Question 1

What do you do with H3O+, H2O (acid catalyst hydration) (alkene)?

Accepted Answer

Adds an H and OH in place of the pi-bond. OH is on the most substituted, and carbocation rearrangement is possible.

Question 2

1. Hg(OAc)2, H2O   2. NaBH4  
Oxymercuration-reduction hydration: alkene

Accepted Answer

Adds OH and H in place of pi-bond, with OH on most substituted. NO carbocation rearrangement.

Question 3

1. BH3    2. H2O2
Hydroboration hydration; alkene

Accepted Answer

Adds OH and H in place of pi-bond, with OH on least substituted.

Question 4

1. OsO4    2. NHSO4 (or Zn)  
Sin-dihydroxylation; alkene

Accepted Answer

Adds 2OH in syn (cis) conformation with NO carbocation rearrangement.

Question 5

1. O3     2. Zn (or NaSO3)
Ozonolysis

Accepted Answer

Cleaves both the pi and sigma bonds and adds an O to each end via double bonds.

Question 6

Addition of X2

Accepted Answer

Adds 2X in anti (trans) formation with NO carbocation rearrangement

Question 7

X2, H2O

Accepted Answer

Adds X and OH in anti (trans) conformation, with OH on most substituted. NO carbocation rearrangement

Question 8

HX addition

Accepted Answer

H and X added with X on most substituted. Can have carbocation rearrangement.

Question 9

1. NaNH2 (base)    2. Carbon skeleton with X attached

Accepted Answer

Adds carbon skeleton to alkyne. Only works when alkyne is on end

Question 10

1. X2    2. NaNH2 (base)

Accepted Answer

Changes alkene to alkyne

Question 11

HX addition to alkyne

Accepted Answer

Adds 2H on least sub, and 2X on most sub. 
Goes from alkyne ->alkane

Question 12

X2 addition to alkyne

Accepted Answer

No stereochemistry needed. 4X added, with 2X on each C. However, like with alkene rxn each 2X is added in anti conformation.

Question 13

H2, Pd
Alkene

Accepted Answer

Adds 2H in syn (cis) conformation in place of pi-bond.

Question 14

HgSO4; H2O, H3O+

Accepted Answer

For alkynes, adds O double bonded to most substituted C. 
Alkyne -> alkane but with O double bond

Question 15

1. HBR2 (BH3)     2. H2O2

Accepted Answer

Adds O double bonded to least substituted.

Question 16

H2, Pd
Alkyne

Accepted Answer

Saturates alkyne to alkane (adds 4H, 2 on each C)

Question 17

H2, Lindler's catalyst

Accepted Answer

Stops reaction after addition of 2H, making it go alkyne to alkene. Alkene is in CIS/Z conformation (rxn proceeds syn)

Question 18

Na, NH3

Accepted Answer

Adds 2H (similar to H2, Lindler's catalyst), but resulting alkene is in trans/E conformation. This is more stable.

Question 19

Transition state

Accepted Answer

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.

Question 20

In an exothermic reaction, the transition state will be:

Accepted Answer

"Early", looks like reactants

Question 21

In an endothermic reaction, the transition state will be:

Accepted Answer

"Late", looks like products

Question 22

In free radical bromination (more specific type of halogenation) (addition of Br from Br2 to a carbon skeleton) , the Br will be placed:

Accepted Answer

On the most substituted C

Question 23

Free radical halogenation

Accepted Answer

Remove an H, replace it with an X. Cl isn't selective, but Br wants to be on most sub.

Question 24

Hemolysis

Accepted Answer

Where 1 e- goes to each atom in a bond

Question 25

Initiation step

Accepted Answer

Beginning of free radical halogenation where more free radicals are created (most often X2 being 2X with each being a free radical)

Question 26

Propegation step 1

Accepted Answer

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.

Question 27

Propegation step 2

Accepted Answer

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.

Question 28

Z

Accepted Answer

Same side

Question 29

E

Accepted Answer

Different side

Term	Definition
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

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Organic Test 2