Indian National Chemistry Olympiad 2020 Theory (3 hours)
Roll No. Exam Centre:
Indian National Chemistry Olympiad 2020 Theory (3 hours)
Date: February 1, 2020 Do not write anything below this line
Question No
1
2
3
4
5
6
Total
Marks
16
22
24
13
29
15
119
Marks Obtained
Signature of Examiner
Instructions for students Write last four digits of your Roll No. at the top of all pages. This examination booklet consists of 29 pages of problems including answer boxes. Kindly check that the booklet has all the pages. If not, report to the invigilator immediately. All answers must be written in the appropriate boxes. Anything written elsewhere will not be considered
for assessment. Adequate space has been provided in the answersheet for you to write/calculate your answers. In case you
need extra space to write, you may request for additional blank sheets from the invigilator (Please draw a box and write the Q. No. in the box on this sheet for evaluation). Remember to write your roll number on the extra sheets and get them attached to your answersheet. Use only a pen to write the answers in the answer boxes. Answers written in pencil (except for graph) will be penalized. You must show the main steps in the calculations. For objective type question, mark X in the correct box. Some of the objective questions may have more than one correct answer. A copy of the Periodic Table of the Elements is provided at the end. Do not leave the examination room until you are directed to do so.
Fundamental Constants
Avogadro number Electronic charge
NA = 6.022 ? 1023 mol1 e = 1.602 ? 1019 C
Molar gas constant
R = 8.314 J K?1 mol1 = 0.08205 Latm K-1mol-1
1 atm = 101325 Pa
Mass of electron Speed of light 1 atomic mass unit
pH = log [H+]
me = 9.109 ? 1031 kg c = 2.998 ? 108 m s1 (1 amu) = 1.660 ? 1027 kg
pKa = log Ka
? Homi Bhabha Centre For Science Education, Tata Institute of Fundamental Research V.N. Purav Marg, Mankhurd, Mumbai 400 088
IUPAC Periodic Table of the Elements
1
1
H
hydrogen
1.008
[1.0078, 1.0082]
2
3
Li
lithium
6.94
[6.938, 6.997]
4
Be
beryllium
9.0122
11
Na
sodium
22.990
12
Mg
magnesium
24.305
[24.304, 24.307]
19
K
potassium
20
Ca
calcium
Key:
atomic number
Symbol
name
conventional atomic weight standard atomic weight
3
21
Sc
scandium
4
22
Ti
titanium
5
23
V
vanadium
39.098
37
Rb
rubidium
40.078(4)
38
Sr
strontium
44.956
39
Y
yttrium
47.867
40
Zr
zirconium
50.942
41
Nb
niobium
85.468
55
Cs
caesium
87.62
56
Ba
barium
88.906
57-71 lanthanoids
91.224(2)
72
Hf
hafnium
92.906
73
Ta
tantalum
132.91
87
Fr
francium
137.33
88
Ra
radium
89-103 actinoids
178.49(2)
104
Rf
rutherfordium
180.95
105
Db
dubnium
6
7
24
Cr
chromium
25
Mn
manganese
51.996
42
Mo
molybdenum
54.938
43
Tc
technetium
95.95
74
W
tungsten
75
Re
rhenium
183.84
106
Sg
seaborgium
186.21
107
Bh
bohrium
8
26
Fe
iron
55.845(2)
44
Ru
ruthenium
101.07(2)
76
Os
osmium
190.23(3)
108
Hs
hassium
18
2
He
helium
13
14
15
16
17
4.0026
9
27
Co
cobalt
58.933
45
Rh
rhodium
10
28
Ni
nickel
58.693
46
Pd
palladium
11
29
Cu
copper
63.546(3)
47
Ag
silver
12
30
Zn
zinc
65.38(2)
48
Cd
cadmium
5
B
boron
10.81
[10.806, 10.821]
6
C
carbon
12.011
[12.009, 12.012]
7
N
nitrogen
14.007
[14.006, 14.008]
8
O
oxygen
15.999
[15.999, 16.000]
9
F
fluorine
18.998
10
Ne
neon
20.180
13
Al
aluminium
26.982
14
Si
silicon
28.085
[28.084, 28.086]
15
P
phosphorus
30.974
16
S
sulfur
32.06
[32.059, 32.076]
17
Cl
chlorine
35.45
[35.446, 35.457]
18
Ar
argon
39.95
[39.792, 39.963]
31
Ga
gallium
69.723
32
Ge
germanium
72.630(8)
33
As
arsenic
74.922
34
Se
selenium
78.971(8)
35
Br
bromine
79.904
[79.901, 79.907]
36
Kr
krypton
83.798(2)
49
In
indium
50
Sn
tin
51
Sb
antimony
52
Te
tellurium
53
I
iodine
54
Xe
xenon
102.91
77
Ir
iridium
192.22
109
Mt
meitnerium
106.42
78
Pt
platinum
107.87
79
Au
gold
195.08
196.97
110
111
Ds Rg
darmstadtium roentgenium
112.41
114.82
80
Hg
mercury
200.59
81
Tl
thallium
204.38
[204.38, 204.39]
112
Cn
copernicium
113
Nh
nihonium
118.71
82
Pb
lead
207.2
114
Fl
flerovium
121.76
83
Bi
bismuth
208.98
115
Mc
moscovium
127.60(3)
84
Po
polonium
116
Lv
livermorium
126.90
85
At
astatine
117
Ts
tennessine
131.29
86
Rn
radon
118
Og
oganesson
57
La
lanthanum
138.91
89
Ac
actinium
58
Ce
cerium
140.12
90
Th
thorium
232.04
59
60
Pr Nd
praseodymium neodymium
140.91
91
Pa
protactinium
144.24
92
U
uranium
231.04
238.03
61
Pm
promethium
93
Np
neptunium
62
Sm
samarium
150.36(2)
94
Pu
plutonium
63
Eu
europium
151.96
95
Am
americium
64
Gd
gadolinium
157.25(3)
96
Cm
curium
For notes and updates to this table, see . This version is dated 1 December 2018. Copyright ? 2018 IUPAC, the International Union of Pure and Applied Chemistry.
65
Tb
terbium
158.93
97
Bk
berkelium
66
Dy
dysprosium
162.50
98
Cf
californium
67
Ho
holmium
164.93
99
Es
einsteinium
68
Er
erbium
167.26
100
Fm
fermium
69
Tm
thulium
168.93
101
Md
mendelevium
70
Yb
ytterbium
173.05
102
No
nobelium
71
Lu
lutetium
174.97
103
Lr
lawrencium
Indian National Chemistry Olympiad 2020
Problem 1
Chemistry of the artificial hair dyes
Dyeing of hair is an ancient art. Earlier, dyes were
obtained
from
plants
such
as indigo, henna, turmeric and amla. In dyeing with
natural henna, an aglycone molecule (structure shown
below), gets converted to orange colored Lawsone in
the presence of an oxidant (such as air).
Oxidant
Roll No
16 marks
Aglycone
Lawsone
Hydroquinone (1,4-dihydroxybenzene) and analogous 1,4-diimine derivatives also undergo similar transformations under the influence of an oxidant. A chemistry teacher at a hair salon once observed the barber preparing a low-cost hair colorant for a customer by mixing a spoonful of brownish-white crystalline solid with H2O2 and shaving cream. A little investigation indicated that the white solid was a chemical called PPD (p-phenylenediamine). On further reading, the teacher realized that PPD was one of the first synthetic chemical launched for hair dyeing market in the year 1907. Later he found that most of the hair colorants being sold in the market, including the so-called "natural/herbal" hair colours had PPD in them. Hair proteins have iso-ionic points (when the number of +ve and ?ve charged species in protein are equal) around pH = 4.86.8. Therefore, some of the hair dyes are also +vely charged species, which hair proteins absorb from neutral-alkaline solutions. In this problem, we will explore how colourless PPD leads to the intense black and numerous other hair colours. Under oxidizing condition, PPD gets converted to corresponding diimine A (C6H6N2).
1.1 Draw the structure of A.
?HBCSE, February 1, 2020
1
Indian National Chemistry Olympiad 2020
Roll No
In aqueous solutions in the pH range 7-10, a small fraction of A molecules exists in monoprotonated form B. 1.2 Draw the structure of B.
B
Usually colour in organic molecules originates due to extended conjugation. Black colour is generated from PPD by two pathways I and II. Pathway I: In many commercial hair colorants, the white powder contains a mixture of several compounds (called couplers) along with PPD. Under oxidizing conditions, PPD reacts with different couplers producing different hair colors, which in specific combination give black colour. For example, oxidized form of PPD (B) reacts with the coupler m-phenylenediamine in aqueous phase to form the thermodynamically stable product C, which on further oxidation produces a blue hair dye D (a cationic species having two rings). 1.3 Write the structure of C and dye D.
C C
D (Hair dye)
When resorcinol is used as a coupler, a yellow-green dye E is produced which reacts with diazomethane to form a compound having molecular formula C20H18N4O2. 1.4 Write the structure of E.
B
?HBCSE, February 1, 2020
E (Yellow-Green colour) 2
Indian National Chemistry Olympiad 2020
Roll No
Pathway II: When PPD is alone subjected to oxidation, the oxidized form A slowly gets polymerized to give an intermediate F, which on further oxidation gives an intense black hair colour.
1.5 Write the part-structure of the polymer (F).
F
The oxidative coupling of 4-dimethylamino aniline with phenol can take a different pathway. A 2 e oxidation of 4-dimethylamino aniline generates a cationic intermediate species G. Reaction of phenol with G, produces a colourless species leuco-indoaniline, which gets easily oxidized to give a dye, indoaniline. Quantum mechanical calculations indicate that conversion from G to leuco-indoaniline can go through a very low energy pathway involving formation of another intermediate species H. Due to - stacking interactions between two rings, H can easily rearrange to produce leuco-indoaniline.
1.6 Draw the most stable resonance form of G and possible structure of H.
[O]
G Rearrangement
leuco-indoaniline
?HBCSE, February 1, 2020
H 3
................
................
In order to avoid copyright disputes, this page is only a partial summary.
To fulfill the demand for quickly locating and searching documents.
It is intelligent file search solution for home and business.
Related download
- chemistry syllabus examinations
- 66092 y18 sy chemistry o level for 2018 seab
- level 5 exemplar comments paper 1b
- ifrn k i u i 2020 d kk 12 le 3 k v 15 feuv 70
- chemistry theory
- 9791 y20 sm 1
- level 1 level 2 gcse 9 1 chemistry
- 1sc0 1ch additional sams combined science paper 3
- indian national chemistry olympiad 2020 theory 3 hours
- cambridge international as a level
Related searches
- national health observances 2020 calendar
- national awareness months 2020 calendar
- national health observances 2020 calendar pdf
- national health observances 2020 at a glance
- national day calendar 2020 free
- national merit semifinalist 2020 list
- national health observances 2020 july
- national days in 2020 printable
- chemistry igcse 2020 past paper
- chemistry igcse 2020 z notes
- act national test dates 2020 21
- national defense strategy 2020 pdf