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Unformatted text preview: 82 ALCOHOLS AND ALKYL HALIDES need to complete the equation by realizing that HF is also formed in the ﬂuorination of alkanes. The
balanced equation is therefore:
4.39 (CF3)4C 12F2 12HF The reaction is free-radical chlorination, and substitution occurs at all possible positions that bear a
replaceable hydrogen. Write the structure of the starting material, and identify the nonequivalent
C C F Cl Cl H CH3 1,2-Dichloro-1,1-difluoropropane The problem states that one of the products is 1,2,3-trichloro-1,1-diﬂuoropropane. This compound
arises by substitution of one of the methyl hydrogens by chlorine. We are told that the other product
is an isomer of 1,2,3-trichloro-1,1-diﬂuoropropane; therefore, it must be formed by replacement of
the hydrogen at C-2.
C C F Cl Cl F H Cl 1,2,3-Trichloro-1,1-difluoropropane 4.40 C C F CH2Cl Cl Cl CH3 1,2,2-Trichloro-1,1-difluoropropane Free-radical chlorination leads to substitution at each carbon that bears a hydrogen. This problem essentially requires you to recognize structures that possess various numbers of nonequivalent hydrogens. The easiest way to determine the number of constitutional isomers that can be formed by chlorination of a particular compound is to replace one hydrogen with chlorine and assign an IUPAC
name to the product. Continue by replacing one hydrogen on each carbon in the compound, and
compare names to identify duplicates.
(a) 2,2-Dimethylpropane is the C5H12 isomer that gives a single monochloride, since all the hydrogens are equivalent.
CH3 CH3 CH3CCH3 Cl2 CH3CCH2Cl light CH3 CH3
2,2-Dimethylpropane (b) 1-Chloro-2,2-dimethylpropane The C5H12 isomer that has three nonequivalent sets of hydrogens is pentane. It yields three
isomeric monochlorides on free-radical chlorination.
1-Chloropentane CH3CH2CH2CH2CH3 Cl2 CH3CHCH2CH2CH3 Pentane Cl
3-Chloropentane Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website ...
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