butanoic acid with K₂Cr₂O₇. – Butanone (B) is a ketone and resists further oxidation under these conditions.
3. Conclusion: – Only butanal (A) fits both the mass spectrometry data and the oxidation behavior.