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Chapter 17 Suggested solutions for Chapter 17 135 Problem 12 State with reasons whether these reactions will be either SN1 or SN2. 0 (b) 0 0 H Br N3 N3 0 0 OMe 0 0 0 (d) 0 n-PrOH 0 n-PrO OPr H Purpose of the problem examples of choice between our two main mechanisms with (c) and (d) differing only in Suggested solution (a) offers a choice between an SN2 reaction at a tertiary carbon atom or an SN 1 reaction next to group. In fact, these are about the only known examples of SN2 reactions at tertiary carbon work because the carbonyl group accelerates SN2 reactions so much. The diagram of the transition shows how the P orbitals on the carbonyl group are parallel to the orbital in the SN2 transition Azide is also an excellent nucleophile being sharp and narrow and not affected much by steric We know the reaction is SN2 because we find inversion of configuration. (-) Br 0 N N Br N 0 0 0 N (-) N N N N N Example (b) is like acetal formation though it is strictly orthoester exchange and should be compared the chemistry described in Chapter 14 and particularly with Problems 4 and 8 at the end of that The replacement of OMe by the primary OH is a nucleophilic substitution at saturated carbon it goes by the SN1 route because of the cation-stabilizing effect of the other two oxygen atoms. 0 0 0 H OMe OMe 0 0 Ho 0 Ho 0 0 H Examples (c) and (d) add the same group (PrO) to the same starting material (an epoxide) under conditions. We can tell that (c) must be SN1 because the nucleophile has added to the rather than the secondary end of the epoxide and that (d) must be SN2 because of the reverse uselectivity. Acid catalysis makes SN1 better by improving the leaving group and base catalysis SN2 better by improving the nucleophile. Notice that inversion of configuration occurs in reactions. This is expected in SN2 and may be the case in SN1 because the underside of the is not blocked by the OH group. If you said that inversion implied SN2 with a loose cationic state, you may well be right as this is also a valid explanation. 0 0 OH H n-PrOH H OH (c) (+) 0 n-PrOH alternative PrOH explanation OH (+) OH OPr