Exam 3

Question 1

Sequential reaction

Answer 1

1. first reaction (to completion)
2. Second reaction (no interaction with the first reaction)

Question 2

Substitution Rxn

Answer 2

Replacing one group on a carbon with another

Question 3

Addition Rxn

Answer 3

combining reactants to form new products (typically one pi bond is split to form two new sigma bonds)

Question 4

Elimination Rxn

Answer 4

A Hydrogen and a Leaving Group are removed and replaced by a pi bond

Question 5

Rearrangement Rxn

Answer 5

a reaction that changes orientation

Question 6

Concerted

Answer 6

All bonds are broken and formed simultaneously - no intermediates with formal charges

Question 7

Step-wise

Answer 7

bonds are broken and formed in specific steps creating intermediates with formal charges that can move

Question 8

Homolydic cleavage

Answer 8

each atom gets one electron, typically seen in a free radical halogenation reaction

Question 9

Heterolydic Cleavage

Answer 9

electrons travel together

Question 10

Free energy equation

Answer 10

∆G= ∆H - (Tx∆S)
∆H = BFE - BDE

Question 11

∆H is positive

Answer 11

Energy is absorbed - Endothermic

Question 12

∆H is negative

Answer 12

Energy is released - exothermic

Question 13

stronger bonds…

Answer 13

requires more energy to break

Question 14

Energy Diagram

Answer 14

one peak represents a concerted reaction, multiple peaks represent each step in the reaction

Question 15

intermediate stability

Answer 15

Carbon with the charge:
primary < secondary < tertiary < benzylic < Allylic (adjacent to a double bond) < Tertiary allylic

Question 16

Nucleophilic addition

Answer 16

nucleophile uses a lone pair of e- to add itself to the molecule and push off another atom

Question 17

Leaving Group

Answer 17

atom or group comfortable with a negative charge

Question 18

Proton Transfer

Answer 18

addition of a hydrogen

Question 19

Pi bond formation

Answer 19

Transfer of a non bonded lone pair to move the positive charge

Question 20

Free radical halogenation

Answer 20

*Stereochemistry for chirol centers
1. homolytic cleavage of a halogen, one removes a hydrogen
2. a second homolytic cleavage of a halogen to bond with the radical carbon intermediate

Question 21

SN2 Reaction

Answer 21

• Concerted substitution
• Backside attack

Factors to consider:
1. Electrophile
methyl > primary > secondary, no tertiary addition
2. Leaving group
I > Br > Cl > F
3. Nucleophile
Must have an available lone pair
4. Solvent
Polar aprotic - no O-H Bonds
DMSO, HMPA, Acetone

Question 22

SN1 Reaction

Answer 22

• Stepwise
• Leaving groups leaves on its own first
• Racemic
• Cation rearrangement for best stability (Hydride/Methyl shift)

Factors to consider:
1. Electrophile
tertiary > secondary > primary > Methyl
2. Leaving group
I > Br > Cl > F
3. Nucleophile
Weak - Compounds with lone pairs but are neutral
4. Solvent
Polar protic - an O-H Bond helps with cation stability

Question 23

Elimination Reaction - E1

Answer 23

• Stepwise
• Hydride shift
• double bond with more substituents is preferable
• Leaving group leaves on its own
• Trans double bond is more stable

Factors to consider:
1. Electrophile
tertiary > secondary > primary > Methyl
2. Leaving group
I > Br > Cl > F
3. Solvent
Polar protic - stabilizes the carbocation intermediate and one of its lone pairs are able to remove the hydrogen to form the double bond
4. Environment
When heat is applied E1 is favored over SN1

Question 24

Alcohol (OH) Leaving group

Answer 24

• poor leaving group
• can be activated with either an acid (SN1/E1) or with TsCl and pyr (SN1, SN2, E1, E2)

Question 25

Activation with Acid (SN1, E1)

Answer 25

1. a hydrogen is taken from the acid to make the OH an H2O 2. H2O is a good leaving group, so it does 3. Hydrogens can shift if needed 4. the H2O in solution can remove the second H which leaves its electrons to the electrophile

Question 26

Activation with TsCl and Pyr (SN1, SN2, E1, E2)

Answer 26

1. TsCl turns the OH into OTs which is a good leaving group, akin to Br 2. from there it can do any of the reactions

Question 27

Elimination reaction - E2

Answer 27

- Concerted - Strong Base - no positive charges allowed - small bases -> less hydrogens -Bulky Bases -> more hydrogens - stereoselective: leaving group and hydrogen must be in the same plane, preferably anti planer Factors to consider: 1. Electrophile tertiary > secondary > primary > Methyl 2. Leaving group I > Br > Cl > F 3. Nucleophile strong base - any salt with an organic compound (NaOCH3, ect.) 4. Solvent not important but typically Conj. Acid of the base

Question 28

E2 Elimination in rings

Answer 28

Normal ring: - Trans diaxial formation (one up one down) Bridged system: - Syn elimination only, both down