Concepts: exercise practice and examination questions

5.1 Exercise practice

The chapter has one exercise set with questions on naming, isomer counting, ozonolysis, combustion, conformations and aromatic substitution. The items below work through the ones that carry the most marks.

Exercise 9.1. Methane gives ethane during chlorination because the chloromethane that is first formed can react with a methyl radical, and two methyl radicals can combine to give ethane. The termination steps of the free radical chain are the source of the ethane, and the same kind of coupling explains the small amounts of other products.

Exercise 9.2. For the naming items the longest chain containing the principal double or triple bond is chosen first, numbering starts from the end that gives that bond the lower number, and substituents are listed alphabetically with their locants. A chain with two double bonds is named as a diene, and a triple bond takes the suffix yne.

Exercise 9.3. C4H8 with one double bond has three structural isomers: but-1-ene, but-2-ene and 2-methylprop-1-ene. C5H8 with one triple bond has three structural isomers: pent-1-yne, pent-2-yne and 3-methylbut-1-yne; the terminal ones show the acetylide test and the internal one does not.

Exercise 9.4. Ozonolysis of pent-2-ene gives ethanal and propanal. 3,4-Dimethylhept-3-ene gives butan-2-one and pentan-2-one. 2-Ethylbut-1-ene gives butan-2-one and methanal. 1-Phenylbut-1-ene gives benzaldehyde and propanal.

Exercise 9.5. If the products are ethanal and pentan-3-one, the alkene is rebuilt by removing the oxygen from each carbonyl and joining the two carbon atoms with a double bond; the structure is 3-methylhex-3-ene, because ethanal contributes two carbons and pentan-3-one contributes five, with the double bond between the two carbonyl carbons.

Exercise 9.8. The combustion equations are written by balancing carbon dioxide and water for each formula. Butane, C4H10, gives four carbon dioxide and five water; pentene, C5H10, gives five carbon dioxide and five water; hexyne, C6H10, gives six carbon dioxide and five water; toluene, C7H8, gives seven carbon dioxide and four water. The oxygen coefficients follow from the balance: thirteen by two for butane, fifteen by two for pentene, seventeen by two for hexyne and nine for toluene.

Exercise 9.9. Hex-2-ene has a cis form, with both alkyl groups on the same side, and a trans form, with them on opposite sides. The trans isomer is more symmetrical and packs better in the liquid, so it has the higher boiling point, while the cis isomer is slightly more polar and has the higher density and dipole moment.

Exercise 9.10. Benzene is extraordinarily stable because its six pi electrons are delocalised over the ring, so the actual molecule is a resonance hybrid of the two Kekule structures and has resonance energy. Any reaction that would break the ring would lose this stabilisation, which is why benzene prefers substitution to addition.

5.2 Final revision points

For the examination, keep the reagents together with the reactions and keep the formulae together with the class of the compound. The combustion calculations come from the general equations in the formula sheet, and the isomer counts come from drawing every skeleton before counting. The aromatic questions are answered from the three aromaticity conditions and from the directing effect of the substituent, and the carcinogenicity point is one short paragraph that can be attached to any question on benzene's stability.

5.3 Two more worked items

Aromatics and substituents. A monosubstituted benzene reacts at the positions decided by the group already present. The methyl group of toluene is electron releasing, so it activates the ring and sends the next electrophile to the ortho and para positions; that is why toluene on Friedel-Crafts methylation gives a mixture of ortho and para xylene. A nitro group does the opposite: it withdraws electron density through its inductive and resonance effects, so nitrobenzene is less reactive than benzene and the next group enters at the meta position. A halogen is a special case, because it withdraws electrons inductively but donates them by resonance; overall the ring becomes less reactive, but the resonance donation still sends the new group to the two neighbouring positions and to the opposite one. These three examples cover the directive-effect questions of the exercise.

Counting and naming. For an isomer-counting question, draw the carbon skeletons first and place the multiple bond in each distinct position, checking each structure against the molecular formula. For a naming question, number the longest chain from the end that gives the principal multiple bond the lower locant, then list the substituents in alphabetical order with their locants. The same discipline answers the questions on the products of ozonolysis, because the carbonyl fragments must be joined at the carbons that carried the double bond, and the resulting structure must be checked against the formula of the original alkene.