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PSE 406: Lecture 5 1 Wood Chemistry Wood Chemistry PSE 406/Chem E 470 Lecture 5 Cellulose

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Wood Chemistry PSE 406/Chem E 470. Lecture 5 Cellulose. Cellulose: the Basics. Linear polymer made up of  -d glucopyranose units linked with  glycosidic bonds. Repeating unit = glucose (cellobiose) - PowerPoint PPT Presentation

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Page 1: Wood Chemistry PSE 406/Chem E 470

PSE 406: Lecture 5 1

Wood Chemistry

Wood ChemistryPSE 406/Chem E 470

Lecture 5

Cellulose

Page 2: Wood Chemistry PSE 406/Chem E 470

PSE 406: Lecture 5 2

Wood Chemistry Cellulose: the Basics

Linear polymer made up of -d glucopyranose units linked with glycosidic bonds.

Repeating unit = glucose (cellobiose) Glucopyranose units in chair form - most thermodynamically stable.

Only 2% in other forms CH2OH and OH groups in equatorial positions stability

O

O O

O

O

O

O

O

CH2OH

OHHO

OHHO

CH2OHOH

CH2OH

HO

OH

CH2OH

HO

O

Cellulose

Cellobiose Unit

Page 3: Wood Chemistry PSE 406/Chem E 470

PSE 406: Lecture 5 3

Wood Chemistry Cellulose: More Basics

Cellulose is elongated and the glucose units in one plane for the following reasons: linkages » The thermodynamic stability of the chair form» The positioning of the groups on the ring: E versus Al

Amylose (starch) occurs as a helix in solid state because of the 1-4 linkage.

O

OH

CH2OH

O

HO

O O

OH

CH2OH

OHO O

OH

CH2OH

OHOAmylose

Page 4: Wood Chemistry PSE 406/Chem E 470

PSE 406: Lecture 5 4

Wood Chemistry Reducing End Groups

Each cellulose chain has 1 reducing end group at the C1 position of the terminal glucopyranose unit

The C4 position of the other terminal unit is an alcohol and therefore not reducing.

Does the reducing end mutarotate?» In fibers, probably not because of hydrogen bonding, etc.» In solution, probably

Reducing EndNon-reducing End

O

OO

O

O

O

OCH2OH

OHHO

OHHO

CH2OHOH

CH2OH

HO

OH

CH2OH

HO

HO

OH

Cellulose

nNotes

Page 5: Wood Chemistry PSE 406/Chem E 470

PSE 406: Lecture 5 5

Wood Chemistry

Cellulose: Molecular Weight

Degree of Polymerization of Cellulose

DP = molecular weight of cellulose

molecular weight of one glucose unit

Determination of molecular weight requires isolation and solubilization of cellulose

Isolation procedures will modify (reduce) molecular weight Various modification procedures used for isolation

» Derivatize with a variety of agents» Metal complexes» Pulping » Solvent systems

Page 6: Wood Chemistry PSE 406/Chem E 470

PSE 406: Lecture 5 6

Wood Chemistry Degree of Polymerization

Material Degree of polymerizationCotton (unopened) 15,300

Aspen (Hardwood) 10,300

Jack Pine (Softwood) 7900

Bacteria 5000

Sulfite pulp, bleached 1255

Kraft pulp 975

Rayon 305

Notes

Page 7: Wood Chemistry PSE 406/Chem E 470

PSE 406: Lecture 5 7

Wood Chemistry

Carbons 6 x 12 = 72Oxygen 6 x 16 = 96Hydrogen 12 x 1 = 12

180 g/mole

Molecular Weight Determination

Determination of the molecular weight of the glucose molecule on the left is quite simple.

Simply count all of the atoms and add up the molecular weight.

O

HO OH

OH

OH

CH2OH

Page 8: Wood Chemistry PSE 406/Chem E 470

PSE 406: Lecture 5 8

Wood Chemistry

Molecular Weight of Mixtures

O

OH

CH2OH

HO OH

HO

180g/mole

O

OO

OH

CH2OH

HO

OH

CH2OH

HO

HOOH

342g/moleO O

OO

OCH2OH

OHHO

OHHO

CH2OHOH

CH2OH

HO OH

HO

504g/mole

O

O O

O

O

O

OCH2OH

OHHO

OHHO

CH2OHOH

CH2OH

HO

OH

CH2OH

HO

HOOH

666g/mole

What if you have a mixture of 4 different chemicals and refer to it as a single compound like cellulose. What is the molecular weight?

Page 9: Wood Chemistry PSE 406/Chem E 470

PSE 406: Lecture 5 9

Wood Chemistry

The simple answer to the question of molecular weight of mixtures is that you use an average of the molecular weights. This is known as the number average molecular weight.

Although this gives a number which is usable, it doesn’t completely describe the system. This is because you can obtain the same number average molecular weight with completely different mixtures. For example, a sample of all medium sized molecules and a mixtures of big and little molecules could give the same value.

Molecular Weight of Mixtures

Number Average Molecular Weight (Mn)

Mn = weight

# molecules=

NxMx

Nx

Page 10: Wood Chemistry PSE 406/Chem E 470

PSE 406: Lecture 5 10

Wood Chemistry Molecular Weight Equations

Weight Average Molecular Weight (Mw)

Mw = CxMx

Cx=

NxMx)(Mx)

NxMx=

NxMx2

NxMx

The second method for determining molecular weight is the weight average method. This method gives values which are influenced by the amount of larger molecules. This equation is developed from the number average equation by replacing the number of molecules Nx by the weight of the molecules Cx. For examples on calculating molecular weight, see the end of this lecture.

Page 11: Wood Chemistry PSE 406/Chem E 470

PSE 406: Lecture 5 11

Wood Chemistry Polydispersity

Polydispersity is the ratio of the weight average molecular weight to the number average molecular weight. This value provides information on the distribution of molecular weights. Larger values indicate that you have a wide range of molecular weights while low values mean a narrow distribution.

Polydispersity

Polydispersity = Mw/Mn

Page 12: Wood Chemistry PSE 406/Chem E 470

PSE 406: Lecture 5 12

Wood Chemistry

Molecular Weight Distribution

Length of chain

We

igh

t o

f p

oly

me

rs in

siz

e in

terv

al

Number average (Mn)

Weight average (Mw)

Page 13: Wood Chemistry PSE 406/Chem E 470

PSE 406: Lecture 5 13

Wood Chemistry

Molecular Weight Determination Methods

Number Average (Mn)» Osmometry» Reducing end group analysis (cellulose)

Weight Average (Mw)» Light Scattering

Others: Mz and Mv» Ultracentifugation» Viscosity Measurements

Page 14: Wood Chemistry PSE 406/Chem E 470

PSE 406: Lecture 5 14

Wood Chemistry

Lecture 5

Example Problems

Page 15: Wood Chemistry PSE 406/Chem E 470

PSE 406: Lecture 5 15

Wood Chemistry

Molecular Weight: Example 1

1g/mole 5g/mole 10 g/mole

Note: These values represent the molecular weight of each of spheres

Page 16: Wood Chemistry PSE 406/Chem E 470

PSE 406: Lecture 5 16

Wood Chemistry

CelluloseMolecular Weight: Example

1

1. Number Average Mn

Mn = NxMx

Nx=

(5)(1) + (5)(5) + (5)(10)5+5+5

= 5.33

2. Weight Average Mw

Mw = CxMx

Cx=

(5)(1) + (5)(5) + (5)(10)

(5)(1)(1) + (5)(5)2 + (5)(10)2

= 7.875

3. Polydispersity

= Mw

Mn= 1.477

Page 17: Wood Chemistry PSE 406/Chem E 470

PSE 406: Lecture 5 17

Wood Chemistry

CelluloseMolecular Weight: Example

2

1g/mole 5g/mole 10 g/mole

Page 18: Wood Chemistry PSE 406/Chem E 470

PSE 406: Lecture 5 18

Wood Chemistry

CelluloseMolecular Weight: Example

2

1. Number Average Mn

Mn = NxMx

Nx=

(25)(1) + (5)(5) + (5)(10)25+5+5

= 2.86

2. Weight Average Mw

Mw = CxMx

Cx=

(25)(1) + (5)(5) + (5)(10)

(25)(1)(1) + (5)(5)2 + (5)(10)2

= 6.50

3. Polydispersity

= Mw

Mn= 2.27

Page 19: Wood Chemistry PSE 406/Chem E 470

PSE 406: Lecture 5 19

Wood Chemistry

CelluloseMolecular Weight: Example

3

1g/mole 5g/mole 10 g/mole

Page 20: Wood Chemistry PSE 406/Chem E 470

PSE 406: Lecture 5 20

Wood Chemistry

CelluloseMolecular Weight: Example

3

1. Number Average Mn

Mn = NxMx

Nx=

(5)(1) + (25)(5) + (5)(10)5+25+5

= 5.14

2. Weight Average Mw

Mw = CxMx

Cx=

(5)(1) + (25)(5) + (5)(10)

(5)(1)(1) + (25)(5)2 + (5)(10)2

= 6.27

3. Polydispersity

= Mw

Mn= 1.22

Page 21: Wood Chemistry PSE 406/Chem E 470

PSE 406: Lecture 5 21

Wood Chemistry

CelluloseMolecular Weight: Example

4

1g/mole 5g/mole 10 g/mole

25 x

Page 22: Wood Chemistry PSE 406/Chem E 470

PSE 406: Lecture 5 22

Wood Chemistry

CelluloseMolecular Weight: Example

4

1. Number Average Mn

Mn = NxMx

Nx=

(5)(1) + (5)(5) + (25)(10)5+5+25

= 8.00

2. Weight Average Mw

Mw = CxMx

Cx=

(5)(1) + (5)(5) + (25)(10)

(5)(1)(1) + (5)(5)2 + (25)(10)2

=9.39

3. Polydispersity

= Mw

Mn= 1.17