Topics in the June 2013 Exam Paper for CHEM1611

Click on the links for resources on each topic.

2013-J-2: Atomic StructureChemical Bonding

2013-J-3: Chemical Bonding

2013-J-4: Chemical BondingThe Shapes of Molecules

2013-J-5: Atomic Structure

2013-J-6: AlkenesAlcohols, Phenols, Ethers and ThiolsCarboxylic Acids and DerivativesAmines

2013-J-7: Introduction to Organic ChemistryStereochemistry

2013-J-8: AlkenesAlcohols, Phenols, Ethers and ThiolsCarboxylic Acids and Derivatives

2013-J-9: Carbohydrates

2013-J-10: Amino Acids, Peptides and Proteins

2013-J-11: Amino Acids, Peptides and Proteins

2013-J-12: DNA and Nucleic Acids

June 2013

2216(a) THE UNIVERSITY OF SYDNEY

CHEM1611 - CHEMISTRY 1A (PHARMACY) FIRST SEMESTER EXAMINATION

CONFIDENTIAL JUNE 2013 TIME ALLOWED: THREE HOURS GIVE THE FOLLOWING INFORMATION IN BLOCK LETTERS

FAMILY NAME

SID NUMBER

OTHER NAMES

TABLE NUMBER

• All questions are to be attempted. There

are 19 pages of examinable material.

• Complete the examination paper in INK.

• Read each question carefully. Report the appropriate answer and show all relevant working in the space provided.

• The total score for this paper is 100. The possible score per page is shown in the adjacent tables.

• Each new short answer question begins with a •.

• Only non-programmable, University- approved calculators may be used.

• Students are warned that credit may not be given, even for a correct answer, where there is insufficient evidence of the working required to obtain the solution.

• Numerical values required for any question, standard electrode reduction potentials, a Periodic Table and some useful formulas may be found on the separate data sheets.

• Pages 14, 16, 22 and 24 are for rough work only.

OFFICIAL USE ONLY Multiple choice section

Marks Pages Max Gained

2-9 30

Short answer section

Marks Page Max Gained Marker

10 4

11 5

12 10

13 5

15 11

17 6

18 6

19 6

20 6

21 6

23 5

Total 70

Check Total

CHEM1611 2013-J-2 June 2013

• In the spaces provided, briefly explain the meaning of the following terms. Marks

4

Effective nuclear charge

Atomic emission spectrum

Ionic bonding

Core electrons

THE REMAINDER OF THIS PAGE IS FOR ROUGH WORKING ONLY.

CHEM1611 2013-J-3 June 2013

• Silicon and carbon are both in Group 14 and form dioxides. Carbon dioxide is a gas at room temperature while silicon dioxide (sand) is a solid with a high melting point. Describe the bonding in these two materials and explain the differences in properties they show.

Marks 3

• Complete the following nuclear equations by filling in the missing particle.

2

147 N + 11p → 116C +

116C → + 0 +

1e

THE REMAINDER OF THIS PAGE IS FOR ROUGH WORKING ONLY.

CHEM1611 2013-J-4 June 2013

• Draw the Lewis structure of the acetate ion (CH3COO–) showing all appropriate resonance structures.

Marks 10

Indicate the hybridisation, molecular geometry and approximate bond angle about each of the carbon atoms in the acetate ion.

–CH3 –COO–

Hybridisation of C

Molecular geometry about C

Approximate bond angles about C

The actual structure of the acetate ion is a weighted combination of all resonance structures. Sketch the σ-bond framework of the acetate ion and indicate the p-orbitals that are involved in the π-bonding of the acetate ion.

How many electrons are involved with the π-bonding?

What is the hybridisation of the oxygen atoms in the acetate ion?

CHEM1611 2013-J-4 June 2013

• The yellow light emitted from an excited sodium atom has a wavelength of 590 nm. What is the energy of one photon of this light and one mole of photons? Specify appropriate units with your answers.

Marks 5

Energy of one photon: of 1 mol of photons:

The yellow light is associated with the longest wavelength transition as the atom returns to the ground state electron configuration. Fill in the following energy level diagram for sodium and indicate the transition associated with the emission of yellow light.

A quantum mechanical model of an atom can explain the emission spectrum of sodium, but the Bohr model of the atom cannot. Why?

CHEM1611 2013-J-6 June 2013

• Complete the following table. Make sure you complete the name of the starting material where indicated.

Marks 11

STARTING MATERIAL REAGENTS/ CONDITIONS

CONSTITUTIONAL FORMULA(S) OF MAJOR ORGANIC PRODUCT(S)

HBr / CCl4 (solvent)

Name:

1. Mg / dry ether

2. CO2

O

O

3 M NaOH

excess (CH3)2NH

Name:

NH2Br

2 M HCl

Cl

O

CHEM1611 2013-J-7 June 2013

• Oseltamivir, marketed under the trade name Tamiflu, is an antiviral drug, which may slow the spread of influenza (flu) virus between cells in the body by stopping the virus from chemically cutting ties with its host cell.

Marks 6

Give the molecular formula of Tamiflu.

List the functional groups present in Tamiflu.

How many stereogenic centres are there in Tamiflu?

How many possible stereoisomers can exist for Tamiflu?

Add the NH2 and H groups to the stereogenic centre indicated below to give the (R)-configuration of that centre.

Tamiflu

CHEM1611 2013-J-8 June 2013

• Show clearly the reagents you would use to carry out the following chemical conversions. Note that more than one step is required and you should indicate all necessary steps and the constitutional formulas of any intermediate compounds.

Marks 6

!

!

CHEM1611 2013-J-9 June 2013

• Consider the following two monosaccharides, (A) and (B).

β-D-altropyranose α-D-xylofuranose

Draw Fischer projections of the open chain forms of (A) and (B).

Marks 6

(A)

(B)

Draw the major organic product of the reaction of D-altropyranose with the following reagents.

Draw the Haworth stereoformula of a non-reducing disaccharide formed from (A) and (B).

CHEM1611 2013-J-10 June 2013

• The amino acid, asparagine, was isolated from asparagus juice in 1806. The uncharged form, Y, is given below.

Draw the constitutional formula of the product(s) formed in the reaction of Y with the following reagents.

Marks 6

Cold, dilute hydrochloric acid

Cold, dilute sodium hydroxide

Hot, 6 M hydrochloric acid

Hot, 6 M sodium hydroxide

THE REMAINDER OF THIS PAGE IS FOR ROUGH WORKING ONLY.

CHEM1611 2013-J-11 June 2013

• Alanine (R = CH3) and lysine (R = CH2CH2CH2CH2NH2) are two common amino acids. Using ala and lys to represent the two amino acids, represent all constitutional isomers of the tripeptide formed from one ala and two lys units.

Marks 6

Comment, giving your reason(s), on whether the tripeptide(s) will be acidic, neutral or basic in character.

The pKa values of lysine are 1.82 (α-COOH), 8.95 (α-NH3⊕) and 10.53 (side chain). What is the value of the isoelectric point of lysine?

pI =

Draw the Fischer projection of the zwitterionic form of lysine.

CHEM1611 2013-J-12 June 2013

• Consider the following structure. Marks

5

H

N

HO

HOOP

OOO

P OO

O

HHO

HO

P OO

O

HHOH

HO

NH

O

O

N

NN

NNH2

N

N

NH2

O

Is it a fragment of DNA or RNA? Give two reasons.

Clearly identify on the above structure one example of each of the following subunits.

nucleic base

nucleoside

nucleotide

2216(b) June 2013 CHEM1611 - CHEMISTRY 1A (PHARMACY)

DATA SHEET Physical constants Avogadro constant, NA = 6.022 × 1023 mol–1 Faraday constant, F = 96485 C mol–1

Planck constant, h = 6.626 × 10–34 J s

Speed of light in vacuum, c = 2.998 × 108 m s–1

Rydberg constant, ER = 2.18 × 10–18 J

Boltzmann constant, kB = 1.381 × 10–23 J K–1

Permittivity of a vacuum, ε0 = 8.854 × 10–12 C2 J–1 m–1

Gas constant, R = 8.314 J K–1 mol–1 = 0.08206 L atm K–1 mol–1

Charge of electron, e = 1.602 × 10–19 C

Mass of electron, me = 9.1094 × 10–31 kg

Mass of proton, mp = 1.6726 × 10–27 kg

Mass of neutron, mn = 1.6749 × 10–27 kg

Properties of matter Volume of 1 mole of ideal gas at 1 atm and 25 °C = 24.5 L

Volume of 1 mole of ideal gas at 1 atm and 0 °C = 22.4 L Density of water at 298 K = 0.997 g cm–3 Conversion factors 1 atm = 760 mmHg = 101.3 kPa = 1.013 bar 1 Ci = 3.70 × 1010 Bq

0 °C = 273 K 1 Hz = 1 s–1 1 L = 10–3 m3 1 tonne = 103 kg

1 Å = 10–10 m 1 W = 1 J s–1

1 eV = 1.602 × 10–19 J

Decimal fractions Decimal multiples Fraction Prefix Symbol Multiple Prefix Symbol

10–3 milli m 103 kilo k 10–6 micro µ 106 mega M 10–9 nano n 109 giga G 10–12 pico p 1012 tera T

2216(b) June 2013 CHEM1611 - CHEMISTRY 1A (PHARMACY)

Standard Reduction Potentials, E° Reaction E° / V Co3+(aq) + e– → Co2+(aq) +1.82 Ce4+(aq) + e– → Ce3+(aq) +1.72 MnO4

–(aq) + 8H+(aq) + 5e– → Mn2+(aq) + 4H2O +1.51 Au3+(aq) + 3e– → Au(s) +1.50 Cl2 + 2e– → 2Cl–(aq) +1.36 O2 + 4H+(aq) + 4e– → 2H2O +1.23 Pt2+(aq) + 2e– → Pt(s) +1.18 MnO2(s) + 4H+(aq) + e– → Mn3+ + 2H2O +0.96 NO3

–(aq) + 4H+(aq) + 3e– → NO(g) + 2H2O +0.96 Pd2+(aq) + 2e– → Pd(s) +0.92 NO3

–(aq) + 10H+(aq) + 8e– → NH4+(aq) + 3H2O +0.88

Ag+(aq) + e– → Ag(s) +0.80 Fe3+(aq) + e– → Fe2+(aq) +0.77 Cu+(aq) + e– → Cu(s) +0.53 Cu2+(aq) + 2e– → Cu(s) +0.34 BiO+(aq) + 2H+(aq) + 3e– → Bi(s) + H2O +0.32 Sn4+(aq) + 2e– → Sn2+(aq) +0.15 2H+(aq) + 2e– → H2(g) 0 (by definition) Fe3+(aq) + 3e– → Fe(s) –0.04 Pb2+(aq) + 2e– → Pb(s) –0.126 Sn2+(aq) + 2e– → Sn(s) –0.136 Ni2+(aq) + 2e– → Ni(s) –0.24 Co2+(aq) + 2e– → Co(s) –0.28 Cd2+(aq) + 2e– → Cd(s) –0.40 Fe2+(aq) + 2e– → Fe(s) –0.44 Cr3+(aq) + 3e– → Cr(s) –0.74 Zn2+(aq) + 2e– → Zn(s) –0.76 2H2O + 2e– → H2(g) + 2OH–(aq) –0.83 Cr2+(aq) + 2e– → Cr(s) –0.89 Al3+(aq) + 3e– → Al(s) –1.68 Sc3+(aq) + 3e– → Sc(s) –2.09 Mg2+(aq) + 2e– → Mg(s) –2.36 Na+(aq) + e– → Na(s) –2.71 Ca2+(aq) + 2e– → Ca(s) –2.87 Li+(aq) + e– → Li(s) –3.04

2216(b) June 2013 CHEM1611 - CHEMISTRY 1A (PHARMACY)

Useful formulas

Quantum Chemistry

E = hν = hc/λ

λ = h/mv

E = –Z2ER(1/n2)

Δx⋅Δ(mv) ≥ h/4π

q = 4πr2 × 5.67 × 10–8 × T4

T λ = 2.898 × 106 K nm

Electrochemistry

ΔG° = –nFE°

Moles of e– = It/F

E = E° – (RT/nF) × lnQ

E° = (RT/nF) × lnK

E = E° – 0.0592n

logQ (at 25 °C)

Acids and Bases

pH = –log[H+] pKw = pH + pOH = 14.00

pKw = pKa + pKb = 14.00

pH = pKa + log{[A–] / [HA]}

Gas Laws

PV = nRT (P + n2a/V2)(V – nb) = nRT

Ek = ½mv2

Radioactivity

t½ = ln2/λ

A = λN

ln(N0/Nt) = λt 14C age = 8033 ln(A0/At) years

Kinetics

t½ = ln2/k

k = Ae–Ea/RT

ln[A] = ln[A]o – kt

2

1 1 2

1 1ln = - ( )ak Ek R T T

Colligative Properties & Solutions

Π = cRT

Psolution = Xsolvent × P°solvent

c = kp

ΔTf = Kfm

ΔTb = Kbm

Thermodynamics & Equilibrium

ΔG° = ΔH° – TΔS°

ΔG = ΔG° + RT lnQ

ΔG° = –RT lnK

ΔunivS° = R lnK

Kp = Kc100( )

nRT Δ

Miscellaneous

A = –log0

II

A = εcl

E = –A2

04erπε

NA

Mathematics

If ax2 + bx + c = 0, then x = 2b b 4ac

2a− ± −

ln x = 2.303 log x

Area of circle = πr2

Surface area of sphere = 4πr2

PER

IOD

IC T

AB

LE

OF T

HE

EL

EM

EN

TS

1

2 3

4 5

6 7

8 9

10 11

12 13

14 15

16 17

18

1 H

YD

RO

GE

N

H

1.008

2 H

EL

IUM

He

4.003 3

LIT

HIU

M

Li

6.941

4 B

ER

YL

LIU

M

Be

9.012

5 B

OR

ON

B

10.81

6 C

AR

BO

N

C

12.01

7 N

ITR

OG

EN

N

14.01

8 O

XY

GE

N

O

16.00

9 FL

UO

RIN

E

F 19.00

10 N

EO

N

Ne

20.18 11

SOD

IUM

Na

22.99

12 M

AG

NE

SIUM

Mg

24.31

13 A

LU

MIN

IUM

Al

26.98

14 SIL

ICO

N

Si 28.09

15 PH

OSPH

OR

US

P 30.97

16 SU

LFU

R

S 32.07

17 C

HL

OR

INE

Cl

35.45

18 A

RG

ON

Ar

39.95 19

POT

ASSIU

M

K

39.10

20 C

AL

CIU

M

Ca

40.08

21 SC

AN

DIU

M

Sc 44.96

22 T

ITA

NIU

M

Ti

47.88

23 V

AN

AD

IUM

V

50.94

24 C

HR

OM

IUM

Cr

52.00

25 M

AN

GA

NE

SE

Mn

54.94

26 IR

ON

Fe 55.85

27 C

OB

AL

T

Co

58.93

28 N

ICK

EL

Ni

58.69

29 C

OPPE

R

Cu

63.55

30 Z

INC

Zn

65.39

31 G

AL

LIU

M

Ga

69.72

32 G

ER

MA

NIU

M

Ge

72.59

33 A

RSE

NIC

As

74.92

34 SE

LE

NIU

M

Se 78.96

35 B

RO

MIN

E

Br

79.90

36 K

RY

PTO

N

Kr

83.80 37

RU

BID

IUM

Rb

85.47

38 ST

RO

NT

IUM

Sr 87.62

39 Y

TT

RIU

M Y

88.91

40 Z

IRC

ON

IUM

Zr

91.22

41 N

IOB

IUM

Nb

92.91

42 M

OL

YB

DE

NU

M

Mo

95.94

43 T

EC

HN

ET

IUM

Tc

[98.91]

44 R

UT

HE

NIU

M

Ru

101.07

45 R

HO

DIU

M

Rh

102.91

46 PA

LL

AD

IUM

Pd 106.4

47 SIL

VE

R

Ag

107.87

48 C

AD

MIU

M

Cd

112.40

49 IN

DIU

M

In 114.82

50 T

IN

Sn 118.69

51 A

NT

IMO

NY

Sb 121.75

52 T

EL

LU

RIU

M

Te

127.60

53 IO

DIN

E

I 126.90

54 X

EN

ON

Xe

131.30 55

CA

ESIU

M

Cs

132.91

56 B

AR

IUM

Ba

137.34

57-71 72

HA

FNIU

M

Hf

178.49

73 T

AN

TA

LU

M

Ta

180.95

74 T

UN

GST

EN

W

183.85

75 R

HE

NIU

M

Re

186.2

76 O

SMIU

M

Os

190.2

77 IR

IDIU

M

Ir 192.22

78 PL

AT

INU

M

Pt 195.09

79 G

OL

D

Au

196.97

80 M

ER

CU

RY

Hg

200.59

81 T

HA

LL

IUM

Tl

204.37

82 L

EA

D

Pb 207.2

83 B

ISMU

TH

Bi

208.98

84 PO

LO

NIU

M

Po [210.0]

85 A

STA

TIN

E

At

[210.0]

86 R

AD

ON

Rn

[222.0] 87

FRA

NC

IUM

Fr [223.0]

88 R

AD

IUM

Ra

[226.0] 89-103 104

RU

TH

ER

FOR

DIU

M

Rf

[263]

105 D

UB

NIU

M

Db

[268]

106 SE

AB

OR

GIU

M

Sg [271]

107 B

OH

RIU

M

Bh

[274]

108 H

ASSIU

M

Hs

[270]

109 M

EIT

NE

RIU

M

Mt

[278]

110 D

AR

MST

AD

TIU

M

Ds

[281]

111 R

OE

NT

GE

NIU

M

Rg

[281]

112 C

OPE

RN

ICIU

M

Cn

[285]

114

FLE

RO

VIU

M

Fl [289]

116

LIV

ER

MO

RIU

M

Lv

[293]

LAN

THA

NO

IDS

57 L

AN

TH

AN

UM

La

138.91

58 C

ER

IUM

Ce

140.12

59 PR

ASE

OD

YM

IUM

Pr 140.91

60 N

EO

DY

MIU

M

Nd

144.24

61 PR

OM

ET

HIU

M

Pm

[144.9]

62 SA

MA

RIU

M

Sm

150.4

63 E

UR

OPIU

M

Eu

151.96

64 G

AD

OL

INIU

M

Gd

157.25

65 T

ER

BIU

M

Tb

158.93

66 D

YSPR

OSIU

M

Dy

162.50

67 H

OL

MIU

M

Ho

164.93

68 E

RB

IUM

Er

167.26

69 T

HU

LIU

M

Tm

168.93

70 Y

TT

ER

BIU

M

Yb

173.04

71 L

UT

ET

IUM

Lu

174.97

AC

TINO

IDS

89 A

CT

INIU

M

Ac

[227.0]

90 T

HO

RIU

M

Th

232.04

91 PR

OT

AC

TIN

IUM

Pa [231.0]

92 U

RA

NIU

M

U

238.03

93 N

EPT

UN

IUM

Np

[237.0]

94 PL

UT

ON

IUM

Pu [239.1]

95 A

ME

RIC

IUM

Am

[243.1]

96 C

UR

IUM

Cm

[247.1]

97 B

ER

KE

LL

IUM

Bk

[247.1]

98 C

AL

IFOR

NIU

M

Cf

[252.1]

99 E

INST

EIN

IUM

Es

[252.1]

100 FE

RM

IUM

Fm

[257.1]

101 M

EN

DE

LE

VIU

M

Md

[256.1]

102 N

OB

EL

IUM

No

[259.1]

103 L

AW

RE

NC

IUM

Lr

[260.1]

2216(b) CHEM1611 June 2013