烃

Alkanes 烷烃 are saturated hydrocarbons (general formula $\text{C}_n\text{H}_{2n+2}$).

Making alkanes

• hydrogenation 氢化: add hydrogen to an alkene 烯烃, using a $\text{Pt}$ or $\text{Ni}$ catalyst and heat.
• cracking 裂化: break a long-chain alkane into shorter ones by heating with $\text{Al}_2\text{O}_3$.

Combustion

In combustion 燃烧 an alkane burns in oxygen:

• complete combustion 完全燃烧 (plenty of oxygen) gives carbon dioxide and water.
• incomplete combustion 不完全燃烧 (not enough oxygen) gives water plus toxic carbon monoxide 一氧化碳 ($\text{CO}$) and soot (carbon).

You are often asked to write the balanced equation. Balance it in a fixed order - carbon first, then hydrogen, and oxygen last - because oxygen is the only element left on just one side:

The 6 carbons fix $6\,\text{CO}_2$; the 14 hydrogens fix $7\,\text{H}_2\text{O}$; counting the oxygens on the right gives $12 + 7 = 19$, so the left needs $19 \div 2 = 9\tfrac{1}{2}$. A half of $\text{O}_2$ is perfectly acceptable in this equation - and if the question asks for whole numbers, simply double everything ($2\,\text{C}_6\text{H}_{14} + 19\,\text{O}_2 \rightarrow 12\,\text{CO}_2 + 14\,\text{H}_2\text{O}$).

Free-radical substitution

Alkanes react with chlorine or bromine by free-radical substitution 自由基取代, in ultraviolet light 紫外线. For ethane and chlorine the mechanism has three steps:

• initiation 引发 — UV light splits the halogen into two free radicals 自由基: $\;\text{Cl}_2 \rightarrow 2\,\text{Cl}\cdot$ This kind of break is homolytic fission 均裂: the bond splits evenly, one electron going to each atom, which is what makes two radicals. (The opposite, heterolytic fission 异裂, sends both electrons to one atom and makes a pair of ions - that is what happens in the polar mechanisms such as electrophilic addition.) Examiners ask for this word by name, so use it.
• propagation 增长 — radicals react and make new radicals:
  $$\text{Cl}\cdot + \text{C}_2\text{H}_6 \rightarrow \text{C}_2\text{H}_5\cdot + \text{HCl} \qquad \text{C}_2\text{H}_5\cdot + \text{Cl}_2 \rightarrow \text{C}_2\text{H}_5\text{Cl} + \text{Cl}\cdot$$
• termination 终止 — two radicals join, ending the chain: $\;\text{Cl}\cdot + \text{C}_2\text{H}_5\cdot \rightarrow \text{C}_2\text{H}_5\text{Cl}$

Why cracking is useful, and why alkanes are unreactive

Cracking turns heavy fractions of crude oil 原油 into more useful, lower-$M_r$ alkanes and alkenes. (A fraction 馏分 is a group of molecules with a similar boiling-point range.)

Alkanes are generally unreactive, especially towards polar reagents. This is because the C–H and C–C bonds are strong and have little polarity 极性, so there is no charge to attract an attacking species.

Environmental effects

Burning alkanes in an internal combustion engine gives off carbon monoxide, oxides of nitrogen and unburnt hydrocarbons. A catalytic converter removes these by turning them into harmless gases.

Alkenes have a C=C double bond (general formula $\text{C}_n\text{H}_{2n}$). The double bond is the functional group, so alkenes are reactive.

Making alkenes

• elimination 消去 of $\text{HX}$ from a halogenoalkane 卤代烷, using $\text{NaOH}$ dissolved in ethanol, with heat.
• dehydration 脱水 of an alcohol 醇, using a hot $\text{Al}_2\text{O}_3$ catalyst or concentrated sulfuric acid.
• cracking of a longer-chain alkane.

Reactions of alkenes

Most reactions are electrophilic addition 亲电加成 across the double bond:

There are also two oxidation reactions with acidified $\text{KMnO}_4$:

• cold, dilute $\text{KMnO}_4$ adds two $\text{–OH}$ groups to give a diol 二醇.
• hot, concentrated $\text{KMnO}_4$ breaks the C=C bond right apart, and what each half turns into tells you exactly where the double bond used to be.

That second reaction is worth learning properly, because it is how the exam asks you to locate a double bond in a large molecule. Oxidation is written with $[\text{O}]$, meaning "an oxygen atom from the oxidising agent". Cut the molecule at the C=C, then look at what each of the two carbons was carrying:

The pattern is simply how many hydrogens that carbon had: none stops at a ketone, one is pushed on to an acid, and two are oxidised all the way to $\text{CO}_2$.

Worked example. Give the products when $(\text{CH}_3)_2\text{C=CHCH}_3$ is heated with hot concentrated acidified $\text{KMnO}_4$. Cut at the C=C and take each carbon separately. The left carbon carries two methyl groups and no hydrogen, so it stops at a ketone: $(\text{CH}_3)_2\text{C=O}$, which is propanone. The right carbon carries one methyl and one H, so it goes on to a carboxylic acid: $\text{CH}_3\text{COOH}$, ethanoic acid. So:

Now run it backwards, which is the way the question is usually set. Given the products, rebuild the alkene by putting the two carbonyl carbons back together as a C=C: a ketone means that carbon had two alkyl groups, an acid means one alkyl and one H, and $\text{CO}_2$ means the chain ended in $\text{=CH}_2$. Getting $\text{CO}_2$ is the strongest clue of all - it can only come from a terminal double bond.

Test for a C=C bond

Shake the compound with orange bromine water 溴水. An alkene decolourises it (turns it colourless) by electrophilic addition. An alkane does not.

Addition polymerisation

In addition polymerisation 加成聚合, many alkene molecules join into one long chain, with no other product. Ethene gives poly(ethene). The long-chain product is a polymer 聚合物.

The mechanism and Markovnikov's rule

In electrophilic addition, the electron-rich C=C attracts an electrophile 亲电试剂 (an electron-pair acceptor). This forms a positive intermediate called a carbocation 碳正离子, which the negative part then attacks. The syllabus uses two examples: bromine with ethene, and hydrogen bromide with propene.

Where the $\delta+$ comes from. $\text{Br}_2$ is not polar on its own. As it comes close to the C=C, the cloud of electrons in the double bond repels the electrons in Br–Br. This makes an induced dipole 诱导偶极: the nearer Br becomes $\delta+$ and the further Br $\delta-$. $\text{HBr}$ already has a permanent dipole, with H $\delta+$.

Drawing it for full marks (Br$_2$ + ethene):

1. the dipole on Br–Br, $\delta+$ on the Br nearer the C=C.
2. a curly arrow from the C=C double bond to the $\delta+$ Br.
3. a curly arrow from the Br–Br bond to the $\delta-$ Br.
4. the carbocation, with a full $+$ on the carbon that did not gain the Br.
5. a curly arrow from a lone pair on $\text{Br}^-$ to that $\text{C}^+$, and the product.

In the first step the Br–Br bond breaks heterolytically: both electrons go to one Br, which leaves as $\text{Br}^-$.

Common mistakes.

• An arrow starting at a carbon atom, not at the middle of the double bond.
• An arrow from the $\delta+$ Br towards the C=C. The electrons move from the double bond to the electrophile.
• The $+$ drawn on the carbon that has just bonded to Br. The other carbon has lost its share of the C=C pair.
• An arrow starting from the minus sign of $\text{Br}^-$. Draw the lone pair and start there.

With bromine water the carbocation meets far more water molecules than $\text{Br}^-$ ions, so a lone pair on water often attacks it instead. The main product is then 2-bromoethanol, not 1,2-dibromoethane. The colour still goes, because the $\text{Br}_2$ is used up either way.

Alkyl groups push electrons towards the positive carbon — this is the inductive effect 诱导效应. So a carbocation with more alkyl groups is more stable: tertiary is more stable than secondary, which is more stable than primary. When $\text{HBr}$ adds to propene, the more stable carbocation forms, so hydrogen adds to the carbon that already has more hydrogens. This pattern is Markovnikov's rule 马氏规则.

Worked example. Predict the major product when $\text{HBr}$ adds to propene, $\text{CH}_3\text{CH=CH}_2$. The $\text{H}^{+}$ adds first, and it adds in whichever way makes the more stable carbocation. Adding the $\text{H}$ to the end carbon puts the positive charge on the middle carbon, giving a secondary carbocation, which is stabilised by electron-releasing alkyl groups on two sides. Adding it to the middle carbon would leave a less stable primary carbocation. The bromide ion then attacks the secondary carbocation, so the major product is 2-bromopropane. That is Markovnikov's rule - but quote the reason (carbocation stability: tertiary > secondary > primary), because the rule on its own is not the explanation.

• Alkanes: free-radical substitution needs UV light — show initiation, propagation and termination with the radical dots.
• Alkenes: electrophilic addition via a carbocation; draw curly arrows from the $\text{C}=\text{C}$.
• Markovnikov: H adds to the carbon with more H's, via the more stable carbocation (alkyl groups are electron-releasing).
• Test for $\text{C}=\text{C}$: bromine water decolourises (orange → colourless) — state the observation.