lecture 7. rearrangement involving carbocations qn: why is rearrangement quite common to...

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Lecture 7

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Lecture 7

R R

TS

R

R R

TS

R R

TS

C R

Antibonding

Bonding

Orbital at migrating centre

Rearrangement Involving Carbocations

Qn: Why is rearrangement quite common to carbocations, and not in case of radicals or anions?

Formation of TS involves combination of bonding orbital (C-R)with the vacant orbital on the carbocation,

singly occupied orbital in case of radical and doubly occupied orbital in case of carbanion.This overlap generates

one bonding and one antibonding scenario.

In case of cation, both the electrons occupy the bonding orbital……hence rearrangement is facile

R3C

Carbocation

CR2

Carbene

YY

R CH2

CH2 RHC CH2

RR

R1 R

R

R1R

R1

CH2

RR

R R

RCH2

R

1)

2) - H

3)

Polymer

4) Product

R3C R3C

Electron Deficient Carbon Carbene

Reactions of Carbocations:

1) Combination with a nucleophile

2) Elimination of a proton

3) Addition to an unsaturated linkage

4) rearrangement

Me

Me

HO

Me

OH

Me

Me

Me

Me O

Me

H

Me

Me

HO

Me

OH

Me

Me

Me

HO

Me

OH2

Me

H H2O-Me

Me

HO

Me

Me

Me

HO

Me

Me

Me

Me Me

Me

MeHO

Me

Me

Me O

Me

Pinacol - Pinacolone Rearrangement ----

Pinacol Pinacolone

Involvs 1, 2 shift of Me :--

Pinacolone

Pinacol

Qn: Why does the methyl migrate to a tertiary cation?

Ans: To achieve greater stabilization by delocalization of charge through electron pair of oxygen atom.

R1

R2

HO

R3

OH

R4

R1

R2

HO

R3

R4

R1OCC

R2

R3

R4

R2OCC

R1

R3

R4

R1

R2

R3

OH

R4

R1

CCOR4R2

R3

R2 CCOR3

R4

R1

4 Possibilities

Ph

HO

Ph

Me

Me

OH

Ph

Ph

Me

Me

OH

Ph

HO

Ph

Me

Me

Ph

PhMe

O

Me

Example : --

OR

( Major )

( less stable )

For differently substituted pinacols

How to decide which will be major?

1) First decide which carbocation is more stable

2) Then consider the relative migratory aptitude of the groups that will undergo 1,2-shift.

HO OH2

HO

ph

HOPh

Ph

PhO

Migratory aptitude aryl>alkyl usually H>alkyl

Aryl has greater migratory aptitude because it assists in the departure of leaving group (anchimeric assistance)

Via bridged carbocation

Electron donating groups at the ortho or para position will enhance the migratory aptitude while

electron withdrawing groups will lower the aptitude for migration.

H3C

H3CHO

CH3

CH3Br

H3C

H3CHO

CH3

CH3

H3C

H3CHO

CH3

CH3NH2

H3C

H3CHO

CH3

CH3N2

H3C

H3CHO

CH3

CH3

Ag

AgBr

HNO2

Product.

Product

H

H3CHO

H3C

HO

CH3

H

HO

Stabilised by hyperconjugation

If Methyl migrates

Less Stable

In case of hydride vs alkyl migration

Hydride generally migrates because of the generation of a more stable carbocation

Other ways of carrying out pinacol-pinacolone type rearrangements

HO

R1

R3

R2

O

R1R3

R2

H

OH

NH2

HNO2

OH

NH2HNO2

??

??

R'

R

OH

R'

R

OH

OH OH

+

Stereochemistry of migration:

The migrating group retains configuration. That means migrating group is never totally detached

This is also proved by cross over experiments Home Assignment

LECTURE 8

NH2

Me

OH

HMeNaNO2

PhHCl

A =HMe

PhMe

OH

NH2

OH

PhMeHMe

NH2

OHMe

Ph

Me H

Ph

Me H

Me O

COMe

Ph

HMe

MeH

N2

A

Me

Me

Me

Nu

Ph

OHMe

Ph

OHMe

H

N2+

NH2

HNO2

H

H

?

Ans.

1 2

Nu (Ph)

2 2

(major product)

211

2

2

Qn: Write down the Fisher projection formula of the major product

Problems on Pinacol-pinacolone rearrangement

O

O

O

OH OHOH

OHO

Mg

Benzene

H

Convert

NH2

OH

N2

OH

H

OH

O

NaNO2

HCl

(major)

N2

OH

OH

NaNO2

HClOH

NH2

CHO

NH2

OH

34

56 1

1

23

4

5

6

1

2

3

4

5 6

( major)

2

R C

O

N

R C

O

NH2

R C

O

N

B

A

R C

O

NHOCOR

R

RC NHBr

O

R N C O

R C

O

N

R C

O

N A

R C

O

N OCOR

R C

O

N Br

R C

O

N

RNH2

R NHCO2H

RCON

RCON

"a"

B = H

"b"

B =

Acyl nitrene

- CO2R'OH

NHCOOR'Carbamate or Urethane

Hofmann :

Lossen :

Migration to electron deficient Nitrogen

Hofman, Curtius,

Lossen, Schmidt

ROCHN C

O

NHR

RCONHOH

NaOBr

RCOOH

H2O

R C N

R C

O

NH2

O

RCON

HN3

RCONHNH2

R CO

N N N

Curtius(Schimdt)

(1/2 eq)

HNO2

Problem

HO R2

N

R1

R2 C

O

NHR1

N

NH

O

H2O

R2 C

O

NHR1

N

R2

R1

NOH2

PPA

R1 R2

NOH2

H

NOH

R1

C

R2

N

OH

H

R1

C

R2

N

OH

Base[ NH(CH2)5CO]n

Nylon 6Caprolactam

Cyclohexanone Oxime

H2SO4

SOCl2

P2O5

BF3

Beckman Rearrangement

Anti group always migrates…no question of migratory aptitude!

Mechanism

R

O

R

O

R

H

R

R1 C

O

O O H

R C

OH

O R

R C

O

OR- H

R

O

OR

R

OH

R

HO R

R O O C

O

R1

Peracid

orPeroxide

O

X

X

O

OH

COOH

1. OH

2. H

Rearrangement Involving Carbanions

Favorskii Rearrangent

Migration to electron Deficient Oxygen

Bayer Villiger Oxidation

COOH

O

OH

OH

O

COOH

O

Cl

O

OH

COOH

Cl

COOH

OH

OH

OH

O

O

Cl

O

Cl

COOH

Cl

*

*

* *

*

*

*

*

+

*

*

*

*

*

*

*

*

*

*

** *

*

+

R1

R2

O

R4

R1

R2 CO2H

R3

R4

OH

R1 CO2H

R2CHR3R4

R1

R2

O

R4

R3

X

OHR1

R2

H

R3

R4

X

O

+

R3

Earlier mechanism

However

Application to acyclic systems

OH OH