Brainscape's Knowledge GenomeTM

Browse over 1 million classes created by top students, professors, publishers, and experts.


See full index

Orgo 1 > Ch 9 - Addition Reactions of Alkenes > Flashcards

Ch 9 - Addition Reactions of Alkenes Flashcards

1
Q

addition reactions

A

common reactions of alkenes characterized by the addition of two groups across a double bond
- a pie bond is broken in the process

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
2
Q

many times addition is simply the reverse of

A

an elimination reaction

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
3
Q

delta G = delta H + (-Tdelta S)

A

delta G = delta H + (-Tdelta S)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
4
Q

sigma bonds are

A

stronger than pie bonds

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
5
Q

hydrohalogenation

A

the treatment of alkenes with HX(X = Cl, Br, I) results in an addition reaction in which H and X are added across the pie bond

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
6
Q

H is generally placed at the vinylic position already bearing the

A

larger number of hydrogen atoms

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
7
Q

the halogen is generally placed at the

A

more substituted position

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
8
Q

Markovnikov Addition

A

the vinylic position bearing more alkyl groups is more substituted and that is where the halogen is placed

	- the regiochemical preference
	- also observed in HCl and HI
	- regioselective
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
9
Q

anti-Markovnikov Addition

A

when the alkyl group which is less substituted is where the halogen gets placed

	- HBr does this sometimes
	- impurity of peroxides(ROOH) even in trace amounts cause this
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
10
Q

purity of reagents is the critical feature of Markovnikov addition

A
  • pure reagents go by Markovnikov addition
    - impure reagents sometimes go by anti-Markovnikov addition
    - ROOH(peroxides) lead to anti-Markovnikov addition
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
11
Q

mechanism for hydrohalogenation

A
  • step 1
    - the piece bond of the alkene is protonated, generating a carbocation intermediate
    - step 2
    - the intermediate is attack by a bromide ion(NA)
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
12
Q

regioselectivity of an ionic addition reaction is determined by the preference for the reaction to proceed through the more stable carbocation intermediate

A
  • tertiary is the most stable carbocation

- lower Ea and total energy needed and therefore a faster reaction

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
13
Q

hydrohalogenation often involved the formation of a chirality center

A

the two enantiomers are produced in equal amounts(racemic mixture)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
14
Q

if the intermediate of a hydrohalogenation can undergo carbocation rearrangement it will

A

the result is in an unequal mixture of products of rearranged and nonrearranged products

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
15
Q

when carbocation rearrangements can occur,

A

they do occur

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
16
Q

hydration

A

the methods for adding water(H and OH) across a double bond

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
17
Q

acid-catalyzed hydration

A

addition of water across a double bond in the presence of an acid

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
18
Q

for most simple alkenes the acid-catalyzed hydration reaction proceeds

A

via Markovnikov addition

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
19
Q

H3O+ is representative of

A

H2O and an acid source(H+)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
20
Q

brackets around the proton source above a reaction arrow indicate the proton source is not consumed in the reaction

A
  • it’s a catalyst

- acid-catalyzed hydration

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
21
Q

with each additional alkyl group the reaction rate will increase by orders of magnitude

A

the OH is placed at the more substituted position

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
22
Q

three step for acid-catalyzed hydration

A
  • the alkene is protonated to generate a carbocation intermediate
    - then attacked by the nucleophile
    - results in an oxonium ion(exhibits an oxygen atom with a +)
    - bc the NA started neutral a proton transfer is necessary at the end
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
23
Q

acid-catalyzed hydration is an equilibrium process

A

the reverse is an E1 acid-catalyzed dehydration reaction(alcohol to an alkene)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
24
Q

the equilibrium of an acid-catalyzed reactions are sensitive to temperature and concentration of water

A
  • can use dilute or concentrated acids to control the amount of water
    - LeChateliers principle(the system responds to minimize stress)
    - dilute acid increases water is used to convert alkene into an alcohol
    - concentrated acid will reduce water and is used to favor the formation of the alkene
    - remove water = favor alkene
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
25
Q

an acid-catalyzed hydration will have an intermediate carbocation which can be NA from either side

When a chirality center is formed the mixture will be

A

a racemic mixture of enantiomers

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
26
Q

if protonation of the alkene leads to carbocation rearrangement then acid-catalyzed hydration is

A

an inefficient method for adding water across the alkene

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
27
Q

oxymercuration-demercuration

A

a hybrid of two resonancy structures
- an intermediate which instead of an ordinary carboncation is forms a bridge with its electrons to the carbocation lcoation

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
28
Q

a mercurium ion has some character of a carbocation and some character of a bridge three member ring

A

reduces carbocation rearrangement while still being subject to a NA

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
29
Q

Hydroboration-oxidation uses

A

anti-Markovnikov addition of water

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
30
Q

hydroboration-oxidation

A

places the OH group at the less substituted carbon

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
31
Q

syn addition

A

when two new chirality centers are formed the addition of water(H and OH) occurs in a way that places the H and OH on the same face of the pie bond
- stereospecific because only two of the four possible stereoisomers are formed

32
Q

syn addition will not form stereoisomers with H and OH on

A

opposite sides of a pie bond

33
Q

hydroboration-oxidation must explain both:

A
  • the regioselectivity(anti-Markovnikov addition)

- stereospecificity(syn addition)

34
Q

BH3 structure is similar to a carbocation without the charge

A
  • has an empty p orbital

- lacks an octet = very reactive

35
Q

diborane

A

when two borane interact forming a dimeric structure

	- possesses a special type of bonding
	- H atoms actually break their bond with the borane to resonate between the two boranes
36
Q

three-center, two-electron bonds

A

resonance structure in which a H atom from 2 borane is partially bonded to two borane atoms with a total of two electrons between the three entities

37
Q

mechanism for hydroboration-oxidation

A

Regioselectivity

stereospecificity of Hydroboration-Oxidation

38
Q

mechanism for hydroboration-oxidation

Regioselectivity

A
  • electronic consideration
    - attack of the pie bond triggers a simultaneous hydride shift
    - pie bond attacks the empty p orbital of boron and one of the vinylic positions can begin to develop a partial positive charge leading to the hydride shift
    - steric considerations
    - since BH2 is bigger than H the BH2 will be placed in the less substituted carbon atom(less steric hindrance and lower energy)
39
Q

mechanism for hydroboration-oxidation

stereospecificity

A
  • syn addition(same face of the double bond) of the H and BH2 groups across the face of the alkene
    - the number of chirality centers formed is key:
    - 0 chirality centers = no stereospecificity
    - 1 chirality center = both enantiomers are obtains because syn addition can take place from either side equally
    - 2 chirality centers = the requirement of syn addition determines which pair of enantiomers are obtains while the other 2 possible enantiomers are not obtained
40
Q

catalytic hydrogenation

A

the addition of a molecular hydrogen(H2) across a double bond in the presence of a metal catalyst
- the net result of the process is to reduce an alkene to an alkane

41
Q

when two chirality centers are present only the pairs with

A

syn addition are formed

42
Q

catalytic hydrogenation is accomplishing by

A

treating an alkene with H2 gas and a metal catalyst

- often at high pressures

43
Q

in any given case the stereochemical outcome is dependent on the number of chirality centers formed in the process

A

must be careful with symmetrical alkenes as they will produce a meso compound instead of a pair of enantiomers

44
Q

heterogeneous catalysts

A

catalyst which do not dissolve in the reaction medium

45
Q

homogeneous catalysts

A

catalysts which are soluble in the reaction medium

46
Q

the most common homogeneous catalyst for hydrogenation is

A

the Wilkinson’s catalyst

47
Q

a syn addition is observed in

A

homogeneous catalysts

48
Q

asymmetric hydrogenation

A

creating only one enantiomer instead of a pair
- achieved by using a chiral Wilkinson’s catalyst which can change the Ea of one enantiomer more than the other allowing the creation of only one(or significantly high ee)

49
Q

halogenation

A

involved the addition of X2(either Br2 or Cl2) across an alkene

	- only practical for chlorine or bromine
	- fluorine is too violent and iodine produces very low yields
50
Q

anti addition

A

addition of halogenation occurs in a way where the halogen atoms end up on opposite sides of the pie bond

51
Q

for most simple alkenes halogenation appears to proceed primarily via

A

an anti addition

52
Q

bromonium ion

A

bridged intermediate

- similar to the mercurinium ion

53
Q

halogenation mechanism

A

NA + LG create a bromonium atom followed by a second NA by a Br

54
Q

the stereochemical outcome for halogenation reactions is dependent on the configuration of the starting alkene

A

can be cis, trans, meso

55
Q

bromination occurring in a non-nucleophilic solvent results in the addition of Br2 across the pie bond

A

the addition of Br2 across the pie bond

56
Q

bromination in the presence of water can allow the bromonium ion that is initially formed to be captured by the water molecule instead of the bromide resulting in the formation of Halohydrin instead

A
  • halohydrin formation
    - bromohydrin
    - if a chlorine is used then chlorohydrin
57
Q

halohydrin formation is typically regioselective

A

the OH generally positioned at the more substituted position

58
Q

a partial positive charge passes through the carbon atom allowing the transition state for a halohydrin formation to bear partial carbocationic character

A

allows the water molecule to attack the more substituted carbon(the carbon can better stabilize the partial positive charge in transition)

59
Q

dihydroxylation

A

the addition of OH and OH across an alkene

60
Q

epoxide

A

three membered cyclic ether

61
Q

peroxy acids are strong oxidizing agents

A

capable of delivering an oxygen atom to an alkene in a single step

62
Q

a protonated epoxide can also be attacked by water from

A

the back side

63
Q

when an alkene is treated with osmium tetroxide(OsO4) a cyclic osmate ester is produced

A
  • concerted process where both oxygen atoms attach to the alkene simultaneously
    - same face of alkene so syn addition
64
Q

OsO4 acts as a catalyst to produce large quantities of

A

diol

65
Q

Syn Dihydroxylation creates a product with

A

two OH groups on the same face via syn addition

66
Q

ozonolysis

A

add across an alkene and completely cleaving the C=C bond

- after the split there are now two separate C=O double bonds

67
Q

stereochemistry and regiochemistry become irrelevant

A

irrelevant for oxidative cleavage

68
Q

For oxidative cleavage, ozone(O3) wil react with an alkene to produce an initial primary ozonide(or molozonide) which

A

undergoes rearrangement to produce a more stable ozonide

69
Q

three questions to consider to predict the products of an addition reaction

A
  • What are the identities of the groups being added across the double bond?
    - what is the expected regioselectivity(Markovnikov or anti-Markovnikov addition)?
    - What is the expected stereospecificity(syn or anti addition)?
70
Q

it is absolutely essential to recognize reagents

A

must understand the proposed mechanisms for each reaction type

71
Q

a proposed mechanism must explain the experimental observations

A

the mechanism for each reaction can serve as the key to remembering the three pieces of information

72
Q

elimination reactions can be used to convert

A

alkyl halides or alcohols into alkenes

73
Q

addition reactions are characterized by

A

two groups adding across a double bond

74
Q

be familiar with all reagents employed for each type of reaction

A

be familiar with all reagents employed for each type of reaction

75
Q

it is possible to combine elimination and addition reactions

A
  • two step sequence(elimination then addition to change to position of a leaving group)
    - in the elimination reaction the product can be either to Zaitsev or Hofmann product
    - a careful choice of reagents makes it possible to control the regiochemical outcome
    - strong base to more substituted position
    - sterically hindered base to less substituted position
76
Q

when changing the location of a leaving group make sure to remember that OH is a terrible leaving group

A
  • this is critical because you may think you can add acid to protonate the group BUT remember if its an E1 process the regiochemical can not be controlled
    - it is not possible to protonate with a strong acid and then use a strong base to achieve an E2 reaction because the strong acid and strong base will neutralize each other
    - a better avenue is to convert the OH group into tosylate which is a much better leaving group which can then used anti-Markovnikov addition of H-OH
77
Q

2 step process to change the position of a pie bond(addition then elimination)

A
  • allows for the regioselectivity of each step to be carefully controlled
    - uses Markovnikov addition with HBr and anti-Markovnikov addition si achieved by using HBr with ROOR(peroxides)
    - the elimination step can be accomplished by using a strong base for Zaitsev products and the Hofmann can use a strong, sterically hindered base

Orgo 1 (10 decks)

Brainscape helps you reach your goals faster, through stronger study habits.
© 2024 Bold Learning Solutions. Terms and Conditions