The partial rate factors for nitration of toluene (relative to one position of benzene) are $f_{o} = 42$ , $f_{m} = 2.5$ and $f_{p} = 58$ . Remember that toluene has two ortho positions, two meta positions and one para position. Calculate the percentage of the para isomer in the mononitration product mixture (to one decimal place).

Numerical answer type, answer in %, tolerance 0.2.

Answer: 39.5 %

The yield of each isomer is proportional to (number of positions) x (partial rate factor). Ortho: 2 x 42 = 84; meta: 2 x 2.5 = 5; para: 1 x 58 = 58; total = 147.0. Para percentage = 58/147.0 x 100 =

39.5%

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Answer: 39.5 %

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