Download Chapter 9 Compounds of carbon

Worked solutions to student book questions
Chapter 9 Compounds of carbon
E1.
Write the structural formula of the heptane and 2,2,4-trimethylpentane molecules.
EA1.
E2.
What is the name of the straight-chain hydrocarbon that has same molecular formula
as 2,2,4-trimethylpentane?
EA2.
C8H18 octane
Q1.
Give the meaning of the following terms:
a homologous series
b structural isomers
c structural formula
d semistructural formula
e saturated
f unsaturated
A1.
a
b
c
d
e
g
Homologous series: A series of organic compounds in which each members
differs by a CH2– group from the previous members. Members of a homologous
series have similar chemical properties.
Structural isomers: Molecules with the same formula but different molecular
structures.
Structural formula: A formula that represents the three-dimensional arrangement
of atoms in a molecules.
Semi-structural formula: A formula that shows the sequence of atoms in a
molecules without indicating the three dimensional arrangement of the atoms
within the molecule.
Saturated: Carbon compounds that only contain single bonds between the carbon
atoms.
Unsaturated: Carbon compound that contain at least one double or triple bond
between adjacent carbon atoms.
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Worked solutions to student book questions
Chapter 9 Compounds of carbon
Q2.
Identify the homologous series to which each of the following belongs:
a C3H8
b C2H4
c C5H10
d C8H18
e CH3(CH2)5CH3
f CH3CH=CHCH2CH3
A2.
a
b
c
d
e
f
alkane
alkene
alkene
alkane
alkane
alkene
Q3.
Draw the structural formula of:
a ethane
b propene
c butane
d methylbutane
e 3-ethyloctane
f 2,3-dimethylhexane
A3.
a
ethane
b
propene
c
butane
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Chapter 9 Compounds of carbon
d
methylbutane
e
3-ethyloctane
f
2,3-dimethylhexane
Q4.
Explain why there is only one compound corresponding to the formula C3H8 while
there are over 70 compounds corresponding to the molecular formula C10H22.
A4.
The number of possible ways of arranging carbon atoms within a molecule increases
as the number of carbon atoms increase.
Q5.
Describe and explain the difference in bonding and structure between alkanes and
alkenes.
A5.
In alkanes, there is a single bond between all the carbon atoms. There is a tetrahedral
arrangement of the bonds formed by each carbon atom. Alkenes contain at least one
double bond between adjacent carbon atoms. The bonds formed around the doublebonded carbon atoms lie in the same plane.
Q6.
Why does the alkene homologous series begin with ethene, C2H4, while the alkanes
start with methane, CH4?
A6.
Alkenes have a double bond between two adjacent carbon atoms. Hence the smallest
alkene is ethene, CH2=CH2.
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Worked solutions to student book questions
Chapter 9 Compounds of carbon
Q7.
Give the systematic names for:
a CH3OH
b HCOOH
c CH3Cl
d CH3NH2
A7.
a
b
c
d
methanol (1 carbon, hydroxyl, –OH, functional group)
methanoic acid (1 carbon, carboxyl (or acid) , –COOH, functional group)
chloromethane (1 carbon , chloro, Cl, functional group)
methanamine or methylamine (1 carbon, amine, NH2, functional group)
Q8.
Write the systematic names of:
a CH3CH2CH2Cl
b CH2ClCH2CH3
c CH3(CH2)3CH2OH
d CH3(CH2)3CHOHCH2CH3
e CH3 CH2CH2NH2
A8.
a
b
c
d
e
1-chloropropane
1-chloropropane
pentan-1-ol
heptan-3-ol
propan-1-amine or 1-propylamine
Q9.
Write the semistructural formulas of:
a 2-hexanol
b 1-chloropentane
c butan-1-amine
d 2-methylhexane
e 4-nonanol
A9.
a
b
c
d
e
CH3CHOHCH2CH2CH2CH3
CH2CH2CH2CH2CH2Cl
CH3CH2CH2CH2NH2
CH3CH(CH3)CH2CH2CH2CH3
CH3CH2CH2CHOHCH2CH2CH2CH2CH3
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Worked solutions to student book questions
Chapter 9 Compounds of carbon
Q10.
Why don’t we use the names 1-chloroethane or 3-propanol?
A10.
Since 1-chloroethane and 2-chloroethane are identical, the molecule is simply named
chloroethane. In the case of 3-propanol, this molecule is identical to 1-propanol and it
is called 1-propanol by convention. (The numbers used are always the smaller values.)
Q11.
The semi-structural formulas of some organic compounds is given below. For each
compound:
i identify the homologous series to which it belongs
ii give its systematic name.
a CH3(CH2)5CH2Cl
b CH3(CH2)2CHCl(CH2)2CH3
c CH3CHOH(CH2)4CH3
d CH3(CH2)4COOH
e CH3(CH2)2CH(NH2)CH3
f (CH3)2CHCH2CH3
g (CH3)2C=CH2
A11.
Formula
CH3(CH2)5CH2Cl
CH3(CH2)2CHCl(CH2)2CH3
CH3CHOH(CH2)4CH3
CH3(CH2)4COOH
CH3(CH2)2CH(NH2)CH3
(CH3)2CHCH2CH3
(CH3)2C=CHCH3
Homologous
series
chloroalkanes
chloroalkanes
alkanols
carboxylic acids
amines
alkanes
alkenes
Name
1-chloroctane
4-chloroheptane
2-heptanol
hexanoic acid
pentan-2-amine or 2-pentylamine
2-methylbutane
2-methylbut-2-ene
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Worked solutions to student book questions
Chapter 9 Compounds of carbon
Q12.
Draw the structural formula and name one structural isomer that has the molecular
formula:
a C5H11Cl
b C4H9OH
c C2H5COOH
d C3H7COOH
e C5H11NH2
A12.
There are a number of possible answers. Some sample answers are provided.
a 1-chloropentane
or
3-methyl-2-chlorobutane
b
1-butanol
c
propanoic acid
d
butanoic acid
e
pentan-1-amine
pentan-1-amine
or
2-butanol
or
methyl propanoic acid
or
pentan-2-amine
pentan-2-amine
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Worked solutions to student book questions
Chapter 9 Compounds of carbon
Chapter review
Q13.
Give the systematic names for:
a CH3(CH2)7CH2Cl
b CH3CH2COOH
A13.
a
b
1-chlorononane
propanoic acid
Q14.
Write the semi-structural formulas of:
a 2-chloroheptane
b butan-2-ol
c decane-3-amine
d butanoic acid
A14.
a
b
c
d
CH3CHCl(CH2)4CH3
CH3CHOHCH2CH3
CH3CH2CHNH2CH2CH2CH2CH2CH2CH2CH3 or CH3CH2CHNH2(CH2)6CH3
CH3(CH2)2COOH
Q15.
Draw the structural formulas of:
a butan-2-ol
b pentanoic acid
c 2-chloropentane
d propene
e hexan-3-amine
f octane
A15.
a
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Worked solutions to student book questions
Chapter 9 Compounds of carbon
b
c
d
e
f
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Chapter 9 Compounds of carbon
Q16.
Write semistructural formulas and systematic names for the following substances:
a
b
c
A16.
a
b
c
CH3CH2COOH; propanoic acid
CH3CHOHCH3; propan-2-ol
HCOOH; methanoic acid
Q17.
Draw and name all possible isomers of:
a C3H7Cl
b C5H10 (an alkene)
c C5H12
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Worked solutions to student book questions
Chapter 9 Compounds of carbon
A17.
a
1-chloropropane, 2-chloropropane
b
1-pentene, 2-pentene, 2-methyl-1-butene, 2-methyl-2-butene
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Chapter 9 Compounds of carbon
c
pentane, 2-methylbutane, 2,2-dimethylpropane
Q18.
Draw the structural formulas of five isomers of C5H11Cl.
A18.
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Worked solutions to student book questions
Chapter 9 Compounds of carbon
Q19.
Table 9.7 gives the molecular masses and boiling points of butane, propan-1-ol and
chloroethane. Explain why the boiling points differ even though the compounds have
similar molecular masses.
Table 9.7
Compound
Butane
Propan-1-ol
Chloroethane
Molecular mass
58
60
65
Boiling point (°C)
–0.5
97.0
12.5
A19.
The differences in boiling points is indicative of the strength of bonds between
molecules. There a weak dispersion forces between butane molecules. In
chloroethane, the bond between chlorine and carbon is polar and there are dipoledipole bonds between chloroethane molecules resulting in it having a higher boiling
point than butane. Hydrogen bonds exist between the propan-1-ol molecules due to
the electronegative nature of the oxygen atom. These bonds are stronger than the
bonds between chloroalkane molecules and the bonds between butane molecules.
Q20.
Explain why alkanols such as methanol and ethanol are soluble in water in all
proportions whereas the alkanols higher in the homologous series such as 1-octanol
are insoluble.
A20.
Hydrogen bonds form between the polar hydroxy function group in alkanol in alkanol
and water molecules. The alkyl group is non polar and does not interact with water
molecules. As the size of the alkyl group increases, the solubility of the alkanol in
water decreases.
Q21.
Write an equation for the ionisation of methanoic acid in water.
A21.
HCOOH(l) + H2O(l) → HCOO–(aq) + H3O+(aq)
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Worked solutions to student book questions
Chapter 9 Compounds of carbon
Q22.
With the aid of a diagram, explain why methylamine is soluble in water.
A22.
Bonds around the N atom in methylamine are polar due to the highly electronegative
nature of the N atom. Methylamine is soluble in water because hydrogen bonds form
between methylamine and water molecules.
Q23.
Why would ethanamine be more soluble in water than butanamine?
A23.
The highly polar amino functional group can form hydrogen bonds with water
molecules but as the non-polar hydrocarbon chain grows longer, as in butanamine, the
solubility is reduced.
Q24.
Write equations for the the reaction between ethanoic acid solution and:
a sodium hydroxide solution
b solid sodium carbonate
c magnesium
A24.
a
b
c
CH3COOH(aq) + NaOH(aq) → CH3COONa(aq) + H2O(l)
2CH3COOH(aq) + Na2CO3(s) → 2CH3COONa(aq) + CO2(g) +H2O(l)
2CH3COOH(aq) + Mg(s) → (CH3COO)2Mg + H2(g)
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Worked solutions to student book questions
Chapter 9 Compounds of carbon
Q25.
Prepare a poster that summarises the rules for the systematic naming of carbon
compounds.
A25.
The poster should include reference to the following.
Alkanes:
• The first part of the name indicates the number of carbon atoms.
• The last part ends in -ane.
• The stem of the name of branched alkanes is derived from longest alkane
hydrocarbon chain. The name and number indicating the position of any akyl
side chains is then added.
Alkenes:
• The names of alkenes end in -ene.
• The stem of the systematic name of alkenes is based on the longest carbon
chain that contains the double bond.
• The position of the double bond is indicated by the number of the first carbon
atom involved in the double bond. By convention, numbering starts from the
end nearest the double bond. The rules for naming of any side chains are
similar to those for the alkanes.
Functional groups:
• Numbers are used to indicate the position of the functional group attached to
the carbon chain
• Chloroalkanes: the name starts with chloro- followed by the name of the
alkane from which it is derived.
• Alkanols are named by dropping the ‘e’ at the end of the hydrocarbon name
and replacing it with ‘ol’.
• Carboxylic acid: The name of carboxylic acids is determined from the total
number of carbon atoms in the molecule, and -oic acid is added at the end of
the name.
• Amines are named by adding -amine to the alkane stem.
Q26.
Prepare notes for a PowerPoint presentation that outlines the reasons why carbon is
able to form so many different compounds.
A26.
The notes might include.
• Carbon forms stable bonds with other carbon atoms, and can form large chains
as well as cyclic structures.
• Carbon forms single, double or triple bonds with other carbon atoms.
• Carbon is able to bond with the atoms of other elements, H, O, N, P, Cl, and
with groups of atoms (functional groups).
• There are different organic compounds that have the same molecular formula
– isomers.
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