Alkanes

Alkanes are saturated hydrocarbons. This means that they contain only carbon and hydrogen atoms and they contain no carbon-carbon double bonds.

The alkane homologous series has the general formula of $C_{n}H_{2n+2}$ . An exception is the cycloalkanes which lose two hydrogens so that the carbon can form another C-C bond; cycloalkanes have the general formula of $C_{n}H_{2n}$

Physical Properties

There are a few things you will need to know about their physical properties in an exam:

Boiling/Melting Point

As the relative molecular mass, or number of carbon atoms in an alkane increases, so does its boiling or melting point:

Alkane	Molecular Mass	Molecular Formula	Boiling Point (K)
Methane	16	CH₄	109
Ethane	30	C₂H₆	185
Propane	44	C₃H₈	231
Butane	58	C₄H₁₀	273
Pentane	72	C₅H₁₂	309

As you should be aware, alkanes are held together by induced dipole-induced dipole (ID-ID) forces. The larger a molecule is, the more electrons it has. This means it can form larger dipoles and its ID-ID forces will be larger. It will therefore take more energy to break the bonds and so the boiling and melting point will be higher.

Some alkanes can be branched instead of straight-chained. Boiling point decreases as a molecule becomes more branched. This is because branched molecules cannot pack so tightly together, so their ID-ID forces must act over larger distances (intensity of the ID-ID forces decreases) and thus require less energy to break. For example, hexane has five isomers:

Isomer	Structural Formula	Boiling Point (K)
Hexane	CH₃CH₂CH₂CH₂CH₂CH₃	342
3-Methylpentane	CH₃CH₂CH(CH₃)CH₂CH₃	337
2-Methylpentane	CH₃CH(CH₃)CH₂CH₂CH₃	333
2,3-Dimethylbutane	CH₃CH(CH₃)CH(CH₃)CH₃	331
2,2-Dimethylbutane	CH₃CH(CH₃)(CH₃)CH₂CH₃	323

Reactions

The alkanes are fairly unreactive, as the C-H bond is non-polar. However they undergo three reactions.

Reactions of alkanes

Take the alkane C₄H₁₀. It will undergo combustion to form CO₂ + H₂O.

It will undergo radical substitution reactions (e.g. with Cl₂ to create C₄H₉Cl + HCl)

It will also undergo cracking reactions to create a smaller alkane and at least one alkene, in this case, C₂H₆ + C₂H₄ or CH₄ + C₃H₆

The most important of these being combustion. Alkanes are very volatile and burn easily when in the presence of plentiful amounts of oxygen and thus are major fuels.

Cracking

Longer chain molecules have higher boiling points and are more difficult to ignite. These long chain molecules can be broken into shorter chain (more useful) molecules through catalytic cracking. Any alkane from $C_{4}$ to $C_{50}$ can be cracked. As it is catalytic cracking it requires a catalyst, previously this catalyst was either $SiO_{2}$ or $Al_{2}O_{3}$ , but now zeolite is preferred. These reactions also need high temperatures (773 K or 450 °C is usually used).

The products from cracking can vary. For example you can create several moles of one alkene and hydrogen from cracking an alkane if done right. However you can also create a mixture of alkenes & alkanes or just a mixture of alkenes.

Example: $C_{8}H_{18}\rightarrow \;4C_{2}H_{4}+H_{2}$

Under conditions of 450 °C and with an $Al_{2}O_{3}$ catalyst present.

Radical substitution

Alkanes are very unreactive due to the non-polar C-H bond, so most reactions are not possible. However, radical substitutions are possible due to the reactivity a radicals. Radical substitution involves a halogen and UV light to initiate homolytic fission.

An example with methane.

Initiation reaction:

$Cl_{2}\rightarrow \;Cl^{\bullet }+Cl^{\bullet }$

Propagation:

$Cl^{\bullet }+CH_{4}\rightarrow HCl+CH_{3}^{\bullet }$

This sets into motion the chain reaction, where one free radical reacts with a stable species to form another stable species and a free radical.

The next propagation steps:

$CH_{3}^{\bullet }+Cl_{2}\rightarrow CH_{3}Cl+Cl^{\bullet }$

Or

$CH_{3}^{\bullet }+CH_{3}Cl\rightarrow C_{2}H_{6}+Cl^{\bullet }$

This goes on until 2 radicals meet and combine to form a stable species. Once all the radicals have done this, the reaction terminates, hence why it is the termination step.

In this reaction we have 3 possible termination steps:

$CH_{3}^{\bullet }+Cl^{\bullet }\rightarrow CH_{3}Cl$

$Cl^{\bullet }+Cl^{\bullet }\rightarrow Cl_{2}$

$CH_{3}^{\bullet }+CH_{3}^{\bullet }\rightarrow C_{2}H_{6}$

Henceforth, the reaction ends.

Note that a free-radical hydrogen is never produced.

Problems

As we know from above, alkanes burn readily in oxygen and give off carbon dioxide, which is a weak greenhouse gas and thus a contributor to global warming.

Alkanes must burn in excess oxygen for complete combustion and the production of carbon dioxide:

CH₄ + 2 O₂ → CO₂ + 2 H₂O

Incomplete combustion comes from a reduced supply of oxygen and produces water and carbon monoxide:

CH₄ + 1½ O₂ → CO + 2 H₂O

Carbon monoxide is poisonous when inhaled. It binds irreversibly to haemoglobin in red blood cells and makes carboxyhaemoglobin. The red blood cells can no longer carry oxygen and this can cause suffocation if enough cells are carrying carboxyhaemoglobin.

UK regulations ensure that all gas equipment must be annually serviced and that all gas installations require adequate ventilation.

Incomplete combustion can also produce soot (carbon particulates):

CH₄ + O₂ → C + 2 H₂O

Alkenes

Alkenes are aliphatic hydrocarbons containing carbon-carbon double bonds and general formula C_nH_2n.

Naming Alkenes

Alkenes are named as if they were alkanes, but the "-ane" suffix is changed to "-ene".

propane: C₃H₈ (CH₃CH₂CH₃)
propene: C₃H₆ (CH₂=CHCH₃)

If it is unclear where the double bond is, then the carbons should be numbered in such a way as to give the first of the two double-bonded carbons the lowest possible number, and that number should precede the "-ene" suffix with a dash, as shown below.

correct: pent-2-ene (CH₃CH=CHCH₂CH₃)
incorrect: pent-3-ene (CH₃CH₂CH=CHCH₃)
The second one is incorrect because flipping the formula horizontally results in a lower number for the alkene.

If there is more than one double bond in an alkene, all of the bonds should be numbered in the name of the molecule. The numbers should go from lowest to highest, and be separated from one another by a comma. The IUPAC numerical prefixes (di-, tri-, tetra-, etc) are used to indicate the number of double bonds.

octa-2,4-diene: CH₃CH=CHCH=CHCH₂CH₂CH₃
deca-1,5-diene: CH₂=CHCH₂CH₂CH=CHCH₂CH₂CH₂CH₃

Note that the numbering of "2,4" above yields a molecule with two double bonds separated by just one single bond. Double bonds in such a condition are called "conjugated", and they represent an enhanced stability of conformation, so they are energetically favoured as reactants in many situations and combinations.

Diastereomerism

Restricted rotation

Because of the characteristics of π-bonds, alkenes have very limited rotation around the double bonds between two atoms. In order for the alkene structure to rotate the π-bond would first have to be broken - which would require about 250 to 300 kJ of energy per mole. For this reason alkenes have different chemical properties based on which side of the bond each atom is located.

For example, but-2-ene exists as two diastereomers:


(Z)-but-2-ene	(E)-But-2-ene

cis-but-2-ene	trans-but-2-ene

E-Z Notation

The C=C bond does not rotate. This is because rotation would break the π bond. Without rotation, the atoms attached to the C=C carbons are fixed in position and we can have two different stereoisomers.

Using cis/trans notation to describe the isomers, cis- means same side and trans- means opposite side. The cis/trans notation sometimes breaks down and is best used only if we can see H atoms on the same, or opposite sides.

The above example is pretty straight-forward. On the left, we have two hydrogen atoms on the same side, so it is cis-but-2-ene. And on the right, we have them on opposite sides, so we have trans-but-2-ene. So in this situation, the cis/trans notation works.

From the example above, how would you use cis and trans? Which is the same side and which is the opposite side? Whenever an alkene has 3 or 4 differing substituents, one must use E-Z nomenclature, coming from the German words, Entgegen (opposite) and Zusammen (same).

E: Entgegen, opposite sides of double bond
Z: Zusammen, same sides (zame zides) of double bond

Let's begin with (Z)-3-methylpent-2-ene. We begin by dividing our alkene into left and right halves. On each side, we assign a substituent as being either a high priority or low priority substituent. The priority is based on the atomic number of the substituents. So on the left side, hydrogen is the lowest priority because its atomic number is 1 and carbon is higher because its atomic number is 6.

On the right side, we have carbon substituents on both the top and bottom, so we go out to the next bond. On to the top, there's another carbon, but on the bottom, a hydrogen. So the top gets high priority and the bottom gets low priority.

Because the high priorities from both sides are on the same side, they are Zusammen (as a mnemonic, think 'Zame Zide').

Now let's look at (E)-3-methylpent-2-ene. On the left, we have the same substituents on the same sides, so the priorities are the same as in the Zusammen version. However, the substituents are reversed on the right side with the high priority substituent on the bottom and the low priority substituent on the top. Because the High and Low priorities are opposite on the left and right, these are Entgegen, or opposite.

The system takes a little getting used to and it's usually easier to name an alkene than it is to write one out given its name. But with a little practice, you'll find that it's quite easy.

Comparison of E-Z with cis-trans


(Z)-but-2-ene	(E)-but-2-ene

cis-but-2-ene	trans-but-2-ene

To a certain extent, the Z configuration can be regarded as the cis- isomer and the E as the trans- isomers. This correspondence is exact only if the two carbon atoms are identically substituted.

In general, cis-trans should only be used if each double-bonded carbon atom has a hydrogen atom (i.e. R-CH=CH-R').

IUPAC Gold book on cis-trans notation.

IUPAC Gold book on E-Z notation.

Properties

Alkenes are molecules with carbons bonded to hydrogens which contain at least one carbon-to-carbon double bond, where the carbon atoms, in addition to an electron pair shared in a sigma (σ) bond, share one pair of electrons in a pi (π) bond between them.

The general formula for an aliphatic alkene is: C_nH_2n -- e.g. C₂H₄ or C₃H₆

Reactions

Preparation

There are several methods for creating alkenes.

Dehydrohalogenation of Halogenoalkanes

Halogenoalkanes are converted into alkenes by dehydrohalogenation: elimination of the elements of hydrogen halide. Dehydrohalogenation involves removal of the halogen atom together with a hydrogen atom from a carbon adjacent to the one bearing the halogen. It is not surprising that the reagent required for the elimination of what amounts to a molecule of acid is a strong base for example: alcoholic KOH.

In some cases this reaction yields a single alkene. and in other cases yield a mixture. 1-chlorobutane, for example, can eliminate hydrogen only from C-2 and hence yields only but-1-ene. 2-chlorobutane, on the other hand, can eliminate hydrogen from either C-l or C-3 and hence yields both but-1-ene and but-2-ene. Where the two alkenes can be formed, but-2-ene is the chief product.

Dehydration of alcohols

An alcohol is converted into an alkene by dehydration: elimination of a molecule of water. Dehydration requires the presence of an acid and the application of heat. It is generally carried out in either of two ways, heating the alcohol with sulfuric or phosphoric acid to temperatures as high as 200, or passing the alcohol vapor over alumina, Al₂O₃ , at 350-400, alumina here serving as a Lewis acid.

Ease of dehydration of alcohols : 3° > 2° > 1°

Where isomeric alkenes can be formed, we again find the tendency for one isomer to predominate. Thus, butan-2-ol, which might yield both but-2-ene and but-1-ene, actually yields almost exclusively the 2-isomer

The formation of but-2-ene from butan-1-ol illustrates a characteristic of dehydration that is not shared by dehydrohalogenalion: the double bond can be formed at a position remote from the carbon originally holding the -OH group. It is chiefly because of the greater certainty as to where the double bond will appear that dehydrohalogeation is often preferred over dehydration as a method of making alkenes.

Electrophilic Addition Reactions

Catalytic addition of hydrogen

Catalytic hydrogenation of alkenes produce the corresponding alkanes. The reaction is carried out under pressure in the presence of a metallic catalyst. Common industrial catalysts are based on platinum, nickel or palladium, but for laboratory syntheses, Raney nickel (formed from an alloy of nickel and aluminium) is often employed.

The catalytic hydrogenation of ethene to yield ethane proceeds thusly:

CH₂=CH₂ + H₂ + catalyst → CH₃-CH₃

Halogenation

Addition of elementary bromine or chlorine in the presence of an organic solvent to alkenes yield vicinal* dibromo- and dichloroalkanes, respectively. (*"Vicinal" means "adjacent" e.g. the halogen atoms become attached to carbon atoms 1 and 2.).

The decoloration of a solution of bromine in water is an analytical test for the presence of alkenes: CH₂=CH₂ + Br₂ → BrCH₂-CH₂Br

The reaction works because the high electron density at the double bond causes a temporary shift of electrons in the Br-Br bond causing a temporary induced dipole. This makes the Br closest to the double bond slightly positive and therefore an electrophile.

Hydrohalogenation

Addition of hydrohalic acids like HCl or HBr to alkenes yield the corresponding haloalkanes.

an example of this type of reaction is: CH₃CH=CH₂ + HBr → CH₃-CHBr-CH₃

If the two carbon atoms at the double bond are linked to a different number of hydrogen atoms, the halogen is found preferentially at the carbon with less hydrogen substituents (Markovnikov's rule).

Markovnikov's Rule

Before we continue discussing reactions, we need to take a detour and discuss a subject that is very important in alkene reactions, "Markovnikov's Rule." This is a simple rule stated in 1869 by the Russian Vladimir Markovnikov (1837-1904), as he was showing the orientation of addition of HBr to alkenes.

His rule states: "When an unsymmetrical alkene reacts with a hydrogen halide to give a halogenoalkane, the hydrogen adds to the carbon of the alkene that has the greater number of hydrogen substituents, and the halogen to the carbon of the alkene with the fewer number of hydrogen substituents" (This rule is often compared to the phrase: "The rich get richer and the poor get poorer." Aka, the Carbon with the most Hydrogens gets another Hydrogen and the one with the least Hydrogens gets the halogen)

This means that the nucleophile of the electrophile-nucleophile pair is bonded to the position most stable for a carbocation, or partial positive charge in the case of a transition state.

Examples

CH₂=CH-CH₃ + H-Br → CH₃-CHBr-CH₃

Here the Br attaches to the middle carbon and not the terminal carbon, because of Markovnikov's rule, and this is called a Markovnikov product.

Why it works

Markovnikov's rule works because of the stability of carbocation intermediates. Carbocations are planar molecules, with a carbon that has three substituents at 120° to each other and a vacant p-orbital that is perpendicular to the trisubstituent plane.

This leads to a stabilising effect called hyperconjugation. Hyperconjugation occurs between an unfilled p orbital and a filled C-H σ bond orbital next to each other†. The result is that the filled C-H σ orbital interacts with the unfilled p-orbital and stabilises the carbocation. Any alkyl group attached to the C⁺ ion can contribute a C-H σ bond orbital to the hyperconjugation. More alkyl groups, therefore, create more hyperconjugation.

In the ethyl carbocation illustrated, one C-H σ bond can be in the same plane as the unfilled p-orbital. The σ bond and the p-orbital can conjugate. If the C⁺ ion had a second or third alkyl group attached, a C-H σ bond from each alkyl group could also conjugate with the p-orbital, increasing the structure's stability.

†It is beyond the scope of A-level, but hyperconjugation can also occur in neutral molecules between an unfilled π bond orbital, or an "antibonding" π* orbital, and a filled C-H σ bond orbital.

The inductive effect is often used to explain the increased stability of secondary carbocations. Alkyl groups are supposed to release partial electron character onto the C⁺ ion, which makes it more stable. This explanation does not account for carbon being more electronegative than hydrogen; Why does replacing H with C allow C⁺ to take more electrons?

Oxidation

Alkenes are oxidised with a large number of oxidising agents. For example, alkaline potassium manganate(VII) will oxidise alkenes and turn from purple to green or brown in the process. Acidified potassium manganate(VII) will oxidise alkenes and turn from purple to pale pink.

Polymerisation

Polymerisation of alkenes is an economically important reaction which yields polymers of high industrial value, such as the plastics polyethylene and polypropylene.

Addition polymers, at this level, are always based on ethene. The C=C bond becomes a C-C bond, and the C-C units link to one another. Whatever atoms are attached to the C=C unit will be attached to the C-C chain.

References