UAS 23-24 FAD1018 Basic Chemistry II

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UNIVERSITI MALAYA UNIVERSITI MALAYA

PEPERIKSAAN ASASI SAINS HAYAT / FIZIKAL EXAMINATION FOR FOUNDATION IN LIFE / PHYSICAL SCIENCES

SESI AKADEMIK 2023/2024 : SEMESTER 2 ACADEMIC SESSION 2023/2024 : SEMESTER 2

FAD 1018 : Kimia Asas 2 Basic Chemistry 2

Mei/Jun 2024 Masa : 3 jam May/June 2024 Time : 3 hours

ARAHAN KEPADA CALON : INSTRUCTIONS TO CANDIDATES :

Calon dikehendaki menjawab semua soalan. Candidate is required to answer all questions.

(Kertas soalan ini mengandungi 15 soalan dalam 15 halaman yang dicetak) (This question paper consists of 15 questions on 15 printed pages)

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FAD 1018

JISIM ATOM RELATIF UNSUR-UNSUR TERPILIH TABLE OF RELATIVE ATOMIC MASS

[Two-column table with headers:] | Nama Name | Simbol Symbol | No. Atom Atomic No. | J.A.R R.A.M | | Nama Name | Simbol Symbol | No. Atom Atomic No. | J.A.R R.A.M |

[Left column rows:] Aluminum | Al | 13 | 26.982 Antimony | Sb | 51 | 121.76 Argon | Ar | 18 | 39.958 Arsenic | As | 33 | 74.922 Astatine | At | 85 | 209.99 Barium | Ba | 56 | 137.33 Beryllium | Be | 4 | 9.0122 Bismuth | Bi | 83 | 208.98 Boron | B | 5 | 10.81 Bromine | Br | 35 | 79.904 Cadmium | Cd | 48 | 112.41 Calcium | Ca | 20 | 40.078 Carbon | C | 6 | 12.011 Cesium | Cs | 55 | 132.91 Chlorine | Cl | 17 | 35.45 Chromium | Cr | 24 | 51.996 Cobalt | Co | 27 | 58.933 Copper | Cu | 29 | 63.546 Hydrogen | H | 1 | 1.0080 Indium | In | 49 | 114.8 Iodine | I | 53 | 126.904 Iridium | Ir | 77 | 192.22 Iron | Fe | 26 | 55.845 Krypton | Kr | 36 | 83.798 Lead | Pb | 82 | 207.2 Lithium | Li | 3 | 6.94 Molybdenum | Mo | 42 | 95.95 Neon | Ne | 10 | 20.180 Nickel | Ni | 28 | 58.693 Niobium | Nb | 41 | 92.906 Nitrogen | N | 7 | 14.007 Osmium | Os | 76 | 190.23

[Right column rows:] Oxygen | O | 8 | 15.999 Palladium | Pd | 46 | 106.42 Phosphorus | P | 15 | 30.974 Platinum | Pt | 78 | 195.08 Polonium | Po | 84 | 208.98 Potassium | K | 19 | 39.098 Radium | Ra | 88 | 226.02 Radon | Rn | 86 | 222.0 Rhenium | Re | 75 | 186.21 Rhodium | Rh | 45 | 102.91 Rubidium | Rb | 37 | 85.468 Ruthenium | Ru | 44 | 101.07 Scandium | Sc | 21 | 44.956 Selenium | Se | 34 | 78.971 Silicon | Si | 14 | 28.085 Silver | Ag | 47 | 107.87 Sodium | Na | 11 | 22.990 Strontium | Sr | 38 | 87.62 Sulfur | S | 16 | 32.06 Tantalum | Ta | 73 | 180.95 Technetium | Tc | 43 | [97] Tellurium | Te | 52 | 127.60 Thallium | Tl | 81 | 204.38 Tin | Sn | 50 | 118.71 Titanium | Ti | 22 | 47.867 Tungsten | W | 74 | 183.84 Vanadium | V | 23 | 50.942 Xenon | Xe | 54 | 131.29 Yttrium | Y | 39 | 88.906 Zinc | Zn | 30 | 65.38 Zirconium | Zr | 40 | 91.224

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FAD 1018

  1. Nyatakan tiga sifat koligatif.

State three colligative properties.

(3 markah/marks)

  1. Lakarkan satu sel galvani yang terdiri daripada elektrod zink (Zn) dan elektrod kuprum (Cu) yang terendam dalam larutan ion logam masing-masing. Labelkan terminal anod dan terminal katod dalam lakaran anda.

Sketch a galvanic cell comprising a zinc electrode (Zn) and a copper electrode (Cu) immersed in their corresponding metal ion solutions. Label the anode and cathode terminals in your sketch.

(3 markah/marks)

  1. Lukis formula struktur untuk butana-2,3-diol. Dengan menggunakan unjuran Fischer, lukis diastereomer yang tidak aktif optik untuk molekul ini. Tunjukkan satah simetri dalaman, jika ada.

Draw the structural formula for butane-2,3-diol. By using Fischer projection, draw an optically inactive diastereomer for this molecule. Denote the internal plane of symmetry, if any.

(3 markah/marks)

  1. Tindak balas ozonolisis bagi 2-metilbut-2-ena menghasilkan dua sebatian, A dan B. Sebatian A telah mencapai keadaan pengoksidaan paling tinggi namun sebatian B masih boleh mengalami tindak balas pengoksidaan. Lukis struktur molekul bagi semua sebatian utama A – E dalam skema tindak balas berikut:

[Reaction scheme diagram. Left: skeletal structure of 2-methylbut-2-ene with an explicit H on the right-hand vinylic carbon, labeled “2-metilbut-2-ena”. Horizontal arrow right labeled “1. O$_3$ / 2. Zn/H$_2$O” leading to “A + B”. Vertical arrow down from A to C. Vertical arrow down from B to D labeled “KMnO$_4$, H$^+$, $\Delta$”. Vertical arrow down from D to E labeled “1. LiAlH$_4$ berlebihan / 2. H$_3$O$^+$”. Structure C is a benzene ring with O$_2$N at the top position, H$_2$N—HN— at the left position, and NO$_2$ at the right position.]

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FAD 1018

The ozonolysis reaction of 2-methylbut-2-ene produces two compounds, A and B. Compound A has reached the highest oxidation state but compound B can still undergo oxidation reactions. Draw the molecular structures of all the main compounds A – E in the following reaction scheme:

[Reaction scheme diagram. Left: skeletal structure of 2-methylbut-2-ene with an explicit H on the right-hand vinylic carbon, labeled “2-methylbut-2-ene”. Horizontal arrow right labeled “1. O$_3$ / 2. Zn/H$_2$O” leading to “A + B”. Vertical arrow down from A to C. Vertical arrow down from B to D labeled “KMnO$_4$, H$^+$, $\Delta$”. Vertical arrow down from D to E labeled “1. excess LiAlH$_4$ / 2. H$_3$O$^+$”. Structure C is a benzene ring with O$_2$N at the top position, H$_2$N—HN— at the left position, and NO$_2$ at the right position.]

(5 markah/marks)

  1. Berikan nama bagi molekul di bawah dan lukis produk yang terbentuk apabila ia beroligomer dengan 3 unit berulang. Nyatakan jenis tindak balas yang terlibat.

[Chemical structure diagram. From left to right: HO–C(=O)–CH$_2$–CH(NH$_2$)–C(=O)–OH.]

Give the name for the molecule below and draw the product formed when it oligomerizes with 3 repeating units. State the type of reaction involved.

[Chemical structure diagram identical to the one above: HO–C(=O)–CH$_2$–CH(NH$_2$)–C(=O)–OH.]

(3 markah/marks)

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FAD 1018

(b) Dalam bidang perubatan, aspirin (asid 2-etanoiloksibenzoik) merupakan ubat anti radang bukan steroid yang digunakan untuk mengurangkan kesakitan. Di dalam badan manusia, aspirin mengalami hidrolisis untuk menghasilkan asid 2-hidroksibenzoik dan asid etanoik seperti ditunjukkan dalam persamaan di bawah.

[Chemical structure diagram showing aspirin (with H₃C-C(=O)-O- attached to a benzene ring with -COOH at ortho position) reacting with H-O-H to give 2-hydroxybenzoic acid (benzene ring with -OH and -COOH at ortho positions) plus ethanoic acid (H₃C-C(=O)-OH). The ΔH = +46.024 kJ mol⁻¹ is shown below the products.]

(i) Hitung tenaga pengaktifan, $E_a$ (dalam kJ mol⁻¹) untuk tindak balas tersebut sekiranya pemalar kadar adalah $5.0 \times 10^{-5}\ \text{s}^{-1}$ pada 25 °C dan $1.5 \times 10^{-3}\ \text{s}^{-1}$ pada 60 °C.

(ii) Lakar dan label gambar rajah profil tenaga untuk tindak balas tersebut.

(iii) Hitung tenaga pengaktifan untuk tindak balas ke belakang.

In the medical field, aspirin (2-ethanoyloxybenzoic acid) is a nonsteroidal anti-inflammatory drug used to reduce pain. In human body it undergoes hydrolysis to produce 2-hydroxybenzoic acid and ethanoic acid as shown in the equation below.

[Chemical structure diagram showing aspirin (with H₃C-C(=O)-O- attached to a benzene ring with -COOH at ortho position) reacting with H-O-H to give 2-hydroxybenzoic acid (benzene ring with -OH and -COOH at ortho positions) plus ethanoic acid (H₃C-C(=O)-OH). The $\Delta H = +46.024\ \text{kJ mol}^{-1}$ is shown below the products.]

(i) Calculate the activation energy, $E_a$ (in kJ mol⁻¹) for the reaction, if the rate constant is $5.0 \times 10^{-5}\ \text{s}^{-1}$ at 25 °C and $1.5 \times 10^{-3}\ \text{s}^{-1}$ at 60 °C.

(ii) Sketch and label the energy profile diagram for the reaction.

(iii) Calculate the activation energy for the reverse reaction.

(9 markah/marks)

  1. (a) Larutan etilamina, CH₃CH₂NH₂ digunakan secara meluas dalam industri kimia dan sintesis organik. Etilamina merupakan suatu komponen asap tembakau. Ciri-ciri berbahaya utama etilamina adalah berpotensi menghakis dan mudah terbakar.

(i) Hitung pH bagi 0.5 M larutan etilamina.

(ii) Hitung pH apabila 0.1 M CH₃CH₂NH₃Cl ditambah kepada 0.5 M larutan etilamina dan nyatakan jenis larutan yang terhasil. ($K_b = 5.6 \times 10^{-4}$ pada suhu 25 °C)

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FAD 1018

Ethylamine, CH₃CH₂NH₂, is widely used in chemical industry and organic synthesis. It is a component of tobacco smoke. The main hazardous properties of ethylamine are its corrosive potential and flammability.

(i) Calculate the pH of 0.50 M ethylamine solution.

(ii) Calculate the pH when 0.10 M CH₃CH₂NH₃Cl is added to the 0.50 M ethylamine solution and state the type of solution produced. ($K_b = 5.6 \times 10^{-4}$ at 25 °C)

(9 markah/marks)

(b) Dalam suatu eksperimen keseimbangan kimia yang mengkaji kesan kepekatan, ion kompleks ferum(III) tiosianat, Fe(SCN)²⁺ terbentuk apabila ion ferum(III), Fe³⁺ bertindak balas dengan ion tiosianat, SCN⁻. Persamaan seimbang bagi tindak balas adalah seperti ditunjukkan di bawah.

$$\text{Fe}^{3+}(\text{ak}) + \text{SCN}^-(\text{ak}) \rightleftharpoons \text{Fe(SCN)}^{2+}(\text{ak})$$ kekuningan          merah darah

Apabila natrium hidroksida, NaOH, ditambah ke dalam larutan, ia menghasilkan pembentukan mendakan kekuningan disebabkan oleh pemendakan ferum(III) hidroksida, Fe(OH)₃. Jika kepekatan ion OH⁻ di dalam larutan tepu ialah $4.571 \times 10^{-5}$ M, dan dengan mengandaikan bahawa Fe(OH)₃ tercerai sepenuhnya dalam air tanpa terdapat sebarang keseimbangan serentak lain yang melibatkan ion Fe²⁺ atau OH⁻, hitung nilai $K_{sp}$ bagi sebatian ini.

In an experiment on chemical equilibrium investigating the effect of concentration, the iron(III) thiocyanate complex ion, Fe(SCN)²⁺ is formed when the iron (III) ion, Fe³⁺ reacts with the thiocyanate ion, SCN⁻. The balanced equation for the reaction is as shown below.

$$\text{Fe}^{3+}(\text{aq}) + \text{SCN}^-(\text{aq}) \rightleftharpoons \text{Fe(SCN)}^{2+}(\text{aq})$$ yellowish          blood red

When sodium hydroxide, NaOH is added to the solution, it results in the formation of a yellowish precipitate due to the precipitation of iron(III) hydroxide, Fe(OH)₃. If the concentration of OH⁻ ions in a saturated solution is $4.571 \times 10^{-5}$ M, and assuming that Fe(OH)₃ dissociates completely in water without any other simultaneous equilibria involving the Fe²⁺ or OH⁻ ions, calculate the value of $K_{sp}$ for this compound.

(6 markah/marks)

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FAD 1018

  1. (a) Alkohol P, dengan formula molekul $\mathrm{C_5H_{12}O}$ mengandungi hanya satu karbon kiral, menjalani tindak balas kimia seperti yang ditunjukkan dalam gambar rajah berikut.

[Reaction flow diagram with rectangular boxes connected by arrows. Top row: box "Sebatian U" ← (arrow labeled "$\mathrm{I_2}$, NaOH") — box "Sebatian T, $\mathrm{C_5H_{10}O}$" — box "Sebatian W" with "+ gas hidrogen" below it. Middle: arrow from T down to P labeled "$\mathrm{KMnO_4}$, $\mathrm{H_2SO_4}$" with Δ below; arrow from P up to W labeled "Na". Center box: "Sebatian P, $\mathrm{C_5H_{12}O}$". Left of P: box "Sebatian S, $\mathrm{C_4H_8O}$" → (arrow labeled "(1) $\mathrm{CH_3MgI}$ / (2) $\mathrm{H_3O^+}$") → P. Right of P: arrow labeled "$\mathrm{H_2SO_4}$, $\mathrm{CH_3COCl}$ / Δ" → box "Sebatian V". Below P: arrow down labeled "Δ | $\mathrm{H_2SO_4}$ pekat" → box "Campuran alkena Q". Bottom left: box "Sebatian R" → (arrow labeled "KOH, aseton") → P.]

(i) Tulis formula struktur bagi sebatian P hingga W. (ii) Tandakan pusat kiral pada struktur P menggunakan asterisk. (iii) Lukis struktur P dengan menggunakan notasi 3D. (iv) Lukis enantiomer bagi P menggunakan unjuran Fischer.

Alcohol P, with the molecular formula $\mathrm{C_5H_{12}O}$ contains only one chiral carbon, undergoes chemical reactions as depicted in the following diagram.

[Reaction flow diagram with rectangular boxes connected by arrows. Top row: box "Compound U" ← (arrow labeled "$\mathrm{I_2}$, NaOH") — box "Compound T, $\mathrm{C_5H_{10}O}$" — box "Compound W" with "+ hydrogen gas" below it. Middle: arrow from T down to P labeled "$\mathrm{KMnO_4}$, $\mathrm{H_2SO_4}$" with Δ below; arrow from P up to W labeled "Na". Center box: "Compound P, $\mathrm{C_5H_{12}O}$". Left of P: box "Compound S, $\mathrm{C_4H_8O}$" → (arrow labeled "(1) $\mathrm{CH_3MgI}$ / (2) $\mathrm{H_3O^+}$") → P. Right of P: arrow labeled "$\mathrm{H_2SO_4}$, $\mathrm{CH_3COCl}$ / Δ" → box "Compound V". Below P: arrow down labeled "Δ | Concentrated $\mathrm{H_2SO_4}$" → box "Mixture of alkenes Q". Bottom left: box "Compound R" → (arrow labeled "KOH, acetone") → P.]

(i) Write the structural formulae for compounds P to W. (ii) Denote the chiral centre on P with an asterisk. (iii) Draw the structure of P using 3D notation. (iv) Draw the enantiomer of P using Fischer projection.

(12 markah/marks)

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FAD 1018

(b) (i) Beri satu contoh polimer sintetik dan polimer semula jadi. (ii) Lukis formula struktur bagi monomer polistirena.

(i) Give an example of one synthetic polymer and one natural polymer. (ii) Draw the structural formula for the monomer of polystyrene.

(3 markah/marks)

  1. (a) Sebatian organik X, $\mathrm{C_8H_8O}$ diperolehi apabila sebatian pemula benzena menjalani tindak balas pengasilan Friedel-Crafts. Sebatian X membentuk mendakan kuning-oren dengan 2,4-dinitrofenilhidrazina dan boleh diturunkan kepada sebatian Y oleh litium aluminium hidrida. Kedua-dua X dan Y bertindak balas dengan iodin beralkali dan menghasilkan mendakan kuning pucat. Berdasarkan kenyataan ini:

(i) Jelaskan pemerhatian bagi setiap uji kaji dan kenal pasti jenis dan kelas bagi sebatian X dan Y. (ii) Lukiskan formula struktur untuk sebatian X dan Y. (iii) Tulis persamaan kimia untuk penurunan X kepada Y oleh litium aluminium hidrida. (iv) Tulis persamaan kimia untuk menunjukkan bagaimana sebatian X diperolehi daripada tindak balas pengasilan Friedel-Crafts benzena.

An organic compound X, $\mathrm{C_8H_8O}$ is obtained when the starting compound benzene undergoes Friedel-Crafts acylation. Compound X forms a yellow-orange precipitate with 2,4-dinitrophenylhydrazine and can be reduced to compound Y by lithium aluminum hydride. Both X and Y react with iodine alkaline yield a pale-yellow precipitate. Based on these statements:

(i) Explain the observations of each of the tests and identify the type and class of compound X and Y. (ii) Draw the structural formulae for compounds X and Y. (iii) Write the chemical equation for the reduction of X to Y by lithium aluminum hydride. (iv) Write the chemical equation to show how compound X is obtained from Friedel-Crafts acylation of benzene.

(9 markah/marks)

(b) (i) Alizarin Kuning ialah pewarna yang digunakan sebagai penunjuk pH. Ia dihasilkan oleh gandingan azo asid 2-hidroksibenzoik dan p-nitroanilina. Tunjukkan semua langkah tindak balas, reagen dan keadaan yang diperlukan bagi sintesis sebatian ini.

Alizarine Yellow is a dye used as a pH indicator. It is produced by azo coupling of 2-hydroxybenzoic acid and p-nitroaniline. Show all the reaction steps, reagents and conditions required to synthesise this compound.

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FAD 1018

(ii) Molekul di bawah adalah asid amino alfa yang boleh didapati dalam makanan seperti telur, protein soya, rumpai laut, ayam belanda, ayam, kambing, keju, dan ikan. Asid amino wujud dalam bentuk ionik yang berbeza bergantung kepada pH larutan.

[Chemical structure diagram: A central alpha-carbon with four substituents arranged in tetrahedral geometry. A hashed wedge bond points upward to a hydrogen atom. A solid wedge bond points downward to an $\mathrm{NH_2}$ group. A standard bond points to the left to an isopropyl group $\mathrm{-CH(CH_3)_2}$. A standard bond points to the right to a carboxylic acid group $\mathrm{-COOH}$ (carbon double-bonded to $\mathrm{O}$ and single-bonded to $\mathrm{OH}$).]

Lukiskan struktur ionik produk yang terbentuk apabila asid amino di atas bertindak balas dengan reagen berikut dalam keadaan akueus: I. NaOH II. HCl

The molecule below is an alpha amino acid that can be found in foods such as eggs, soy protein, seaweed, turkey, chicken, lamb, cheese, and fish. The amino acid exists in different ionic forms depending on the pH of the solution.

[Chemical structure diagram: Identical to the diagram above — a central alpha-carbon with a hashed wedge to $\mathrm{H}$ (up), a solid wedge to $\mathrm{NH_2}$ (down), a bond to an isopropyl group $\mathrm{-CH(CH_3)_2}$ (left), and a bond to a carboxylic acid group $\mathrm{-COOH}$ (right).]

Draw the ionic structure of the product formed when the above amino acid is reacted with the following reagents in aqueous conditions: I. NaOH II. HCl

(6 markah/marks)

TAMAT END

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Verbatim transcription via Kimi K2.6 vision subagents.