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CH341—Organic Chemistry I > Exam 3 Reactions > Flashcards

Exam 3 Reactions Flashcards

1
Q

Dehydrohalogenation SM

A

alkyl halide
usually 3°, or 2° with stronger conditions

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2
Q

E1 DEHYDROHALOGENATION MECHANISM

A
  1. halogen leaves
  2. C+ rearrangement
  3. weak base deprotonates beta H; H e- make alkene
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3
Q

E1 dehydrohalogenation solvent

A

good ionizers - water, alcohol

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4
Q

E1 dehydrohalogenation rate law

A

rate = k[RX]

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5
Q

base used in E1 dehydrohalogenation
small or bulky?

A

weak
small yields Zaitsev product
bulky yields Hofmann product

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6
Q

E1 dehydrohalogenation byproducts

A

Sn1

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7
Q

how to favor E1 over Sn1?

A

heat

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8
Q

E2 dehydrohalogenation SM configuration requirement

A

anticoplanar
if cyclohexane, X and H must both be axial

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9
Q

E2 dehydrohalogenation SM

A

alkyl halide
usually 2° or 3°

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10
Q

E2 dehydrohalogenation rate law

A

rate = k[RX][B]

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11
Q

best base characteristics for E2 dehydrohalogenation

A
  • strong
  • nonpolarizable
  • matched with solvent
  • small (for Zaitsev product)
  • large (for Hofmann product)
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12
Q

how to favor E2 dehydrohalogenation over Sn2

A
  • polar protic solvent
  • heat
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13
Q

E2 dehydrohalogenation stereospecificity

A

2 “wedges” end up one one side of alkene (longways)

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14
Q

Dehydration of alcohol SM

A

an alcohol

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15
Q

E1 DEHYDRATION MECHANISM

A
  1. hydroxy attacks H of acid to make a good LG
  2. C-O bond breaks and water leaves
  3. C+ rearrangement
  4. water attacks & removes beta H - H e- form alkene
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16
Q

E1 dehydration SM and acid

A

2° and 3° alcohol required
phosphoric/sulfuric acid

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17
Q

E2 DEHYDRATION MECHANISM

A
  1. acid attacks hydroxy to make good LG
  2. water leaves, and water attacks beta H - H e- make alkene
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18
Q

E2 dehydration SM & acid

A

1° alcohol
concentrated H2SO4

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19
Q

Hydrohalogenation SM

A

alkene

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20
Q

HYDROHALOGENATION MECHANISM

A
  1. alkene attacks H (or other electrophile), splitting it from X
  2. C+ rearrangement
  3. X attacks C+
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21
Q

hydrohalogenation product

A

alkyl halide
pair of enantiomers

22
Q

hydrohalogenation variation

A

organic peroxide rxn
HBr
RO-OR solvent
radical mechanism

23
Q

hydrohalogenation of alkynes SM and their products

A

terminal alkyne yields vinyl halides or geminal dihalides (depends on molar ratio)

internal alkyne yields 4 products, no Markovnikov application

24
Q

HX addition to alkynes
1HX:1SM =

A

vinyl halide alkene

25
Q

HX addition to alkynes
2HX:1SM =

A

geminal dihalide alkane

26
Q

ACIDIC WATER HYDRATION MECHANISM

A
  1. alkene attacks H of acid
  2. C+ rearrangement
  3. water attacks C+, giving protonated alcohol
  4. another water deprotonates alcohol & regenerates acid
27
Q

OXYMERCURATION-DEMERCURATION HYDRATION MECHANISM

A
  1. oxymercuration—alkene attacks Hg, Hg attacks C, forming mercurinium ion ring
  2. water attacks more stable partial + C, and Hg moves to the other C
  3. another water deprotonates alcohol, forming organomercurial alcohol
  4. demercuration—addition of NaBH4 drives SN2-type reaction, where Hg is replaced with H
28
Q

why is mercuration used?

A

to get a non-rearranged hydrated product

29
Q

species involved in oxymercuration-demercuration

A
  • SM: alkene
  • Hg(OAc)2
  • THF-H2O solvent
  • NaBH4 hydride
  • product: non-rearranged alcohol
30
Q

How is alkyoxymercuration different from oxymercuration? (2)

A
  1. Hg(O2CCF3) is used
  2. OR is attached to product, not OH; results in ether, not alcohol
31
Q

Mercuric ion-catalyzed hydration of alkyne product

A

ketone

32
Q

MERCURIC HYDRATION OF ALKYNE MECHANISM

A
  1. alkyne attacks Hg; mercury attaches to less sub. C
  2. water attacks C+, forming orthomercurial alcohol
  3. another water deprotonates water, giving alcohol
  4. alkene attacks proton from acid; Hg is replaced by H
  5. keto-enol tautomerization occurs
33
Q

type of Hg salts used in mercuric hydration of alkynes

A

Hg (II) - HgO, HgSO4, Hg(OAc)2

34
Q

hydroboration TS

A

4-membered ring, with Boron aligned with less sub. C

35
Q

stereoselectivity of hydroboration

A

syn - H and Boron add to same face of alkene

Boron adds to less hindered face

if both faces are equally hindered, 2 products result

36
Q

species involved in hydroboration of alkenes

A
  • BH3 (or B2H6 dimer) - THF
  • H2O2, OH-
37
Q

boron reagent used in hydroboration of alkynes

A

Sia2BH

38
Q

rearrangement involved with hydroboration of alkynes

A

enol-aldehyde tautomerization

39
Q

HALOGENATION OF ALKENE MECHANISM

A
  1. halonium ion formation—alkene attacks X, X-X bond polarizes, X- breaks away
  2. X- opens 3-membered ring with backside attack
40
Q

type of X-X addition to alkenes

A

anti

41
Q

produces PVC

A

halogenation of alkenes

42
Q

halogenation of symmetrical trans substrate gives…

of asymmetrical substrate gives…

A

meso product

enantiomers

43
Q

halogenation of cis substrate gives…

A

pair of enantiomers

44
Q

halogenation solvent

A

CCl4

45
Q

halogenation of alkynes occurs in ________ conditions

A

anhydrous

46
Q

halohydrin SM and product

A

SM: alkene
product: OH and X on alkane

47
Q

halohydrin solvent

A

H2O

48
Q

Nu and El for halohydrin formation

A

OH- is Nu
X+ is El

49
Q

type of halohydrin addition

A

trans (anti)

50
Q

HALOHYDRIN MECHANISM

A
  1. alkene attacks X-X, polarizing bond; X+ attacks one C; X- detaches
  2. water opens 3-membered ring from backside
  3. another water deprotonates alcohol
51
Q

Halohydrin – is OH or X on more sub. C?

A

Markovnikov – OH is on more sub. C

52
Q

2 variations of halohydrin formation

A
  1. ROH solvent gives ether, not OH
  2. salt in H2O gives mixed product, ex) Cl and Br on different products

CH341—Organic Chemistry I (9 decks)

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