The least sterically hindered position gives a 1245

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Unformatted text preview: group has been added to the benzene ring. The proton at 7.81 with only one small (meta) coupling must be between 9 10 Solutions Manual to accompany Organic Chemistry 2e the nitro group and the other ring and is marked on the two possible structures. Ha O2N HNO3 N H or H2SO4 N H N H O2N Hb You could argue that NH is ortho,para directing and so the second structure is more likely. But this is a risky argument as the reaction is carried out in strong acid solution where the nitrogen will mostly be protonated. It is safer to use the predicted δH from tables. Here we get: Proton ortho meta para predicted δH Ha NO2 = +0.95 CH2 = –0.14 NH = –0.25 7.73 Hb NO2 = +0.95 NH = –0.75 CH2 = –0.06 7.31 There’s not much difference but Ha at 7.73 is closer to the observed 7.81, so it looks as though the small amount of unprotonated amine directs the reaction. PROBLEM 8 What are the two possible isomeric products of this reaction? Which structure do you expect to predominate? What would be the bromination product from each? MeO A + B Br2 both C12H16O Br2 OH ? ? Purpose of the problem Getting you to think about alternative products and possible reactions on compounds that haven’t been made (yet). Suggested solution The reaction is a Friedel Crafts cyclisation, as you could have deduced by the simple loss of water. The resulting cation could cyclise in two ways, arbitrarily called A and B. Steric hindrance suggests that A would be the more likely product. 11 Solutions for Chapter 21 – Electrophilic Aromatic Substitution MeO MeO MeO H MeO H A OH2 MeO MeO B Bromination will go either ortho or para to the...
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This document was uploaded on 02/10/2014.

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