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Organic Chemistry 1 > Chapter 6 > Flashcards

Chapter 6 Flashcards

1
Q

Alkyl halides

A

Halogen, X, is directly bonded to sp3 carbon

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2
Q

Vinyl halides

A

X is bonded to sp2 carbon of alkene

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3
Q

Aryl halides

A

X is bonded to sp2 carbon on phenyl ring

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4
Q

CH2X2

A

methylene halide

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5
Q

CHX3

A

haloform

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6
Q

CX4

A

carbon tetra halide

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7
Q

Geminal dihalide

A

two halogen atoms are bonded to the same carbon

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8
Q

Vicinal dihalide

A

two halogen atoms are bonded to adjacent carbons

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9
Q

Allylic Halogination

A
  • Allylic radical is resonance stabilized
  • Bromination occurs with good yield at the allylic position (sp3 C next to C=C)
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10
Q

SN2 Reaction

A
  • Bimolecular nucleophilic substitution
  • Concerted reaction: new bond forming and old bond

breaking at same time

  • Rate is first order in each reactant
  • Inversion at carbon
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11
Q

SN2 Energy Diagram

A
  • The SN2 reaction is a one-step reaction
  • Transition state is highest in energy
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12
Q

What does SN2 stand for?

A

SN2 stands for: substitution, nucleophilic, bimolecular

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13
Q

Bimolecular

A

TWO species are involved in rate determining step

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14
Q

SN2 Mechanism Favored When:

A

Substrate is PRIMARY and UNHINDERED

GOOD (strongly basic) Nucleophile

GOOD (weakly basic) Leaving Group

POLAR, APROTIC Solvent

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15
Q

Kinetics of SN2 Reaction

A

Second order reaction: Rate = k [:Nu] [substrate]

Suggests that C-Nu bond forms as C-X bond breaks

Requires backside attack of nucleophile

Trigonal bipyramidal transition state

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16
Q

SN2 and Stereochemistry

A
  • Backside attack of nucleophile results in INVERSION of configuration
  • The stereochemistry is always reversed, but the R/S does not always switch. Depends on the priority of the nucleophile and substituents on the stereocenter.
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17
Q

SN2 Orbital Inversion

A

Inversion of Configuration Required by Orbitals

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18
Q

Relative Reactivity of Halides via SN2

A

methyl > primary > secondary >> tertiary (doesn’t happen)

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19
Q

Steric Effects of the Substrate on SN2 Reactions

A

As substitution increases, steric crowding increases.

Approach of the nucleophile is increasingly hindered.

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20
Q

Nucleophilicity across a period

A

Nucleophilicity increases

across a period (R to L).

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21
Q

Nucleophilicity in groups

A

Nucleophilicity increases down a group.

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22
Q

Nucelophilicity in acids vs bases

A

Neg. charged species a stronger nucleophile than its conjugate acid.

Why? Ability to donate electrons.

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23
Q

What makes a good leaving group in SN2 reactions?

A

More stable anion = better leaving group

24
Q

Leaving Groups

A

Good leaving groups: I -, Br -, sulfonates

Modest leaving group: Cl -

Poor leaving groups: F -, - OH, CH3CO2 -, PhO -

25
Are strong bases good leaving groups in SN2 reactions?
No, strong bases are poor leaving groups
26
The SN2 Mechanism – Solvent Effects
SN2 reactions favored (faster) in polar, aprotic solvents SN2 reactions disfavored (slower) in protic solvents: water, alcohols, acetic acid
27
Solvent Effects: Protic Problems
Polar protic solvents have acidic hydrogens (O—H or N—H) which can solvate the nucleophile reducing their nucleophilicity.
28
Nucleophile strength and rate of the reaction
## Footnote - Stronger nucleophile = faster reaction
29
Stability of leavin group and rate of the reaction
More stable leaving group = faster reaction
30
What are the steps in an SN1 reaction?
1) formation of carbocation by loss of leaving group 2) attack (substitution) of nucleophile
31
When is SN1 favorable?
Substrate is tertiary, allylic / benzylic or sterically hindered PROTIC / MILDLY ACIDIC Solvent
32
Rate determining step in SN1
Formation of carbocation is the rate determining step
33
Racemization in SN1
The lobes of the empty p-orbital are on both sides of the trigonal plane Nucleophile can attack the carbocation from either side – leads to racemization of optically active S.M.
34
Hydrogen Shift
Hydrogens adjacent to a 1° or 2° halide can shift over to form a more stable cation
35
Methyl Shift
Since a primary carbocation cannot form, the methyl group on the adjacent carbon can move (along with both bonding electrons) to the primary carbon displacing the bromide and forming a tertiary carbocation.
36
Relative Reactivity of Halides via SN1
tertiary \> secondary \> primary \>\> methyl
37
Do nucleophiles have an effect on the rate of SN1 reactions?
Carbocation formation is rate determining: Thus, different nucleophiles do not change rate in SN1 Poor nucleophiles better to avoid side reactions
38
Leaving groups in SN1 reactions
Better leaving groups lead to faster carbocation formation I - \> Br - ~ - OTs \> Cl – Typically need good leaving groups for SN1
39
In what type of solvent do SN1 reactions work best?
SN1 reactions work best in POLAR PROTIC solvents: H2O, CH3CO2H, CF3CH2OH, CH3OH, etc.
40
Elimination reactions
The halogen and one hydrogen are eliminated from the molecule resulting in a double bond
41
E1 Reaction
Unimolecular elimination Two groups lost: a hydrogen and the leaving group Nucleophile can act as a base The E1 and SN1 reactions have similar conditions so a mixture of products often observed
42
Step 1 of E1 reaction
Step 1: Leaving group leaves, forming a carbocation
43
Step 2 of E1
Step 2: Base abstracts H+ from adjacent carbon forming the double bond
44
E1 and SN1 are competing reactions because their first steps both for a carbocation. What conditions favor elimination over substitution?
High temperature and weak/bulky bases will favor elimination over substitution.
45
In E1, which step is the rate-determining step?
The first step (just like SN1)
46
Zaitsev's Rule
More substituted double bonds are more stable In elimination reactions, the major product of the reaction is the more substituted double bond: Zaitsev’s Rule
47
The E2 reaction
Elimination, bimolecular Requires a strong base Concerted reaction: the proton is abstracted, the double bond forms and the leaving group leaves, all in one step
48
E1 Mechanism
•Order of reactivity for alkyl halides: 3° \> 2 ° \> 1° Mixture may form, but Zaitsev product predominates
49
E2 Stereochemistry
- The halide and the proton to be abstracted must be anti-coplanar (q=180º) to each other for the elimination to occur - The orbitals of the hydrogen atom and the halide must be aligned so the s bond overlaps with s\* - The anti-coplanar arrangement minimizes any steric hindrance between the base and the leaving group
50
E2 Reactions on Cyclohexanes
- An anti-coplanar conformation (180°) can only be achieved when both the hydrogen and the halogen occupy trans diaxial positions. - The chair must flip to the conformation with the axial halide in order for the elimination to take place.
51
How does the strength of the nucleophile determine order?
Strong nucleophiles or bases promote bimolecular reactions (E2 or Sn2)
52
What type of Alkyl halides usually undergo SN2?
1° alkyl halides usually undergo SN2
53
What type of reactions do tertiary alkyl halides usually undergo?
3° alkyl halides mixture of SN1, E1 or E2. They cannot undergo SN2
54
What type of reactions do secondary alkyl halides usually undergo?
2° alkyl halides: mix of substitution and elimination. Look at nucleophile / base strength
55
\_\_\_\_\_ temperature favors elimination
High temperature favors elimination
56
\_\_\_\_\_\_ bases favor elimination
Bulky bases favor elimination NaOC(CH3)3
57
Common polar, aprotic solvents
acetone, DMF, DMSO, acetonitrile, HMPA

Organic Chemistry 1 (10 decks)

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