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335 Midterm 1 Flashcards

1
Q

what do high temperatures (delta) or intense light (hv) provide for the molecule

A

gives the molecule a lot of energy

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2
Q

what is the difference between heterolytic and homolytic bond cleavage

A
  • heterolytic bond cleavage: both electrons remain with one atom
  • homolytic bond cleavage: each atom gets one electron
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3
Q

what are the three steps for radical chlorination

A
  1. initiation step
  2. propagation steps
  3. termination steps
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4
Q

why is excess alkane used to get monochlorination

A

using an excess of the starting alkane minimizes dichlorination, trichlorination, etc.

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5
Q

what are the relative stabilities of radicals

A

most stable
1. tertiary radical
2. secondary radical
3. primary radical
4. methyl radical
least stable

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6
Q

what are the pitfalls of radical chlorination

A

main problems:
- selectivity for secondary over primary
- stability of radicals determines…
- alkane reactivity
- product selectivity
- potential for dichlorination, trichlorination, etc.

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7
Q

why is it easier to make a secondary radical over a primary radical

A

it’s more stable. thus, the reactivity (how easy it is to break C-H bond) follows the trend of stability of the radical intermediate

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8
Q

what are the relative rates of radical formation by a chlorine radical at room temperature

A

greatest rate of formation
tertiary (5.0)
secondary (3.8)
primary (1.0)
lowest rate of formation

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9
Q

what are the relative stabilities of alkyl radicals

A

most stable
tertiary radical
secondary radical
primary radical
methyl radical
least stable

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10
Q

what does the stability of radicals determine

A
  • alkane reactivity
  • product selectivity (the preference of a reaction to occur in a particular manner, yielding a specific product)
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11
Q

relative rates of bromination vs. chlorination

A

bromination:
greatest rate of formation
- tertiary (1600)
- secondary (82)
- primary (1)
lowest rate of formation

chlorination:
greatest rate of formation
- tertiary (5.0)
- secondary (3.8)
- primary (1.0)
lowest rate of formation

**bromination works way better

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12
Q

what is the relativeity-selectivity principle

A

a bromine radical is much more selective than a chlorine radical, making radical bromination of an alkane a much more useful reaction.

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13
Q

contrast the additions of HBr in the presence of peroxide vs. without it

A

R + HBr —no peroxide—> Br attached to most substituted carbon
R + HBr —peroxide—> Br attached at the end

suggests different rxn mechanisms

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14
Q

does the free radical addition of HX work with other compounds other than HBr? Why or why not?

A

no. it works only for HBr and not for HCl or HI.
- both propagation steos for HBr addition are exothermic (so chain mechanism works well for HBr)
- HCl and HI have one endothermic and exothermic step, so chain mechanism does not work well for HCl or HI.

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15
Q

what are the relative stabilities of radicals

A

most stable
allylic
tertiary
secondary
primary
methyl
vinylic
least stable

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16
Q

explain how an allylic radical is stabilized by resonance

A

the p orbital on the central carbon can overlap equally well with a p orbital on either neighboring carbon, resulting in two equivalent resonance structures.

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17
Q

can strongly basic leaving groups be displaced

A

no. OH (for example) is not a good leaving group because…
- weak base
- poor nucleophile
- negative charge stabilized by electronegative bromine

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18
Q

how can you make something more reactive

A
  • to convert a poor leaving group into a good leaving group you can use protination.
  • only weakly basic nucleophiles can be used because strongly basic nucleophiles would react with the proton.
  • for example, alcohols must be activated before they can react
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19
Q

how can you convert alcohols into alkyl halides

A

primary and secondary alcohols require heat (tertiary alcohols do not).

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20
Q

how can you convert alcohols into alkyl halides

A

least reactive
methyl
primary
secondary
tertiary
most reactive

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21
Q

why is it important to convert alcohols into alkyl halides

A

alcohols are readily available but are generally unreactive. alkyl halides are less available but reactive and can be used to synthesize a wide variety of compounds via SN2 reactions.

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22
Q

what are other methods of converting alcohols into alkyl halides

A

reacting alkens with an OH at the end with another reactant and with pyridine. Reactants:
- PBr3 (adds Br to the product)
- PCl3 (adds Cl to the product)
- SOCl2 (adds Cl to the product)

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23
Q

explain how carbon can be an electrophile or a nucleophile

A
  • carbon is an electrophile when it is attached to an electron-withdrawing group (e.g. a halide (F, Cl, I, Br))
  • carbon is a nucleophile when it is attached to a metal (e.g. Li, MgX)
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24
Q

come back to missed lecture from snow break

A
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25
Q

suffix for naming an alkyne

A

-yne. numbered giving triple bond the lowest number

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26
Q

how do you number an alkyne with a double bond

A

add -en and -yne
ex. 2-hexEN-4-yne

*when he same number is obtained for both the double and triple bonds, the double bond gets the lower number.

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27
Q

what is the structure of alkynes

A
  • the triple bond is composed of a sigma bond and two pi bonds
  • two orthogonal p-orbitals (like a giant X)
  • 180 degrees
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28
Q

what are the characteristics of a triple bond

A
  • bond angle of 180 degrees (linear)
  • short CC bond
  • strong bond
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29
Q

explain how alkynes undergo electrophilic addition rxns

A

since alkynes are pi-electron rich molecules (i.e. nucleophilic), they react readily with electrophiles such as HCl and HBr.

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30
Q

addition of an electrophile to an alkene vs. to an alkyne

A

alkene - produces an alkyl cation product
alkyne - produces a vinylic cation

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31
Q

what are the relative stabilities of carbocations

A

most stable
tertiary carbocation
secondary carbocation
secondary vinylic cation (similar to the next)
primary carbocation
primary vinylic cation (similar to the next)
methyl cation
least stable

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32
Q

explain how the addition to a terminal alkyne is regioselective

A

the halogen goes on the element with the least amount of hydrogens (most substituted)

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33
Q

why is the terminal alkyne regioselective

A

the formation of the more stable transition state accounts for the regioselectivity. it helps improve stability.

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34
Q

are alkyl halides electrophiles or nucleophiles

A

electrophiles. RCH2–X where X is an electronegative atom or an electron withdrawing group. The bond is polar

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35
Q

electrophile definition

A

they have an electronegative (electron withdrawing) atom or group that is attached to an sp^3 carbon

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36
Q

why do alkyl halides react with nucleophiles

A

because alkyl halides are electrophiles, they react with nucleophiles

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37
Q

what are the two possible outcomes when alkyl halides and nucleophiles interact

A

a substitution rxn (the EN group is replaced by another group)
an elimination rxn (the EN group is eliminated along with a hydrogen)

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38
Q

what can the rate law tell us

A

tells us that bothe molecules are involved in the transition state of the RDS

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39
Q

what are the kinetics of a reaction

A

the factors that affect the rate of the reaction - help determine the mechanism

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40
Q

explain SN2 rxn

A

substitution nucleophilic bimolecular
- involves a back-side attack of the nucleophile

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41
Q

why is SN2 bimolecular

A
  • the bond to the leaving group breaks at the same time that the bond to the nucleophile forms
  • since both the Nu and leaving group are involved, its bimolecular
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42
Q

relative rates of an SN2 rxn

A

most reactive
methyl halide
primary alkyl halide
secondary alkyl halide
tertiary alkyl halide
least reactive

as sterics increase, the speed of an SN2 reaction decreases.

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43
Q

why is tertiary alkyl halide least reactive in a SN2 rxn

A

as the sterics get bigger, it becomes harder for the Nu to access the carbon.

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44
Q

inversed configuration

A

A process in which the configuration of an atom is changed (ex. R to S, or a substitution occurs and the compound contains a new molecule)
- if the halogen is bonded to an asymmetric center, the product will have the inverted configuration

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45
Q

rank the best leaving groups

A

the weakest bases are the best leaving groups (the better the leaving group, the faster the reaction)

most reactive
RI
RBr
RCl
RF
least reactive

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46
Q

what is a sulfonate ester

A
  • a good leaving group
  • aka -OTs
  • it turns a hydroxyl into a good leaving group for SN2 rxns as it is highly reactive. This is because it’s negative charge can be delocalized over the three oxygen atoms and one sulfur
  • significant resonance delocalization energy (three good, equivalent resonance structures)
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47
Q

how can primary and secondary alcohols be activated by being converted into sulfonate esters

A

reaction with pyridine (pyridine as the solvent can act as a base to neutralize the acid by-product that is also formed)

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48
Q

nucleophilicity vs. basicity

A

nucleophilicity roughly parallels basicity (but not always)
- nucleophilicity usually increases going down a column of the periodic table

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49
Q

base strength vs. nucleophile strength

A

anything with a minus is typically a better base and a better nucleophile than the same atom that is neutral.

the strongest base is the best nucleophile

50
Q

polar solvents: protic vs aprotic solvents

A

polar solvents contain bonds between atoms with very different ENs, which in turn create dipole moments.

protic polar solvents –> have a hydrogen attached to an oxygen (e.g. H2O)

aprotic polar solvents –> do not have a hydrogen attached to an oxygen atom (e.g. DMF)

51
Q

non polar solvents

A

contain bonds between atoms with similar ENs, such as carbon and hydrogen. negatively charged species cannot dissolve in non polar solvents

52
Q

why do protic polar solvents make the strongest bases the poorest nucleophiles

A

F- is the best nucleophile in an apotic polar solvent whereas I- is the best nucleophile in a protic polar solvent.

because strong bases form strong ion-dipole interactions, these interactions must be broken before the nucleophile can react.

53
Q

how does steric hinderance affect nucleophilicity

A

steric hinderance decreases nucleophilicity. atoms larger in size makes it difficul for the Nu to approach the back side of the carbon, thus nucleophilicity is reduced.

54
Q

are SN2 reactions reversible

A

most are irreversible because a weak base cannot displace a strong base.

55
Q

SN1 reaction

A

substitution nucleophilic unimolecular

56
Q

what is in the transition state of the RDS for an SN1 rxn

A

only the alkyl halide

57
Q

example of an SN1 rxn

A

Tertiary alcohols. They react fastest because they give the most stable carbocation intermediates.

58
Q

example of SN2 rxn

A

Primary or secondary alcohols. They are preferred to prevent steric congestion caused by the simultaneous binding of the nucleophile and release of the leaving group. This reaction mechanism is faster because it omits the formation of a carbocation intermediate.

59
Q

solvolysis rxn

A

the solvent is also the nucleophile (only one reactant and the Nu is on the arrow)

60
Q

what happens to the product of the SN1 rxn if the leaving group is bonded to an asymmetric / chiral center

A

produces a pair of enantiomers (one product with inverted configuration, the other with retained configuration)

61
Q

why does SN1 (at times) produce enantiomers

A

because the rxn goes through an achiral intermediate, the Nu has equal probability of additing to either face of the naked PLANAR carbocation, giving a racemic product.

most SN1 rxns lead to partial racemization.

62
Q

explain how an inverted product is formed vs. racemic mixture

A

first the nucleophile back side attacks the carbon on the front face. the leaving group is still attached on the other side of the molecule. this produces an INVERSION. when the leaving group leaves, the Nu has equal access to the carbon from both sides, creating a racemic mixture.

63
Q

effect of solvation and reactivity in SN1 rxn

A

the electron rich oxygen atoms of solvent molecules orient around the positively charged carbocation and thereby stabilize it.

64
Q

primary alkyl halides / methyl halides with a nucleophile intermediate

A

they cannot form carbocations (SN2)

65
Q

secondary alkyl halides with a nucleophile intermediate

A

carbocation formation is too slow to make up for the large concentration of the nucleophile in a solvolysis rxn (SN2)

66
Q

tertiary alkyl halides with a nucleophile intermediate

A

they cannot undergo backside attack, carbocation is formed (SN1)

67
Q

SN2 vs. SN1
how many steps?
RDS?
how does rate increase or decrease?
inversion?
leaving group?
alkyl halide reactants?
how does the Nu affect the rate of the rxn?

A

SN2
one step mechanism
bimolecular RDS
rate decreases with increasing steric hinderance
product has the inverted configuration relative to the reactant
I- > Br- > Cl- > F-
alkyl halide reactants: methyl, primary, secondary
the better the Nu, the faster the rate of the rxn

SN1
two or three step mechanism
unimolecular RDS
rate decreases with decreasing stability of the carbocation
products have both retained and inverted config.
I- > Br- > Cl- > F-
alkyl halide reactants: tertiary
the strength of the Nu has no affect on rxn rate

68
Q

explain how alkyl halides undergo elimination rxns

A

a halogen is removed from one carbon and a hydrogen is removed from an adjacent carbon. a double bond is formed between the two carbons that the atoms were removed.

69
Q

what occurs in the TS of the RDS for an E2 rxn

A

the alkyl halide and the base (aka elimination is bimolecular because the C-H and C-X bonds break simultaneously)

70
Q

beta carbon

A

the carbon with the hydrogen being removed

71
Q

alpha carbon

A

the carbon with the hydrogen staying

72
Q

how many steps does E2 elimination occur in

A

one step (its concerted as there are no charged intermediates)

73
Q

explain how E2 is regioselective

A

the major product is the most stable alkene. connects with zaitsev’s rule

74
Q

zaitsev’s rule

A

the most stable alkene is obtained by removing a hydrogen from the beta carbon that is bonded to the fewest hydrogens

75
Q

why is anti elimination preferred in E2 rxns

A
  • anti elimination requires the molecule to be in a staggered conformation
  • anti periplanar attack of the base achieves the best overlap of interacting orbitals
  • anti elimination avoids the repulsion of the electron-rich base wth the electron rich leaving group
76
Q

why is periplanar important

A

because the sigma bonds in the reactants must overlap with each other perfectly to form pi bonds, which can be made easier when they are already in the same plane

77
Q

syn vs. anti periplanar geometry

A

syn:
- eclipsed, higher energy

anti:
- staggered, lower energy

78
Q

explain how an E2 rxn is stereoselective

A

the alkene with the largest / bulkiest groups are on opposite sides of the double bond because it is more stable. AKA E is prefered over Z

79
Q

how does the stereopchemistry of the starting material affect the major product of an E2 rxn

A

when only one hydrogen is bonded to the beta carbon, the major product of an E2 rxn depends on the structure of the alkyl halide as it must achieve an anti comformation

80
Q

explain how stereochemistry affects what side the atoms are on

A

if both atoms are going back, they are on the same side of a flat molecule (cis)

81
Q

explain an E1 rxn

A

the C-X bond breaks first to afford a carbocation, followed by base removal of a proton to give the alkene

82
Q

what is in the TS of the RDS for an E1 rxn

A

only the alkyl halide (makes it unimolecular)

83
Q

explain how an E1 rxn is regioselective

A

the major product is the most stable alkene (following zaitsev’s rule)

84
Q

can the stereochemistry of the starting material affect the major product of an E1 rxn?

A

no. however when only one hydrogen is bonded to the beta carbon, the major product of an E1 rxn is still the more stable alkene because the carbocation intermediate is planar and thus ‘loses’ its stereochemical information.

85
Q

relative reactivities of alkyl halides in E2 and E1 rxns

A

most reactive
RI
RBr
RCl
RF
least reactive

86
Q

which alkyl halides undergo E2 rxns

A

primary, secondary, AND tertiary alkyl halides

87
Q

which alkyl halides undergo E1 rxns

A

tertiary

88
Q

when do tertiary alkyl halides go through E2 vs. E1 rxns

A
  • E2 –> favored by a high concentration of a strong base (prolly have a minus sign)
  • E1 –> favored by a low concentration of a weak base (prolly just have a lone pair)
89
Q

compare SN2 vs. E2

A

SN2
- only the inverted product is formed

E2
- both E and Z stereoisomers are formed (with more of the stereoisomer with the largest groups on opposite sides of the double bond) unless the beta carbon from which the hydrogen is removed is bond to only one hydrogen, in which case only one stereoisomer is formed. the stereoisomer configuration depends on the configuration of the reactant.

90
Q

compare SN1 vs. E1

A

SN1
- both stereoisomers (R and S) are formed (generally with more inverted product than with retained

E1
- both E and Z stereoisomers are formed (with more of the stereoisomer with the largest groups on opposite sides of the double bond)

91
Q

explain E2 elimination from a six-membered ring

A

both groups being eliminated must be in an anti orientation (meaning in a six membered ring, they most both be in the axial position). **remember the ring may have to be flipped (k eq being on the arrow going either way)

92
Q

explain E1 elimination from a six-membered ring

A

the hydrogen and halogen atoms do not have to be in axial positions because the rxn is not concerted.

93
Q

what products are obtained from primary alkyl halides

A

primarily substitution (SN2) unless there is steric hinderance in the alkyl halide or Nu in which case elimination (E2) is favored

CANNOT undergo SN1 / E1

94
Q

what products are obtained from secondary alkyl halides

A

both substitution (SN2) and elimination (E2); the stronger and bulkier the base and the higher the temperature, the greater the percentage of elimination.

CANNOT undergo SN1 / E1

95
Q

what products are obtained from tertiary alkyl halides

A

only elimination (E2)

both substitution (SN1) and elimination (E1) with substitution are favored.

96
Q

benzyl vs. phenyl groups

A

benzyl group:
- six membered ring with a CH2 attached then R

phenyl group:
- sex membered ring and an R group attached

97
Q

what rxn does benzylic and allylic halides undergo

A

unless benzylic and allylic halides are tertiary, they readily undergo SN2 rxns. They use strong nucleophiles.

97
Q

when / why do benzylic and allylic halides undergo SN1 rxns

A

benzylic and allylic halides also undergo SN1 rxns because they form relatvely stable carbocations. They use weak bases

98
Q

how are the benzylic and allylic halides stabilized in E2 rxns

A

the product is relatively stable because the new double bond is conjugated. They use strong bases

99
Q

how are the benzylic and allylic halides stabilized in E1 rxns

A

benzylic and allylic halides also undergo E1 rxns because they form good resonance-stabilized carbocations. They use weak bases

100
Q

how are resonance forms involved in benzylic and allylic carbocations in SN1 and E1 rxns

A

resonance forms of benzylic and allylic carbocations. the positive charge is delocalized over the entire pi system in both cations.

101
Q

what is the stability for benzylic and allylic carbocations in SN1 and E1 rxns

A

most stable
tertiary
secondary (about the same as the next)
benzylic (about the same as the next)
allylic
primary
methyl
least stable

102
Q

can vinylic and aryl halides undergo SN2 rxns

A

it is impossible for an sp^2 hybridized carbon to undergo an SN2 rxn as back side attack of the nucleophile cannot occur. Basically the Nu is repelled by the pi electron cloud

103
Q

can vinylic and aryl halides undergo SN1 or E1 rxns

A

it is impossible for an sp^2 hybridized carbon to undergo a unimolecular rxn as the carbocation is too unstable to be formed.

104
Q

can a vinylic halide undergo an E2 rxn

A

it is possible for an sp^2 hybridized carbon to undergo an E2 rxn but a strong base (such as -NH2) is required.

105
Q

what is an organometallic compound

A

part organic, part metal. it contains a carbon-metal bond. Ex. C-Pb

106
Q

what are organolithium compounds

A

organolithium compounds require two equivalents of Li metal (2 Li) and one equivalent of alkyl / aryl halide. organolithium compounds are very strong bases and excellent Nu’s

107
Q

what are organomagnesium compounds

A

organomagnesium compounds are called grignard reagants. organomagnesium compounds require one equivalent of Mg metal and one equivalent of alkyl/alkenyl/aryl halide. organomagnesium compounds are very strong bases and excellent Nu’s

108
Q

what are grignard reagants

A

RMgX

  • an organomagnesium compound
  • ethers are the usual solvents in grignard rxns due to their unreactive nature
  • the solvent provides electrons so the magnesium can complete its octet.
109
Q

why do organolithium and organomagnesium compounds react as if they were carbanions

A
  • carbon is more electronegative than Li or Mg
  • since organometallic reagants can be thought of as carbanions, they are great Nu’s
110
Q

what happens when organolithium and organomagnesium compounds react with a proton of an acid group

A
  • when the grignard reagent reacts with a proton source (ex. H2O or CH3OH), it forms an alkane.
  • organolithium and organomagnesium compounds react violently with water and alcohols so you must exclude protic molecules or add them to a rxn very slowly.
111
Q

what are organocuprates

A
  • organocuprates are also called gilman reagants
  • organocuprates undergo coupling rxns to link any two alkyl, aryl or vinyl groups together
112
Q

how is the configuration of the double bond retained for organocuprates

A
  • organocuprates can be used to prepare compounds that cannot be prepared by using nucleophilic substitution rxns
  • organocuprates substitute the halogen on an alkene or arene sp^2 carbon with one of the “R” groups attached to the Cu atom.
113
Q

what R group of an alkyl halide and the organocubrate can be used

A
  • the R group of the alkyl halide (and thus the organocuprate) can be primary, methyl, aryl, vinylic, or acllylic
  • the R group cannot be secondary or tertiary for steric reasons
114
Q

oxidation vs. reduction rxns

A

oxidation:
- C-H bond broken and C-Cl bond formed

reduction:
- C-Cl bond broken and C-H bond formed

115
Q

what is the solvent for a free radical addition of HBr to a terminal alkene

A

HBr

peroxide

116
Q

what is the solvent for the SN1 mechanism of an alcohol to an alkyl halide

A

HI

117
Q

what is the solvent for a free radical bromination

A

Br2

hv

118
Q

what is the solvent for the SN1 mechanism of an alkyl halide to an alcohol

A

H2O

119
Q

what is the solvent for keto / enol tautomerization

A

H2SO4

H2O

120
Q

what is the solvent for a free radical chlorination

A

hv

ochem (2 decks)

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