17th Chemistry Olympiad of the Baltic States

Riga, Latvia, 2009

Theoretical examination

Answers

BChO17, Riga, Latvia, 2009 Theoretical examination

Answers

2 of 8

Problem 1. Hydrolysis of pyrophosphate (10 pts).

a) 3

2 71 2

2 2 7

[HP O ][H ][H P O ]

K

42 7

2 32 7

[P O ][H ][HP O ]

K

2 3 42 2 7 2 7 2 7[H P O ] [HP O ] [P O ]c

b) 2

42 7

2 1 2

[H ] [H ]1 [P O ]cK K K

2121

21272

]H[]H[]OP[

KKKKK

c

00979,0)10(10101010

101027712,695,812,6

95,812,6

11

7

*7 s102,0

0,00979s0010,0

kk

c) 000484,0)10(10101010

101026612,695,812,6

95,812,6

6

0997,0)10(10101010

101028812,695,812,6

95,812,6

8

1516

*6 s1093,4s102,0000484,0

kk

1218

*8 s100,1s102,00997,0

kk

BChO17, Riga, Latvia, 2009 Theoretical examination

Answers

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Problem 2. Iodine and polyiodides (12 pts).

a) A KI B HIO3 C I2O5 D KIO3 E HI F + G CuI + I2 H AgI Equations: 2KI + KNO3 + H2SO4 I2 + KNO2 + K2SO4 + H2O 3I2 + 10HNO3 6HIO3 + 10NO + 2H2O HIO3 + KOH KIO3 + H2O 2HIO3 I2O5 + H2O 5CO + I2O5 5CO2 + I2 2I2 + N2H4 4HI + N2 HI + KOH KI + H2O 4KI + 2CuSO4 2CuI + I2 + 2K2SO4 KI + AgNO3 AgI + KNO3 b)

Linear I I I c) Possible geometries:

V-shaped and L-shaped: Linear pentaiodides also are possible.

d) R4NI2x+1 + 2xS2O32 R4NI + 2xI + xS4O62 n(S2O32)= 0.01023 L0.112 mol/L = 0.001146 mol if x=1 n(R4NI2x+1)= 0.001146/2=0.0005729 mol ; M= 0.219 g/0.0005729 mol =382.3 g/mol M too small for R4NI3. Only I3 gives 381 g/mol.

BChO17, Riga, Latvia, 2009 Theoretical examination

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if x=2 n(R4NI2x+1)= 0.001146/4=0.0005729 mol ; M= 0.219 g/0.0002864 mol =764.5 g/mol (in general M= 382.3x) R4NI5 M=(14n+1)4+14+1275 =764.5 n=2 Molecular formula (C2H5)4NI5. Higher values of x doesnt fit any plausible R.

Problem 3. Small rings fighting diabetes (10 pts).

NHOOC

COOMe

N

COOMe

ClOCO

N

COOMe

N

COOMe

N

COOMe

N

COOH

(COCl)2

AlCl3

1. MeMgCl 1 mol

2. H+

1. NaOH

2. H+

2.

S+ O

H

S+

CH2-O

Ar Ar

Ar Ar

S+

O

-Ar Ar

base

BChO17, Riga, Latvia, 2009 Theoretical examination

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Problem 4. Uses and misuses of nitrogen compounds (10 pts).

NH2 NH2

ON

N

N

OH

OH OH

NH

NH

NH

O

O O

N

N

N

Cl

Cl Cl

N

N

N

O

O O

ClCl

Cl

N

N

N

NH2

NH2 NH2

N

N

N

N+

OMe

MeO

Cl N

N

N

N

OMe

MeO

NNMP =

1. CH3OH

2. NMP

+

POCl3 Cl2

175 C

+

BA1 A2

C1 C2

B

D

E FG

NH3

NH3

CH3Cl

2. Replace chloride with a nonnuclephilic anion (BF4-, PF6-)

BChO17, Riga, Latvia, 2009 Theoretical examination

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6 of 8

3.

N

N

N

OMe

MeO O R

ON

N

N

N+

OMe

MeOO

RO

N

N

N

N+

OMe

MeO

Cl

R OH

O

R'NH2

N

N

N

OMe

MeO OHR N

H

OR'+

Problem 5. Kinetics and equilibrium (8 pts).

a) exp exp 0.00264G H SKRT RT R

298 4730 33.5exp 0.002648.314 298 8.314

K

273 4730 33.5exp 0.002218.314 273 8.314

K

b) 1

298 12

0.0375 s 14,2s0.00264

k

1273 12

0.0021s 0,95s0.00221

k

a1

2 1 2

1 1ln Ekk R T T

1 2 1a

1 2 2

lnTT kE RT T k

1 1 11

298 273 0.0375K 8.314 J mol K ln 78kJ mol25 0.0021

E

1 1 12

298 273 14.2K 8.314 J mol K ln 73kJ mol25 0.95

E

BChO17, Riga, Latvia, 2009 Theoretical examination

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Problem 6. Chanel 5 (10 pts).

a) A. B.

b)

c) C. (R)-2-methylundecanal

d) (S) enantiomer is used. After imine D is formed, metoxy group sterricaly bloks methyl addition to the imine in enolic form from the Si face. Methyl adds to the imines unhindered face and after hydrolysis (R)-methylundecanal is formed.

BChO17, Riga, Latvia, 2009 Theoretical examination

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e) D. E.

F.