Unit notes and practice questions

Grade 10 Chemistry unit 6: Hydrocarbons and their natural sources

Everything here is for unit 6 of the Grade 10 Chemistry textbook: what the chapter covers, the short notes, practice questions with their answers, and the lesson video where one exists.

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Cards: 10
Practice questions on unit 6

What unit 6 covers

1.Saturated Hydrocarbons: Alkanes (CnH2n+2)
2.Unsaturated Hydrocarbons: Alkenes, Alkynes
3.Aromatic Hydrocarbons: Benzene
4.Natural Sources of Hydrocarbons

Short notes for unit 6

The cards filed under this chapter on the Grade 10 Chemistry page. Open one at a time.

Alkane Structural Isomers FormulaFormula

N = 2^{n-4} + 1

N = 2^{n-4} + 1

$N$: number of structural isomers of the alkane
$n$: number of carbon atoms, valid for 4 to 7 carbons

When you use it

Use this shortcut to find the exact number of structural isomers for alkanes containing between 4 and 7 carbon atoms.

Watch out

This formula applies strictly for n between 4 and 7. It fails for methane, ethane, propane, and alkanes with 8 or more carbons.

Drafted from Grade 10 Chemistry, page 237 onwards, then checked twice before it went up

Complete Combustion of AlkanesFormula

\mathrm{C}_n\mathrm{H}_{2n+2} + \left(\frac{3n+1}{2}\right)\mathrm{O}_2 \rightarrow n\,\mathrm{CO}_2 + (n+1)\,\mathrm{H}_2\mathrm{O}

\mathrm{C}_n\mathrm{H}_{2n+2} + \left(\frac{3n+1}{2}\right)\mathrm{O}_2 \rightarrow n\,\mathrm{CO}_2 + (n+1)\,\mathrm{H}_2\mathrm{O}

$n$: number of carbon atoms in the alkane

When you use it

Use this general equation to balance any complete combustion reaction of an alkane producing carbon dioxide and water.

Watch out

When n is an even number, multiply the entire equation by 2 to clear the fractional coefficient for oxygen.

Drafted from Grade 10 Chemistry, page 237 onwards, then checked twice before it went up

Hydrocarbon Homologous FamiliesReference tableAlkane · Alkene · Cycloalkane

Class	General Formula	Saturation	Characteristic Feature
Alkane	$C_n H_{2n+2}$	Saturated	Single $C-C$ bonds only
Alkene	$C_n H_{2n}$	Unsaturated	At least one $C=C$ bond
Cycloalkane	$C_n H_{2n}$	Saturated	Single-bonded carbon ring

When you use it

Use this to classify open-chain and cyclic hydrocarbons, deduce molecular formulas from carbon counts, or check degree of saturation.

Watch out

Cycloalkanes share the general formula CnH2n with alkenes, but cycloalkanes are saturated rings containing only single bonds.

Drafted from Grade 10 Chemistry, page 237 onwards, then checked twice before it went up

Laboratory Preparation of AlkanesReference tableHydrogenation · Wurtz synthesis · Decarboxylation

Reaction Type	Reactants	Reagent or Catalyst	Product
Hydrogenation	Alkene $+ \mathrm{H}_2$	$\mathrm{Pt}$ , heat	Alkane
Wurtz synthesis	$2\mathrm{R-X} + 2\mathrm{Na}$	Heat	$\mathrm{R-R} + 2\mathrm{NaX}$
Decarboxylation	$\mathrm{RCOONa} + \mathrm{NaOH}$	$\mathrm{CaO}$ , heat	$\mathrm{R-H} + \mathrm{Na}_2\mathrm{CO}_3$

When you use it

Use this summary to identify reagents and products for standard synthetic routes to alkanes in the laboratory.

Watch out

Decarboxylation yields an alkane with one fewer carbon atom than the parent carboxylate salt, because one carbon leaves as sodium carbonate.

Drafted from Grade 10 Chemistry, page 237 onwards, then checked twice before it went up

The first ten alkanesReference tableMethane · Ethane · Propane · +7

Name	Carbon atoms	Molecular formula
Methane	1	$\mathrm{CH_4}$
Ethane	2	$\mathrm{C_2H_6}$
Propane	3	$\mathrm{C_3H_8}$
Butane	4	$\mathrm{C_4H_{10}}$
Pentane	5	$\mathrm{C_5H_{12}}$
Hexane	6	$\mathrm{C_6H_{14}}$
Heptane	7	$\mathrm{C_7H_{16}}$
Octane	8	$\mathrm{C_8H_{18}}$
Nonane	9	$\mathrm{C_9H_{20}}$
Decane	10	$\mathrm{C_{10}H_{22}}$

When you use it

Reach for this when a question names an alkane and asks for its formula, or gives a formula and asks for the name. These rows are the straight chains, because from butane on the formula alone no longer names one compound: C4H10 is butane, and it is also methylpropane.

Watch out

Losing the plus two. A student doubles the carbon count and stops, so hexane comes out as C6H12, which is a different compound carrying a double bond or a ring. Each end of the chain has one spare bond where the chain stops, so every alkane holds two hydrogens more than twice its carbons, and hexane is C6H14. Check any answer by counting: the hydrogen number is always even, and always two above the doubled carbon count.

Addition reactions of alkenesReference tableHydrogenation · Halogenation · Hydration · +2

Reaction	What is added, and the conditions	Product
Hydrogenation	$\mathrm{H_2}$ over heated $\mathrm{Ni}$ , or over $\mathrm{Pt}$ or $\mathrm{Pd}$ at room temperature	an alkane, $\mathrm{RCH_2-CH_3}$
Halogenation	$\mathrm{Br_2}$ at room temperature, with no catalyst	$\mathrm{RCHBr-CH_2Br}$ , and the orange colour fades
Hydration	$\mathrm{H_2O}$ with an acid catalyst	a secondary alcohol, $\mathrm{RCH(OH)-CH_3}$
Adding a hydrogen halide	$\mathrm{HBr}$ or $\mathrm{HCl}$ at room temperature	$\mathrm{RCHBr-CH_3}$ , the H joining the carbon that already carries more hydrogens
Polymerization	many $\mathrm{CH_2=CH_2}$ molecules with a catalyst	polyethene, $\mathrm{(-CH_2-CH_2-)_n}$

When you use it

Use when something is added across a C=C: hydrogen, bromine, water or a hydrogen halide. Addition breaks only the weaker pi bond of the double bond and puts two new single bonds in its place, which is why an alkene reacts in the cold where an alkane does not.

Watch out

Students put the OH on the wrong carbon. Water added to propene gives propan-2-ol, CH3CH(OH)CH3, and not propan-1-ol: the H joins the CH2 end that already carries two hydrogens, so the OH is left on the middle carbon. The same trap catches HBr, where propene gives 2-bromopropane. Ethene and but-2-ene are the exception, because both of their carbons carry the same number of hydrogens and only one product is possible.

The bromine water test for unsaturationMethodAn alkene, such as ethene · An alkyne, such as ethyne · An alkane, out of direct sunlight · +2

What is in the tube	What you see	What it tells you
An alkene, such as ethene	The orange brown colour goes at once, in the cold and with no catalyst	Unsaturated. The bromine adds across the double bond, and the Grade 10 equation is $\mathrm{C_2H_4 + Br_2 \rightarrow C_2H_4Br_2}$
An alkyne, such as ethyne	The colour goes as well, and twice as much bromine is used up	Unsaturated too. $\mathrm{C_2H_2}$ takes $\mathrm{2Br_2}$ and gives $\mathrm{C_2H_2Br_4}$ , so the test cannot tell a double bond from a triple one
An alkane, out of direct sunlight	The colour stays orange brown	Saturated
An alkane, in sunlight	The colour fades slowly and hydrogen bromide is given off	A false positive from substitution, so keep the tube away from the window
Phenol, or an aldehyde	The colour goes, and phenol also leaves a white precipitate	The sample reacts with bromine without being unsaturated, so the test does not name the group on its own

When you use it

Use to tell a saturated hydrocarbon from an unsaturated one. Add bromine water drop by drop and shake: if the orange brown colour goes at once in the cold, the sample holds a double or a triple bond.

Watch out

Two slips, and both cost marks. Writing that the solution turns clear instead of colourless: bromine water is already clear, meaning you can see through it, so the answer is orange brown going to colourless. Then reading a colourless tube as proof of a C=C. An alkyne does exactly the same thing, and uses twice the bromine doing it, so the only safe conclusion is that the compound is unsaturated.

Substitution or addition: how a hydrocarbon reactsRuleAlkanes, C_nH_2n+2, saturated · Alkenes C_nH_2n and alkynes C_nH_2n-2, unsaturated · Complete combustion, any hydrocarbon · +1

Family	How it reacts	Worked example
Alkanes, $\mathrm{C_nH_{2n+2}}$ , saturated	Substitution, because there is no bond to open. A chlorine atom takes the place of a hydrogen and $\mathrm{HCl}$ leaves. Sunlight starts a chain, so the real products are a mixture of $\mathrm{CH_3Cl}$ (chloromethane), $\mathrm{CH_2Cl_2}$ , $\mathrm{CHCl_3}$ and $\mathrm{CCl_4}$	$\mathrm{CH_4 + Cl_2 \rightarrow CH_3Cl + HCl}$ , the first step
Alkenes $\mathrm{C_nH_{2n}}$ and alkynes $\mathrm{C_nH_{2n-2}}$ , unsaturated	Addition. The multiple bond opens and both added atoms join the one molecule, so nothing is released	$\mathrm{CH_2=CH_2 + H_2 \rightarrow CH_3-CH_3}$ , over heated $\mathrm{Ni}$
Complete combustion, any hydrocarbon	Burning in plenty of oxygen gives carbon dioxide and water. A limited supply gives carbon monoxide or soot instead	$\mathrm{CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O}$
Esterification, a different family	A carboxylic acid condenses with an alcohol. Neither one is a hydrocarbon, the reaction is reversible so it is written with a two way arrow, and the concentrated $\mathrm{H_2SO_4}$ is a catalyst and a dehydrating agent rather than a reactant	$\mathrm{CH_3COOH + C_2H_5OH \rightarrow CH_3COOC_2H_5 + H_2O}$

When you use it

Use to decide which reaction a hydrocarbon does before you write a single equation. An alkane has no bond to open, so it substitutes and pushes out a small molecule; an alkene or an alkyne opens its multiple bond and takes everything in.

Watch out

Students mix the two mechanisms in one equation and write CH2=CH2 with Cl2 giving CH3CH2Cl and HCl, opening the double bond and releasing HCl as well. Count the atoms and the mistake shows at once: the left side holds C2H4Cl2 and the right side C2H6Cl2, two hydrogens that came from nowhere. An addition joins everything into one product and releases nothing, while a substitution always kicks out a small molecule such as HCl. So the alkene answer is CH2ClCH2Cl on its own, and the alkane answer is CH3Cl with HCl beside it.

Naming a branched-chain alkaneRuleFind the parent chain · Number the chain · Give every substituent its locant · +4

Step	What to do	Example
Find the parent chain	Take the longest continuous chain of carbon atoms, counting around corners rather than along the row it happens to be drawn in. If two chains tie on length, take the one carrying more substituents.	3-ethylhexane
Number the chain	Number the carbons from the end that reaches a branch first, so the first substituent gets the lowest number it can have.	2-methylhexane
Give every substituent its locant	Each substituent takes the number of the parent carbon it is bonded to. Two on the same carbon each keep their own number, so that number is written twice.	2,2-dimethylpropane
Order the substituents	Write them alphabetically in front of the parent name. Ignore di-, tri- and tetra- and the italic prefixes n-, sec- and tert- when alphabetising, so n-propyl files under p. The prefixes iso-, neo- and cyclo- are part of the name and do count.	4-ethyl-2,2-dimethylhexane
Punctuate the name	Put a hyphen (-) between a number and a word and a comma (,) between two numbers, then close with the parent chain and its suffix.	2,3-dimethylbutane
Break a tie in the numbering	When both ends give the same first locant, compare the two full sets of locants term by term and keep the set that is lower at the first point of difference. If the two sets are identical, the lower number goes to the substituent that comes first alphabetically.	3-ethyl-4-methylhexane
Name a branched substituent	Treat it as a chain of its own, number its carbon 1 at the atom bonded to the parent chain, and put its whole name in parentheses.	4-(1-methylethyl)heptane

When you use it

Use whenever you have to turn a drawn branched alkane into its IUPAC name, or read a name back into a structure. Every name is built in one fixed order: the locant that says where a substituent sits, the prefix that names it, the parent chain, and the family suffix, which is -ane for an alkane.

Watch out

Two slips cost the most marks. The first is taking the chain that happens to be drawn straight across the page as the parent chain. A five-carbon row with a propyl branch on carbon 3 is not 3-propylpentane, because the chain that turns down the branch is six carbons long, so the name is 3-ethylhexane. The second is alphabetising by the multiplier. Dimethyl files under m, so the correct name is 4-ethyl-2,2-dimethylhexane and 2,2-dimethyl-4-ethylhexane is wrong.

Primary, secondary, tertiary and quaternary carbonsRuleNone · Primary · Secondary · +2

Class	Other carbons bonded to it	Where you meet it
None	0	the carbon in methane, CH4, which is bonded to no other carbon
Primary	1	the two end carbons of butane, CH3-CH2-CH2-CH3, and both carbons in ethane
Secondary	2	the two middle carbons of butane
Tertiary	3	the central carbon of 2-methylpropane, (CH3)3CH, which holds three methyl groups and one hydrogen
Quaternary	4	the central carbon of neopentane, C(CH3)4, which carries no hydrogen at all

When you use it

Use when a question asks you to classify a carbon atom, or when a reaction depends on how substituted that carbon is. Count only the other carbon atoms bonded directly to it: one makes it primary, two secondary, three tertiary and four quaternary, and a hydrogen takes the class of the carbon it sits on.

Watch out

Students count hydrogens instead of carbons. A CH3 carbon carries three hydrogens, so they call it tertiary, when it is primary: it touches only one other carbon. Read the neighbours rather than the hydrogens, because the carbon with the most hydrogens is the least substituted and the carbon with none is the quaternary one. Two follow-on traps come after that. An oxygen, nitrogen or halogen on the carbon is never counted, so the carbon in (CH3)3C-OH holds three carbons and is tertiary. And amines break the pattern entirely, because there the class is counted on the nitrogen, so a primary amine RNH2 has nothing to do with a primary carbon.

Free-radical halogenation of an alkaneMethodChain initiation · Chain propagation, step 1 · Chain propagation, step 2 · +1

Stage	What happens	With methane and chlorine
Chain initiation	Ultraviolet light or strong heat splits the halogen molecule homolytically, one bonding electron going to each atom. It is the halogen bond that breaks, because it is the weakest bond present, and the free radical it leaves is neutral with one unpaired electron, which is what makes it so reactive.	Cl2 gives two Cl radicals
Chain propagation, step 1	A chlorine radical pulls a hydrogen atom off the alkane and leaves an alkyl radical behind.	CH4 + Cl radical gives CH3 radical + HCl
Chain propagation, step 2	The alkyl radical attacks a fresh halogen molecule, takes one atom from it and hands back a chlorine radical. That returned radical is why one initiation event can drive a long run of cycles, and why this is called a chain reaction.	CH3 radical + Cl2 gives CH3Cl + Cl radical
Chain termination	Two radicals meet and combine, and that chain stops. Finding a trace of ethane among the products of methane chlorination is the evidence that methyl radicals were present.	CH3 radical + CH3 radical gives ethane, CH3-CH3

When you use it

Use when a question asks how an alkane reacts with chlorine or bromine, or asks you to write the mechanism out. A substitution reaction replaces an atom or group in a molecule with a different atom or group, and an alkane will only do it once ultraviolet light or strong heat is supplied, through three stages in this order: chain initiation, chain propagation, chain termination.

Watch out

Students write CH4 + Cl2 giving CH3Cl + HCl and stop there, as though the product were pure chloromethane. A chlorine radical attacks CH3Cl as readily as it attacks CH4, so the real product is a mixture of CH3Cl, CH2Cl2, CHCl3 and CCl4, plus a trace of ethane from termination, and a large excess of methane is what pushes the mixture towards the mono-substituted product. The second slip is in the initiation step. Many write CH4 giving a methyl radical and a hydrogen radical, but the bond that breaks is Cl-Cl, the weakest bond in the mixture, and the radicals must be drawn neutral with a single dot, never as Cl+ and Cl-.

Physical properties of the alkanesReference tableState at 25 degrees C · Polarity · Solubility · +4

Property	What it does	Why
State at 25 degrees C	Methane to butane (C1 to C4) are gases, pentane to heptadecane (C5 to C17) are liquids, and octadecane (C18) and longer are waxy solids.	The liquid to solid line is set by melting point: heptadecane melts at about 22 degrees C and octadecane at about 28 degrees C.
Polarity	Alkanes are nonpolar and carry no significant dipole.	Carbon and hydrogen sit close together on the Pauling scale, 2.55 and 2.20 in current tables and 2.5 and 2.1 in older ones, so a C-H bond has almost no charge separation.
Solubility	Insoluble in water, and freely soluble in nonpolar solvents such as diethyl ether, benzene or carbon tetrachloride.	Like dissolves like.
Density	Liquid alkanes run about 0.62 to 0.79 g/mL and solid paraffin wax about 0.9 g/mL, all below water's 1.00 g/mL, so petrol, kerosene and paraffin oil spread as a film on the surface.	Density climbs with chain length and then levels off near 0.8 g/mL for the liquids, so it never reaches the density of water.
Boiling point	Rises steadily as the chain gets longer, by roughly 20 to 30 degrees C for each added CH2 at the short-chain end and less as the chain grows.	A bigger molecule has a larger, more easily distorted electron cloud, so the London dispersion forces between neighbouring molecules are stronger and more heat is needed to separate them.
Melting point	Rises with chain length overall, but zigzags on the way up instead of climbing in a straight line.	Even-numbered chains pack into the crystal lattice better than odd-numbered ones.
Branching	Lowers the boiling point. Pentane boils at 36 degrees C, 2-methylbutane at 28 degrees C and 2,2-dimethylpropane at 9.5 degrees C, and all three are C5H12.	A branched isomer is more compact and closer to spherical, so neighbouring molecules touch over less surface and the dispersion forces between them are weaker.

When you use it

Use when a question asks you to compare two alkanes, or to explain why one boils higher than another, floats on water or refuses to dissolve in it. Alkanes are nonpolar molecules held to each other only by London dispersion forces, and every property in this table follows from that.

Watch out

Students learn that branching lowers the boiling point and then apply it to the melting point as well. Melting depends on how neatly molecules stack in a crystal, and a compact symmetric branched molecule stacks better than a long floppy chain: 2,2-dimethylpropane melts at about -17 degrees C while straight-chain pentane melts at about -130 degrees C, even though the branched one boils 27 degrees C lower. Branching lowers the boiling point and often raises the melting point, so check which property the question is asking about before you reach for the rule.

Practice questions on unit 6

10 multiple-choice questions written against this chapter, each with its answer and the reason. Try one before you open the answer.

Question 1

Which scientist disproved the Vitalism theory in 1828 by synthesising urea from ammonium cyanate in the laboratory?

A.Charles-Adolphe Wurtz
B.Friedrich Wohler
C.John Dalton
D.Dmitri Mendeleev

Show the answer and why

Answer: B Friedrich Wohler

Why

In 1828, the German chemist Friedrich Wohler showed that the organic compound urea could be prepared by heating an aqueous solution of the inorganic compound ammonium cyanate.

Question 2

What is the general molecular formula for open-chain saturated hydrocarbons (alkanes)?

A.CnH2n
B.CnH2n-2
C.CnH2n+2
D.CnHn

Show the answer and why

Answer: C CnH2n+2

Why

Alkanes are saturated hydrocarbons containing only single bonds, and their general formula is C_nH_2n+2.

Question 3

In the alkane homologous series, each consecutive member differs from the preceding one by which constant group?

A.-CH3-
B.-CH2-
C.-CH-
D.-C2H4-

Show the answer and why

Answer: B -CH2-

Why

In a homologous series, each successive member differs from the previous member by a constant -CH₂- (methylene) unit, which has a molecular mass of 14.

Question 4

At room temperature (25 °C and 1 atm), which of the following alkanes exists in the gaseous state?

A.Propane (C3H8)
B.Pentane (C5H12)
C.Hexane (C6H14)
D.Octane (C8H18)

Show the answer and why

Answer: A Propane (C3H8)

Why

The first four unbranched alkanes (C₁ to C₄: methane, ethane, propane, and butane) are gases at room temperature. Alkanes from C₅ to C₁₇ are liquids.

Question 5

In a laboratory experiment in Addis Ababa, a student heats sodium acetate (CH₃COONa) with soda lime (NaOH and CaO). Which gas is collected?

A.Ethene
B.Ethane
C.Methane
D.Propane

Show the answer and why

Answer: C Methane

Why

Heating the sodium salt of a carboxylic acid with soda lime is a decarboxylation reaction. Sodium acetate produces methane: CH₃COONa + NaOH (with CaO and heat) → CH₄ + Na₂CO₃.

Question 6

How is a carbon atom classified if it is directly connected to three other carbon atoms in an alkane chain?

A.Primary (1°) carbon
B.Secondary (2°) carbon
C.Tertiary (3°) carbon
D.Quaternary (4°) carbon

Show the answer and why

Answer: C Tertiary (3°) carbon

Why

A carbon atom bonded directly to one carbon is primary (1°), to two carbons is secondary (2°), to three carbons is tertiary (3°), and to four carbons is quaternary (4°).

Question 7

What alkane is formed when methyl chloride reacts with sodium metal in a Wurtz synthesis reaction?

A.Methane
B.Ethane
C.Propane
D.Butane

Show the answer and why

Answer: B Ethane

Why

The Wurtz reaction couples two alkyl groups: 2CH₃Cl + 2Na → CH₃-CH₃ (ethane) + 2NaCl.

Question 8

According to the textbook classification, which of the following groups of hydrocarbons has no functional group?

A.Alkenes
B.Alkynes
C.Alkanes
D.Haloalkanes

Show the answer and why

Answer: C Alkanes

Why

Alkanes do not have a functional group because they contain only unreactive single C-C and C-H bonds, whereas alkenes have carbon-carbon double bonds and alkynes have carbon-carbon triple bonds.

Question 9

A farmer in Hawassa uses a biogas stove. Based on the balanced combustion equation, how many moles of oxygen (O₂) are needed to completely burn one mole of methane (CH₄)?

A.1 mole
B.2 moles
C.3 moles
D.4 moles

Show the answer and why

Answer: B 2 moles

Why

The balanced equation for the complete combustion of methane is CH₄ + 2O₂ → CO₂ + 2H₂O + energy. Thus, 2 moles of O₂ are consumed per mole of CH₄.

Question 10

Using the textbook formula 2^n-4 + 1 for calculating the number of isomers of alkanes with 4 to 7 carbon atoms, how many structural isomers does heptane (C₇H₁₆) have?