Lecture 8: Stereochemistry

Mechanism of organic reactions: E1

and E2 Reactions
Dr Mohammed Jasim Mohammed Al Yasiri

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 Elimination Reactions, E1 and E2:

We have seen that alkyl halides may react with basic nucleophiles such as NaOH
via substitution reactions .
H H
.. H .. .. .. .. ..
:O H + C Br : H C Br : C + : Br :
..
H
.. ..O .. H ..O ..
H
H HH H
transition state

Also recall our study of the preparation of alkenes. When a 2° or 3° alkyl halide
is treated with a strong base such as NaOH, dehydrohalogenation occurs
producing an alkene – an elimination (E2) reaction.

KOH in ethanol + KBr + H2O

Br
-HBr

bromocyclohexane + KOH → cyclohexene (80 % yield)

Substitution and elimination reactions are often in competition. We shall

consider the determining factors after studying the mechanisms of elimination.

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Dr Mohammed Jasim Mohammed Al Yasiri
 E2 Reaction Mechanism

There are 2 kinds of elimination reactions, E1 and E2.

E2 = Elimination, Bimolecular (2nd order). Rate = k [RX] [Nu:-]
❑ E2 reactions occur when a 2° or 3° alkyl halide is treated with a strong
base such as OH-, OR-, NH2-, H-, etc.

H
 C C + Br- + HO H
OH- + C C
b
Br
The Nu: - removes an H+ from a b-carbon &
the halogen leaves forming an alkene.

❑ All strong bases, like OH-, are good nucleophiles. In 2° and 3° alkyl halides
the -carbon in the alkyl halide is hindered. In such cases, a strong base
will ‘abstract’ (remove) a hydrogen ion (H+) from a b-carbon, before it hits
the -carbon. Thus strong bases cause elimination (E2) in 2° and 3° alkyl
halides and cause substitution (SN2) in unhindered methyl° and 1° alkyl
halides.

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Dr Mohammed Jasim Mohammed Al Yasiri
 E2 Reaction Mechanism

In E2 reactions, the Base to H  bond formation, the C to H  bond breaking,

the C to C  bond formation, and the C to Br  bond breaking all occur
simultaneously. No carbocation intermediate forms.

B:- d
+
B
H R H R R R
C C R C C C C + B H + X-
R R R R
R X R R
X-
d

Zaitsev’s Rule:
Recall that in elimination of HX from alkenes, the more highly substituted
(more stable) alkene product predominates.

Br CH3CH2O-Na+
CH3CH2CHCH3 CH3CH CHCH3 + CH3CH2CH CH2
EtOH 2-butene 1-butene

major product minor product

( > 80%) ( < 20%)
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Dr Mohammed Jasim Mohammed Al Yasiri
 E2 Reactions are ‘antiperiplanar’

❑ E2 reactions, do not always follow Zaitsev’s rule.

❑ E2 eliminations occur with anti-periplanar geometry, i.e., periplanar

means that all 4 reacting atoms - H, C, C, & X - all lie in the same
plane. Anti means that H and X (the eliminated atoms) are on opposite
sides of the molecules.

❑ Look at the mechanism again and note the opposite side & same plane
orientation of the mechanism:

B:- d
+
B
H R H R R R
C C R C C C C + B H + X-
R R R R
R X R R
X-
d

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Dr Mohammed Jasim Mohammed Al Yasiri
 Antiperiplanar E2 Reactions in Cyclic Alkyl Halides

❑ When E2 reactions occur in open chain alkyl halides, the Zaitsev

product is usually the major product. Single bonds can rotate to the
proper alignment to allow the antiperiplanar elimination.
❑ In cyclic structures, however, single bonds cannot rotate. We need to
be mindful of the stereochemistry in cyclic alkyl halides undergoing E2
reactions.
See the following example.

❑ Trans –1-chloro-2-methylcyclopentane undergoes E2 elimination with

NaOH. Draw and name the major product .

H H3C H
H3C H Na+ OH- + NaCl
H H3C
H H + HOH
H
H
H E2 H
Cl
Little or no Zaitsev (more stable)
Non Zaitsev product product is formed.
is major product.

3-methylcyclopentene 1-methylcyclopentene

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Dr Mohammed Jasim Mohammed Al Yasiri
 E1 Reactions

❑ Just as SN2 reactions are analogous to E2 reactions, so S N1 reactions have

an analog, E1 reaction.
❑ E1 = Elimination, unimolecular (1st order); Rate = k  [RX]

CH3 CH3
slow H
rapid CH3
CH3 C Br CH3 C+ -
C C + B H + Br
- Br-
CH3 H C H
CH3 H
H B:-

❑ E1 eliminations, like SN1 substitutions, begin with unimolecular

dissociation, but the dissociation is followed by loss of a proton from the
b-carbon (attached to the C+) rather than by substitution.
❑ E1 & SN1 normally occur in competition, whenever an alkyl halide is
treated in a protic solvent with a nonbasic, poor nucleophile.
❑ Note: The best E1 substrates are also the best S N1 substrates, and
mixtures of products are usually obtained.

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Dr Mohammed Jasim Mohammed Al Yasiri
 E1 Reactions

❑ As with E2 reactions, E1 reactions also produce the more highly

substituted alkene (Zaitsev’s rule). However, unlike E2 reactions where no
C+ is produced, C+ rearrangements can occur in E1 reactions.
❑ e.g., t-butyl chloride + H2O (in EtOH) at 65 C → t-butanol + 2-methylpropene

CH3 H2O, EtOH CH3 CH3 H

CH3 C Cl CH3 C OH + C C
65ºC
CH3 CH3 CH3 H
36%
64%
E1
S N1
product
product

❑ In most unimolecular reactions, S N1 is favored over E1, especially at low

temperature. Such reactions with mixed products are not often used in
synthetic chemistry.
❑ If the E1 product is desired, it is better to use a strong base and force the
E2 reaction.
❑ Note that increasing the strength of the nucleophile favors S N1 over E1.
Can you postulate an explanation?

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Dr Mohammed Jasim Mohammed Al Yasiri
 Predicting Reaction Mechanisms

1. Non basic, good nucleophiles, like Br- and I- will cause substitution not
elimination. In 3° substrates, only SN1 is possible. In Me° and 1°
substrates, SN2 is faster. For 2° substrates, the mechanism of
substitution depends upon the solvent.
2. Strong bases, like OH- and OR-, are also good nucleophiles.
Substitution and elimination compete. In 3° and 2° alkyl halides, E2 is
faster. In 1° and Me° alkyl halides, SN2 occurs.
3. Weakly basic, weak nucleophiles, like H2O, EtOH, CH3COOH, etc.,
cannot react unless a C+ forms. This only occurs with 2° or 3°
substrates. Once the C+ forms, both SN1 and E1 occur in competition.
The substitution product is usually predominant.
4. High temperatures increase the yield of elimination product over
substitution product.
5. Polar solvents, both protic and aprotic, like H2O and CH3CN,
respectively, favor unimolecular reactions (SN1 and E1) by stabilizing
the C+ intermediate. Polar aprotic solvents enhance bimolecular
reactions (SN2 and E2) by activating the nucleophile.

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Dr Mohammed Jasim Mohammed Al Yasiri
 Predicting Reaction Mechanisms

- - - -
good Nu good Nu good Nu very poor Nu
alkyl nonbasic strong base strong bulky base nonbasic
halide e.g., bromide e.g., ethoxide e.g., t-butoxide e.g., acetic acid
- - -
(substrate) Br C2H5O (CH3)3CO CH3COOH

Me SN2 SN2 SN2 no reaction

1° SN2 SN2 E2 (SN2) no reaction

2 SN2 E2 E2 SN1, E1

3 SN1 E2 E2 SN1, E1

Strong bulky bases like t-butoxide are hindered. They have difficulty hitting
the -carbon in a 1° alkyl halide. As a result, they favor E2 over SN2
products.

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 Predicting Reaction Mechanisms

The nucleophiles in the table on slide 10 are extremes. Some

nucleophiles have basicity and nucleophilicity in between these extremes.
The reaction mechanisms that they will predominate can be interpolated
with good success.
Predict the predominant reaction mechanisms the following table.
-
v. gd. Nu-
……….. v. gd. Nu-
……….. ………..
fair Nu HI
alkyl ………. .base
moderate ……….. base
moderate ………..
weak base alcohol HBr
halide e.g., cyanide e.g., alkyl sulfide e.g., carboxylate (substrate)
- - - - HCl
(substrate) CN RS , also HS RCOO
4.7
pkb = ……… 6.0 / 7.0
pkb = ……… 9
pkb = ………

Me SN2 SN2 SN2 Me SN2

1 SN2 SN2 SN2 1 SN2
2 SN2 SN2 E2 2 SN1
3 E2 E2 E2 3 SN1

HCl, HBr and HI are assumed to be in aqueous solution, a protic solvent.

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Dr Mohammed Jasim Mohammed Al Yasiri
 Alkylation of Alkynides

Recall the preparation of long alkynes.

1. A terminal alkyne (pKa = 25) is deprotonated with a very strong base…

R-C  C-H + NaNH2 → R-C  C:- Na+ + NH3

2. An alkynide anion is a good Nu: - which can substitute (replace) halogen

atoms in methyl or 1° alkyl halides producing longer terminal alkynes 

R-C  C: - Na+ + CH3CH2-X → R-C  C-CH2CH3 + NaX

❑ The reaction is straightforward with Me and 1 alkyl halides and proceeds

via an SN2 mechanism

❑ Alkynide anions are also strong bases (pKb = -11) as well as good Nu:-’s,
so E2 competes with SN2 for 2 and 3 alkyl halides
CH3(CH2)3CC:- Na+ + CH3-CH(Br)-CH3 → CH3(CH2)3CCCH(CH3)2 (7% SN2)
+ CH3(CH2)3CCH + CH3CH=CH2 (93% E2)

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Dr Mohammed Jasim Mohammed Al Yasiri
 Preparation of Alkyl Halides from Alcohols

Alkyl halides can be prepared from alcohols by reaction with HX, i.e., the
substitution of a halide on a protonated alcohol.

3º - H2O
.. + ..
(Lucas Test) (CH3)3C Cl : + H2O
(CH ) C OH + H Cl (CH3)3C OH (CH3)3C + ..
SN1 33 .. .. 2
.. -
Rapid. 3° C+ is stabilized by protic sovent (H 2O) : Cl :
..

❑ OH- is a poor leaving group, i.e., is not displaced directly by nucleophiles.

Reaction in acid media protonates the OH group producing a better leaving
group (H2O). 2 and 3 alcohols react by SN1 but Me° and 1 alcohols react
by SN2.
1º
.. +
SN2 CH3CH2 OH + H Cl CH3CH2 OH2 CH3CH2Cl + H2O
.. D .. -
: Cl :
..
Very slow. Protic solvent inhibits the nucleophile.

❑ Draw the mechanism of the reaction of isopropyl alcohol with HBr.

❑ What products form if concentrated H2SO4 is used in place of aq. HCl?

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Dr Mohammed Jasim Mohammed Al Yasiri
 Preparation of Alkyl Halides from Alcohols

Alternative to using hydrohalic acids (HCl, HBr, HI), alcohols can be

converted to alkyl halides by reaction with PBr 3 which transforms OH- into
a better leaving group allowing substitution (S N2) to occur without
rearrangement.
Br
:P
Br
1º
.. Br + PBr2
CH3(CH2)4CH2Br
SN2 CH3(CH2)4CH2 OH CH3(CH2)4CH2 O
.. ..
ether H
-
Br

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Dr Mohammed Jasim Mohammed Al Yasiri
 Preparation of Alkenes from Alkyl Halides

On Slide 2 we noted that 2° and 3° alkyl halides can be dehydrohalogenated

with a strong base such as OH- producing an alkene.

KOH in ethanol + KBr + H2O

Br
-HBr

bromocyclohexane + KOH → cyclohexene (80 % yield)

Clearly, this is an E2 reaction.

❑ Predict the mechanism that occurs with a Me° or 1° alkyl halide.

❑ Predict the products and mechanism that occur with isopentyl chloride and
KOH

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Dr Mohammed Jasim Mohammed Al Yasiri
 Summary of SN /Elimination Reactions

Alkyl Halide Substrate Reactivity:

methyl 1º 2º 3º

H H CH3 CH3

H C Br CH3 C Br CH3 C Br CH3 C Br

H H H CH3

unhindered substrates favor S N2 hindered substrates. S N2 increasingly unfavorable, E2 is OK

do not form a stable C + form increasingly stable C +
do not react by S N1 or E1 favors SN1 and E1. E2 is OK

E2 reactions possible with strong bases

E2 reactions possible with strong bulky bases (t-butoxide)

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 Summary of SN /Elimination Reactions

Reactivity of Nucleophiles:
HS- CN- I- CH3O- HO- NH3 Cl- H2 O

125,000 125,000 100,000 25,000 16,000 1000 700 1

good nucleophiles which are good nucleophiles which are

weak bases favor SN reactions also strong bases favor elimination

Note that poor nucleophiles that are also weak bases (H2O, ROH, CH3COOH,
etc.) do not undergo any reaction unless a C + is formed first. If a C+ can form
(as with a 2º, 3º, any benzylic, or any allylic halides), then E1 and S N1
generally occur together.
Leaving Group Activity:

pKb = 23 pKb = 22 pKb = 21 pKb = 11 pKb = -1.7 pKb = -2 pKb = -21

I- Br - Cl- F- HO- RO- H2N-
30,000 10,000 200 1 0 0 0

good leaving groups favor both poor leaving groups make both
substitution and elimination reactions substitution and elimination
reactions unfavorable
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Dr Mohammed Jasim Mohammed Al Yasiri
 Summary of SN /Elimination Reactions

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Dr Mohammed Jasim Mohammed Al Yasiri
 Summary of SN /Elimination Reactions

Question #1: Substrate: is this carbon primary, secondary, or tertiary?

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Dr Mohammed Jasim Mohammed Al Yasiri
 Summary of SN /Elimination Reactions

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Dr Mohammed Jasim Mohammed Al Yasiri
 Summary of SN /Elimination Reactions

Question #3: What’s The Solvent?

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Dr Mohammed Jasim Mohammed Al Yasiri
Summary of SN /Elimination Reactions

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 Practice reactions, SN2, E2, SN1, E1

1- Give the major organic product of the following reactions. Also, state the mechanism
through which each reaction proceeds (e.g. SN2). (Do not draw out the mechanism.)

SN2

2- Draw a complete mechanism for the following reaction. Be sure to include all intermediates, formal
charges, and pushing arrows.

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Dr Mohammed Jasim Mohammed Al Yasiri