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SolutionsNomenclature

1. Name these halides

  1. CH3CH2CH2Br
  2. (CH3)2CHCl
  3. CH3CHBrCH2CH3
  4. (CH3)3CI
  5. CH2Cl2
  6. CHCl3
  1. 1-Bromopropane (propyl bromide).
  2. 2-Chloropropane (isopropyl chloride).
  3. 2-Bromobutane.
  4. 2-Iodo-2-methylpropane (tert-butyl iodide).
  5. Dichloromethane (methylene chloride).
  6. Trichloromethane (chloroform).

2. Write the formula

  1. 1-Bromo-2-methylpropane
  2. 1,2-Dichloroethane
  3. 3-Bromo-2-chloropentane
  4. Iodocyclohexane
  1. (CH3)2CHCH2Br
  2. ClCH2CH2Cl
  3. CH3CHClCHBrCH2CH3
  4. C6H11I (a cyclohexane bearing an iodine)

3. Primary, secondary and tertiary

Classify these alkyl halides:

  1. 1-Chloro-2-methylpropane
  2. 2-Bromobutane
  3. 2-Chloro-2-methylbutane
  4. Chloromethane
  5. Bromocyclohexane
  1. Primary (although branched at the adjacent carbon).
  2. Secondary.
  3. Tertiary.
  4. Methyl halide.
  5. Secondary.

4. The order of prefixes

Why is CF3CH2CHClCH(CH3)CH2CH2CH(CH3)2 named 3-chloro-1,1,1-trifluoro-4,7-dimethyloctane and not 1,1,1-trifluoro-3-chloro-4,7-dimethyloctane?

Substituents are cited in alphabetical order, ignoring multiplying prefixes (di-, tri-…): chloro comes before fluoro and methyl. Numbering starts at the CF3 group because this gives the lowest locants (1,1,1,3,4,7).

5. Alkyl, aryl or alkenyl?

  1. Bromobenzene
  2. Vinyl chloride, CH2=CHCl
  3. Benzyl bromide, C6H5CH2Br
  4. 3-Chlorocyclohexene
  1. Aryl: bromine is bonded to a carbon of the aromatic ring.
  2. Alkenyl: chlorine is bonded to an sp2 carbon of the double bond.
  3. Alkyl: bromine is bonded to an sp3 carbon (a benzylic halide).
  4. Alkyl: chlorine is bonded to an sp3 carbon next to the double bond (an allylic halide).

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