methyl RX example primary RX example secondary RX example secondary RX example tertiary RX example

(1)

C

_β

H

D CH

₃

C

_α

Br C

_β

H C

_α

C

_β

H

Br C

_β

H

₃

C D

H

T H

D T

Example 2 - two possible perspectives (deuterium is an isotope of hydrogen that can be distinguished)

C

_β

H

CH

₃

D

C

_α

Br D H

C

_α

C

_β

H

Br D H H

₃

C D

(1S,2S)- 1,2-dideuterio-1-bromopropane reaction

conditions reaction

conditions

? ?

(1S,2R)- 1,2-dideuterio-1-bromopropane

(1S,2S,3R)-1,3-dideuterio-2-bromo-1-tritiobutane (1S,2R,3R)-1,3-dideuterio-2-bromo-1-tritiobutane Example 3 - two possible perspectives (deuterium is an isotope of hydrogen that can be distinguished)

C

_β

H

D CH

₃

C

_α

Br C

_β

H C

_α

C

_β

D

Br C

β

H

₃

C H

H

CH

₂

CH

₃

H

CH

₃

CH

₂

CH

₃

(2S,3R,4S)-2-deuterio-4-methyl-3-bromohexane

Example 4 - two possible perspectives (deuterium is an isotope of hydrogen that can be distinguished)

(2S,3S,4S)-2-deuterio-4-methyl-3-bromohexane

C

_β

H

D CH

₃

C

_α

Br

C

β

C H

₂

C

_α

C

_β

D

Br C

_β

CH

₂

H

₃

C H

H

CH

₂

CH

₃

H

CH

₃

CH

₂

CH

₃

(2S,3R,4S)-2-deuterio-4-methyl-3-bromohexane

Example 5 - two possible perspectives (deuterium is an isotope of hydrogen that can be distinguished)

(2S,3S,4S)-2-deuterio-4-methyl-3-bromohexane H

H D

H

3

C H

₃

C T

C

_α

Br D H

S-bromodeuterioprotiotritiomethane

C

_α

T

Br D H

Example 1 - two possible perspectives (deuterium and tritium are isotopes of hydrogen that can be distinguished)

methyl RX example

reaction conditions

?

reaction conditions

?

reaction conditions

?

reaction conditions

?

reaction conditions

?

reaction conditions

?

reaction conditions

?

reaction conditions

? primary RX example

secondary RX example

tertiary RX example secondary RX example

Beginning RX patterns to knowfor S

_N

and E Reactions - horizontal and vertical templates for practice

simplify structures by making D and T = H

S-bromodeuterioprotiotritiomethane

simplify structures by making D and T = H

simplify structures by

making D and T = H

(2)

Practice arrow pushing reactions using acid/base chemistry. Model the example reaction.

Basic solutions

O

H H₃C O

HC C

N N N

O C H₃C H₃C

H₃C

H Na

K Na

Na

B H

H H

H

Na pK_a1=16

O H

H

pKa2=16 O H

H

O H

∆G = pK_a1 -pK_a2 = 16 - 16 = 0

K_eq = Ka1

Ka2

= 10^-16 10^-16

= 1

pKa1=16 O H

H

pK_a2= 16

∆G = pK_a1 -pK_a2 =

K_eq = Ka1

Ka2

=

pKa1=16 O H

H

pKa2= 5

∆G = pKa1 -pKa2 =

Keq = Ka1

K_a2

=

pKa1=16 O H

H

pK_a2= 19

∆G = pKa1 -pKa2 =

Keq = Ka1

K_a2

= H3C

C O

O Na

pK_a1=16 O H

H

pK_a2= 9

∆G = pKa1 -pK_a2 =

K_eq = K_a1

K_a2

=

pK_a1=16 O H

H

pKa2= 25

∆G = pKa1 -pK_a2 =

K_eq = K_a1

K_a2

= N

C

Na

pKa1=16 O H

H

pKa2= 5

∆G = pKa1 -pKa2 =

Keq = Ka1

Ka2

=

pKa1=16 O H

H

pK_a2= ?

∆G = pKa1 -pKa2 =

Keq = Ka1

Ka2

=

pK_a1=16 O H

H

pK_a2= 37

∆G = pKa1 -pK_a2 =

K_eq = K_a1

K_a2

= Example

hydroxide methoxide

(alkoxides)

potassium t-butoxide ethanoate

(carboxylates)

cyanide terminal acetylides

sodium borohydride (also lithium aluminiumhydride) azide

hydride (always a base)

(3)

Acidic solutions – model the example reaction.

O

H H₃C O

O C H3C H₃C

H3C Na

pKa1= -2

O H

H pK_a2=16

O H

H

∆G = pKa1 -pK_a2 = -2 - 16 = -18

Keq = Ka1

K_a2

= 10² 10^-16

= 10⁺¹⁸

pKa2= -3

∆G = pKa1 -pK_a2 =

Keq = Ka1

K_a2

=

∆G = pK_a1 -pK_a2 =

K_eq = K_a1

Ka2

=

pKa2= -7

∆G = pKa1 -pKa2 =

K_eq = K_a1

Ka2

=

pKa2= -3

∆G = pKa1 -pKa2 =

K_eq = K_a1

Ka2

=

pKa2= 9

∆G = pKa1 -pKa2 =

K_eq = K_a1

Ka2

=

pK_a2= -8

∆G = pKa1 -pKa2 =

K_eq = K_a1

Ka2

= Example

hydroxide methoxide

(alkoxides)

t-butyl alcohol (2-methylpropan-2-ol)

alkene (use the pi electrons)

H O

H H H

pKa1= -2 H

H O

H H

pK_a2= -3

H3C C O

CH3

propan-2-one

O H

SO3H pKa1= -10 O

H3C

pK_a2= -3

∆G = pKa1 -pKa2 =

Keq = Ka1

K_a2

= methoxide

(alkoxides) H

O H

SO₃H pK_a1= -10

O H

SO₃H pK_a1= -10 sulfuric acid

O H

SO₃H pK_a1= -10 O

H3C

dimethyl ether CH3

N H3C

dimethyl ether CH3

CH₃

O H

SO3H pKa1= -10

O H

SO₃H pK_a1= -10

S

N

and E reactions (S

N

2 vs E2) and (S

N

1 vs E1) Methyl RX compounds – always S

N

2 reactions

O H

Na

Example reaction

hydroxide nucleophile

T

C

_α

Br D H

C O H

T

D H

Br

absolute configuration = S absolute configuration = R leaving

group

mech. = S

_N

2

(4)

O H₃C

Na methoxide nucleophile

T

C_α Br D H

H₃C C O

O Na ethanoate nucleophile

T

C_α Br D H

O C H₃C H₃C

H₃C K potassium t-butoxide nucleophile, usually a base

T

C_α Br D H

N C

Na cyanide nucleophile

T

C_α Br D H

HC C

Na terminal acetylide

nucleophile

T

C_α Br D H

N N N

Na azide nucleophile

T

C_α Br D H

B H

H H

H

Na sodium borohydride

nucleophile

T

C_α Br D H

(5)

S

N

2 vs E2 Competition at primary RX compounds

Cβ

H

CH₃ D

Cα Br D H O

H Na

Example S_N2 reaction

hydroxide nucleophile

C_α O H

C_β

D

Br

absolute configuration = 1S,2S absolute configuration = 1R,2S

leaving group mech. = S_N2

hydroxide base

H

CH₃ D H

Cβ

H

CH₃ D

Cα Br D H O

H

Na

Example E2 reaction

Br

absolute configuration = 1S,2S

leaving group mech. = E2

C C D

D H₃C

H E stereochemistry C1-C2 single bond rotation

Cβ

D H

CH₃

Cα Br D H

hydroxide base

O H

Na

mech. = E2

C C CH₃

D H

H Z stereochemistry

Br

leaving group

C_β H

CH₃ D

Cα Br D H

mech. = S_N2

Cβ

H

CH₃ D

C_α Br D H

mech. = E2

C1-C2 single bond rotation

Cβ

D H

CH₃

C_α Br D

H mech. = E2

methoxide nucleophile

methoxide base

O H₃C

Na O H₃C

Na

O H₃C

Na methoxide

base

(6)

Cβ

H

CH₃ D

C_α Br D H H₃C

C O

O Na

mech. = S_N2

C_β H

CH₃ D

Cα Br D H

mech. = E2

Cβ

D H

CH₃

Cα Br D

H mech. = E2

ethanoate nucleophile

H₃C C O

O Na ethanoate

base

H₃C C O

O Na ethanoate

base

Cβ

H

CH₃ D

C_α Br D H O

C H₃C H₃C

H₃C K

mech. = S_N2

Cβ

H

CH₃ D

Cα Br D H

absolute configuration = 1S,2S mech. = E2

Cβ

D H

CH₃

Cα Br D

H mech. = E2

potassium t-butoxide nucleophile

O C H₃C H₃C

H₃C K

O C H₃C H₃C

H₃C K

potassium t-butoxide is sterically large and very basic and makes a poor nucleophile, mainly a base in our course

potassium t-butoxide base

(7)

Cβ

H

CH₃ D

C_α Br D H N

C

Na

mech. = S_N2

C_β H

CH₃ D

Cα Br D H

Cβ

D H

CH₃

Cα Br D

H mech. = E2

cyanide nucleophile

N C

Na cyanide

base

N C

Na cyanide

base

S_N2 = major product

E2 = minor product

Cβ

H

CH₃ D

C_α Br D H HC

C

Na

mech. = S_N2

Cβ

H

CH₃ D

Cα Br D H

Cβ

D H

CH₃

Cα Br D

H mech. = E2

E2 = minor product

E2 = minor product terminal acetylide

nucleophile

HC C

base

HC C

base

(8)

Cβ

H

CH₃ D

C_α Br D H N

N N

Na

mech. = S_N2

C_β H

CH₃ D

Cα Br D H

Cβ

D H

CH₃

Cα Br D

H mech. = E2

E2 = minor product

E2 = minor product azide

nucleophile

N N N

Na azide

base

N N N

Na azide

base

Cβ

H

CH₃ D

C_α Br D H B

H H H

H

Na

mech. = S_N2

Cβ

H

CH₃ D

Cα Br D H

Cβ

D H

CH₃

Cα Br D

H mech. = E2

E2 = minor product

E2 = minor product borohydride

nucleophile

B H

H H

H

Na borohydride

base

B H

H H

H

Na borohydride

base

(9)

Secondary RX compounds

Cβ

H D

CH₃

C_α Br Cβ

H H

D T O

H Na

Example S_N2 reaction

Cα

O H

C_β

C

Br

absolute configuration = 1S,2S,3R

leaving group mech. = S_N2

hydroxide base

H D CH₃ H

O H

Na

Example E2 reaction

Br

C C CH₃

CHDT D

H Z stereochemistry C2-C3 single bond rotation

H D

T S_N2 = minor product

absolute configuration = 1S,2R,3R

Cβ

H D

CH₃

C_α Br Cβ

H H

D T

E2 = major product

C1-C2 single bond rotation hydroxide

base O H

Na

Br

C C H

CHDT H₃C

H E stereochemistry Cβ

D CH₃

H

Cα Br Cβ

H H

D T

E2 = major product

hydroxide base

O H

Na

Br

Z stereochemistry Cβ

D CH3

H

Cα Br C_β H H

D T

E2 = major product C

C H

CHDCH₃

D TMS

hydroxide base

O H

Na

Br

E stereochemistry Cβ

D CH₃

H

C_α Br Cβ

H D

T H

C H

CHDCH₃

T H

hydroxide base

O H

Na

Br

Z stereochemistry Cβ

D CH3

H

Cα Br Cβ

H T

H D

C H

CHDCH₃

H D

(10)

Use the above template to write out the mechanisms for the following nucleophile/bases.

O H

HC C

N N N

Na Na

Na N

C

Na

hydroxide nucleophile hydroxide

base

cyanide nucleophile

O H

₃

C

Na methoxide

base methoxide nucleophile H

₃

C

C O

O Na ethanoate nucleophile

ethanoate base

potassium t-butoxide is sterically large and very basic and makes a poor nucleophile, mainly a base in our course

O C H

₃

C H

₃

C

H

₃

C K

potassium t-butoxide nucleophile cyanide

base

terminal acetylide nucleophile terminal acetylide

base azide nucleophile

azide base

For the following nucleophile/bases S

_N

2 is the major product and E2 is the minor product at secondary RX centers in our course.

For the following nucleophile/bases E2 is the major product and S

_N

2 is the minor product at secondary RX centers in our course.

B H

H H

H

Na borohydride nucleophile borohydride

base

(11)

Tertiary RX compounds – always E2 with strong base/nucleophiles in our course

O H

Na

No S_N2 reactions at tertiary RX centers

absolute configuration = 1S,2S,3R

hydroxide base

O H

Na

Example E2 reaction - all E2 reactions via anti C_β-H and C_α-X conformations

Br

C C CH₃

CH D

CH₃ Z stereochemistry C2-C3 single bond rotation

E2 = product

hydroxide base

O H

Na

Br

E stereochemistry

hydroxide base

O H

Na

Br

Z stereochemistry

hydroxide base

O H

Na

Br

No E/Z stereochemistry C_β

H D

CH₃

C_α Br

Cβ

CH₂ H

CH₃ CH₂CH₃ H

No S_N2 reactions at tertiary RX centers

Cβ

H D

CH₃

C_α Br

C_β CH₂ H

CH₃ CH₂CH₃ H

C_β D

CH₃ H

Cα Br

Cβ

CH₂ H

CH₃ CH₂CH₃ H

E2 = product

C₂H₅

CH₃

C C H

CH H₃C

CH₃

C₂H₅

CH₃

Cβ

D

CH₃ H

C_α Br

C_β CH₂ H

CH₃ CH₂CH₃ H

Cβ

D

CH₃ H

Cα Br

Cβ

CH₂ H

CH₃ CH₂CH₃ H

C C H₃C

CHDCH₃

H3C

CH₂CH₃

C H₂C

CHDCH₃

C H

CH₃ CH₂CH₃

(12)

O

H HC

C

N N N

Na Na

Na N

C

Na

hydroxide nucleophile hydroxide

base cyanide

nucleophile O

H₃C

Na methoxide

base methoxide nucleophile

H₃C C O

O Na ethanoate nucleophile

ethanoate base

potassium t-butoxide is sterically large and very basic and makes a poor nucleophile, mainly a base in our course

O C H₃C H₃C

H₃C K

potassium t-butoxide nucleophile

cyanide base

terminal acetylide nucleophile terminal acetylide

base

azide nucleophile

azide base For the following nucleophile/bases E2 is the only product at tertiary RX centers in our course.

B H

H H

H

Na borohydride nucleophile borohydride

base

(13)

Example Reaction Conditions with RX Compounds (methyl, primary, secondary, tertiary)

O

H H₃C O

H₃C C O

O N

C HC

C N

N N

O C H₃C H₃C

H3C

O

H H₃C O

H₃C C O

O

H

Na Na Na K Na Na Na

H

H H

Na B

H H H

H

Na

basic hydride (always a base) nucleophilic

hydride sodium

azide sodium

azide potassium

t-butoxide (sterically bulky base) sodium

ethanoate (carboxylate) sodium

methoxide (alkoxide) sodium

hydroxide

water methanol

(alcohols)

ethanoic acid (carboxylic acids)

Strong base/nucleophiles = SN2 / E2 Reactions (conjugate acid pKa in parentheses)

Weak base/nucleophiles = S_N1 / E1 Reactions

X H₃C

X H₂C

X CH

X C R

R

R R R

R

only S_N2

NA

S_N2 > E2

E2 > SN2

S_N2 > E2 S_N2 > E2 E2 > S_N2

E2 > S_N2

S_N2 > E2 S_N2 > E2 S_N2 > E2 S_N2 > E2 NA

E2 > SN2 S_N2 > E2 S_N2 > E2 E2 > S_N2 S_N2 > E2 S_N2 > E2 NA

only E2

only

E2 only

E2

only

E2 only

E2

only E2

only

E2 NA

methyl RX

primayr RX

secondary RX

tertiary RX

pKa=16 pK_a=16 pK_a=5 pK_a=19 pKa=25 pKa=9 pKa=5 pK_a=37

No

Reaction No

Reaction

No Reaction

S_N1 > E1 S_N1 > E1 S_N1 > E1

S_N2 1. bimolecular

2. always backside attack = inversion of configuration 3. less basic is better for SN2 4. steric hindrance at C_α or C_β or in nucleophile slows S_N2

E2

1. bimolecular

2. anti C_β-H / C_α-X attack 3. more basic is better for E2 4. steric hindrance at C_α or C_β or in nucleophile makes E2 more competitive

SN1

1. unimolecular in RX 2. first step forms carbocation 3. top and bottom attack at carbocation racemization of configuration 4. generally lose "H⁺" from oxygen 4. generally S_N1 > E1

E1

1. unimolecular in RX 2. first step forms carbocation 3. generally loss of C_β-H via syn and anti attack, can make E & Z alkenes

4. generally S_N1 > E1

3 Carbocation Reactions

1. add nucleophile (top or bottom) ≈ S_N1 2. lose beta H (top or bottom) ≈ E1 3. rearrangement (not covered in 314) X

H₃C

X H₂C

X CH

X C R

R

R R R

R methyl RX

primayr RX

secondary RX

tertiary RX

(14)

C

_α

C

_β1

C

Br C

_β2

H

_b

H

_a

H

H H

C

_α

C

_β1

H

_a

Br C

_β2

H

₃

C

H

_b

H

Rotation of C

_α

-C

_β1

brings H

_a

or H

_b

anti to C-Br, which allows S

_N

2 and two different E2 possibilities: H

_a

(Z) and H

_b

(E). Since C

_β2

is a simple methyl, there is no C

_β2

substituent to inhibit either of these reactions.

C

_α

C

_β1

H

_b

Br C

_β2

H

_a

H

₃

C H

H When methyl on C

_β

1 is anti

to C-Br, no S

_N

2 is possible and no E2 is possible from C

_β

1. S

_N

2 possible

E2 from C

_β1

(2Z- butene) E2 from C

_β2

(1-butene)

S

_N

2 possible

E2 from C

_β1

(2E- butene) E2 from C

_β2

(1-butene) No S

_N

2 possible and no

E2 from C

_β1

, but E2 from C

_β2

(1-butene) is possible.

E2 possible here

Use these ideas to understand cyclohexane reactivity.

H H

H

H H

Br H H

H H

H

H H

H

H Br H

H H

H

H H

No S

_N

2 is possible (1,3 diaxial positions block approach of nucleophile), and no E2 is possible because ring carbons are anti.

No S

_N

2 or E2 when "X"

is in equatorial position.

S

_N

2 possible if C

_α

is not tertiary and there is no anti C

_β

"R" group.

E2 possible with anti C

_β

-H.

Both S

_N

2 and E2 are possible in this conformation with leaving group in axial position.

H H

H

H H

Br H C

H H

H

H H

H

H H Br

H H

H

₃

C

H H

No S

_N

2 or E2 when "X"

is in equatorial position.

No S

_N

2 possible if there is an anti C

_β

"R" group.

E2 possible with anti C

_β

-H.

Only E2 is possible in this conformation. Leaving group is in axial position.

H

H H full rotation

is possible in chain

only partial rotation is possible in ring

full rotation is possible in chain

only partial rotation is possible in ring

H H H

equatorial leaving group

axial leaving group

No S

_N

2 is possible (1,3 diaxial positions block approach of nucleophile), and no E2 is possible because ring carbons are anti.

No E2 possible,

no anti C

_β

-H.

(15)

S

_N

1 and E1 Factors

1.

We write nucleophiles (H-Nu:) and bases (H-B:) differently. A proton is included and they are written as neutral molecules. Those most often used by us are solvent molecules of polar, protic solvents (water, alcohols and liquid carboxylic acids).

Weak nucleophile bases

used in our course: ^H ^O ^H ^R ^O ^H ^C

R O

O H

water alcohols carboxylic acids

2. Role of RX compound – RX needs to stabilize a carbocation carbon.

a. “R” groups on the C

α

carbon are inductively donating and help stabilize positive charge. Tertiary is better than secondary and we will not propose primary or methyl carbocations.

C H

H

very poor carbocation

C R

H

poor carbocation

C R

R

H

C R

R

relatively good carbocation

= donating inductive effect OK

carbocation

b. Hyperconjugation stabilizes positive charge. Tertiary is better than secondary and we will not propose primary or methyl carbocations.

C C

H

C C

H sigma resonance?

carbocation Carbocation stabilized by

the electrons in an parallel adjacent sigma bond.

c. Resonance is very good at stabilizing charge (positive, negative, free radicals or neutral conjugate pi systems).

i. Pi bonds allow delocalization of electrons via parallel overlap of adjacent 2p orbitals. Most often these pi electrons are from an alkene or aromatic pi system.

carbocation next to a pi bond (alkene or aromatic)

C C

C

C C

C

(16)

ii. Lone pairs adjacent to empty orbitals can share their electron density to help stabilize positive charge (they also delocalize into neutral pi bonds, like enols and amides). Most often nitrogen or oxygen is the donor atom. (Under very different conditions, carbon can do so when present as a carbanion.)

carbocation next to a lone pair (usually nitrogen or oxygen)

X C X C

X = "N" or "O" X = "N" or "O"

iii. Really poor carbocations – We usually won’t propose these, though there are rare occasions where we do invoke some of them.

C C

R

H

Phenyl carbocation has an empty sp

²

orbital (33% "s").

R

Vinyl carbocation has an empty 2p orbital, but on a sp hybridized carbon atom. sp carbocation is more electronegative because 50% "s".

C C

R

Terminal alkyne carbocation, has an empty sp orbital (50% "s").

3. Attack of nucleophile/base

a. Attack as H-Nu: at C

α

carbon (top and bottom)

Nu H

H Cα

R

If R = R, then we cannot detect top/bottom attack.

top

bottom

H Cα

R2

R₁

If R

₁

≠ R

₂

, then we can detect top/bottom attack because opposite configurations will form at the C

_α

carbon, which are enantiomers. We assume racemization, but is not always true.

H Cα

R2

R₁

If R

₁

≠ R

₂

, and R

₁

and/or R

₂

have chiral centers, then we can detect top/bottom attack because opposite configurations will form at the C

_α

carbon and will form diastereomers in combination with the other chiral centers.

* = has chiral center

*

C R

R

H

There are no chiral centers, but top/bottom attack is seen in cis/trans products which are diastereomers.

b. Attack as H-B: at any C

β

-H: Anything goes (can rotate C

β

-H up or down relative to the

carbocation 2p orbital). Alkene mixtures are expected. More substituted tends to be more stable.

C_β H

R₁ R₂

C_α R₃

R₄

C_β C_α R₃

R₄

Rotate C

_β

-H 180

^o

.

R₁ R₂

H

E and Z possible stereochemistry.

Cα

Cβ

R₁

R

R₃

R

Cα

Cβ

R₂

R

R₃

R B

H

B H

(17)

4. Role of solvent – Polar, protic solvent stabilizes charges of both types in first step of the S

N

1 and E1 reaction possibilities (ionization of the C-X bond).

O R

H

O R H

O R

H O

R H

Oxygen end of solvent molecules stabilizes the cations.

O R

H O

R H

O R H O R

H

Hydrogen end of solvent molecules stabilizes the anions.

5. Leaving groups – Same as in S

N

2 reactions, all are very good.

6. Special cases of S

N

2 and S

N

1 and E1 reactions.

a. Assume mainly S

N

product over E product when alcohols (ROH) are mixed with HX acids (H-Cl, H-Br and H-I). The acid protonates the alcohol and makes a good leaving group (water). The nucleophile is the “halide” counter anion. S

N

2 reactions are assumed at methyl and primary alcohols and S

N

1 reactions are assumed at secondary, tertiary, allylic and benzylic alcohols.

O

R H H X

(X = Cl, Br, I)

_X

R = methyl = S

_N

2

R

R = primary = S

_N

2 R = secondary = S

_N

1 R = tertiary = S

_N

1 R = allylic = S

_N

1 R = benzylic = S

_N

1

b. Assume mainly E1 reactions when alcohols are mixed with concentrated sulfuric acid (H

₂

SO

₄

) when heated (∆). As in 6a, the alcohol is protonated to make a good leaving group (water). A carbocation forms (assumed for primary, secondary and tertiary). When an E1 reaction occurs the alkene is distilled from the mixture, which continually shifts the equilibrium to make more.

O

R H H O

R = methyl = not possible R = primary = difficult, E1 R = secondary = moderate, E1 R = tertiary = relatively easy, E1

S

O

O OH

∆ = heat Alkenes, with lower boiling points, possible rearrangement, more substituted alkenes tend to be the major products formed. This is the only reaction where we will propose E1 as the major product.

alcohols,

have higher

boiling points

(18)

Essential clues to make educated guesses about what is occurring.

Reagents Reaction Conditions

Products

X

R Nu H

B H

Nu

H H

Nu R

B

H H

B X

R

Nu R X

X R

X C

_α

C

_β

C

_α

C

_β

B H

X

R = carbon portion methyl = CH

₃

-X primary = RCH

₂

-X secondary = R

₂

CH-X tertiary = R

₃

C-X allylic = CH

₂

=CHCH

₂

-X benzylic = C

₆

H

₅

CH

₂

-X

Nu = B

strong electron pair donors = S

_N

2 / E2 anions, neutral nitrogen, neutral sulfur H

Nu = B H

weak electron pair donors = S

_N

1 / E1 neutral solvent H

₂

O, ROH, RCO

₂

H

Nu R

substitution products

C

_α

C

_β

elimination products

X leaving group

Typical R-X Structures X

H

₃

C methyl X (unique)

CH

₂

R X

primary X (general)

CH R

R X secondary X (general)

C R

R

R X tertiary X (general)

CH H

₂

C CH

₂

X

allylic X (and variations)

CH

₂

X

benzylic X (and variations) Typical leaving groups (for our course = X) – stable anions or neutral molecules

chloride Cl

bromide Br

iodide

I O S

O

O tosylate

O H

H water

These can be leaving groups in basic, neutral or acidic conditions

water is a leaving

group from a

protonated alcohol

in acid conditions

(19)

S

N

/ E Worksheet

Nu B

Nu H

B H

H C_α

X H

H

C_β C_α

X H

H H

Cβ

C_α X

H C_β

H H

C_β C_α

X C_β C_β H

H H Nu

B

Nu B

Nu H H B

H C_α

X H

H

C_β C_α

X H

H H

C_β C_α

X H

C_β

H H

C_β C_α

X C_β Cβ

H H

H H Nu

B H

H Nu B H

Possibilities: SN2 E2 SN1 E1

Possibilities: S_N2 E2 S_N1 E1

examples

X H3C

CH2

X C CH3

H CH2

H3C

C C CH3

H CH2

H3C

X

CH2CH3

H

C C CH₃ H CH2

H3C

X

CH2CH2CH3

CH₃

X H3C

CH₂ X C CH₃ H CH2

H3C

C C CH3

H CH2

H3C

X

CH₂CH₃ H

C C CH3

H CH2

H3C

X

CH₂CH₂CH₃ CH3