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Chemistry · Ch 14 — Haloalkanes and Haloarenes

Nomenclature

14.3.1

Nomenclature

Haloalkanes can be named by the common system -- naming the alkyl group followed by 'halide' (e.g. methyl iodide, ethyl bromide, tert-butyl chloride) -- or by the IUPAC system, applying the general nomenclature rules (Unit 11) with the halogen cited as a 'halo' prefix and the lowest locants given to the substituents. Vinylic and allylic examples follow the same rule set once the parent alkene chain is n …

Table 14.1Common and IUPAC names of haloalkanes and polyhalogen compounds
S.NoStructural formulaCommon nameIUPAC name
1CH3Imethyl iodideIodomethane
2CH3CH2Brethyl bromideBromoethane
3CH3CH2CH2Fn-propyl fluoride1-Fluoropropane
4CH3-CHF-CH3iso-propyl fluoride2-Fluoropropane
5CH3-CH2-CH2-CH2-Cln-butyl chloride1-Chlorobutane
6CH3-CH(CH3)-CH2-Cliso-butyl chloride1-Chloro-2-methylpropane
7CH3-CH(Cl)-CH2-CH3sec-butyl chloride2-Chlorobutane
8CH3-C(Cl)(CH3)-CH3tert-butyl chloride2-Chloro-2-methylpropane
9(CH3)3C-CH2-Brneo-pentyl bromide1-Bromo-2,2-dimethylpropane
10CH2=CH-Clvinyl chlorideChloroethene
11CH2=CH-CH2-Brallyl bromide3-Bromopropene
12ClCH2-CH2ClEthylene chloride (ethylene dichloride)1,2-Dichloroethane
Misc evaluate-yourself-1Evaluate Yourself 1 -- IUPAC names of three drawn structures

Worked out. Book's practice box (no printed solution). i) CH2=C(CH3)-CH2-Cl; ii) the trisubstituted alkene CH3-CH=C(CH3)-CH(I)-CH3 drawn with the C1-methyl and the C4 iodo-chain on opposite sides of the double bond; iii) CH3-CH=CH-CH(F)-CH3 drawn trans (the C1-methyl and the C4 fluoro-chain on opposite sides). Working through them: (i) numbering from the =CH2 end gives the double bond the lower locant 1, so the name is 3-chloro-2-methylprop-1-ene; (ii) the higher-priority groups (C1-methyl on C2; the iodo-bearing C4 chain on C3) sit on opposite sides, so it is (E)-4-iodo-3-methylpent-2-ene; (iii) by the same CIP logic it is (E)-4-fluoropent-2-ene (own solutio …

Misc evaluate-yourself-2Evaluate Yourself 2 -- structures from IUPAC names

Worked out. Book's practice box (no printed solution): draw i) 1-bromo-4-ethylcyclohexane (a cyclohexane ring with Br on C1 and an ethyl group on C4, a 1,4/para relationship around the ring); ii) 1,4-dichlorobut-2-ene (ClCH2-CH=CH-CH2Cl, a symmetric four-carbon chain with the double bond in the middle and Cl on both end carbons); iii) 2-chloro-3-methylpentane (CH3-CHCl-CH(CH3)-CH2-CH3) (own solutions, not printed in the textbook …