Organic Chem Exam 3 Study Guide
Organic Chem Exam 3 Study Guide CEM 251
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This 6 page Study Guide was uploaded by Sarah Struble on Friday November 20, 2015. The Study Guide belongs to CEM 251 at Michigan State University taught by C. Vasileiou in Fall 2015. Since its upload, it has received 131 views. For similar materials see Organic Chemistry I in Chemistry at Michigan State University.
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Date Created: 11/20/15
tof6 CEM 251 Study Guide SN2CH3Xgt1 gt gtgtgt3 W StrongNVeak Nuc Strong Nuc aprotic solution not allowed SN13 gtgtgtgt1 methyl not allowed 0 is not stable enough 2 can do BOTH SN1 and SN2 gt if weak Nuc with polar protic solvent H20 alcohol SN1 Alkenes Alkenes have at least one CC double bond R t Lox 0103 ltKlt Q R least stable of 3 mono substituted smallest possible disubstituted LEAST stable lt F lt R R trisubstituted MOST stable internal more stable terminal Naming Alkenes double bonds should be the lowest possible number a 4 b r 2pentene 9 4 l c I 3 4bromo 2pentene 20f6 1cyclohexene R2hydroxyl3hexene 4 ethyl 1 cyclohexene 39 cis 2butene 3 cis 6 trans 36 decadiene trans 2butene E or Z Steps look at atomsgroups on one side of the 00 bond and prioritize them like we prioritized when figuring out 8 or R ohirality H is always last then look at the atomsgroups on the other C of the double bond and prioritize those If the priorities are opposite on either side of the double bond it is E if they are the same it is Z 9 3 9L CD E Z opposite priorities together priorities E3methyl 2pentene Z3 methyl 2pentene Eliminations Alkenes Happen with a strong basebad Nuo ex OK DBN DBU Types of Elimination 1 E2 one step no intermediates 2 E1 two steps oarbooation intermediate E2 Reaction 1 2 3 uses polar aprotio solutions bases like DBU or DBN 30f6 Key Points Always form the most stable double bond The base can only attack the H that is pointing in the opposite direction as the leaving group ie if the leaving group is a wedge the base can only attack a dashed H H ll WANw disubstituted minor pdt MAJOR pdt l Jt n v mi 4 a only one product because there is 05 only one way the reaction can occur 41 V l l only ONE H is opposite oF Cl so it is the only one that can be used W Nquot Ha HC Hb quota 0K W lt U r L Minor Product mega r 4of6 E2 Reactions carbocation intermediate polar protic solutions ie H2O with H catalyst 3 gt 2 gtgt l methyl H0 MO i V Cr Addition Reactions 1 Addition of H X Hydrohalogenation Markomikov s Rule X goes on more substituted C H goes on less substituted C 221 Orr0 2 Addition of X2 Cl2 Br2 l2 Halogenation u Ct Ut LL 3 Addition of X2H2O OH goes to more substituted X goes to less substituted Er Brbr v HlO V0 4 Addition of H OH Hydration Markomikov s H goes to less substituted C one step reaction with H20 and H or two step reaction 1 HgOAc2 H2O 2 NaBH4 0H rT liizes J G 50f6 AntiMarkomikov s two step reaction 1 BH3 2 H202 Whenever you see BH3 the H will go to the more substituted carbon and the OH will go to less b393 H Mb 2 0H 5 Addition of RO H Markomikov s rue OR goes to more substituted C OW gCAc 2 BOH 2 NaBH4 6 Addition of H2 Hydrogenation H2 and PdC or Pt02 syn addition The H s will always ben pointing in same direction as each other 7 Addition of O syn addition Epoxidation uses mCPBA Q mCPBA 3 60f6 8 Dihydroxylation syn addition two step 1 0304 o 2 NaHSOB added OH s are either both wedged or both dashed I39M OH
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