CHEM 232 final exam Questions and Answers Already Passed Latest Update

CHEM 232 final exam Questions and Answers Already Passed Latest Update 3 steps in each radical mechanism - Answers 1. Initation 2. propagation 3. termination Homolytic fragmentation - Answers electrons in bonding pair move together. Creates anion with missing pair and cation with complete octet Heterolytic fragmentation - Answers electrons in bond move independently. highly reactive. creates 2 radical products Sn2 - Answers 1. One Step 2. transition state of RDS has 2 species coming together 3. nucleophile does backside attack in coaxial way: inversion of stereochemistry 4. n--- sigma*, sigma type interaction 5. less substituted carbon = more accessible Sn1 - Answers 1. two steps 2. transition state of RDS has 1 species 3. proceeds without stereospecificity or selectivity (racemic) the transition state of an exothermic rxn will be more like the products or reactants? - Answers reactants Late transition state will resemble products or reactants more? - Answers products stereospecific rxn - Answers rxn when stereochemistry of reactant determines stereochemistry of product w/o any other option stereoselective rxn - Answers rxn when theres a choice of pathway, but product stereoisomer is formed preferentially because its rxn pathway is more favorable than other available E2 - Answers 1. stereospecific 2. rxn path must be continuous overlap among orbitals 3. sigma bonds are mad/broken and pi bonds are made 4. -X and -H oriented anti-periplanar (preferred) 5. -X and -H oriented syn-periplanar 6. stereochemistry of reactant determines stereochemistry of product E2 in cyclic compounds - Answers 1. anti-periplanarity demands the groups going to be eliminated occupy axial positions E2 in acyclic substrates - Answers 1. must always undergo stereospecific antipariplanar rxn acyclic compounds can undergo bond rotations so possible to find more than one elimination path E1 - Answers 1. no antipariplanar requirement 2. carbocation intermediate undergoes rotation about single bond [1, 2R] - Answers elementary step for carbocation rearrangements. carbocations can rearrange themselves to become more stable