Lecture 3 Isomerism. STRUCTURE CHANGE (ISOMERIZATION) and CONFORMATION CHANGE COMPOSITION C k H l O...

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Lecture 3 Isomerism

Transcript of Lecture 3 Isomerism. STRUCTURE CHANGE (ISOMERIZATION) and CONFORMATION CHANGE COMPOSITION C k H l O...

Lecture 3

Isomerism

STRUCTURE CHANGE (ISOMERIZATION) and CONFORMATION CHANGE

COMPOSITIONCkHlOm

STRUCTURE 1 STRUCTURE 2 STRUCTURE 3Reaction 1→2 Reaction 2→3

HC

H

O

C O

HH

C O

H

H

C O

H

H

3n-6=12-6=6

Reactivity

ReactivityTS1→2

TS2→3

STABILITY 2 STABILITY 3

1

2

3

ΔE = f(x1, x2, x3, x4, x5)

ΔE = f(x6)

H

C

H

O C O

H

H

C O

H

H

12 3

4

5

1 2

35

41

253

4

C O

H

H

x6

A molecule may change its structure without bound breaking and with bound breaking. The former one leads to different

conformers the latter change produces isomers. If σ-bounds are broken then we may have structural isomers and if π-bounds are

broken then we may have geometrical isomers.

Geometrical Isomers Structural IsomersConformers

CO2 + H2O

C + H2 + O2

A̶ 2706.6

A̶ 2709.8

A̶ 3.2A̶ 3.8

A̶ 2877

A̶ 125

A̶ 2875

A̶ 3000

A̶ 134

9

C C

H

CH3

H3C

H

C C

CH3

H

H3C

H

C C

CH3

H

CH3

H

HH

C C

H

CH3

H3C

H

H

H

C C

HCH3

H

H3C

HH

CO2+H2O CO2+H2O

Δ‡H=

ΔcH=

7. 4 Relative Stabilities of Alkenes

CH2

H

CH2H3C

H

H3C

H

CH3

H

H3C

CH3

H

1 .0

1 .7

-3 0 .3 -2 8 .6 -2 7 .6

CH3

CH2CH2

CH3

H

Figure 7.3 Enthalpies (DH) of hydrogenation (in units of Kcal/mol) for the isomeric n-butenes.

Figure 7.3 shows the heats or enthalpies of hydrogenation of three isomeric butenes. The enthalpies of hydrogenation are a measure of the molecular stabilities of unsaturated

hydrocarbons.

Two things should be noted: 1) Disubstituted carbon-carbon double bonds (at the middle and right hand

side of Figure 7.3) are more stable than mono-substituted olefinic double bonds.

2) The trans isomer (at the extreme right in Figure 7.3) is more stable than the cis isomer.

In Chapter 4, we saw that the stabilities of isomeric saturated hydrocarbons (CnH2n+2) can be assessed from their heat or enthalpy of combustion (c.f. Figure 4.2). In this chapter, we see that the heat or enthalpy of hydrogenation may be used to measure the stability of unsaturated hydrocarbons (CnH2n), as shown in Figure 7.3.

The results of these two methods can be put together, as illustrated for cis - and trans - 2 - butene and n - butane in Figure 7.4. As far as stabilities are concerned, the greater the substitution, the more stable the alkene. Of the disubstituted cases, 1,1 - disubstituted ethylene is more stable than 1,2-disubstituted ethylene. For the latter, the trans - isomer is more stable than the cis - isomer. These stabilities are presented in Figure 7.5 in terms of enthalpies of hydrogenation in a schematic fashion.

Figure 7.4 Comparison of enthalpies of formation, enthalpies of

hydrogenation and enthalpies of combustion for selected

hydrocarbons.

Me

Me

Me

MeMe

Et

Me

H Et

Et

H

HH

Et

Et

HH

Et

H

EtH

n -Bu

H

H

H

Figure 7.5 A schematic illustration of the relative stabilities of olefinic double bonds with the extent of substitution as measured by the enthalpies of

hydrogenation.

One may wonder if an alkyl substitution on an olefinic double bond can be carried out that will make it more stable. The answer lies in the concept of hyperconjugation.

We saw in Chapter 1 that conjugation of two carbon-carbon double bonds leads to stabilization (c.f. [1.42] and Figure 1.28). We can now assess the magnitude of such conjugative

stabilization to be of a few Kcal/mol. The example in Figure 7.6 shows the stabilization for butadiene to be 3.5 Kcal/mol.

2 C H2=C H–C H2–C H3 + 2 H2C H3–C H2–C H2–C H3

+C H2=C H–C H2=C H2 + 2 H2

-3.5 kcal/mol

-57.1 kcal/mol-60.6 kcal/mol

2 C H3–C H2–C H2–C H3

Figure 7.6 Conjugative stabilization (-3.5 Kcal/mol) of 1, 3 - butadiene with respect to two moles of 1-butene as measured by the difference in enthalpies of hydrogenation.

The structural isomerism requires a large initial energy input even though the energy is recovered at the end of the reaction.

Clearly this is not a spontaneous isomerization.„Tautomers” represent a special structural isomers which earn be formed quite easily without large energy investment. The two isomers differ from each other only in the position of a proton.

C C

C C

C C

C +CH3

+H C C

C H

CH3

0 Appr. 650 kJ/mol 9kJ

The ketoo-enol tautomerization is formed easily with acid catalysis.

CC

O

CC

O

O

H

C

C

OH

O

H

H

O

H

C

C

OH

O H

H

In general, most chemical reaction starts with a complex formation which is called the REACTANT COMPLEX. Usually the in

the pen-ultimate step a PRODUCT COMPLEX is formed. These two coplexes separated by a transition state (TS) as shown

below.

A + B → A···B → [TS]± → C···D → C + DReactants Reactant

ComplexTransition

StateProductComplex

Products

A schematic energy change is shown below

A+B A···B TS C···D C+DReaction

Coordinata

Ener

gy o

r Fre

e En

ergy

Product complex

Reactant complex

This situation dominantly occur in biological reactions which involve specific enzyme.

The first step is the ENZYME-SUBSTRATE complex formation which is converted via a TS to an ENZYME-PRODUCT comlex. The assiciated energy

profile and mechanistic equation is analogons to the shown below.

The following figure shows a bacterial enzyme which is able to oxidise CH4 to H3C-OH. The figure also shows the ENZYME-PRODUCT comlex. Since the H

atoms are not shown H3C-OH is shown in the the complex as C-O.

Methane MonooxygenaseCH4 + NADH + H+ + O2 → CH3OH + H2O + NAD+

A given molecular structure, such as C4H6 has numerous isomers, as few of them are illustrated below.

They may have numerous molecular fragments some of them a complete organic molecules while some of them contein two valent carbons, called carbenes.

CH3 CH3

H C C H CH

H

CH

H

CH4

+

C C

+

C

Nevertheless the potential energy surface associated with the particular chemical composition (like C4H6) does contain all these entities as local minimum.

1 2 3( , , ,... )DE f x x x x

k l m nC H N O

3 6D N

N k l m n

The function of the total PEHS is:

If the number of atoms have the following form:

The degrees of freedom (D) of the system given of N number of atoms:

where the N is:

The number of stretches, bends and the torsion add up to 3N-6.

1 2 3 2 3 2 2 3 6( , , ,... , ... )

reaction conformationalsubspace subspace

N N NE f x x x x x x

F + C Cl

H

H H

H

Et

Cl

MeH

Et

Cl

Me

H

2

1

3

H

2

1

3

S-isomer

clockwisecounterclockwise

R-isomer

H

Me

Et

Cl H

Me

Et

Cl

R S

R S

R S

RS

enantiomers

diastereomers

enantiomers

Me

Et

H

ClH

Cl

Me

Et

Cl

HCl

H

Me

Cl H

Et

H Cl

Me

H Cl

Et

Cl H

mirror

Me

Et

H

HCl

Cl

Me

Et

Cl

ClH

H

Me

Cl H

Et

Cl H

Me

H Cl

Et

H Cl

mirror

Me

Me

H

FH

F

Me

Me

F

HF

H

Me

F H

Me

H F

Me

H F

Me

F H

mirror

Me

Me

H

HF

F

Me

Me

F

FH

H

Me

F H

Me

F H

Me

H F

Me

H F

mirror

mirror

mirror

RR

SS

RS

SR

=

enantiomeric pair meso-structure

Cl

HH

Cl

H

HH

ClH

Cl

H

HCl

HH

H

Cl

H

meso SS RR

cis trans

H

Cl

H

Cl

H

H

Cl Cl

a

e

ee transcis

HH

Cl

Cle

e

HCl

Cl

H

e

HCl

Cl

H

e

aa

transcis

H

H

H

H

Cl ClCl Cl

e e

ee

RR

SS

H

Cl

H

Cl

H HCl Cl

Me

ET

HR HS

Me

ET

D HS

Me

ET

HR D

R-2-Deuteriobutane

S-2-Deuteriobutane

E[R(a)] = E[S(a)] [5.10] E[R(g+)] = E[S(g-)] [5.11a] E[R(g-)] = E[S(g+)] [5.11b]

Y

R H· H

Y

R · R H

Y·Y

R H·

H

Y

R · R H

Y

·Y

R H·

0o 60o 120o 180o 240o 300o 360o

-360o -300o -240o -180o -120o -60o -0o

Y

H R· R

Y

H · H R

Y·Y

H R·

R

Y

H · H R

Y

·Y

H R·

clockwise

counter clockwise

R-enatiomer

S-enatiomer

R(g+) S(g+)

E

D

D

DD

R(g-) S(g -)

C Z

Y

RH

CZ

Y

RH

S-configuration R-configuration