5.12_Pset_5_Solutions

5.12_Pset_5_Solutions - Massachusetts Institute of...

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Massachusetts Institute of Technology Organic Chemistry 5.12 April 9, 2011 By Yu-Pu Wang and Phil Hamzik Problem Set 5 Answer Key (1) Predict the products of the reactions below. (Multiple products are possible in some cases.) Explicitly show stereochemistry in all products, where appropriate. Designate each product as being the result of S N 1, S N 2, cross-coupling, E1, E2, E1cB, or other. If you predict “no reaction” occurs, indicate as NR. ANSWER: (a) Cl KI DMSO NR S N 2 cannot occur at tertiary carbon, and S N 1 requires polar protic solvent to occur at any appreciable rate. (b) Cl NaOMe DMSO major minor Strong base with tertiary C-X gives E2; Zaitsev's rule predicts more substituted alkene to be major product. (c) H 3 C CH 3 CH 3 Br NaO t -Bu DMSO H 3 C CH 3 CH 3 H 3 C CH 3 major minor Br H CH 3 H CH 3 Et t -BuO (d) H 3 C CH 3 Br NaCN DMF H 3 C CH 3 C N Cyanide is a good nucleophile, but a weak base (pK a of HCN ~ 9), so get S N 2; note inversion of stereochemistry (e) CH 3 Br H 3 C H 3 C NaCN DMF NR Neopentyl halides are inert to S N 2 due to steric hindrance. (f) H 3 C 18 OH CH 3 pyridine H 3 C 18 OM s CH 3 NaOH H 2 O O H major minor M sCl Recall that for secondary C-X, strong base gives predominantly E2; by Zaitsev's rule, the favored elimination This is the only staggered conformation that positions the β -hydrogen on the secondary β -carbon anti-periplanar to the C-Br bond.
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product is the more substituted product; within that regioisomer, the trans alkene is favored for steric reasons. The less substituted alkene and the substitution products are formed only in very small amounts. (2) The rate of reaction of an amine increases significantly with the degree of substitution. For example, methyl amine reacts with iodomethane much faster than does ammonia.
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