Question 1. Identify the sugars below by filling in the table below (except shaded areas). Use the page or a separate sheet

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1 Question 1. Identify the sugars below by filling in the table below (except shaded areas). Use the page or a separate sheet Sugar name aworth projection(s) of the corresponding pyranose form (Six membered rings) A D-galactose (show only β-form) B C β-dglucuronic acid N-acetyl-β-Dglucosamine C 2 C C 2 N Fisher projection C C 2 C C Preferred ring conformation (name only) 4 C 1 (name only) 4 C 1 2 C N C 3 Polysaccharide containing the sugar (name Fill one) Pectins, Guar, locust bean gum or other galactomannans Xanthan yaluronan (hyaluronic acid) eparin Chitin Chitosan yaluronan (Not heparin, which is α-..) 3 C

2 D 3,6-anhydro- D-galactose C 2 (name only) 1 C 4 (3,6-anhydro flips the galactose ring) Carrageenans (Not agar) E α-l-guluronic acid C C (structure and name) - C Alginate C 1 C 4

3 Question 2. a) Show the chemical structure of cellulose chains (without substituents) (Type of monomers, aworth structure, linkage types) Monomers: nly 4-linked β-d-glucose, i.e linkages are only β-1,4 aworth: C 2 C 2 C 2 C 2 b) The figure shows a Ramachandran plot for cellobiose. What do the curves and symbols in the plot refer to? Indicate the stable conformation of cellobiose that corresponds to the plot. The plot refers to the energy map calculated by systematically varying the torsion angles Ψ and Φ (see figure), which are the glycosidic bond angles,

4 and calculating interactions between the residues: Cellobiose (most stable structure): The two rings are in 4 C 1 conformation, and rotated about 180 relative to each other as shown below (Note: -bonds shown below are not part of the question) 4 6 C C c) What is the basis for the crystallinity observed in cellulose? - Rigid, extended chains as described above - Intramolecular -bonds (along the chain) - Intermolecular -bonds link chains together Question 3. PMC is a family of cellulose derivatives where some of the hydroxyl groups are substituted with methyl (-C 3 ) and hydroxypropyl (-C 2 C()C 3 ) groups. Some PMC types dissolve in cold water to give clear solutions, but precipitate or form gels by heating above a certain temperature.

5 d) Which types of chain-chain interactions lead to the precipitation of PMC at high temperature? ydrophobic interactions (intermolecular) between non-polar substituents e) Which enthalpic () and entropic factors (S) contribute to this behaviour of PMC? Precipitation (ordering) above a certain T happens when both Δ (for order-todisorder) and ΔS (for order-disorder) are negative (or positive for opposite direction). The negative Δ corresponds to breaking non-covalent bonds (favours order), but negative ΔS (=ΔS M + ΔS w ) must be linked to the entropy of water since ΔS M is positive for order-to-disorder. rdering releases water which obtain higher entropy in the bulk phase. f) Would the intrinsic viscosity of PMC (below the precipitation temperature) be essentially unchanged or very different if determined at high (0.1 M) and low (0.01 M) ionic strengths, respectively? Explain briefly. PMC has no charges (not polyelectrolyte) and the intrinsic viscosity should not depend the ionic strength. (It could be argued that high salt favours hydrophobic associations, hence compacting the polymer, leading to a decrease in IV). Question 4. Pullulan, hyaluronan and xanthan have persistence lengths of 1.3 nm, 7 nm, and 120 nm, respectively, in 0.1 M NaCl. a) What is the persistence length (definition and brief comment)? The persistence length is defined according to a vector model (n vectors of length l). It is the dot product of:

6 a) The unity vector in direction 1 ( l i / l 1 ) b) The average end-to-end vector ( R = l i i The PL is a stiffness parameter used in the wormlike chain model b) Discuss the structural basis for the differences in persistence length of the three polymers. Pullulan contains flexible α-1,6 linkages (maltotriose repeating units) results in a relatively little extended randomly coiled polymer (consequently low R G and low IV compared to the others (at the same M) yaluronan has alternating B-1,3 and β-1,4 linkages. The absence of 1,6 linkages leads to a more extended chain Xanthan has a cellulosic backbone (with trisaccharide side chains on every second Glc), and adopts a double-stranded structure with very high stiffness. c) Explain in plain words (max ½ page) a method or strategy to determine the persistence length from experimental data - btain fractions differing in molecular weight (fractionate or degrade to different extents) - Determine for each fraction the molecular weight and either R G or IV - Compare data to those calculated from the WCM (wormlike chain model) by systematically varying the PL (q) and the mass-per-unit-length (M L ) and find best fit.

7 Question 5. An alginate has the following properties: M w = Da M n = Da F G = 0.72 F GG = 0.68 F MGM = F GGM = a) Define M w M w = n i=1 n i=1 n i M i 2 n i M i = n w i M i i=1 n w i i=1 = n c i M i i=1 n c i i=1 n i : moles of component i w i : weight (grams) of component i c i : Concentration (g/l) of component i M i : molecular weight of component i b) Which experimental methods could be used to find M w and M n, respectively?

8 M w : Light scattering M n : smometry or end-group analysis c) Determine the average G-block length (N G>1 ) for the alginate N G>1 = (F G -F MGM )/F GGM = 7.7 d) The alginate is degraded randomly by means of dilute acid at elevated temperature. After 30 min M w = Calculate M w and M n after 60 min. - Use the equation 1/M w,t = 1/M w,t=0 + kt/2m 0 to find k/2m 0 using M w data for 0 and 30 min - Then calculate M w for 60 min - Since M w /M n 2 for random degradation (also for the t = 0 sample), M n is found as M w,t=60 /2 t Mw Mn k/2m E

9 e) Alginates degrade both at low p and high p. Which are the main reaction mechanisms in the two cases? Low p: Acid hydrolysis Glycoside (acetal) R + + R - Protonation of glycosidic oxygen - Cleavage of C1 bond (rate limiting) producing a resonance stabilised carbocation - Addition of water leads to normal reducing end Slow (rate limiting step) -R Carbocation (carbonium ion) Partial double bond to C1-5 => planar structure igh p: Alkaline β-elimination (not hydrolysis) - C5 proton slightly acidic: abstracted by - - Resonace stabilised carbanion intermediate (C4-C5-C6) - Elimination of substituent at C4 - End product (non-reducing end) unsaturated derivative. Normal reducing end f) When subjected to random depolymerisation agents (e.g. 2 2 /Fe ++ ) polysaccharides such as xanthan and scleroglucan tend to loose solution viscosity much more slowly than water soluble cellulose derivatives (e.g. CMC) or synthetic polymers such as PAAM (partially hydrolysed polyacrylamide), even if the rates of chain cleavage are essentially the same in all cases. Explain this phenomenon. 2

10 PAAM: > Every PAAM or CMC cleavage leads to chain fragmentation (decrease in M w and IV) because they are single stranded Xanthan and scleroglucan are double- and triple stranded. They are stabilised by intermolecular, non-covalent bonds: ================= Cleavage of one of two (or three) strands may initially have no effect because the probability of having another break nearby is low. ence, the structure remains intact.

11 TBT4135 Exam Dec 2012 SLUTIN Note: Solution given as keywords only, i.e. main points that should be mentioned to obtain an A.

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