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New 21st Century Chemistry

Suggested answers to in-text activities and unit-end exercises 1 © Jing Kung. All rights reserved.

Topic 16 Unit 56

Suggested answers to in-text activities and unit-end exercises

Topic 16 Unit 56

In-text activities

Checkpoint (page 162)

a)

b) As the concentration of fluoride ions increases, more In(aq) (red-coloured indicator) reacts.

c) 1.50 mg dm–3

d) At zero absorbance, the concentraton of fluoride ions is 7.50 mg dm–3.

Number of moles of fluoride ions in 25.0 cm3 = x dm3

= 9.87 x 10–6

mol

∴ the value of X is 9.87 x 10–6

.


1 a)

b) The broad absorption at 2 500 – 3 300 cm–1

indicates the presence of an O–H bond (in

acid).

25.0

1 000

mol–1

7.50 mg dm–3

19.0 g mol–1



Topic 16 Unit 56

The absorption at 1 680 – 1 750 cm–1

indicates the presence of a C=O bond.

Thus, acrylic acid was formed.

2 Suppose we have 100 g of either compound X or Y, so there are 40.0 g of carbon, 6.70 g of

hydrogen and 53.3 g of oxygen.

Carbon Hydrogen Oxygen

Mass of element in the

compound 40.0 g 1.70 g 53.3 g

Number of moles of

atoms that combine

= 3.33 mol

= 6.70 mol

= 3.33 mol

Simplest ratio of atoms = 1.00 = 2.01 = 1.00

∴ the empirical formula of either compound is CH2O.

Let (CH2O)n be the molecular formula of either compound.

Relative molecular mass = n(12.0 + 2 x 1.0 + 16.0)

= 30n

i.e. 30n = 60.0

n = 2

∴ the molecular formula of either compound is C2H4O2.

The absorption at 1 700 cm–1


The absorption at 3 350 cm–1

indicates the presence of an O–H bond.

Thus, it can be deduced that compound Y is a carboxylic acid while compound X is either an

aldehyde, a ketone or an ester.

The structure of compound Y is .

As compound X contains two oxygen atoms, thus it should be an ester.

The structure of compound X is .

3 a) The absorption at 3 230 – 3 670 cm–1

indicates the presence of an O–H bond.

Thus, compound Y is an alcohol. As compound Y is formed from propanal, it can be

deduced that compound Y is propan-1-ol.

Compound A is an ester formed by the reaction between compounds X and Y. Thus,

compound X is propanoic acid (formed from propanal).

The structural formulae of the compounds are as follows:

3.33 mol

3.33 mol

6.70 mol

3.33 mol

3.33 mol

3.33 mol

53.3 g

16.0 g mol–1

6.70 g

1.0 g mol–1

40.0 g

12.0 g mol–1



Topic 16 Unit 56

Compound A

Compound X

Compound Y CH3CH2CH2OH

b) K2Cr2O7 / H3O+

Propanal compound X (propanoic acid)

heat 1 LiAlH4 / ethoxyethane

Propanal compound Y (propan-1-ol)

2 H3O+

c) As a catalyst


1 a) The peak at m/e = 72

b) The peak at m/e = 43

CH3CO+ ion

.

c)

CH3CO

+ + •CH2CH3

molecular ion m/e = 43

m/e = 72

2 a)

m/e Ion

100 [H2C=CHCH2OOCCH3]•+

57 H2C=CHCH2O+

43 CH3CO+

b)



Topic 16 Unit 56


1 The difference between 44 and 28 is 16. Thus, the peak at m/e = 28 is formed from the loss of

an oxygen atom from the molecular ion. The peak at m/e = 28 is due to the CO+ ion.

Hence the gas is carbon dioxide.

2 a) The difference between 46 and 29 is 17. Thus, the peak at m/e = 29 is probably due to the

loss of a hydroxyl radical.

b)

m/e Ion

28 CO+

29 CHO+

45 CHO2+

c)


1 a) Any one of the following:

• Mix phosphorus pentachloride with each compound.

Only CH3CH2CH2OH gives steamy fumes.

• Warm each compound with acidified aqueous solution of potassium dichromate.

Only CH3CH2CH2OH turns the orange dichromate solution green.

b) i) The IR spectrum of compound A should have the following absorptions:

1 000 – 1 300 cm–1

, indicating the presence of a C–O bond;

1 680 – 1 750 cm–1

, indicating the presence of a C=O bond.

The IR spectrum of compound B should have the following absorption:

3 230 – 3 670 cm–1

, indicating the presence of an O–H bond.

ii) Confirm the identity of a sample by comparing its spectrum with those of known

compounds.

c) i) The molecular ion peak of CH3COOCH3 is at m/e = 74. The base peak is at m/e = 43,

due to the CH3CO+ ion.

The difference between 74 and 43 is 31. The peak at m/e = 43 is probably due to the

loss of a •OCH3 fragment.



Topic 16 Unit 56

ii)

CH3CO

+ + •OCH3

m/e = 74 m/e = 43

2 a) The difference between 74 and 57 is 17. Thus, the peak at m/e = 57 is probably due to the

loss of a hydroxyl radical.

b) The difference between 74 and 45 is 29. Thus, the peak at m/e = 45 is probably due to the

loss of an ethyl radical.

c) An ethyl ion

d) O–H bond

e) In the mass spectrum, the peak at m/e = 57 suggest the loss of a hydroxyl radical from the

molecular ion. Thus, a hydroxyl group is likely to be present in compound X.

The peak at m/e = 29 is probably due to the CH3CH2+ ion. This suggests the presence of an

ethyl group in compound X.

The infrared absorption at 1 680 – 1 750 cm–1

suggests the presence of a C=O bond.

Hence, a possible structure of compound X is .

Unit-end exercises (pages 198 – 210)

Answers for the HKCEE and HKALE questions are not provided.

1

Physical property measured Analytical method based on measurement of

the property

Mass gravimetric analysis

Volume volumetric analysis

Absorption of electromagnetic radiation colorimetry / infrared spectroscopy

Mass-to-charge ratio mass spectrometry

2 a) Propan-2-ol



Topic 16 Unit 56

Propanone

b) X: C=O bond

Y: O–H bond

c) Spectrum 1 is the spectrum of propanone.

Spectrum 2 is the spectrum of propan-2-ol.

3 a) A — sample inlet

B — ionization chamber

C — ion detector

D — magnetic field

b) i) The peak at the m/e which equals the relative molecular mass of the sample

ii) The strongest detector signal which is set to 100%

4 a) Let CxHy be the molecular formula of A.

The peak at m/e = 42 is the molecular ion peak.

Relative molecular mass of A = 42.0 = 12.0x + 1.0y

x = 3 and y = 6

∴ the molecular formula of A is C3H6.

b) The difference between 42 and 27 is 15, so it is probable that a methyl radical is lost from

the molecular ion, producing the peak at m/e = 27.

The structural formula of hydrocarbon A is CH3CH=CH2.

5 a) The peak at m/e = 91 corresponds to the C6H5CH2+ ion.

The peak at m/e = 65 corresponds to the C5H5+ ion.

b)



Topic 16 Unit 56

6 a) A colorimeter is an instrument for measuring the amount of light absorbed by a sample

when a beam of light passes through the sample.

Prepare a series of standard solutions of the species concerned. Using light of the selected

colour (that the species absorbs the most), record the absorbance of each solution. Plot the

absorbance against the concentration of the solutions.

Shine the light of the selected colour through the sample solution under test. Record the

absorbance.

From the calibration curve, read off the concentration of the species concerned in the

sample solution according to the absorbance recorded.

b) Concentration of H2O2 in rainwater sample = 1.55 x 10–6 mol dm–3 x

= 1.05 x 10–6

mol dm–3

a) Concentration of the stock solution = 3dm1000

250.0

0.100g

= 0.400 g dm–3

Mass of curcumin in 10.0 cm3 of stock solution = 0.400 g dm–3 x dm3

= 4.00 x 10–3

g

Volume of standard 3 containing 4.00 x 10–3 g of curcumin =

= 4.00 x 10–1

dm3

= 400 cm3

Volume of water added to 10.0 cm3 of stock solution = (400 – 10.0) cm

3

= 390 cm3

b) Curcumin concentration in the solution = 8.90 x 10–3 g dm–3

4.00 x 10–3

g

1.00 x 10–2

g dm–3

10.0

1 000

0.278

0.410



Topic 16 Unit 56

Mass of curcumin in 250.0 cm3 solution = 8.90 x 10–3 g dm–3 x dm3

= 2.23 x 10–3

g

= 2.23 mg

Curcumin content in the food sample =

= 0.0910 mg per gram of food sample

8 a) Ethyl ethanoate CH3COOCH2CH3

Butan-1-amine CH3CH2CH2CH2NH2

b) In the spectrum of compound A, the absorption at 3 350 – 3 500 cm–1

is due to the N–H

bond. Hence compound A is butan-1-amine.

In the spectrum of compound B, the absorptions at 1 000 – 1 300 cm–1

and 1 680 – 1 750

cm–1

are due to the C–O bond and the C=O bond. Hence compound B is ethyl ethanoate.

9 a) The spectrum of X shows a strong absorption at about 1 700 cm–1

, indicating the presence

of a C=O bond.

The broad absorption bond at about 3 000 cm–1

is characteristic of an O–H bond.

b) X is a carboxylic acid.

Let CnH2n+1COOH be the molecular formula of X.

Molar mass of X = 60.0 g mol–1

= 12n + 2n + 1 + 12.0 + 2 x 16.0 + 1.0

n = 1

Thus, the structural formula of X is .

10 a) i) Ester functional group

Carbon-carbon double bond

ii)

iii) C12H14O2

2.23 mg

24.5 g

250.0

1 000



Topic 16 Unit 56

b) Stereoisomers have their atoms linked in the same way, but they differ in the spatial

arrangement of their atoms.

Structural formula of compound B:

In compound A, the methyl groups are on opposite sides of the double bond.

In compound B, the methyl groups are on the same side of the double bond.

c)

d) i) The absorption at 1 000 – 1 300 cm–1

is due to the C–O bond.

The absorption at 1 680 – 1 750 cm–1

is due to the C=O bond.

ii) 2 500 – 3 300 cm–1

/ 3 230 – 3 670 cm–1

O–H bond is not present in compound A.

11 a)


indicates the presence of an O–H bond

(alcohol).



Topic 16 Unit 56

As compound X has a pair of enantiomers. Thus, it should have a chiral carbon.

Hence the structure of compound X is as follows:

12 —

13 a) Pentan-2-one CH3CH2CH2COCH3

Pentan-3-one CH3CH2COCH2CH3

b) The peak at m/e = 86

c) The difference between 86 and 71 is 15, so it is probable that a methyl radial is lost from

the molecular ion, producing the peak at m/e = 71.

The CH3CO+ ion is responsible for the peak at m/e = 43.

d) The compound is pentan-2-one.

The compound is not pentan-3-one because its mass spectrum should have a peak at m/e =

57 due to the CH3CH2CO+ ion.

The peak at m/e = 71 is produced by this fragmentation:

CH3CH2CH2CO+ + •CH3

molecular ion

m/e = 86

The peak at m/e = 43 is produced by this fragmentation:

CH3CO+ + •CH2CH2CH3

14 a) CH3CH2CH2CH2CH3



Topic 16 Unit 56

b) The base peak is at m/e = 57. This shows that alkane A can give a stable C4H9+ ion during

fragmentation.

This rule out the structure CH3CH2CH2CH2CH3 because the C4H9+ ion formed from it by

losing a methyl radical is a primary carbocation. The primary carbocation has a low

stability and the intensity for this ion should be low.

CH3CH2CH2CH2+ + •CH3

Alkane A is not . The molecular ion of can give a stable

secondary C4H9+ ion by losing a methyl radical, and also a stable secondary C3H7

+ ion by

losing an ethyl radical.

is consistent with the mass spectrum. Its molecular ion can undergo

fragmentation readily to give a stable tertiary C4H9+ ion.

15 a) i) The difference between 134 and 105 is 29, so it is probably that an ethyl radical is lost

from the molecular ion, producing the peak at m/e = 105.

Thus, the ion is responsible for the peak at m/e = 105.

The C6H5+ is responsible for the peak at m/e = 77.



Topic 16 Unit 56

ii)

b) i)

ii)

c) i)

ii) Geometrical isomerism

iii) Restricted rotation about a carbon-carbon double bond.

Two different atoms / groups on each carbon atom of the carbon-carbon double bond.

16 —

17 a) Suppose we have 100 g of compound A, so there are 69.8 g of carbon, 11.6 g of hydrogen

and 18.6 g of oxygen.


Mass of element

in the compound 69.8 g 11.6 g 18.6 g

Number of moles

of atoms that

combine

= 5.82 mol

= 11.6 mol

= 1.16 mol

Simplest ratio of

atoms = 5.02 = 10.0 = 1.00

1.16 mol

1.16 mol

11.6 mol

1.16 mol

5.82 mol

1.16 mol

18.6 g

16.0 g mol–1

11.6 g

1.00 g mol–1

69.8 g

12.0 g mol–1



Topic 16 Unit 56

∴ the empirical formula of compound A is C5H10O.

b) The molecular on peak is at m/e = 86. Thus, the relative molecular mass of compound A is

86.0.

Let (C5H10O)n be the molecular formula of compound A.

Relative molecular mass of A = n(5 x 12.0 + 10 x 1.0 + 16.0)

= 86n

i.e. 86n = 86.0

n = 1

∴ the molecular formula of compound A is C5H10O.

c) Compound A gives an orange precipitate with 2,4-dinitrophenylhydrazine. Thus,

compound A is either an aldehyde or a ketone.

Compound A gives no observable change with acidified aqueous solution of potassium

dichromate. Thus, compound A is a ketone, not an aldehyde.

Possible structural formulae of compound A:

d) i) The C2H5+ ion is responsible for the peak at m/e = 29.

The CH3CH2CO+ ion is responsible for the peak at m/e = 57.

ii) The structure of compound A is as follows:

18 The molecular ion peak of each alcohol is at m/e = 74.

The major peak at m/e (M – 45) is due to the C2H5+ ion.

The major peak at m/e (M – 43) is due to the loss of a C3H7 radical from the molecular ion.

It can be deduced that alcohol B has a propyl group. Thus, its structural formula is

CH3CH2CH2CH2OH.

It can be deduced that alcohol A has an ethyl group. Thus, its structural formula is

CH3CH2CH(OH)CH3.



Topic 16 Unit 56

19 a) Suppose we have 100 g of compound X, so there are 77.8 g of carbon, 7.40 g of hydrogen

and 14.8 g of oxygen.


Mass of element


Number of moles

of atoms that

combine

= 6.84 mol

= 7.40 mol

= 0.925 mol

Simplest ratio of

atoms = 7.01 = 8.00 = 1.00

∴ the empirical formula of compound X is C7H8O.

i) The m/e of the molecular ion of compound X is 108.

ii) C6H5+ ion


is due to an O–H bond (acid).

In the mass spectrum of compound X, the molecular ion peak is at m/e = 108. Thus, its

relative molecular mass is 108.

Let (C7H8O)n be the molecular formula of compound X.

Relative molecular mass of X = n(7 x 12.0 + 8 x 1.0 + 16.0)

= 108n

i.e. 108n = 108

n = 1

∴ the molecular formula of compound X is C7H8O.

As compound X can undergo oxidation to give an acid, it can be deduced that compound X

is an alcohol or an aldehyde.

As the mass spectrum of compound X has a peak at m/e = 77 due to the C6H5+ ion, it can

be deduced that compound X contains a phenyl group.

Beside the phenyl group (–C6H5), the formula of the other group should be C2H3O. Thus,

compound X probably has a CH2OH group attached to the phenyl group.

The structure of compound X is as follows:

0.925 mol

0.925 mol

7.40 mol

0.925 mol

6.48 mol

0.925 mol

14.8 g

16.0 g mol–1

7.40 g

1.00 g mol–1

77.8 g

12.0 g mol–1



Topic 16 Unit 56

The structure of compound Y is as follows:

20 Suppose we have 100 g of compound A, so there are 66.7 g of carbon, 11.1 g of hydrogen and

the rest is oxygen (i.e. 22.2 g).


Mass of element


Number of moles

of atoms that

combine

= 5.56 mol

= 11.1 mol

= 1.39 mol

Simplest ratio of

atoms = 4.00 = 7.99 = 1.00

∴ the empirical formula of compound A is C4H8O.

In the mass spectrum of compound A, the molecular ion peak is at m/e = 72. Thus, its relative

molecular mass is 72.0.

Let (C4H8O)n be the molecular formula of compound A.

Relative molecular mass of A = n(4 x 12.0 + 8 x 1.0 + 16.0)

= 72n

i.e. 72n = 72.0

n = 1

∴ the molecular formula of compound A is C4H8O.

The difference between 72 and 57 is 15. Thus, the peak at m/e = 57 is probably due to the loss

of a methyl radical from the molecular ion.

The peak at m/e = 43 may be due to the CH3CO+ ion or the C3H7

+ ion.

The strong IR absorption at 1 720 cm–1

is due to a C=O bond.

Hence the structure of compound A

may be CH3CH2COCH3 or CH3CH2CH2CHO.

21 The absorption at 1 680 – 1 750 cm–1


The molecular ion peak of compound X is at m/e = 58. Thus, its relative molecular mass is

58.0.

Other than the oxygen atom (relative atomic mass = 16.0), there is probably 3 carbon atoms

and 6 hydrogen atoms in compound X.

Hence compound X is probably an aldehyde or a ketone.

The peak at m/e = 29 in the mass spectrum is due to the C2H5+ ion. Hence compound X

11.1 mol

1.39 mol

1.39 mol

1.39 mol

5.56 mol

1.39 mol

22.2 g

16.0 g mol–1

11.1 g

1.00 g mol–1

66.7 g

12.0 g mol–1



Topic 16 Unit 56

probably contains an ethyl group.

Thus, the structure of compound X is CH3CH2CHO.

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Suggested answers to in-text activities and exercisesintranet.stmgss.edu.hk/~ted/ccy/new_edition/SuggestedAnswers_56_E... · Suggested answers to in-text activities and unit-end exercises