
Concept explainers
(a)
Interpretation:
The structure of an L- aldopentose needs to be drawn.
Concept introduction:
Sugar molecules can be named as D or L sugars according to their most oxidized carbon at the top of the Fisher projection. Aldopentose is a five-carbon

Answer to Problem 20.27P
Explanation of Solution
Sugar molecules can be named as D or L sugars according to their most oxidized carbon at the top of the Fisher projection. The absolute configuration of a molecule can be explained using D and L configuration. In D- sugars, the OH group on the bottom chiral center points to the right while in L- sugars, the OH group on the bottom chiral center points to the left. Aldopentose is a five-carbon aldehyde sugar. Xylose, Arabinose and Ribose are few examples.
(b)
Interpretation:
The structure of a D- tetrose needs to be drawn.
Concept introduction:
Sugar molecules can name as D or L sugars according to their most oxidized carbon at the top of the Fisher projection. Aldotetrose is a four-carbon aldehyde sugar.

Answer to Problem 20.27P
Explanation of Solution
Sugar molecules can be named as D or L sugars according to their most oxidized carbon at the top of the Fisher projection. The absolute configuration of a molecule can be explained using D and L configuration. In D- sugars, the OH group on the bottom chiral center points to the right while if L-sugars, the OH group on the bottom chiral center points to the left. Aldotetrose is a four-carbon aldehyde sugar.
(c)
Interpretation:
The structure of a five-carbon alditol needs to be drawn.
Concept introduction:
Alditols are polyols. Alditols can be produced by reducing an aldehyde or a

Answer to Problem 20.27P
Explanation of Solution
Alditols are polyols. Alditols can be produced by reducing an aldehyde or a ketone group in a monosaccharide. In this reduction process, an aldehyde group or a ketone group converts into −CH2OH group. Xylitol is an example of an alditol.
For an example:
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Chapter 20 Solutions
EBK GENERAL, ORGANIC, & BIOLOGICAL CHEM
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- Assign all the carbonsarrow_forwardC 5 4 3 CI 2 the Righ B A 5 4 3 The Lich. OH 10 4 5 3 1 LOOP- -147.52 T 77.17 -45.36 200 190 180 170 160 150 140 130 120 110 100 90 80 70 60 50 40 30 20 10 ppm B -126.25 77.03 200 190 180 170 160 150 140 130 120 110 100 90 80 70 60 50 40 30 20 10 ppm 200 190 180 170 160 150 140 130 120 110 100 90 80 TO LL <-50.00 70 60 50 40 30 20 10 ppm 45.06 30.18 -26.45 22.36 --0.00 45.07 7.5 1.93 2.05 -30.24 -22.36 C A 7 8 5 ° 4 3 7.5 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 1.0 ppm 9 8 5 4 3 ཡི་ OH 10 2 7.5 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 5 4 3 2 that th 7 I 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 115 2.21 4.00 1.0 ppm 6.96 2.76 5.01 1.0 ppm 6.30 1.00arrow_forwardCurved arrows were used to generate the significant resonance structure and labeled the most significant contribute. What are the errors in these resonance mechanisms. Draw out the correct resonance mechanisms with an brief explanation.arrow_forward
- What are the: нсе * Moles of Hice while given: a) 10.0 ml 2.7M ? 6) 10.ome 12M ?arrow_forwardYou are asked to use curved arrows to generate the significant resonance structures for the following series of compounds and to label the most significant contributor. Identify the errors that would occur if you do not expand the Lewis structures or double-check the mechanisms. Also provide the correct answers.arrow_forwardhow to get limiting reactant and % yield based off this data Compound Mass 6) Volume(mL Ben zaphone-5008 ne Acetic Acid 1. Sam L 2-propanot 8.00 Benzopin- a col 030445 Benzopin a Colone 0.06743 Results Compound Melting Point (°c) Benzopin acol 172°c - 175.8 °c Benzoping to lone 1797-180.9arrow_forward
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