Final Exam

Question 1

S_N1 Reaction

Answer 1

• S_N1- primary substrates are favored (less bulky)
• Weaker Nuc^-
• Polar protic solvent
• Nuc^- is in the solvent (they are the same, common for S_N1)
• 1^st order rxn
• rate = k[substrate]
• The substrate that can make the more stable carbocation will be faster (resonance!)

Question 2

S_N2 Reaction

Answer 2

• S_N2- primary substrates are favored (less bulky)
• Me groups
• Stronger nucleophiles
• Solvents: polar aprotic

Question 3

E1 Reactions

Answer 3

• If the leaving group is OH and the solvents is H2O and some kind of acid- dehydration reaction and E1 is favored
• 2^o or 3^o

2+ step rxn:

1. kick off LG
2. pull off a proton (i.e. H on CH₃)
3. leftover electrons form a double bond

Question 4

E2 Reactions

Answer 4

• Pull off L.G. and Proton (H⁺)
• Strong bases

You can think about your substrate as being the acid and your nucleophile as being the base

1. draw out base formation, then put your LG and H in that configuration (anti-periplanar)
2. Rotate: Dashes end up on the same side of alkene, wedges end up on same side of alkene.
3. Pull off group that’s anti-periplanar. Both need to be axial

Question 5

Racemates

Answer 5

50:50 mixture of (R) and (S) products

Question 6

Polar Protic Solvents

Answer 6

H₂O

CH₃OH

CH₃CH₂OH

CH₃COOH

i-PrOH

CH₃CH₂CH₂NH₂

Question 7

Polar Aprotic Solvents

Answer 7

DMSO

Acetone

Acetonitrile

DMF

HMPA

Question 8

Zaitsev’s Rule

Answer 8

The alkene formed in greated amount is the one that corresponds to removal of the hydrogen from the beta-carbon having the fewest hydrogen substituents.

Question 9

Alkene Hydrohalogenation

Answer 9

• An addition of hydrogen and halogen across a π bond.
• Step 1: alkene (nuc^-) is protonated by the electrophile, yielding a carbocation and ion.
• Step 2: Ion then acts as a nuc^- and adds to the strongly electrophilic carbocation to yield the product

Question 10

Hydride Shift

Question 11

Cahn Ingold Prelog Rules

(Chirality)

Answer 11

Prioritize by:

1. Highest atomic # (O > N > C > H)
2. Then by isotope atomic weight
3. Then by substituents, one atom at a time
4. If atoms are equal, look at first point of difference
5. Double bonds count twice, triple bonds count 3 times

Question 12

Hydroboration [of Alkenes}

Answer 12

Boron adds to the less substituted end of the alkene (“anti-Marovnikoff”) .

Question 13

Regiochemistry

Answer 13

Site of elimination on the substrate

Question 14

Zaitzev Product

Answer 14

forms the more substituted alkene

Question 15

Hoffman Product

Answer 15

Elimination forms the less substituted alkene

Question 16

C=C Bond Length

Answer 16

1.34 Å

Question 17

Assigning E vs Z

Answer 17

• Rank substituents on each of the double bond
• Z - both high priority substituents on same side
• E - both high priority substituents on opposite sides

Question 18

C-C BDE

Answer 18

1.54 Å

Question 19

H-C_=_C-H

Answer 19

C_=_C BDE: +190 kcal/mol

Question 20

Me-Me Bond Length

Answer 20

1.54 Å

Question 21

H₂C=CH₂Bond Length

Answer 21

1.34 Å

Question 22

HC=CH Bond Length

Answer 22

1.20 Å

Question 23

C-H Bond Lengths

Answer 23

H₃CCH₂-H

H₂C=CH₂

H-C_=_C-H

1.10 Å is close enough