Halogenoalkanes, alcohols and spectra

Swap one hydrogen in an alkane for a halogen or an –OH group and the molecule suddenly has a polar bond that other species can attack. This topic follows what happens next: how halogenoalkanes are substituted or lose HX to give alkenes, how alcohols are oxidised, halogenated and dehydrated, how the products are separated and purified in the lab, and how mass spectra and infrared spectra tell you which molecule you have made.

Halogenoalkanes: names and classes

A halogenoalkane is an alkane in which one or more hydrogen atoms have been replaced by halogen atoms. The name is built on the longest carbon chain, with a prefix for each halogen: fluoro-, chloro-, bromo- or iodo-. Number the chain from the end that gives the substituents the lowest numbers, and write the prefixes in alphabetical order (bromo before chloro before methyl). Multiplying prefixes such as di- and tri- are ignored when you put the prefixes in order.

• CH₃CH₂CH₂Br: 1-bromopropane
• CH₃CHClCH₃: 2-chloropropane
• (CH₃)₃CBr: 2-bromo-2-methylpropane
• CH₃CHBrCHClCH₃: 2-bromo-3-chlorobutane (b comes before c, and numbering from either end gives 2 and 3, so bromo, first alphabetically, takes the lower number)
• CH₂BrCH₂Br: 1,2-dibromoethane

Halogenoalkanes are classified by looking at the carbon atom that carries the halogen and counting how many other carbon atoms are bonded to it. This matters because, as you will see, the class changes both the mechanism and the speed of the reactions.

Halogenoalkanes are polar molecules but they cannot form hydrogen bonds with water, so they do not mix with it. Their boiling temperatures rise with chain length and from chloro to bromo to iodo, because bigger molecules with more electrons have stronger London forces between them. Branched isomers boil lower than straight-chain ones because their molecules touch over a smaller area.

Why halogenoalkanes react: the polar C–X bond

Every halogen is more electronegative than carbon, so the bonding pair in the C–X bond is pulled towards the halogen. The carbon becomes slightly positive (δ+) and the halogen slightly negative (δ−). That electron-poor carbon is a target for nucleophiles.

In a nucleophilic substitution the nucleophile replaces the halogen, which leaves as a halide ion, X⁻. The C–X bond breaks heterolytically: both electrons of the bond go to the halogen.

Bond strength decides the rate

You might expect the most polar bond, C–F, to react fastest because its carbon is the most positive. Experiments show the opposite. The step that controls the rate is breaking the C–X bond, and the bonds get weaker down the group: as the halogen atom gets bigger, the shared pair of electrons is further from its nucleus and is held less strongly.

Nucleophilic substitution reactions

The same pattern works for every nucleophile: the lone pair attacks the δ+ carbon, the C–X bond breaks and the halide ion leaves. What changes is the product, and the solvent and conditions chosen to make that product.

With aqueous hydroxide ions: making alcohols

Heating a halogenoalkane under reflux with aqueous potassium or sodium hydroxide replaces the halogen with –OH:

CH₃CH₂CH₂Br + OH⁻ → CH₃CH₂CH₂OH + Br⁻   or   CH₃CH₂CH₂Br + KOH → CH₃CH₂CH₂OH + KBr

The hydroxide ion carries a full negative charge, so it is attracted far more strongly to the δ+ carbon than a neutral water molecule is. That is why OH⁻ substitutes much faster than water does. For a primary halogenoalkane the whole change happens in one step:

With cyanide ions: making nitriles

Warming a halogenoalkane under reflux with potassium cyanide dissolved in ethanol gives a nitrile. The cyanide ion attacks through the lone pair on its carbon atom, so the reaction adds one carbon to the chain, which makes it useful in synthesis:

CH₃CH₂CH₂Br + CN⁻ → CH₃CH₂CH₂CN + Br⁻   (1-bromopropane gives butanenitrile: four carbons, counting the one in –C≡N)

The mechanism is the same as with hydroxide: an arrow from the lone pair on the carbon of :C≡N⁻ to the δ+ carbon, and an arrow from the C–Br bond to Br. Potassium cyanide is extremely toxic, so it is handled in a fume cupboard and never acidified (that would release HCN gas).

With ammonia: making amines

Ammonia's lone pair on nitrogen lets it act as a nucleophile too. The halogenoalkane is heated with an excess of concentrated ammonia in ethanol in a sealed tube. Ammonia is a gas that would simply boil out of an open flask, so sealing the tube keeps its concentration high; the pressure inside rises as it is heated.

CH₃CH₂CH₂Br + 2NH₃ → CH₃CH₂CH₂NH₂ + NH₄Br

The amine product still has a lone pair on nitrogen, so it is a nucleophile as well. It can attack another halogenoalkane molecule to form a secondary amine, then a tertiary amine, and finally a quaternary ammonium salt. Using a large excess of ammonia makes it far more likely that a halogenoalkane molecule meets an ammonia molecule than an amine molecule, which keeps these further substitutions, and the loss of yield they cause, to a minimum.

Comparing rates of hydrolysis

Hydrolysis means splitting a molecule by reaction with water. Water is a weak nucleophile, but warm halogenoalkanes do react with it slowly:

R–X + H₂O → R–OH + H⁺ + X⁻

If silver nitrate solution is present, each halide ion released is caught straight away as an insoluble silver halide, Ag⁺(aq) + X⁻(aq) → AgX(s). The silver ions do not attack the halogenoalkane themselves; they just reveal how fast the halide ions are being set free. So the sooner a precipitate appears, the faster the hydrolysis. Ethanol is added as a solvent that mixes with both the halogenoalkane and the aqueous silver nitrate, so the reactants are in one phase.

S_N2 and S_N1: primary versus tertiary

Nucleophilic substitution can happen in two different ways, and which one a halogenoalkane uses depends on its class. The names say what happens: Substitution, Nucleophilic, and the number of species involved in the slowest (rate-determining) step.

• S_N2 (primary halogenoalkanes): one step. The nucleophile attacks as the C–X bond breaks, passing through a transition state containing both the halogenoalkane and the nucleophile. The rate depends on both concentrations: rate = k[RX][OH⁻].
• S_N1 (tertiary halogenoalkanes): two steps. First the C–X bond breaks on its own to give a carbocation and a halide ion (slow). Then the nucleophile attacks the carbocation (fast). Only the halogenoalkane is in the slow step: rate = k[RX].
• Secondary halogenoalkanes can react by either route, and often use a mixture of both.

Why does structure decide the route? Two effects work together.

• Carbocation stability: alkyl groups push electron density towards a neighbouring positive carbon (a positive inductive effect), spreading out its charge. A tertiary carbocation has three alkyl groups doing this and is stable enough to form; a primary carbocation has only one and is too unstable, so primary halogenoalkanes cannot take the S_N1 route.
• Crowding: in S_N2 the nucleophile must reach the carbon from the side opposite the halogen. Three bulky methyl groups on a tertiary carbon block that approach, while the small hydrogen atoms on a primary carbon leave it open.

Elimination: making alkenes

Change the solvent and the same reagent does something different. Heated under reflux with potassium hydroxide dissolved in ethanol, a halogenoalkane loses a hydrogen halide and forms an alkene. Here the hydroxide ion acts as a base: it removes an H⁺ from a carbon next to the C–X carbon, a C=C double bond forms between those two carbons, and the halide ion leaves.

CH₃CH₂CH₂Br + KOH → CH₃CH=CH₂ + KBr + H₂O

In practice both reactions compete and you usually get a mixture; the conditions just tip the balance. If the halogen is not on the end of an unsymmetrical chain, the hydrogen can be removed from a carbon on either side, so more than one alkene forms. 2-Bromobutane gives three products, because but-2-ene has E and Z isomers:

Alcohols: names, classes and shape

Alcohols contain the hydroxyl group, –OH, and the open-chain ones with one –OH have the general formula C_nH_2n+1OH. Their names end in -ol, with a number showing where the –OH is: butan-2-ol is CH₃CH(OH)CH₂CH₃. With two or three –OH groups, keep the "e" of the alkane name and add -diol or -triol: ethane-1,2-diol (HOCH₂CH₂OH) and propane-1,2,3-triol (glycerol). When the molecule has a group that takes the suffix, such as a carboxylic acid, the –OH becomes the prefix hydroxy-: CH₃CH(OH)COOH is 2-hydroxypropanoic acid.

Alcohols are classified exactly like halogenoalkanes, by counting the carbon atoms bonded to the carbon that carries the –OH:

• Primary: propan-1-ol, CH₃CH₂CH₂OH (one carbon attached; methanol, with none, behaves like a primary alcohol)
• Secondary: propan-2-ol, CH₃CH(OH)CH₃ (two carbons attached)
• Tertiary: 2-methylpropan-2-ol, (CH₃)₃COH (three carbons attached)

Around each carbon there are four bonding pairs, so the bonds are arranged tetrahedrally at about 109.5°. The oxygen has two bonding pairs and two lone pairs. Lone pairs repel more strongly than bonding pairs, so they squeeze the C–O–H angle down to about 104.5°, giving a bent shape like water.

The O–H group lets alcohol molecules hydrogen bond to one another, so alcohols boil far higher than alkanes of similar mass, and the small ones mix completely with water. Branching lowers the boiling temperature, as for halogenoalkanes: butan-1-ol boils at 118 °C but its more compact isomer 2-methylpropan-2-ol boils at 82 °C.

Reactions of alcohols

Combustion

Alcohols burn in plenty of air with a clean, almost colourless flame to give carbon dioxide and water, which is why ethanol is used as a fuel:

C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l)

Halogenation: swapping –OH for a halogen

The –OH group can be replaced by a halogen, turning the alcohol into a halogenoalkane. Different reagents are used for each halogen.

The reaction with PCl₅ gives off steamy fumes of hydrogen chloride, which turn damp blue litmus red and form white smoke with ammonia. This is a test for an –OH group. It only works on a dry sample, because water and carboxylic acids contain –OH and give the same fumes.

Oxidation with acidified potassium dichromate(VI)

Potassium dichromate(VI), K₂Cr₂O₇, acidified with dilute sulfuric acid oxidises primary and secondary alcohols. As it does so, the orange dichromate(VI) ion, Cr₂O₇²⁻, is reduced to the green chromium(III) ion, Cr³⁺. Equations use [O] to stand for oxygen supplied by the oxidising agent.

Oxidation removes two hydrogen atoms: the one on the –OH and one from the carbon carrying the –OH. A C=O bond forms in their place. This is why the class of alcohol matters so much.

A tertiary alcohol has no hydrogen atom on the carbon bonded to the –OH, so a C=O bond cannot form without breaking a C–C bond, and dichromate cannot do that. A ketone is not oxidised further for the same reason. An aldehyde still has a hydrogen on its C=O carbon, so it is easily oxidised on to a carboxylic acid.

So how do you stop at the aldehyde? Use boiling temperatures. Propanal (49 °C) has no O–H group, so its molecules cannot hydrogen bond to each other, and it boils far lower than propan-1-ol (97 °C) or propanoic acid (141 °C). Gentle warming in a distillation apparatus lets the aldehyde escape the flask as soon as it forms, before it meets more oxidising agent. Heating under reflux does the opposite: everything is returned to the flask until it has all been oxidised to the acid.

Telling the products apart

The Fehling's and Benedict's tests work because an aldehyde can be oxidised and a ketone cannot. The Cu²⁺ ions in the alkaline blue solution oxidise the aldehyde and are themselves reduced to copper(I):

CH₃CHO + 2Cu²⁺ + 5OH⁻ → CH₃COO⁻ + Cu₂O + 3H₂O

Dehydration: making alkenes

Heating an alcohol with concentrated phosphoric(V) acid, H₃PO₄, removes a molecule of water and gives an alkene. The acid is a catalyst and dehydrating agent, and the reaction is an acid-catalysed elimination. The alkene, with its much lower boiling temperature, can be distilled off as it forms.

CH₃CH₂CH₂OH → CH₃CH=CH₂ + H₂O

As with halogenoalkanes, an –OH group in the middle of an unsymmetrical chain can lose a hydrogen from either neighbouring carbon. Butan-2-ol gives but-1-ene and both E- and Z-but-2-ene. Making alkenes this way from ethanol produced by fermentation gives a route to polymers such as poly(ethene) that does not depend on crude oil.

Practical techniques for organic liquids

Heating under reflux

Many organic reactions are slow and their reactants are volatile and flammable. Heating under reflux lets you heat a mixture for a long time without losing anything: a vertical condenser turns the vapours back into liquid, which drips back into the flask.

• Never seal the top of the condenser. Heating a closed system builds up pressure until the apparatus bursts.
• Anti-bumping granules give the liquid many small places to form bubbles, so it boils smoothly instead of in sudden violent bursts.
• Heat electrically (a heating mantle or water bath), not with a Bunsen flame, because most organic liquids catch fire easily.

Distillation

Distillation separates a liquid from a mixture by boiling it off and condensing the vapour somewhere else. It is used to remove a product as it forms (the aldehyde above) and to purify a liquid at the end.

• The thermometer bulb must be level with the side arm, so it measures the vapour leaving the flask, not the liquid boiling below.
• Collect only the fraction that distils within a degree or two of the product's boiling temperature.
• Water enters the condenser at the lower end so the jacket stays full of cold water.
• When drawing apparatus, don't draw lines across the joints between flask, still head and condenser (that would mean they were sealed off), and draw the condenser's water jacket as a closed outer tube with two open water connections.

Separating, washing and drying

• Separating funnel: two liquids that don't mix form two layers. Run off the lower, denser layer through the tap and keep whichever layer holds your product. If you are not sure which layer is which, add a little water and see which one gets bigger.
• Washing: shake the organic layer with sodium hydrogencarbonate solution to neutralise any acid left in it. CO₂ forms, so invert the funnel and open the tap often to release the pressure. Washing with water or saturated sodium chloride solution removes water-soluble impurities.
• Drying: add an anhydrous salt such as anhydrous sodium sulfate, magnesium sulfate or calcium chloride, swirl, and leave until the liquid is clear and the solid no longer clumps. A drying agent must absorb water without dissolving in or reacting with the product. Then decant or filter the liquid off.
• Solvent extraction: shaking a mixture with a solvent that dissolves the product better (iodine moves from water into cyclohexane, for example) concentrates it in one layer, which can then be separated and distilled.
• Boiling temperature: a pure liquid boils at a fixed, known temperature, so measuring it is a check on purity and identity. It is not proof on its own, since different liquids can share a boiling temperature, and it changes with atmospheric pressure.

Mass spectra of organic compounds

In a mass spectrometer, a sample molecule M is hit by high-energy electrons, which knock one electron out of it. That leaves a positive ion with an unpaired electron, the molecular ion:

M(g) → [M]^+•(g) + e⁻    for example   CH₃CH₂CH₂OH → [CH₃CH₂CH₂OH]^+• + e⁻

The dot shows the unpaired electron and is often left out. The ion has a charge of +1, so its m/z value equals its mass. The molecular ion peak, normally the peak with the highest m/z, therefore gives the relative molecular mass of the compound.

Fragmentation

Many molecular ions have so much excess energy that a covalent bond breaks. One piece keeps the positive charge and the other becomes an uncharged free radical:

[M]^+• → X⁺ + Y^•

Only charged particles are deflected and detected, so the spectrum shows X⁺ but never Y^•. Fragment ions that are relatively stable, such as carbocations stabilised by alkyl groups and acylium ions [RCO]⁺, form most often and give the tallest peaks. The tallest peak of all is called the base peak.

Structural isomers have the same molecular ion but break up in different places, so their fragments tell them apart. Pentan-2-one and pentan-3-one both have M_r = 86. Only pentan-2-one has a CH₃–C=O group, so only it gives a large peak at m/z = 43:

[CH₃COCH₂CH₂CH₃]^+• → [CH₃CO]⁺ + ^•CH₂CH₂CH₃   (m/z = 43)

Pentan-3-one has an ethyl group on each side of the C=O, so it gives [CH₃CH₂CO]⁺ at m/z = 57 and CH₃CH₂⁺ at m/z = 29 instead:

[CH₃CH₂COCH₂CH₃]^+• → [CH₃CH₂CO]⁺ + ^•CH₂CH₃   (m/z = 57)

Infrared spectroscopy

Covalent bonds are not rigid: they constantly stretch and bend. Each kind of bond vibrates at its own natural frequency, which depends on the masses of the atoms and the strength of the bond. When infrared radiation of exactly that frequency shines on the molecule, the bond absorbs it and vibrates more strongly. An infrared spectrometer passes a range of IR frequencies through a sample and records how much of each gets through. Dips (troughs) in the transmittance show which bonds are present.

Positions are given as wavenumbers, the number of waves per centimetre, \( \tilde{\nu} = 1/\lambda \), in cm⁻¹. A bigger wavenumber means a higher frequency. Strong bonds and bonds to light hydrogen atoms absorb at high wavenumbers.

These ranges are approximate; in the exam, always quote the range from the data booklet you are given. Above about 1500 cm⁻¹ the troughs are few and easy to assign, so this is the region for identifying functional groups. Below 1500 cm⁻¹ lies the fingerprint region: a complicated pattern of bending and skeletal vibrations that is unique to each compound. A computer can match it against a database of known spectra to identify the substance exactly.

Infrared absorption and the greenhouse effect

The Earth's surface, warmed by sunlight, radiates energy back out as infrared. Bonds in molecules of carbon dioxide, water vapour and methane absorb some of this IR in exactly the way described above, then re-emit it in all directions, some back towards the ground. This traps energy in the lower atmosphere, so these are greenhouse gases. Nitrogen and oxygen, which make up 99% of dry air, cannot absorb IR because their vibrations never change their polarity. Carbon dioxide is non-polar overall, but its asymmetric stretching and bending vibrations do create a temporary dipole, so it absorbs strongly.

Quick checks

Interactive diagrams