Chapter 9 Compounds of Carbon s1

Worked solutions to textbook questions 1

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 semi-structural formula

e saturated

f unsaturated

A1.

a Homologous series: A series of organic compounds in which each members differs by a CH2– group from the previous members. Members of an homologous series have similar chemical properties.

b Structural isomers: Molecules with the same formula but different molecular structures

c Structural formula: A formula that represents the three-dimensional arrangement of atoms in a molecules.

d 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.

e Saturated: Carbon compounds that only contain single bonds between the carbon atoms

g Unsaturated: Carbon compound that contain at least one double or triple bond between adjacent carbon atoms.

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 alkane

b alkene

c alkene

d alkane

e alkane

f 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

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 by the double bonded carbon atoms are planar.

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

Q7.

Give the systematic names for:

a CH3OH

b HCOOH

c CH3Cl

d CH3NH2

A7.

a methanol (1 carbon, hydroxy, –OH, functional group)

b methanoic acid (1 carbon, carboxy (or acid) , –COOH, functional group)

c chloromethane (1 carbon , chlorine, Cl, functional group)

d aminomethane or methylamine (1 carbon, amine, NH2, functional group)

Q8.

Write the systematic name of:

a CH3CH2CH2Cl

b CH2ClCH2CH3

c CH3(CH2)3CH2OH

d CH3(CH2)3CHOHCH2CH3

e CH3 CH2CH2NH2

A8.

a 1-chloropropane

b 1-chloropropane

c pentan-1-ol

d heptan-3-ol

e 1-aminopropane or 1-propylamine

Q9.

Write the semi-structural formula of:

a 2-hexanol

b 1-chloropentane

c 1-aminobutane

d 2-methylhexane

e 4-nonanol

A9.

a CH3CHOHCH2CH2CH2CH3

b CH2ClCH2CH2CH2CH3

c CH2NH2CH2CH2CH2

d CH3CH(CH3)CH2CH2CH2CH3

e CH3CH2CH2CHOHCH2CH2CH2CH2CH3

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 formula 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 / Homologous series / Name
CH3(CH2)5CH2Cl / chloroalkanes / 1-chloroctane
CH3 (CH2) 2CHCl(CH2) 2CH3 / chloroalkanes / 4-chloroheptane
CH3CHOH(CH2)4CH3 / alkanols / 2-heptanol
CH3 (CH2) 4COOH / carboxylic acids / hexanoic acid
CH3(CH2)2CH(NH2)CH3 / amines / 2-pentylamine
(CH3) 2CHCH2CH3 / alkanes / 1-methylbutane
(CH3) 2C=CHCH3 / alkenes / 2-methylbut-2-ene

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 or 2-butanol

c propanoic acid

d butanoic acid or methyl propanoic acid

e 1-aminopentane (or 1-pentylamine) or 2-amonopentane

(or 2-pentylamine)

Chapter review

Q13.

Give the systematic name for:

a CH3(CH2)7CH2Cl

b CH3CH2COOH

A13.

a 1-chlorononane

b propanoic acid

Q14.

Write the semi-structural formula of:

a 2-chloroheptane

b butan-2-ol

c 3-aminodecane or 3-decylamine

d butanoic acid

A14.

a CH3CHCl(CH2)4CH3

b CH3CHOHCH2CH3

c CH3CH2CHNH2CH2CH2CH2CH2CH2CH2CH3

d CH3(CH2)2COOH

Q15.

Draw the structural formula of:

a butan-2-ol

b pentanoic acid

c 2-chloropentane

d propene

e 3-aminohexane

f octane

A15.

Q16.

Write semi-structural formulas and systematic names for the following substances:

A16.

a CH3CH2COOH; propanoic acid

b CH3CHOHCH3; propan-2-ol

c HCOOH; methanoic acid

Q17.

Draw and name all possible isomers of:

a C3H7Cl

b C5H10 (an alkene)

c C5H12

A17.

a 1-chloropropane, 2-chloropropane

b 1-pentene, 2-pentene, 2-methyl-1-butene, 2-methyl-2-butene

c pentane, 2-methylbutane, 2,2-dimethylpropane

Q18.

Draw the structural formulas of five isomers of C5H11Cl.

A18.

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 / Molecular mass / Boiling point (°C)
Butane / 58 / –0.5
Propan-1-ol / 60 / 97.0
Chloroethane / 65 / 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 dipole-dipole 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)

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 electronegative nature of the N atom. Methylamine is soluble in water because hydrogen bonds form between methylamine and water molecules.

Q23.

Write an ionic equation for the reaction that occurs when:

a methylamine is mixed with water

b ethylamine is reacted with a dilute acid

A23.

a CH3NH2(l) + H2O(l)  CH3NH3+(aq) + OH–(aq)

methylammonium ion

b CH3CH2NH2(l) + H3O+(aq)  CH3CH2NH3+(aq) + H2O(l)

ethylammonium ion

Q24.

Write equations for the the reaction between ethanoic acid solution and:

a sodium hydroxide solution

b solid sodium carbonate

c magnesium

A24.

a CH3COOH(aq) + NaOH(aq) ® CH3COONa(aq) + H2O(l)

sodium ethanoate

b 2CH3COOH(aq) + Na2CO3(s) ® 2CH3COONa(aq) + CO2(g) +H2O(l)

c 2CH3COOH(aq) + Mg(s) ® (CH3COO)2Mg + H2(g)

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.

Heinemann Chemistry 2 (4th edition)