This is not unreasonable as the t butyl group

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Unformatted text preview: n how these compounds are formed and suggest the order in which the two substituents are added to form compound C. O O Cl AlCl3 + benzene + A B C O Purpose of the problem Detailed analysis of a revealing example of the Friedel- Crafts reaction. Suggested solution The expected reaction to give A is a simple Friedel- Crafts acylation with the usual acylium ion intermediate. O O O Cl H A Product B must arise from a t- butyl cation and the only way that might be formes is by loss of carbon monoxide from the original acylium ion. Such a reaction happens only when the resulting carbocation is reasonably stable. O H B Solutions for Chapter 21 – Electrophilic Aromatic Substitution The main product C comes from the addition of both these electrophiles, but which adds first? The ketone in A is deactivating and meta directing but the t- butyl group in B is activating and para- directing so it must be added first. O + AlCl3 Cl C O That answers the question but you might like to go further. Both A and C are formed by the alkylation of benzene as the first step. The decomposition of the acylium ion is evidently faster than the acylation of benzene. However, when B reacts further, it is mainly by acylation as only a small amount of di- t- butyl benzene is formed. Evidently the decomposition of the acylium ion is slower than the acylation of B! This is not unreasonable as the t- butyl group accelerates electrophilic attack—but it is a dramatic demonstration of that acceleration. PROBLEM 7 Nitration of this heterocyclic compound with the usual HNO3/H2SO4 mixture gives a single nitration product with the 1H NMR spectrum shown below. Suggest which product is formed and why. C8H8N2O2 N H δH 3.04 (2H, t, J 7 Hz) 3.68 (2H, t, J 7 Hz) 6.45 (1H, d, J 8 Hz) 7.28 (1H, broad s) 7.81 (1H, d, J 1 Hz) 7.90 (1H, dd, J 8, 1 Hz) Purpose of the problem Revision of NMR and an attempt to convince you that the methods of chapter 21 can be applied to molecules you’ve not met before. Suggested solution The two 2H triplets and the broad NH signal show that the heterocyclic ring is intact. One nitro...
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