Question

Chemistry 305. Problem Set 15/16. October 28, 2015 Set 15/16. October 28, 2019. Name Classify each of the following reactions as product. reactions as Sn1, E1, Sn2, E2 (one term each), Explain you heach), Explain your choice and give the reaction DBN THE EtOH HC CH30 WIN NaSH THE (CH3)3C-0 (CH3)3COH

Answer 1

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D01 AS DBN acts as base only (not as ño), elimination reactions are possible. After leaving a carbocation formed is 3°, so the son proceeds via El path . THE (E1) Y Orxides per te nar EDH A (61) After leaving I a 3 - carbocation is formed. So, only r Sulor El path are possible. As the heat is (nainor) mentaoned, the new farcurs el-path (Etoh is solvent) NASH Sh @ we THA (S.2) @NI (H30,000 (CH3), con After leaving IO 1 - carboeation is formed. So, the ron proceeds through either S2 or E2. Aud & polar aprotic solvent, thr). But it is strong a but weak base, so, son proceeds via sx2-path. After leaving go, carbocation is formed. so, the rem proceeds through either s.2 or €2. Bulky nature of reagent makes it a bad luce but strong base. So, the ron proceeds through E2 After leaving Bro, so carbocation is formed. so, the path is either Sulor El. But Buco is weak Me bet strong base. So, non procedds through El.path (€27 (CH₂). Coo مل $1

a Naci DMSO (502) ao behaves as he only not as a base. Du so is polar aprotic solvent reacts to far ours SN2 or er. So, the son proceeds through Se 2 path. HP04 Sa H₂PO4 protonates atst go to makot, which is a good leaving group, since, no. fu is available and heating condition is mentioned, the son follows El path ker cctand Br is a strong Nee but weak base. so must be Salon 302. Though 3° stable cartocation is formed yet, acetone polar aprotic solvent does not stabilise it. So, proceeds via Saia - path L . 24,0 behaves as strong Nu as well as strong base Solvent used is meon Laveurs sula El. But here 9 dorinate Over Sel. D)