classes of alkenes - prof. anil mishra's website ‘s look at an e1 reaction mechanism using...

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Saturated compounds (alkanes): Have the maximum number of hydrogenatoms attached to each carbon atom.

Unsaturated compounds: Have fewer hydrogen atoms attached tothe carbon chain than alkanes.

• Containing double bond are alkenes.CnH2n

• Containing triple bonds are alkynes.CnH2n-2

Alkenes and Alkynes

Classes of Alkenes

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Structure and Bonding

Orbital Description

Sigma bonds around C are sp2 hybridized.Angles are approximately 120 degrees.No nonbonding electrons.Molecule is planar around the double bond.Pi bond is formed by the sideways overlap of parallel p orbitals perpendicular to the plane of the molecule.

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Bond Lengths and Angles

Hybrid orbitals have more s character.Pi overlap brings carbon atoms closer.Bond angle with pi orbitals increases.

Angle C=C-H is 121.7°Angle H-C-H is 116. 6° =>

Pi BondSideways overlap of parallel p orbitals.No rotation is possible without breaking the pi bond (63 kcal/mole).Cis isomer cannot become trans without a chemical reaction occurring.

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IUPAC Nomenclature

Parent is longest chain containing the double bond.-ane changes to -ene. (or -diene, -triene)Number the chain so that the double bond has the lowest possible number.In a ring, the double bond is assumed to be between carbon 1 and carbon 2.

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Using the IUPAC alkane names:

Alkene names change the end to -ene.

Alkyne names change the end to -yne

Naming Alkenes & Alkynes

Give the locationfor double and

triple bond

CH2= CH ─ CH2─ CH3 1-butene

CH3─ CH=CH─ CH3 2-butene

CH3|

CH3─ CH=C─CH3 2-methyl-2-butene

CH3─ C≡C ─ CH3 2-butyne


1 2 3 4

1 2 3 4

4 3 2 1

1 2 3 4

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CH3─ CH2─ C≡C ─ CH3 2-pentyne

CH3

CH3─ CH2─ C=CH ─ CH3 3-methyl-2-pentene

CH2 – CH3

CH3─ CH2─ C=CH ─ CH3 3-ethyl-2-pentene


5 4 3 2 1

5 4 3 2 1

5 4 3 2 1

Cis & Trans Stereoisomers

C = CHH

CH3H3CC = C

H

H CH3

H3C

cis-2-Butene trans-2-Butene

Same molecular formula and same connectivity of their atoms but a different arrangement of their atoms in space.

mp & bp of cis < mp & bp of trans

C4H8 C4H8

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Preparation of Alkenes

Elimination

Addition

C X YC + C C

X Y

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Synthesis of AlkenesDehydrohalogenation

Alkenes are commonly made by elimination of HX from alkyl halide(dehydrohalogenation)

Uses heat and KOH

HH

BrH

KOH

CH3CH2OH

H

HKBr H2O++

Synthesis of AlkenesDehydrohalogenation

Synthesis of AlkenesDehydration

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Synthesis of AlkenesDehydration

elimination of H-OH from an alcohol (dehydration) require strong acids (sulfuric acid, 50 ºC)

CH3

OHH2SO4, H2O

THF, 50 oC

CH3

H2O+

Elimination Reactions

C C

X Y

C C + X Y

Dehydrohalogenation (-HX) and Dehydration (-H2O) are the main types of elimination reactions.

Dehydrohalogenation (-HX)

strong

base

X = Cl, Br, I

+ " "C C

X

H XC C

H

E1 mechanism

1)

++ Br

_slow

+

2)..

:

+fast

O..+O

C C

Br

C C

H

C C

HC C

H

H H

H

H

H

water helpsto stabilizecarbocation

This reaction is done in strong base at low concentration, such as 0.01 M NaOH in water)

E2 mechanism

..:..

__

+

+ Br_

..:

concerted mechanism

H O

C C

Br

H

H OH

C C

This reaction is done in strong base at high concentration, such as 1 M NaOH in water.

_

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The E2 MechanismEnergy Diagram for the E2 Mechanism

Effect of the Substrate on E2 Reactivity

Effect of the LG on E2 Reactivity

The E1Mechanism

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Let ‘s look at an E1 reaction mechanismusing the rxn shown below.

The E1 Mechanism

The first step of the E1 and SN1 mechanisms are identical.

First Step: Same as SN1

The E1 Mechanism Energy Diagram for E1

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SN2

SN1

E1

E2

CH

Br

HH

H3C OC H

HH3CO

H

C BrH3C OH CH3CO+

C BrH3C OH +

Br+H3C O

What about dehydrohalogenationsinvolving RX with hydrogen atoms on different β carbon atoms

The Zaitsev Rule

The Zaitsev Rule- A Regioselective Rxn

Cl

KOC(CH3)3

+

Two possible organic products.Which is the major product?

The Zaitsev Rule (Z-rule)

According to the Z-rule, the major productin a dehydrohalgenation is the …..

most stable product.

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Let’s ask a different question about the products.

Cl

KOC(CH3)3

+

Which is the most stable product?

Elimination Reactions

A compound with an electronegative atom bonded to an sp3 carbon, when approached by a nucleophile/base can undergo either a substitution reaction OR an elimination reaction

+ C C

H

Br

OH- C C + Br- + HO H

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.

The Nu:- removes an H+ from a β-carbon & the halogen leaves forming an alkene.

α

β

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

Elimination Reactions The E2 Mechanism

E2: Deprotonation is First Step

The E2 Mechanism

Energy Diagram

The E2 Mechanism

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The E2 Mechanism

Reaction proceeds by Anti Elimination

The E2 Mechanism

What about dehydrohalogenations involving RX with hydrogen atoms on different β carbon atoms

The Zaitsev (Saytzeff) Rule Z Rule

In reactions of removal of hydrogen halides from alkyl halides or the removal of water from alcohols, the hydrogen which is lost will come from the more highly-branched β-carbon.

A. N. Zaitsev -- 1875γ β α βC C C C

H

H

H H

X

H

H

HHCH3

Less branchedMore branched


Cl

KOC(CH3)3

+

Two possible organic products.Which is the major product?


According to the Z-rule, the major product in a dehydrohalgenation is the …..

most stable product.


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Let’s ask a different question about the products.

Cl

KOC(CH3)3

+

Which is the most stable product?


• When alkyl halides have two or more different β carbons, more than one alkene product is formed.

• When this happens, one of the products usually predominates.• The major product is the more stable product—the one with the

more substituted double bond.• This phenomenon is called the Zaitsev rule.


• When a mixture of stereoisomers is possible from a dehydrohalogenation, the major product is the more stable stereoisomer.

• A reaction is stereoselective when it forms predominantly or exclusively one stereoisomer when two or more are possible.

• The E2 reaction is stereoselective because one stereoisomer is formed preferentially.


The E2 Reaction: Regioselectivity

2-bromobutane has two structurally different β-carbons from which to abstract a hydrogen

E2 reactions give more stable alkene if possible

Zaitsev’s rule (Saytzeff Rule): The more substituted alkene will be formed in elimination reactions


Zaitsev’s rule does not apply when the base is bulkyE2 Rxn is kinetically-controlled


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Zaitsev’s rule does not apply when the leaving group is poorE2-carbanion mechanism operative


Zaitsev’s rule may not apply when conjugated dienes might be formed

CH2 CHCH2CHCHCH3

CH3HO

Cl

CH2

CH2

CHCH

CHCH2CH

CHCHCH3

CH3

CCH3

CH3+

major product


The major product of an E2 reaction is the more substituted alkene unless:

the base is large (i.e. bulky)the leaving group is poor (i.e. F–)

the less substituted β–carbon is allylic or benzylic (ie. more stable)


If the elimination reaction removes two substituents from the same side of the molecule it is syn eliminationIf the elimination reaction removes two substituents from opposite sides of the molecule it is anti elimination

The E2 Reaction: Stereochemistry

The E2 Reaction is stereoselective, but not stereospecific if 2 β H’s are available on carbon bearing eliminated HThe H leading to more stable E isomer is selected to be extracted from β carbon regardless of streochemat α carbon


In an E2 reaction, the bonds to the eliminated substituents must be in the same planeIn this course E2 eliminations will all go via anti-periplanar conformationProduct analysis possible by drawing Newman projections if only 1 β H is available


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When only one hydrogen is on the β carbon predominantly anti elimination leads to high stereospecificity

(2S,3S)-2-bromo-3-phenylbutane (E)-2-phenyl-2-butene

(2S,3R)-2-bromo-3-phenylbutane (Z)-2-phenyl-2-butene

The E2 Reaction: Stereochemistry Effect of the Substrate on E2 Reactivity

Effect of the LG on E2 Reactivity The E1 Reaction

“E1” stands for “Elimination unimolecular”

The E1 reaction is a two-step reaction

The first step is rate-determining

The E1 Mechanism

First Step: Same as SN1

The E1 Mechanism

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The E1 Pathway Competes with SN1

The E1 Mechanism

• The dehydrohalogenation of (CH3)3CI with H2O to form (CH3)C=CH2 can be used to illustrate the second general mechanism of elimination, the E1 mechanism.

• An E1 reaction exhibits first-order kinetics:

Mechanisms of Elimination—E1

rate = k[(CH3)3CI]• The E1 reaction proceed via a two-step mechanism: the bond to

the leaving group breaks first before the π bond is formed. The slow step is unimolecular, involving only the alkyl halide.

• The E1 and E2 mechanisms both involve the same number of bonds broken and formed. The only difference is timing. In an E1, the leaving group comes off before the β proton is removed, and the reaction occurs in two steps. In an E2 reaction, the leavinggroup comes off as the β proton is removed, and the reaction occurs in one step.

Mechanisms of Elimination—E1 The E1 Reaction

E1 reaction involves a carbocation

Therefore rearrangements must be considered

Several Products are Possible

The E1 Mechanism The E1 Reaction: Stereochemistry

With C+ both syn and anti elimination can occur, so E1 reaction forms both E and Z products regardless of whether β-carbon is bonded to one or two H’sProduct stability leads to stereoselectivity but not stereospecificity

classes of alkenes - prof. anil mishra's website ‘s look at an e1 reaction mechanism using...

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