Chem 232 final exam Questions
and Answers 100% PASS
3 steps in each radical mechanism - CORRECT ANSWER-1. Initation
2. propagation
3. termination
Homolytic fragmentation - CORRECT ANSWER-electrons in bonding pair move
together. Creates anion with missing pair and cation with complete octet
Heterolytic fragmentation - CORRECT ANSWER-electrons in bond move
independently. highly reactive. creates 2 radical products
Sn2 - CORRECT ANSWER-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 - CORRECT ANSWER-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? - CORRECT ANSWER-reactants
Late transition state will resemble products or reactants more? - CORRECT
ANSWER-products
stereospecific rxn - CORRECT ANSWER-rxn when stereochemistry of reactant
determines stereochemistry of product w/o any other option
stereoselective rxn - CORRECT ANSWER-rxn when theres a choice of pathway,
but product stereoisomer is formed preferentially because its rxn pathway is more
favorable than other available
E2 - CORRECT ANSWER-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 - CORRECT ANSWER-1. anti-periplanarity demands the
groups going to be eliminated occupy axial positions
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and Answers 100% PASS
3 steps in each radical mechanism - CORRECT ANSWER-1. Initation
2. propagation
3. termination
Homolytic fragmentation - CORRECT ANSWER-electrons in bonding pair move
together. Creates anion with missing pair and cation with complete octet
Heterolytic fragmentation - CORRECT ANSWER-electrons in bond move
independently. highly reactive. creates 2 radical products
Sn2 - CORRECT ANSWER-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 - CORRECT ANSWER-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? - CORRECT ANSWER-reactants
Late transition state will resemble products or reactants more? - CORRECT
ANSWER-products
stereospecific rxn - CORRECT ANSWER-rxn when stereochemistry of reactant
determines stereochemistry of product w/o any other option
stereoselective rxn - CORRECT ANSWER-rxn when theres a choice of pathway,
but product stereoisomer is formed preferentially because its rxn pathway is more
favorable than other available
E2 - CORRECT ANSWER-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 - CORRECT ANSWER-1. anti-periplanarity demands the
groups going to be eliminated occupy axial positions
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