final exam uws chemistry with answers 2013

7/24/2019 Final Exam UWS Chemistry with Answers 2013

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Part 1. The questions in this part to be answered on the scan sheet provided.

Each question is worth one mark.

Questions 1 and 2 refer to the following information:

The concentration of gaseous nitrogen dioxide (NO2(g))was measured as NO2(g)

decomposes to form NO(g) and O2(g) according to the reaction:

NO2(g) NO(g) +

O2(g)

The following plot of 1/[NO2] versus time is shown below and the line of best fit yielded a

straight line through all the data points.

0

10

20

30

40

50

60

70

80

0 200 400 600 800 1000

1/[NO2](M

-1)

Time (sec)

B

A


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Time (sec)

Rate constant is the value of the slope (no sign) = 0.06904


The rate constant of the decomposition of nitrogen dioxide was measured at various

temperatures.

The plot of ln(rate constant) versus 1/temperature is shown below and the line of best fit

has been drawn through all data points.

5 5

6

6.5

7

7.5

8

8.5

9

9.5

ln(rateconst

ant)

x = 0.0000900 K-1

y = -1.240


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The reaction of Br2(g) and NO(g) forms BrNO (g) according to the equation:

Br2(g) + 2 NO(g) 2 BrNO(g)

The following mechanism isproposedfor this reaction:

Step 1 Slow Br2(g) + NO (g) Br2NO (g)

Step 2 Fast Br2NO (g) + NO (g) 2 BrNO (g)

5. For this mechanism to be consistent with kinetic data, the experimental rate lawwould have been?

(d) Rate = k [Br2] [NO]

The rate law is defined by the slow stepin the reaction. The rate law will be the productof the rate constant and the concentrations of the reactants in the slow step

6. The intermediate(s) in thisproposedreaction mechanism is/are

(c) Br2NO

Species that are formed in the initial steps of the mechanism that are not products

7. What is the IUPAC name of the following molecule


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8. Which of the carbon atom(s) labeled in the molecule below is/are chiral?

C

H

CH

CH3

CH

CH2

C

H2

CC

H2

CH3

CH3CH

3

OH

1

2 3

(c) 2 only

Chiral carbon is a carbon with four bonds, each bond connected to a different group,

alkene are not chiral carbons (not 3) carbon 1 has two ethyl (CH2CH3) attached

9. Which of the following molecules has a stereoisomer

1

C CH

CH

F

F

CH3

CH3

2CH

3C

H2

CH

2

C

H2

CH

CH2

3 C C

CH3

CH

CH

F

CH3

CH3

CH3

4CH

3CH

2

CH

CCH

2

CH3

CH3


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11. The following molecule has an enantiomer:

CH2

CCH

2 Cl

H

CH3

CH2

CH3

Which two of the following are enantiomers of this molecule?

1.

H

CCH

2 Cl

CH

2

CH3

C

H2

CH3

2.

H

CCH

2 Cl

CH

2

CH2 CH3

CH3

3.

CH2

CH

CH2

Cl

CH2

CH3

CH3

4.

Cl

CH

CH2

CH2

CH3

CH

2

CH3

(c) 2 and 3

An enantiomer is a mirror image of the molec le This is determined b rotating the


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The least stable conformers are the eclipsed conformers (b and c) where the groups onadjacent carbons are overlapping each other. These are least stable becauseatoms are closest together which yields repulsions. (a and d are staggered

conformers where the groups on adjacent carbons lie in the space inbetween)

b is the least stable because the largest group on each carbon are eclipsed

13. Which of the following is the best definition of a diastereomer?

(c) A diastereomer is one of a pair of stereoisomers which are not mirror images ofeach other.

14. An electrophile is

(d) a reagent that reacts with molecules containing a partialve charge or electron richbond

15. The reaction of 2-bromohexane with excess concentrated ammonia solution willproduce BOTH substitution and elimination products because

(a) ammonia is a base as well as a nucleophile

Substitution reaction needs a nucleophile as the reagent whereas elimination needs abase as the reagent. Nucleophiles are also basic so can do both reactions

16 Th f ll i ti i l hili b tit ti ti f t (th d t


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18. Which is/are the product(s) of the following reaction

HO-

heatCH3

CH C

H2

CH

2

CNH

C

H2

CH

2

CH3

CH3

O

(c)

CH3

C

H CH

2

CH

2

C

CH3

O

OH NH2

CH

2

CH

2

CH3

+

Substrate is amide; only reaction is nucleophilic substitution; HO-is a nucleophile; breakbond between C-N to form leaving group; OH substitutes leaving group to formcarboxylic acid

19. Which is the product of the following reaction

CH

CH2

CH

2

C

CH2

CH2

CH C

H2

CH3

OHPCC

Leaving group


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CH3CH

2

CH C

H CH3

Br

CH3

+ NH

CH3 CH3

(c)CH

3 CH

2

CH C

H CH3

N

CH3

CH3

CH3

Alkyl halide + amine; alkyl halide can do elimination or substitution; amine is anucleophile=> nucleophilic substitution; neutral nucleophileremove H from NHand replace the Br leaving group with nucleophile - H


CH3

CH

2

CH C

CH3

O

CH3

+ CH

3CH

2

CH

2

NH2


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CH3CH

2

C

H2

CH2

COH

OSOCl

2

(b) CH3C

H2

C

H2

C

H2

C

Cl

O

Substitution of OH with Cl


H2(50 atm)

Pt, heat

CH3 C

H

CH

CH

2

CH

2

CH3

(b) CH

2

CH

2

CH

2

CH

2CH

3CH

3

S b t t i lk > ti t b dditi t i H2 hi h lit i t 2 H t


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CH3CH

2

C

O

OC

CH

2

O

CH3NH

2

C

H2

CH3

+

(d) CH3CH

2

C

NH

CH

2

CH3

O

+

CH3

CH

2

C

OH

O

This reaction is an acid anhydride + amine. This is a nucleophilic substitution reaction toform an amide. The leaving group must be identified and exchanged with the aminenucleophile.

The bond in the acid anhydride that breaks to form the leaving group is adjacent to C=O(either one) and forms the leaving group (highlighted in red).

The leaving group is removed and forms one product

The nucleophile is neutral so the H atoms on the NH2group must be removed beforeattaching the the carbon which originally had the leaving group


H


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CH3CH

CCH

2

C

H2

CH

2

CH3

CH3

HBr

(c) CH3CH

2

CCH

2

CH

2

CH

2

CH3

CH3

Br

Starting material is alkene so reaction will be addition reaction and requires an electrophile

formed from the reagent. So HBr is split into H+and Br-as the two groups added tothe two carbons of the double bond.

Apply Markovnikovs rule => H+(electrophile) goes to the carbon with the most Hs on it

28. What REAGENT would you use to perform the following reactions?

CH3

CH

CH

CH C

H2

CH3

CH3

CH3

CH C

H CH C

H2

CH3

Cl

Cl

CH3


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30. What REAGENT(s) would you use to perform the following reactions?

CH2

CH2

CH

2

CH

2

CH

CH2

OHCH2

CH2

CH

2

CH

2

CH

CH

(a) H2SO4and heat

Alkenes are formed by elimination. Elimination from alcohol uses strong acid.

31. The specific heat capacity of Hg(l) is 0.14 J g1K1. The heat required to raise the

temperature of 1.00 mol Hg(l) from 25C to 250C is:

(c) 6.3 kJ

Heat = Cpx mass x T = 0.14 x 200.59 x (25025) = 6318 J = 6.34 kJ (200.59 is atomic

weight of Hg = mass of 1 mol of Hg)

32. The molar enthalpy of vaporisation, vapH, of chloroform, CHCl3, is 29.4 kJ mol-1at

its normal boiling point, 62C. What is the molar entropy change of vaporisation,

vapS of chloroform.

(d) +88 J K-1mol-1


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34. Consider these reactions and their standard enthalpy changes of reaction:

N2(g) +

O2(g) NO(g)

rH

o

= +90.0 kJ

2 NO(g) + O2(g) 2 NO2(g) rHo = -112 kJ

Calculate the rHoat the same temperature for the reaction:

N2(g) + 2 O2(g) 2 NO2(g)

(b) +68 kJ

Use Hesss Law to calculate enthalpy change.Determine the combination of the two equations which results in the third:need to remove NO(g) from equation 2; multiplying equation 1 by 2

N2(g) + O2(g) 2 NO(g) rHo = +180.0 kJ

2 NO(g) + O2(g) 2 NO2(g) rHo = -112 kJ

rH0 = 180 + -112 = 68 kJ

X 2


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Part 2. Questions in this part to be answered in the spaces provided in thisexamination paper.

1. The following rate data were collected at 298 K for the reaction:

2 NO(g) + O2(g) 2 NO2(g)

Experiment Initial concentration (molL-1) Initial rate

NO O2 (molL-1 s-1 )

1 1.26 x 10-2 1.25 x 10-2 1.41 x 10-2

2 2.52 x 10-2 2.50 x 10-2 11.3 x 10-2

3 2.52 x 10-2 1.25 x 10-2 5.64 x 10-2

Deduce the experimental rate law for this reaction including the order with respect to each

reactant and then determine the rate constant.

Show your working clearly, paying attention to significant figures and units. (5 marks)

With respect to NO ([O2] is same in both), use expt 1 and 3

Concentration [NO] is doubled (

), rate x4 (

) => 2ndorder

With respect to O2 ([NO] is same in both), use expt 2 and 3


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2. Draw the structure of the compound with the following systematic name: (5 marks)

Z-4-bromo-3-butylhept-3-en-2-ol

C C

CH2

CH CH3

OH

Br

CH2

CH2

CH3

CH2

CH2

CH3

hept = 7 carbon stem

en = double bond

ol = alcohol functional group

Z = higher ranked group on same side

3. (2 marks)

Explain why 2-fluorobutanoic acid is a stronger acid than 2-chlorobutanoic acid

CH3

CH

2

CH C

OH

O

F

CH3

CH

2

CH C

OH

O

Cl

2-fluorobutanoic acid 2-chlorobutanoic acid

1

2

34

5

6

7


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4. (5 marks)

A chemist has to synthesise ethyl iso-pentanoate as an ingredient in an artificial

banana flavour. The chemist decides to use iso-pentyl chloride as starting material.

Hence the chemist designed a possible synthesis of ethyl iso-pentanoate from iso-

pentyl chloride in a three step process as shown below.

O

O

EDC

A BCl

(i) What is the structure of intermediary compounds A and B? (2 marks)

A

OH

B

OH

O

iso-pentyl chloride ethyl iso-pentanoate

from alcohol

from acid

oxidation


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5. (8 marks)

(a)

The following nucleophilic substitution reaction of S-2-chloropentane with hydroxide

ion in acidic solution yields 2-pentanol as the product.

Experimental data on the rate of reaction showed that the rate law depended on

both [S-2-chloropentane] and [HO-]

CH3 C

H2

CH

2

CCH

3

H Cl

+ OH-

(i) From this data, circle which is the probable mechanism: SN2Rate law is dependent on both substrate and nucleophile => SN2 (1

mark)

(ii) Draw this mechanism for this reaction which explains the above observation. The

product(s) drawn in the mechanism must show the configuration at the chiral

carbon. (4 marks)

N b h ki f


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(b)

There are two possible products of the addition reaction of 3-methyl-2-pentene with

HBr, namely 3-bromo-3-methylpentane and 2-bromo-3-methylpentane. However,

Markovnikovs rule states that the major product is 3-bromo-3-methylpentane as

shown below.

CH3

C

H2

CC

H

CH3

CH3

HBrCH

3C

H2

CC

H2

CH3

CH3

Br Major product

Draw the accepted mechanism for this reaction, explaining why only this product is

formed. (3 marks)

H+

H

+


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6. The equation for the formation of methane at 298 K is given by: (5 marks)

C (graphite) + 2 H2(g) CH4(g)

(a) Calculate the value of rG for this reaction at (298 K)

if rH is74.9 kJ mol1and rS is0.0807 kJ

K1mol1.

rG = rHT . rS

=74.9(298) . 0.0807 make sure you check the units (rH and rS must be same

unit kJ or J)

=50.85 kJ mol-1

(b) Explain the significance of the magnitude and sign of rG for this reactionunder the conditions of part (a) above.

Large and ve => spontaneous reaction


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END of EXAMINATION


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