heat of reaction estimation

7/30/2019 Heat of Reaction Estimation

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Chapter 1 Concepts and

Models in Organic Chemistry

General discussion of some overall concepts

in organic chemistry.

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Chapter 1 Problems

Problems each worth 5 pts.

1, 4, 6, 14, 16, 21


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What is a model?

A representation of the real

Details of the model are only as useful as

they are helpful to explain observations.

When a model does not explain

observations it must be modified or thrown

out.

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Modern Models

How do we arrive at Models

Imagination

Creativity

Desire to explain what we observe

Often quantitative in order to explain

observations from instruments.


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5

Examples of Models.

Lewis dot diagrams

Tinker Toy Models

Dreiding Models

CPK space filling models

Quantum Mechanical Models

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How do we know what amolecule look like?

Imagination

Not very quantitative

Helps sharpen understanding

Graphic representation of quantitative

models.


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Ethane Model

Different views

Each has a use.

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Molecular dimensions.

How do we get the molecular dimensions of

molecules?

X-ray

Electron and Neutron diffraction

Only on solids

Give nuclei positions of heavy atoms

Microwave

Usually low MW molecules


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Bond Lengths and Angles

Represent the inter-nuclei distances as

points in space

The electron density extends out from these

points.Molecule rC-H rC-X


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Covalent Radii

Functions of charge.

C4+ 0.15A while C4- 2.60A

Depends on bonding

C bonded to four groups 0.772A

C double bonded 0.67A

Effective Radii for groups of atoms. CH3 2.0A

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Atomic Volume and Surface

Atomic volumes can be calculated from

values in Table 1.3 (p.8)

Butane

2xCH3 = 2x17.12 = 34.24

2x CH2 = 2x10.23 = 20.46

total=54.7cm3/mole

Theory cal = 99.91 A3 = 60.14 cm3/mole


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13

Heats of Formation

o is the difference in enthalpy between

the compound and the elements in their

standard state.

Obtained from heats of combustion data.

m C(graphite) + n/2 H2(gas ) CmHn Hof

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Heats of Combustion

Ho when the compound is burned with

excess O2

CmHn + (m+ n/4) O2 m CO2 + n/2 H2O Ho

= heat of compustio


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General caseC(graphite) + O2(gas) CO2(gas)

Ho

r= Ho

f(CO2)

H2(gas) + 1/2 O2(gas) H2O (liquid)

Ho

r= Ho

f(H2O)

Combining these quations.

Ho

f(CmHn) = mHo

f(CO2) + n/2 Ho

f(H2O) - Ho

(combustion) (CmHn)

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Corrections

Change of phase

heat of vaporization for liquids

heat of sublimation for solids

Temperature corrections


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Example From BookO

O

+ 8 O2 6 CO2 + 4 H2O Ho

= -735.9 Kcal/mol

Ho

f(cyclohexanedione) = 4 Hof(H2O) + 6 H

of(CO2) - H

ocombustion (cyclohexanedio

Hof(cyclohexanedione) = 4 (68.32) + 6 (94.05) - (-735.9) = -101.68 Kcal/mol

Correction of solid to gas Ho

sub = +21.46 Kc al/mol

Hof(cyclohexanedione) = -101.68 Kcal/mol + 21.46 = -80.22 Kcal/mol

Hof(H2O) = -68.32 Kcal/mol H

of(CO2) = -94.05 Kc al/mol

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Differences BetweenCompounds

You can use heats of reactions if you go to a

common product.

Tran more stable than cis by 1 Kcal/mol

CH3

H

CH3

H

CH3CH2CH2CH3

H

CH3

CH3

H

-27.6 Kcal/mol-28.6 Kcal/mol


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Principle Additivity

The property of a molecule can be estimated

from the the parts (bonds or groups).

If you know the contribution from a group

in one molecule you can use it in another

calculation.

Hof(CH2) = -35.1 - -30.4 Kcal/mol = -4.7 Kcal/mol

CH3CH2CH2CH3 CH3CH2CH2CH2CH3

Ho

f= -30.4 Kcal/mol Ho

f= -35.1 Kcal/mol

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Table 1.5 Bond Increments toHof

How well does this method work?

CH4 cal -15.32 found -17.89 Kcal/mol

Ethane

6 C-H and 1 C-C

6 (-3.83) + 1 (2.73) = -20.25 Kcal/mol

Found -20.24 (not bad)


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Problem

Calculated values

They are the same

Obs. -32.4 Kcal/mol and -30.4 Kcal/molrespectively.

CH3CHCH3

CH3

CH3CH2CH2CH3

10 (-3.83)

3 (2.73)= -30.11 Kcal/mol

10 (-3.83)

3 (2.73)= -30.11 Kcal/mol

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Group Increments(Table 1.7)

Group contributions allow for different

bond types.

Obs. -32.4 Kcal/mol and -30.4 Kcal/mol

respectively (better?)

CH3CHCH3

CH3

CH3CH2CH2CH3

3 CH3 = 3 (-10.8)

1 CH = -1.9= -34.3 Kcal/mol

2 CH3 = 2 (-10.8)

2 CH2 = 2 (-4.5)

= -29.6 Kcal/mol


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Limitation

Group increments do not consider strain

contribution

CH3CHCH2CH2CH3

CH3

3 CH3 = 3 (-10.8)

2 CH2 = 2 (-4.5)

1 CH = 1 (-1.9)=-42.04 Kcal/mol (exp. -41.66)

CH2CH3

HH

CH3

CH3 H

1 gauscheinteratction 0.8

Correction -42.04 + 0.8 = -41.24 (better)

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Bond Dissociation Energies

The energy needed to break the bond.

Homolytic

Hetrolytic

A B A B Hor

A B A B Hor


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Using Bond Dissociation

Energies

Example

This calculation does not say anything

about how fast the reaction occurs.

Ho

= 2(-104. Kcal/mol) + 104.2 + 88 = -15.8 Kcal/mol

A checkHor= 2 H

of (CH4) - H

of(CH3CH3) - H

of(H2) = 2(-17.9) - (-20.2) - 0 = -15.6 Kcal/mol

CH3CH3 + H2 2 CH4

H2 2 H DHo

= 104.2 Kcal/mol

DHo

= 88 Kcal/molCH3CH3 2 CH3

2 CH3 + 2 H 2 CH4 2 DHo

= 2 (-104 Kcal/mol)

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Bonding Models

What holds molecules together?

Groups of atoms find they are a lower energy

when they are associated with each other. They

are bonded.

Electrostatic attractions.

Covalent bond

Na+

Cl- NaCl energy

H H H H


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Models

Lewis Model

For some reason atoms (first row) are in a low

energy state if the have 8 electrons around them.

Why?

Electron transfer between atoms of different

electronegativity

F-1

O-2

N-3Ne

F Cs Cs+1

F-1

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Covalent Bond

Both atoms have the same EN

Bonding only involves outer shell electrons

(valence shell electron)

F F F F

P 1s2 2s2 2p6 3s2 3p3

core valence electrons


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Dipole Moment

Vector sum of the dipole moments of the

bonds.

Dipole moments useful in understandingproperties.

= charge (esu) X distance (cm) = Debye (D)

= 0.1 (4.8 x 10-10

esu) 1.5 A(1 x 10-8

cm/A) = 0.72 x 10-18

esu cm = 0.72 Debye (D)

A B+0.1 esu -0.1 esu

1.5A

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Which Molecules Have DipoleMoments?

Cl HCH3

Cl

H

H

CH3

CH3

H

H

CH3CH2CH2OH

Cl

Cl

Cl

OO

O


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Bond Polarity

Bond polarity comes from unequal sharing

of bonding electrons.

Polar Bond = Covalent part + ionic part

molecule =covalent + ionic

% Ionic character = 2

(1 + 2)

(100)

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Electronegativity

Unequal sharing of electrons between atoms

of a bond.

Pauling Electronegativity Scale

Compares bond dissociation energies of like

atoms with those of unlike atoms.

H2 and Cl2 (104.2 + 58)/2 = 81.1 Kcal/mol HCl 103.2 (larger due to ionic character)


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Table 1.8

Other EN scales

Pauling,p Mullikin,m (I+A)/2

Allen,spec Nagel, Benson,

x Which EN scale to use depends on

application.

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Bonding

Valence Bond Theory (VB)

Bring together of atoms which share electrons.

Molecular Orbital Theory (MO)

Bring together nuclei which are held together

by electrons.


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H2 Molecule

Two Hydrogen atoms

We can write the equations for a 1s orbital.

Wave functions

H1s H1s

1 = c a(1) b(2)

2 = c a(2) b(1)

VB = c a(2) b(1) + c a(2) b(1)

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LCAO Method

LinearCombinations ofAtomic Orbitals

1 = c1 a(1) + c2 b(1)

2 = c1 a(2) + c2 b(2)

MO = 12 = c12 a(1) a(2) + c2

2 b(1) b(2) + c1c2[a(1)b(2) + a(2)b(1)


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Comparison

VB is simpler and is incorporated into the

MO wave function

LCAO method has ionic contribution

As terms are added to either model they

both give better answers and they become

very much alike.

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Energy Levels

The MO model generates the orbital and the

energies of the orbitals.

Energy

H H-H H

AO AOMO

s s


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Methane

Atomic Orbital view.

1s2 2s22p2

This model predicts

CH2 with a bond angle

of 90 o

Energy

s

s

p

AO

Carbon

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Observation

Methane is CH4.

Need new model with 4 unpaired electrons.

Two possible solutions

unpair one of the s electrons and put it in a p

orbital.

Completely reorganize the orbital.


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Hybridization

A model that can be used to explain what

we observe.

Based on AOs

Linear Combination of an s and three p orbitals

Higher in energy than an s but lower than a p.

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Energy

s

sp3

Energy p

Carbon

s

Hybridized


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LCAO

Four AOs must yield 4 MOs

There is an error in the book

sp3 = 1/2( C2s + C2px + C2py + C2pz)

sp3 = 1/2( C2s - C2px - C2py + C2pz)

sp3 = 1/2( C2s + C2px - C2py - C2pz)

sp3 = 1/2( C2s - C2px + C2py - C2pz)

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Shape and Orientation

The electron distribution must have the

characteristics of both s (spherical) and p

(node) orbitals.

The four new electron distributions

(orbitals) will orient as far away from each

other as possible. Toward the corners of atetrahedron.


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Photoelectron Spectroscopy

Measures the energy of electrons ejected

from a molecule when hit by a photon.

Binding energy = photon energy - emitted

photon energy

h = T + EB

Methane shows >290 ev (C1s) and two morebands, 23.0 ev and 12.7 ev. (Two types of

orbitals)

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Orbital Symmetry

The MOs must be symmetric or anti-

symmetric with respect to the AOs (basis

set).

The sigma bond resulting from the C sp3

and H 1s do not have the symmetry of the s

an p orbital.


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New Approach

Consider all the AOs of C and 4 Hs.

Select symmetric or anti-symmetric linear

combination.

If we ignore the C1s electrons there should

be 4 bonding and 4 non-bonding orbitals.

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Orbitals and Energies

-13.30708 Kcal/mol

-28.87874 Kcal/mol


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VSEPR Theory

Valence Shell Electron PairRepulsion

Theory

Electrons in orbitals will take on a geometry

that has the electrons as far away from each

other as possible.

2 sp 180 deg.

3 sp2 120 deg.

4 sp3 109.5 deg.

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Methyl Chloride

Predicted bond angles

H-C-H 109.5o

H-C-Cl 109.5o

Experimental

H-C-H 110o 52

H-C-Cl 108o 0

Why?


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VSEPR Explanation

Cl more EN than Carbon. Draws electon

density toward Cl.

C-H bond shorter.

The electron density is shifted toward the Cl

so the Hs repel each other more than in

methane making H-C-H bond angle greater. The H are attracted to the Cl so the H-C-Cl

bond angle is smaller

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MO Explanation

The HOMOs show significant bonding

between the Cl and C-H bond


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Variable Hybridization

Because of the EN of chlorine the methyl

chloride molecule will be less symmetric.

The C-Cl bond will have more p character

which results in the C-H bond having more

s character.

C-H shorter and H-C-H angle larger.

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Hybridization Parameter

For a bond spn is the hybridization

parameter.

2 = n

For an sp3 the fractional s and p character

are1

1+ 2i% s = % p =2

i1+ 2i


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Interorbital Angles

Based on hybridization model

Angle A-C-B

Angle A-C-A

1 + a b cos ab = 0

1 + a2 cos aa = 0

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Methyl Chloride (again)

Two bond angles H-C-H and H-C-Cl

Relationship between these two angles

If we know one we can calculate the other.

If H-C-Cl is 108 o then

3 sin2ab = 2 (1 - cos a a

cos aa = 1 -3 sin

2ab

2

cos aa = 1 -3 (sin (108))

2

2

aa = 110o


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Carbon Orbital

Orbital to H.

sp2.86

The other C orbital to Cl

sp3.5

a2 =

- 1

cos aa

= 2.86

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1 + 2.86+

1

1 + b2

= 1

b2 = 3.50

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Curved Bonds

Consider CH2Cl2 Measured Cl-C-Cl 111o 47; H-C-H 112 o 0

Cal. C to Cl orbital sp2.69; C to H sp3.37 or a

bond angle of 107.

The bond angle does not fit experimental.


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Are Bonds Straight Lines

Between Nuclei?

Electron density may lie outside the line

between two nuclei.

Internuclear bond angle - defined by nuclear

positions

Interorbital bond angle - defind by the

direction the orbital leaves the atom.

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Bent Bonds

Best example cyclopropane

C-H hybridization sp2.36 which would make

the C-C sp2.69 which would predict an

interorbital angle of 111.8 o; > 60 o.

C

C

C

H

H

Interatomic 115.1


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Measurable Properties

Hybridization parameter can not be

measured directly but it can be correlated to

some measurable parameters.

J13C-H (cps) = 500/ (1 - 2)

rC-H = 1.1597 - 10-4 J13C-H

log k rel = 0.129 J13C-H - 15.9

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, Model for Ethene

Carbon is sp2 hybridized

H-C bond is shorter than normal because

bond has more s character

C-C bond has two parts; from overlap of

two C sp2 hybrids and a from overlap of

two p orbital. H-C-H 120 o and the same for H-C-C.

C-C shorter, more s character


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Energy Diagram

PEsp

2

Hybrid Orbitalsof Carbon

p

Hybrid Orbitalsof Carbon

sp2

p

MO's

to H 1s to H 1s

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Bent Bond Model

Pauling

Carbons are sp3 with two sp3 from each

carbon overlapping to for two bent bonds.

Predicts planar but bond angles of 109.5 o

C

H

H

C

H

H


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Acidity of Hydrocarbons

Hydrocarbons are weak acids.

Ethane 10-42, ethene 10-36.5, ethyne 10-25

R H + B R -

+ HB+

Ka =[R-] [HB+]

[RH] [B]

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Why?

The anion stability follows the s character

of the orbital that the resultant anion exist

in.

sp3 anion less stable than sp2 which is less

stable than sp. Therefore, alkanes less acidic

than alkenes which are less acidic thanalkynes


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Bent Bond Rationalization

The bent bond pulls the electron pair of the

anion closer to the center of the internuclear

axis of the C-C bond.

Less electron pair -- electron pair repulsion

HHH

H H

H

H H

H

0O

60O?

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Which Model is Correct?

Neither!

Both models are useful and must be used

with care.

Usually they converge on the same solution

when considered in great detail.

heat of reaction estimation

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