Indian National Chemistry Olympiad Theory 2012

HBCSE, 28th January 2012

1

B, C

HH H

N N N

H

N+++ +

Problem 1 18 marks

Heterocyclic compounds

1.1

1.2

1.3

B.

1.4 a)

b)

Name of Student Roll No.

F > G > E

NH

NH

NO2

O3S

or

HO3S

NO2

I H

Indian National Chemistry Olympiad Theory 2012

HBCSE, 28th January 2012

2

X

or

or

1.5

1.6

C.

1.7

1.8 Yes

1.9

TS (gives an oxime)

U V

NH

O

H

NNH

O HH

NH

(C9H11NO) (C9H9N) (C9H7N)

1 mole 2 moles

ON H2

N H2

H5C2 CHO+

..

Indian National Chemistry Olympiad Theory 2012

HBCSE, 28th January 2012

3

Problem 2 12 marks

Analysis of alloys

A.

2.1

2.2

2.3

2.4

2.5

B.

2.6

2.7

2.8

2.9

C.

2.10

NO3−(aq) + 10H+ (aq) + 8e− NH4

+(aq) + 3H2O

Fraction of the total current = 36.1%.

Pb2+ (aq) + 2H2O PbO2 ↓ + 4H+ (aq) + 4e−

mmol of PbO2 = 0.0132

Amount of zinc = 0.184 g Percentage = 26.8 % of zinc

2Cu2+ (aq) + 4I− (aq) Cu2I2 ↓ + I2(aq)↑

2S2O3

2− (aq) + I2(aq) S4O62− (aq) + 2I−(aq)

0.484 g of copper

70%

0.469 g of copper Relative percentage deviation = 3.1%

Cr2O72− (aq) + 14H+ (aq) + 6I− (aq) 2Cr3+ (aq) + 7H2O + 3I2↑

Molarity of thiosulphate = 0.0485 M

Indian National Chemistry Olympiad Theory 2012

HBCSE, 28th January 2012

4

Problem 3 18 Marks

Halogens

3.1 (c )

3.2 (a)

(c)

3.3 No

3.4 (a)

(b) (iii)

(c)

3.5

3.6

3.7

Solvent Solubility Solvent Solubility

Carbon tetrachloride 19 n-Hexane 13

Diethyl ether 337 Toluene 182

Solvent λmax Color

Carbon tetrachloride 520 - 540 nm Bright violet

Diethyl ether 460 - 480 nm Deep brown

F2 + Br2 → 2BrF (A) 3BrF → Br2(g) + BrF3 (B) or 5BrF → 2Br2(g) + BrF5 (B)

BrF3 + SbF5 [BrF2]+[SbF6]

−

or BrF5 + SbF5 [BrF4]

+[SbF6]−

KI7 /I7−

X

X

X

X

X

Indian National Chemistry Olympiad Theory 2012

HBCSE, 28th January 2012

5

3.8

3.9 a)

b)

c)

d)

3.10

3.11

Lewis Acid Lewis Base

Reactants SbF5 F−

Products BrF2+ / BrF4

+ SbF6−

S

12

Tetrahedral

XeO64−

N

S

FF

F. ..

..

.

.. .

..

.

Sb

SbFF

FF

F

F

F

F

F

FF

Indian National Chemistry Olympiad Theory 2012

HBCSE, 28th January 2012

6

X

Problem 4 23 marks

Chemistry of carbocations

4.1

4.2

4.3 i) Yes

ii) No

iii) Yes

4.4

4.5

4.6 (iii)

4.7 (ii)

4.8

4.9

sp2

IV > III > I > II

C

+

D

+

Br

X

X

X

X

F [C15H15N2O]+ G [C15H15O] H

Indian National Chemistry Olympiad Theory 2012

HBCSE, 28th January 2012

7

or

(carbocation)L M N (C13 H21O)

X

X

X

4.10

.

4.11

4.12

4.13 (i)

4.14 (i) or (ii) (As per the given structure of P in 4.12)

4.15

and

O O

CH3

O

O

H

O

H

P

O

J (C17H24O)

OH OHOH

Indian National Chemistry Olympiad Theory 2012

HBCSE, 28th January 2012

8

Propagation steps: (ii) & (iii) Termination step: (iv)

]Cl[

]H[

k

I

k

k][H

][Cl2k][H ][Clk]][Cl[Hk2I0td

]d[Cl

].][Cl[Hk]][H[Clk0td

]d[H

2

22

1

3

abs

2

1

232122abs

2221

=

−−+⇒=

−⇒=

•

••••

•••

].[Hk

I2Kor ][H][Ik

dt

d[HCl]21/2

3

1/2abs12

21

abs'=

.051 Yield Quantum =

Problem 5 18 marks

The ‘light’ side of chemistry

5.1

5.2

5.3

5.4 The correct statement/s is/are

i)

ii)

5.5

B.

5.6 a) .

b)

c)

5.7 a)

b)

∆H - ∆H* = NAh (νAOH − νAO− )

∆UAOH = NAhν AOH

∆UAO− = NAhνAO

−

∆pka = − 3.5

∆pka= NAh (νAO

− − νAOH) /2.303RT

X

X

∆U* = ∆UAO− + ∆U − ∆UAOH

Indian National Chemistry Olympiad Theory 2012

HBCSE, 28th January 2012

9

1.0]NH[Pr

]NH[Pr

3

2 =+

0.03 COOH]RN[H

]COORN[H

3

2 =−−−−

+

−

Problem 6 11 marks

Acids, bases and buffers

A.

6.1

6.2

6.3

B.

6.4

6.5 around point B around point D

6.6

6.7 (iii)

CH 2

CH2

CH2

CH2

CH 2

CH2

CH 2

CH2

CH 2

COOH2N

COOH3N

H3N COOH+

+

X X

08.10 pH =

pH ≈ 6

appropriate indicator is –Methyl red. (4.2-6.2)

X