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Organic Test 2

TermDefinition
What do you do with H3O+, H2O (acid catalyst hydration) (alkene)? Adds an H and OH in place of the pi-bond. OH is on the most substituted, and carbocation rearrangement is possible.
1. Hg(OAc)2, H2O 2. NaBH4 Oxymercuration-reduction hydration: alkene Adds OH and H in place of pi-bond, with OH on most substituted. NO carbocation rearrangement.
1. BH3 2. H2O2 Hydroboration hydration; alkene Adds OH and H in place of pi-bond, with OH on least substituted.
1. OsO4 2. NHSO4 (or Zn) Sin-dihydroxylation; alkene Adds 2OH in syn (cis) conformation with NO carbocation rearrangement.
1. O3 2. Zn (or NaSO3) Ozonolysis Cleaves both the pi and sigma bonds and adds an O to each end via double bonds.
Addition of X2 Adds 2X in anti (trans) formation with NO carbocation rearrangement
X2, H2O Adds X and OH in anti (trans) conformation, with OH on most substituted. NO carbocation rearrangement
HX addition H and X added with X on most substituted. Can have carbocation rearrangement.
1. NaNH2 (base) 2. Carbon skeleton with X attached Adds carbon skeleton to alkyne. Only works when alkyne is on end
1. X2 2. NaNH2 (base) Changes alkene to alkyne
HX addition to alkyne Adds 2H on least sub, and 2X on most sub. Goes from alkyne ->alkane
X2 addition to alkyne No stereochemistry needed. 4X added, with 2X on each C. However, like with alkene rxn each 2X is added in anti conformation.
H2, Pd Alkene Adds 2H in syn (cis) conformation in place of pi-bond.
HgSO4; H2O, H3O+ For alkynes, adds O double bonded to most substituted C. Alkyne -> alkane but with O double bond
1. HBR2 (BH3) 2. H2O2 Adds O double bonded to least substituted.
H2, Pd Alkyne Saturates alkyne to alkane (adds 4H, 2 on each C)
H2, Lindler's catalyst Stops reaction after addition of 2H, making it go alkyne to alkene. Alkene is in CIS/Z conformation (rxn proceeds syn)
Na, NH3 Adds 2H (similar to H2, Lindler's catalyst), but resulting alkene is in trans/E conformation. This is more stable.
Transition state Is at crest of reaction energy (where activation energy is on potential energy diagram). Resembles either the reactants or products; whichever is closest to it in energy.
In an exothermic reaction, the transition state will be: "Early", looks like reactants
In an endothermic reaction, the transition state will be: "Late", looks like products
In free radical bromination (more specific type of halogenation) (addition of Br from Br2 to a carbon skeleton) , the Br will be placed: On the most substituted C
Free radical halogenation Remove an H, replace it with an X. Cl isn't selective, but Br wants to be on most sub.
Hemolysis Where 1 e- goes to each atom in a bond
Initiation step Beginning of free radical halogenation where more free radicals are created (most often X2 being 2X with each being a free radical)
Propegation step 1 2nd step of free radical halogenation. Where a free radical X joins with an H from carbon skeleton, leaving the carbon skeleton with a free radical where the H was.
Propegation step 2 Last step of free radical halogenation. Another molecule of X2 splits, with an X going to bond with free radical of carbon skeleton and the other becoming a free radical.
Z Same side
E Different side
Created by: RunningMads
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