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Suggested solution the basic mechanisms for the

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Unformatted text preview: say, HNO3 and H2SO4, each of these compounds forms a single nitration product. What is its structure? Explain your answer with at least a partial mechanism. CO2H NHAc O Me Purpose of the problem Revision of the basic nitration mechanism and extension to compounds where selectivity is an issue. Suggested solution The basic mechanisms for the formation of NO2+ and its reaction with benzene appear on p. 476 of the textbook. Benzoic acid has an electron- withdrawing substituent so it reacts in the meta position. The second compound is activated in all positions by the weakly electron donating alkyl groups (all positions are either ortho or para to one of these groups) but will react at one of the positions more remote from the alkyl groups because of steric hindrance. CO2H CO2H O2N NO2 The remaining two compounds have competing ortho,para- directing substituents but in each case the one with the lone pair of electrons (N or O) is a more powerful director than the simple alkyl group. In the first case nitrogen directs ortho but in the second oxygen activates both ortho and para and steric hindrance makes the para position marginally more reactive. 3 4 Solutions Manual to accompany Organic Chemistry 2e NHAc NHAc NO2 NO2 Me Me O2N O O NO2 PROBLEM 3 How reactive are the different sites in toluene? Nitration of toluene produces the three possible products in the ratios shown. What would be the ratios if all the sites were equally reactive? What is the actual relative reactivity of the three sites? You could express this as x:y:1 or as a:b:c where a+b+c = 100. Comment on the ratio you deduce NO2 Me HNO3 Me Me NO +2 + H2SO4 Me + NO2 59% yield 4% yield 37% yield Purpose of the problem A more quantitative assessment of relative reactivities. Suggested solution As there are two ortho and two meta sites, the ratio if all were equally reactive would be 2:2:1 o:m:p. The observed reactivity is 30:2:37 or 15:1:18 or 43:3:54 depending on how you expressed it. The ortho and para positions are roughly equally reactive because the methyl group is electron- donating....
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