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Organic Chemistry 1 > Chapter 6 > Flashcards

Chapter 6 Flashcards

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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
Q

Are strong bases good leaving groups in SN2 reactions?

A

No, strong bases are poor leaving groups

26
Q

The SN2 Mechanism – Solvent Effects

A

SN2 reactions favored (faster) in polar, aprotic solvents

SN2 reactions disfavored (slower) in protic solvents: water, alcohols, acetic acid

27
Q

Solvent Effects: Protic Problems

A

Polar protic solvents have acidic hydrogens (O—H or N—H) which can solvate the nucleophile reducing their nucleophilicity.

28
Q

Nucleophile strength and rate of the reaction

A

  • Stronger nucleophile = faster reaction
29
Q

Stability of leavin group and rate of the reaction

A

More stable leaving group = faster reaction

30
Q

What are the steps in an SN1 reaction?

A

1) formation of carbocation by loss of leaving group
2) attack (substitution) of nucleophile

31
Q

When is SN1 favorable?

A

Substrate is tertiary, allylic / benzylic or sterically hindered

PROTIC / MILDLY ACIDIC Solvent

32
Q

Rate determining step in SN1

A

Formation of carbocation is the rate determining step

33
Q

Racemization in SN1

A

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
Q

Hydrogen Shift

A

Hydrogens adjacent to a 1° or 2° halide can shift over to form a more stable cation

35
Q

Methyl Shift

A

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
Q

Relative Reactivity of Halides via SN1

A

tertiary > secondary > primary >> methyl

37
Q

Do nucleophiles have an effect on the rate of SN1 reactions?

A

Carbocation formation is rate determining:

Thus, different nucleophiles do not change rate in SN1

Poor nucleophiles better to avoid side reactions

38
Q

Leaving groups in SN1 reactions

A

Better leaving groups lead to faster carbocation formation

I - > Br - ~ - OTs > Cl –

Typically need good leaving groups for SN1

39
Q

In what type of solvent do SN1 reactions work best?

A

SN1 reactions work best in POLAR PROTIC solvents:

H2O, CH3CO2H, CF3CH2OH, CH3OH, etc.

40
Q

Elimination reactions

A

The halogen and one hydrogen are eliminated from the molecule resulting in a double bond

41
Q

E1 Reaction

A

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
Q

Step 1 of E1 reaction

A

Step 1: Leaving group leaves, forming a carbocation

43
Q

Step 2 of E1

A

Step 2: Base abstracts H+ from adjacent carbon forming the double bond

44
Q

E1 and SN1 are competing reactions because their first steps both for a carbocation. What conditions favor elimination over substitution?

A

High temperature and weak/bulky bases will favor elimination over substitution.

45
Q

In E1, which step is the rate-determining step?

A

The first step (just like SN1)

46
Q

Zaitsev’s Rule

A

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
Q

The E2 reaction

A

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
Q

E1 Mechanism

A

•Order of reactivity for alkyl halides:
3° > 2 ° > 1°

Mixture may form, but Zaitsev product
predominates

49
Q

E2 Stereochemistry

A
  • 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
Q

E2 Reactions on Cyclohexanes

A
  • 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
Q

How does the strength of the nucleophile determine order?

A

Strong nucleophiles or bases promote bimolecular reactions (E2 or Sn2)

52
Q

What type of Alkyl halides usually undergo SN2?

A

1° alkyl halides usually undergo SN2

53
Q

What type of reactions do tertiary alkyl halides usually undergo?

A

3° alkyl halides mixture of SN1, E1 or E2. They cannot undergo SN2

54
Q

What type of reactions do secondary alkyl halides usually undergo?

A

2° alkyl halides: mix of substitution and elimination. Look at nucleophile / base strength

55
Q

_____ temperature favors elimination

A

High temperature favors elimination

56
Q

______ bases favor elimination

A

Bulky bases favor elimination

NaOC(CH3)3

57
Q

Common polar, aprotic solvents

A

acetone, DMF, DMSO, acetonitrile, HMPA

Organic Chemistry 1 (10 decks)

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