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Interpretation:
D-Sugar of 2-ketohexosesstructures should be drawn.
Concept introduction:
Ketohexoses: A six-carbon monosaccharidewhich is a
The orientation of the carbon atoms in a molecule is arranged vertically from top to bottom.
The C1 carbon is at the top of the orientation, the hydrogen and hydroxyl groups are placed in right and left side of the orientation. D and L form are depends on the arrangement of the hydroxyl group. If the hydroxyl group is in the right side of the molecule in the penultimate (next-to-last) carbon is called as D sugar, if the If the hydroxyl group is in the left side of the molecule in the penultimate (next-to-last) carbon is called as L sugar.
To find: structures of D-sugar of 2-ketohexoses
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Chapter 24 Solutions
ORGANIC CHEMISTRY-PRINT COMPANION (LL)
- Experiment 1 Data Table 1: Conservation of Mass - Initial Mass Data Table 1 Data Table 2 Data Table 3 Data Table 4 Panel 1 Photo 1 Data Table 5 Reaction Mass of test tube and 5.0% HC₂H₂O2 (g) # (A) (B) Mass of NaHCO, (g) Mass of balloon and NaHCO, (g) (C) 0.10 1 0829 14.38g 0.20 2 0.929 14.29g 0.35 1.00g 3 14.25g 0.50 1.14g 14.29 Experiment 1 Data Table 2: Moles of HC2H3O2 Reaction Volume of Mass of Moles of HC₂H₂O₂ 5.0% Vinegar (g) (ML) 5.0 0.25 0042 mol 2 5.0 0.25 0042 mol 3 5.0 0.25 0042 mol 5.0 0.25 0042 mol Experiment 1 Data Table 3: Moles of NaHCO3 Reaction Mass of NaHCO (g) 10g 20g 35g 50g Experiment 1 Data Table 4: Theoretical Yield of CO₂ Reaction # 1 2 3 Experiment 1 Total mass before reaction (g) (D=A+C) 15.29 15.21g 15.25g 15.349 Exercise 1 Data Table 1 Data Table 2 Data Table 3 Data Table 4 Panel 1 Photo 1 Data Table 5 Exercise 1- Data Table 1 Data Table 2 DataTable 3 Data Table 4 Panel 1 Photo 1 Data Table 5 Exercise 1- Moles of NaHCO 0012 mol 0025 mol 0044 mol 0062 mol…arrow_forwardThe chemical reaction you investigated is a two-step reaction. What type of reaction occurs in each step? How did you determine your answer?arrow_forwardWhat is the relationship between the limiting reactant and theoretical yield of CO2?arrow_forward
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- 1. Show that the change in entropy for a fixed amount of ideal gas held at a constant temperature undergoing a volume change is given by the simple equation AS = NkB In Hint: Start with the equation M dS = du + (Œ) dv - Ž (#) an, dU du+av-dN; j=1 Why doesn't the equation for the entropy of an ideal gas depend on the strength of the intermolecular forces for the gas?arrow_forward2. Make an ice cube at 1 bar pressure by freezing an amount of liquid water that is 2 cm x 2 cm x 2 cm in volume. The density of liquid water at 0 °C is 1.000 g cm³ and the density of ice at 0 °C is 0.915 g cm³. Note that this difference in density is the reason your water pipes burst if they freeze and why you shouldn't forget to take your bottle of pop out of the freezer if you put it in there to try and cool it down faster. A. What is the work of expansion upon freezing? B. Is work done on the system or by the system?arrow_forwardI have a excitation/emission spectra of a quinine standard solution here, and I'm having trouble interpreting it. the red line is emission the blue line is excitation. i'm having trouble interpreting properly. just want to know if there is any evidence of raman or rayleigh peaks in the spectra.arrow_forward
- Give the major product of the following reaction. excess 1. OH, H₂O 1.OH H CH3CH2CH21 H 2. A.-H₂O Draw the molecule on the canvas by choosing buttons from the Tools (for bonds), Atoms, and Advanced Template toolbars. The single bond is active by default.arrow_forward2. Use Hess's law to calculate the AH (in kJ) for: rxn CIF(g) + F2(g) → CIF 3 (1) using the following information: 2CIF(g) + O2(g) → Cl₂O(g) + OF 2(g) AH = 167.5 kJ ΔΗ 2F2 (g) + O2(g) → 2 OF 2(g) 2C1F3 (1) + 202(g) → Cl₂O(g) + 3 OF 2(g) о = = -43.5 kJ AH = 394.1kJarrow_forwardci Draw the major product(s) of the following reactions: (3 pts) CH3 HNO3/H2SO4 HNO3/ H2SO4 OCH3 (1 pts)arrow_forward
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