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Apresentação do PowerPoint - COST FP1205costfp1205.com/wp-content/uploads/2017/schools/... ·...

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Jamerson Oliveira Advisor: Prof. Marie-Pierre Laborie, PhD Co-Advisor: Prof. Dr. Vincent Roucoules
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Page 1: Apresentação do PowerPoint - COST FP1205costfp1205.com/wp-content/uploads/2017/schools/... · 𝑂𝐻 =8.5 10−24 5.3 1020=4.5 10−3 / Title: Apresentação do PowerPoint Author:

Jamerson Oliveira

Advisor: Prof. Marie-Pierre Laborie, PhD

Co-Advisor: Prof. Dr. Vincent Roucoules

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Introduction

Smart Materials

2 ebay.com; eng-cs.syr.edu; Hakalahti et al. 2016.

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Introduction

Click Chemistry

Diels-Alder reaction

3 medchem101.com; summer.catsgarden.net; Liu & Chen, 2007.

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Introduction

Cellulose nanocrystals – Surface Modification

4

CNC 2,4 TDI

OH terminated furan

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Introduction

Cellulose nanocrystals – Surface Modification

5

CNC 2,4 TDI

OH terminated maleimide

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Introduction

Cellulose nanocrystals – Click!

6

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Goals

Main Goal of the Work

- Assess the potential of Diels Alder chemistry to design thermoresponsive

cellulose nanocrystal materials.

Goal of the Presentation

- Overview the work focusing on the importance of knowledge about CNCs

surface for the grafting onto.

7

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Research Question 1: Which are the conditions and thermodynamic/kinetic parameters

for the DA and retro-DA reaction of the selected molecules?

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DA and retro-DA reaction

9

Endo and Exo

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UV-Vis – N-hydroxymaleimide

10

𝐴𝑁 = 𝜀𝑁𝑏𝑐𝑁

𝜀𝑁𝑏 = 406.1

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UV-Vis – Furfuryl alcohol

11

No correlation observed!!

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UV-Vis – Adduct

12

300 350 400 450

0,0

0,2

0,4

0,6

0,8

1,0

Absorb

ance

Wavelength (nm)

C (mg/ml)

1.15

1

0.8

0.6

0.4

0.2

0.1

0.08

0.06

𝐴𝐴 = 𝜀𝐴𝑏𝑐𝐴

𝜀𝐴𝑏 = 41.72

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UV-Vis – Reaction

13

𝐴𝑇 = (𝜀𝐴𝑐𝐴 + 𝜀𝑁𝑐𝑁)𝑏 = 41.72𝑐𝐴 + 406.1𝑐𝑁

• What to expect from the reaction?

• According to Beer’s law:

Stoichiometry 1:1 implies:

𝑐𝐴 + 𝑐𝑁 = 0,0053 𝑐𝐴 = 0,0053 − 𝑐𝑁

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UV-Vis – Reaction

14

NHM solution

Reaction

in which:

𝑐𝑁0 = 0,0053 𝑚𝑜𝑙/𝑚𝑙 FAL solution

𝑐𝐴 + 𝑐𝑁 = 0,0053

k𝐶𝑁0 = 0,0113 min-1

k = 2,13 mol-1.L.min-1

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H NMR – Adduct Characterization

15

H3a H5

H6

H8, H9

H7a

H7

H10

H2

ENDO PRODUCT

Proton Coupling Small coupling

constant

Larger Coupling

constant Chemical Shift

H2 Singlet - - 10.68 s

H6 Doublet of doublet J = 1.62 ; J = 1.61 J = 5.75 ; J = 5.78 6.42 dd, J = 5.8,

1.6

H5 Doublet J = 5.79 - 6.31 d, J = 5.8

H7 Doublet of doublet J = 1.59 ; J = 1.59 J = 5.40 ; J = 5.40 5.27 dd, J = 5.4,

1.6

H10 Triplet

H8, H9

Doublet of

doublets J = 12.9 ; J = 12.9 J = 29.8 ; J = 29.9

3.98 dd, J = 29.9,

12.9

H7a

Doublet of

doublets J = 5.43 ; J = 5.40 J = 7.47 ; J = 7.44

3.57 dd, J = 7.5,

5.4

H3a Doublet J = 7.47 - 3.33 d, J = 7.5

Froidevaux, 2015. SI

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H NMR – Adduct Characterization

16

H3a H5

H6

H8, H9

H7a

H7

H10

H2

EXO PRODUCT

Proton Coupling Small coupling

constant

Larger Coupling

constant Chemical Shift

H2 Singlet - - 10.76 s

H6 Multiplet - - -

H5 Multiplet - - -

H7 Doublet J = 1.38 - 5.01 d, J = 1.38

H10 Triplet - - -

H8, H9

Doublet of

doublets

J = 12.38 ; J =

12.44

J = 102.86 ; J =

107.41

3.81 dd, J = 12.41

J = 105.14

H7a Doublet J = 6.45 - 2.88 d, J = 6.45

H3a Doublet J = 6.42 - 2.75 d, J = 6.42

Froidevaux, 2015. SI

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NMR kinetic measurements

DA reaction

17

5 hours

23 hours

7 min

33 hours

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Research Question 2: How can we tailor the grafting of the selected molecules onto the

surface of CNCs?

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CNCs Characterization - AFM

19

Batch Length (nm) Width (nm) L/W (nm)

1 203 5 44

2 216 5 44

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CNC Characterization – EA, pH and Zeta Potential

20

Elemental Analysis

Sample C H S DS

CNC 1 40.07 6.18 0.71 0.04

CNC 2 39.25 6.11 0.93 0.05

CNC 3 39.45 6.25 0.91 0.05 pH and Zeta potential

Sample pH Zeta

potential

CNC 1 6.6 - 14.9

CNC 2 6.6 - 16.1

CNC 3 6.5 -17.9

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CNCs Characterization – XRD

21

Sample CrI (%)

CNC 1 90.3

CNC 2 89.9

CNC 3 90.4

5 10 15 20 25 30 35 40

Inte

nsity (

a.u

.)

2Theta (degrees)

CNC 1

CNC 2

CNC 3

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CNC Characterization – OH determination

22 Gu et al., 2013.

Sample B110 (nm) B1𝟏 0 (nm)

CNC 1 5.9 5.1

CNC 2 5.9 5.1

CNC 3 6.2 5.1

Mean values 6.0 ± 0.2 5.1 ± 0.0

𝑂𝐻 𝑚𝑜𝑙 𝑝𝑒𝑟 𝑔𝑟𝑎𝑚 = 4.5 𝑥 10−3 𝑚𝑜𝑙/𝑔

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Carbamation

23 Zoppe et al., 2009.

1 1.3

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24

FTIR and EA – Carbamated CNC

4000 3500 3000 2500 2000 1500 1000

Ab

so

rban

ce (

a.u

.)

Wavenumber (cm-1)

CNC2C

CNC2C2

Sample DS

CNC2C 0,50

CNC2C2 0,44

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Surface Grafting with Maleimide Moieties

25 Zoppe et al., 2009. ; Shang et al., 2013.

24 h

75 ºC

Anhydrous toluene

5 days

50 ºC Acetone

TEA

TEA

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26

FTIR – Maleimide grafted CNC

4000 3500 3000 2500 2000 1500 1000

Ab

so

rban

ce (

a.u

.)

Wavenumber (cm-1)

CNC3CM

4000 3500 3000 2500 2000 1500 1000A

bsorb

ance (

a.u

.)

Wavenumber (cm-1)

CNC2C

CNC2C2

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Conclusion

Preliminary results show successful grafting of the moieties onto CNCs

surface.

Further characterization will be performed in order to access more

quantitative results.

Characterization of the CNCs is essential to understand and predict the

behavior of the grafting process.

27

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Acknowledgments

28

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References

Hakalahti M. et al., Direct Interfacial Modification of Nanocellulose Films for Thermoresponsive

Membrane Templates, ACS Appl. Mater. Interfaces, 2016, 8 (5), 2923–2927.

Liu Y-L & Chen Y-W, Thermally Reversible Cross-Linked Polyamides with High Toughness and

Self-Repairing Ability from Maleimide and Furan-Functionalized Aromatic Polyamides. Macromol.

Chem. Phys. 2007, 208, 224–232.

Gu J. et al., Quantification of cellulose nanowhiskers sulfate esterification levels. Carbohydrate

Polymers, 2013, 92, 1809– 1816.

Zoppe J. O. et al., Reinforcing Poly(ε-caprolactone) Nanofibers with Cellulose Nanocrystals, ACS

Appl. Mater. Interfaces , 1 (9), 1996–2004.

Shang W. et al., Hydrophobic modification of cellulose nanocrystal via covalently grafting of castor

oil. Cellulose, 2013, 20, 179–190.

http://www.ebay.com

http://eng-cs.syr.edu/

http://medchem101.com/

http://summer.catsgarden.net/

29

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CNCs Characterization – OH determination

31

Sample B110 (nm) B1𝟏 0 (nm)

CNC 1 5.9 5.1

CNC 2 5.9 5.1

CNC 3 6.2 5.1

Mean values 6.0 ± 0.2 5.1 ± 0.0

𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 = 5.1 + 6.0 𝑥 2 𝑥 1.058 = 23.49 𝑛𝑚2

𝑂𝐻 𝑝𝑒𝑟 𝑎𝑟𝑒𝑎 = 120 ÷ 23.49 = 5.1 /𝑛𝑚2

𝑂𝐻 𝑚𝑜𝑙 𝑝𝑒𝑟 𝑎𝑟𝑒𝑎 = 1 ÷ 6𝑥1023 𝑥 5.1 = 8.5 𝑥 10−24 𝑚𝑜𝑙/𝑛𝑚2

𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑎 𝑐𝑟𝑦𝑠𝑡𝑎𝑙 𝑝𝑒𝑟 𝑔𝑟𝑎𝑚 =1 𝑥 210 𝑥 5 𝑥 4 𝑥 1021

1.5 𝑥 52 𝑥 210= 5.3 𝑥 1020

𝑛𝑚2/𝑔

𝑂𝐻 𝑚𝑜𝑙 𝑝𝑒𝑟 𝑔𝑟𝑎𝑚 = 8.5 𝑥 10−24 𝑥 5.3 𝑥 1020 = 4.5 𝑥 10−3 𝑚𝑜𝑙/𝑔


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