Establish the molecular formula of one of the dibromoalkaces containing 85.11% bromine. Which isomers correspond to this molecular formula. Name them.
Let the formula of an unknown dibromoalkane be RBr2, where R is an unknown alkyl radical.
Determine the molecular weight of the unknown dibromoalkane:
M (RBr2) = M (Br2) * 100 / w (Br2) = 160 * 100 / 85.11 = 188 (amu).
Determine the molecular weight of the unknown alkyl radical:
M (R) = M (RBr2) – M (Br2) = 188 – 160 = 28 (amu).
This molecular weight of the radical corresponds to two CH2 groups:
M (CH2CH2) = 12 * 2 + 1 * 4 = 28 (amu)
or one CH3 group and one CH group:
M (CH3CH) = 12 * 2 + 1 * 4 = 28 (amu).
Thus, the formula for these isomeric dibromoalkanes: C2H4Br2
and it corresponds to BrCH2CH2Br (1,2-dibromoethane) and CH3CHBr2 (1,1-dibromoethane).
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