chapter 10

Question 1

hydrohalogenation

Answer 1

• addition of HX (alkyl halide) (X = Cl, Br, I)
• carbocation intermediate and possible rearrangement
• Markovnikov’s rule
• H bonds to LESS sub. C to form more stable cation
• syn and anti addition

2 steps:

1. H get deprotonated from HX –> carbocation forms
2. X atom bonds

Question 2

hydration

Answer 2

• addition of H20(OH and H - alcohol) or ROH(ether)
• H20 with added H2SO4 makes H3O
• carbocation intermediate and possible rearrangement
• Markovnikov’s rule
• H bonds to LESS sub. C to form more stable cation
• syn and anti addition

three steps:

1. protonation of H30 –> carbocation
2. H2O bonds to carbocation
3. H2O looses H from H2O

Question 3

halogenation

Answer 3

• addition of X2 (Cl or Br)
• bridged halonium ion intermediates
• NO rearrangement possible
• anti addition

two steps:

1. addition of X+ to alkene to form bridged halonium ion
2. nucleophilic attack by X-

Question 4

halohydrin formation

Answer 4

• addition of OH and X(Cl or Br)
• bridged halonium ion intermediates
• NO rearrangement possible
• X bonds to LESS substituted Carbon
• anti addition

3 steps:

1. addition of X+ to alkene to form bridged halonium ion
2. nucleophilic attack of H2O
3. H2O looses proton
   * *NBS in DMSO and H20 adds Br and OH in same fashion

Question 5

hydroboration-oxidation

Answer 5

• addition of H20 (OH and H - alcohol)
• (1) BH3 or 9-BBN/ (2) H202, OH-
• NO rearrangements possible
• OH bonds to less substituted Carbon

2 steps:

1. addition of BH2 and H –> BH2 good LG
2. OH is bonded
   - syn addition of H20 is the result

*follow anti-markonikov’s rule

Question 6

Does cis or trans have a higher boiling point?

Answer 6

cis has a higher boiling point b/c it is more polar

Question 7

syn addition vs anti addition

Answer 7

syn addition - X and Y add to the same side (hydroboration-oxidation)

anti addition - X and Y add from opposite sides (halogenation and halohydrin formation)

Question 8

Degrees of unsaturation

Answer 8

CnH2n
maximum: 2n+2

maximum-actual/2

Oxygen = 0
Halide = +1
Nitrogen = -1

Question 9

E vs Z

Answer 9

E - two higher priority groups on opposite sides

Z - two higher priority groups on same side

Question 10

ethene
methylene group
vinyl group
allyl group

Answer 10

ethene: CH2=CH2

alkyl groups:

methylene: CH2=R
vinyl: CH2=CHR
allyl: CH2=CHCH2-R