2,5-Dimethylfuran is a liquid biofuel that can be synthesized from 5-hydroxymethylfurfural (HMF). As shown below, HMF can be synthesized from D-fructose by treatment with sulfuric acid. FOH H2SO4 HO OH DMSO HO но OH 48 h, 110 °C 68% 5-Hydroxymethylfurfural (HFM) D-Fructose 2,5-Dimethylfuran The mechanism for the formation of HMF from D-fructose is believed to involve the following series of dehydration reactions: он - H,0 HO OH - H20 - H20 но OH но HO OH HO ОН HO D-Fructose A в HO 5-Hydroxymethylfurfural Draw the complete, detailed mechanism for each of these dehydration reactions.

2,5-Dimethylfuran is a liquid biofuel that can be synthesized from 5-hydroxymethylfurfural (HMF). As shown below, HMF can be synthesized from D-fructose by treatment with sulfuric acid. FOH H2SO4 HO OH DMSO HO но OH 48 h, 110 °C 68% 5-Hydroxymethylfurfural (HFM) D-Fructose 2,5-Dimethylfuran The mechanism for the formation of HMF from D-fructose is believed to involve the following series of dehydration reactions: он - H,0 HO OH - H20 - H20 но OH но HO OH HO ОН HO D-Fructose A в HO 5-Hydroxymethylfurfural Draw the complete, detailed mechanism for each of these dehydration reactions.

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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The reaction of liquid methyl chloride (CH3Cl) with water produces methanol and hydrogen chloride gas at room temperature, despite the fact that ΔHo rxn = 28.3 kJ/mol. Using thermodynamic arguments, propose an explanation as to why methanol forms. CH3Cl (l) + H2O (l) → CH3OH (l) + HCl (g)
Calculate the ∆Hof of C6H12O6(s) from the following data: ∆H combustion of C6H12O6(s) = -2816kJ/mol ∆Hof of CO2 (g) = -393.5 kJ/mol ∆Hof of H2O(l) = -285.9 kJ/mol Equation: C6H12O6(s) + O2(g) → CO2(g) + H2O(g)
In biological cells that have a plentiful supply of oxygen, glucose is oxidized completely to CO2 and H2O by a process called aerobic oxidation. Muscle cells may be deprived of O2 during vigorous exercise and, in that case, one molecule of glucose is converted to two molecules of lactic acid (CH3CH(OH)COOH) by a process called anaerobic glycolysis. (a) When 0.3212 g of glucose was burned at 298 K in a bomb calorimeter of calorimeter constant 641 J K−1 the temperature rose by 7.793 K. Calculate (i) the standard molar enthalpy of combustion, (ii) the standard internal energy of combustion, and (iii) the standard enthalpy of formation of glucose. (b) What is the biological advantage (in kilojoules per mole of energy released as heat) of complete aerobic oxidation compared with anaerobic glycolysis to lactic acid?
Calculate the enthalpy change for the reaction C(s) + 2H2 (g)--->CH4(g) from the following data. C(s) + O2 (g) ---> CO2(g). ΔH = – 393.5 kJ/mol H2(g) + 0.5 O2 (g) ---> H2O(l). ΔH = – 285.8 kJ/mol CH4(g) + 2 O2 (g)---> CO2(g) + 2 H2O(l). ΔH = – 890.2 kJ/mol
C(s) + O2(g) -> CO2(g) CH4(g) -> C(s) + 2 H2(g) ∆Hrxn= +74.6 kJ/mol H2(g) + ½ O2(g) ->H2O(l) ∆Hrxn = −285.8 kJ/mol CH4(g) + 2 O2(g) -> CO2(g) + 2 H2O(l) ∆Hrxn = −890.5 kJ/mol 4) Let’s calculate heat of combustion for propane, C3H8, at thermodynamic standard state – first, we need the balanced chemical equation: C3H8(g) + O2(g) -> CO2(g) + H20(l) Use some of the reactions above and the reaction below for the formation of propane (C3H8) to determine the heat of combustion for propane (i.e DH for balanced equation you wrote above). C3H8(g) -> 3 C(s) + 4 H2(g) ΔHrxn = +103.9 kJ/mol
Formula: C3H6OMelting Point: -94.0°CBoiling Point: 56.0°CDensity of liquid: 0.791 g/mLHeat of Fusion: 98.14 J/gHeat of vaporization: 538.9 J/gSpecific heat capacity (solid): 1.653 J/g°C Specific heat capacity (liquid): 2.161 J/g°C Specific heat capacity (gas): 1.291 J/g°C A 791.0 g sample of acetone sitting in the lab has a temperature of 21.7°C. How much heat would be involved in lowering the temperature of the acetone to -200.0°C? (answer in Joules)
The heat of fusion ΔHf of dichloromethane CH2Cl2 is 6.2 /kJmol. Calculate the change in entropy ΔS when 796.g of dichloromethane melts at −95.1°C.
Formula: C3H6OMelting Point: -94.0°CBoiling Point: 56.0°CDensity of liquid: 0.791 g/mLHeat of Fusion: 98.14 J/gHeat of vaporization: 538.9 J/gSpecific heat capacity (solid): 1.653 J/g°C Specific heat capacity (liquid): 2.161 J/g°C Specific heat capacity (gas): 1.291 J/g°C A 523.70 g sample of acetone gas has a temperature of 56.0°C. How much heat is involved in converting it to a solid that has a temperature of -94.0°C? (Answer in Joules)
Use the following data 3N2H4(g) → 4NH3(g) + N2(g) ∆Sº=240.6 J/(K·mol) Sº(N2(g))=216.9 J/(K· mol) Sº(NH3(g))=227.8 J/(K· mol) obtained at 700.0 K to find Sº of N2H4(g).
The complete combustion of 1.20g of cinnamaldehyde (C6H8O, one of the compounds in cinnamon) in a bomb calorimeter (Ccalorimeter = 3.00 kJ/°C) produced an increase in temperature of 13.9°C Calculate the molar enthalpy of combustion of cinnamaldehyde (triangleHcomb) (In kilojoules per mole of cinnamaldehyde)
Calculate the ΔH°f of C6H12O6(s) from the following data: Show the complete and step by step solution ΔH combustion of C6H12O6(s) = -2816kJ/mol ΔH°f of CO2 (g) = -393.5 kJ/mol ΔH°f of H2O (l) = -285.9 kJ/mol Equation: C6H12O6(s) + O2(g) -> CO2(g) + H2O(l)
Naphthalene, better known as “mothballs,” has bp=218∘C and ΔHvap=43.3kJ/mol. What is the entropy of vaporization, ΔSvap in J/(K⋅mol) for naphthalene?