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Concept explainers
(a)
Interpretation: The term monosaccharide, disaccharide, reducing sugar, anomers, enantiomer and aldohexoses whether applies to both the members of the given pair or not has to be stated.
Concept introduction: Monosaccharides are those type of sugars which cannot be hydrolysed further.
Disaccharides are those type of sugars which have two units of monosaccharides.
The enantiomers are the molecules that form non-superimposable mirror images of each other.
Carbon atom which behaves as a stereocenter in the cyclic form of the hemiacetal compound is known as anomeric carbon. The molecule containing the anomeric carbon is called anomer.
The monosaccharides that have
The sugars that have a free carbonyl group are called reducing sugars.
(b)
Interpretation: The term monosaccharide, disaccharide, reducing sugar, anomers, enantiomer and aldohexoses whether applies to both the members of the given pair or not has to be stated.
Concept introduction: Monosaccharides are those type of sugars which cannot be hydrolysed further.
Disaccharides are those type of sugars which have two units of monosaccharides.
The enantiomers are the molecules that form non-superimposable mirror images of each other.
Carbon atom which behaves as a stereocenter in the cyclic form of the hemiacetal compound is known as anomeric carbon. The molecule containing the anomeric carbon is called anomer.
The monosaccharides that have aldehyde as a functional group and a long chain of six carbon atoms are called aldohexoses.
The sugars that have a free carbonyl group are called reducing sugars.
(c)
Interpretation: The term monosaccharide, disaccharide, reducing sugar, anomers, enantiomer and aldohexoses whether applies to both the members of the given pair or not has to be stated.
Concept introduction: Monosaccharides are those type of sugars which cannot be hydrolysed further.
Disaccharides are those type of sugars which have two units of monosaccharides.
The enantiomers are the molecules that form non-superimposable mirror images of each other.
Carbon atom which behaves as a stereocenter in the cyclic form of the hemiacetal compound is known as anomeric carbon. The molecule containing the anomeric carbon is called anomer.
The monosaccharides that have aldehyde as a functional group and a long chain of six carbon atoms are called aldohexoses.
The sugars that have a free carbonyl group are called reducing sugars.
(d)
Interpretation: The term monosaccharide, disaccharide, reducing sugar, anomers, enantiomer and aldohexoses whether applies to both the members of the given pair or not has to be stated.
Concept introduction: Monosaccharides are those type of sugars which cannot be hydrolysed further.
Disaccharides are those type of sugars which have two units of monosaccharides.
The enantiomers are the molecules that form non-superimposable mirror images of each other.
Carbon atom which behaves as a stereocenter in the cyclic form of the hemiacetal compound is known as anomeric carbon. The molecule containing the anomeric carbon is called anomer.
The monosaccharides that have aldehyde as a functional group and a long chain of six carbon atoms are called aldohexoses.
The sugars that have a free carbonyl group are called reducing sugars.
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Chapter 7 Solutions
Organic And Biological Chemistry
- 5) Confidence interval. Berglund and Wichardt investigated the quantitative determination of Cr in high-alloy steels using a potentiometric titration of Cr(VI). Before the titration, samples of the steel were dissolved in acid and the chromium oxidized to Cr(VI) using peroxydisulfate. Shown here are the results (as %w/w Cr) for the analysis of a reference steel. 16.968, 16.922, 16.840, 16.883, 16.887, 16.977, 16.857, 16.728 Calculate the mean, the standard deviation, and the 95% confidence interval about the mean. What does this confidence interval mean?arrow_forwardIn the Nitrous Acid Test for Amines, what is the observable result for primary amines? Group of answer choices nitrogen gas bubbles form a soluble nitrite salt yellow oily layer of nitrosoaminearrow_forward3. a. Use the MS to propose at least two possible molecular formulas. For an unknown compound: 101. 27.0 29.0 41.0 50.0 52.0 55.0 57.0 100 57.5 58.0 58.5 62.0 63.0 64.0 65.0 74.0 40 75.0 76.0 20 20 40 60 80 100 120 140 160 180 200 220 m/z 99.5 68564810898409581251883040 115.0 116.0 77404799 17417M 117.0 12.9 118.0 33.5 119.0 36 133 0 1.2 157.0 2.1 159.0 16 169.0 219 170.0 17 171.0 21.6 172.0 17 181.0 1.3 183.0 197.0 100.0 198.0 200. 784 Relative Intensity 2 2 8 ō (ppm) 6 2arrow_forward
- Solve the structure and assign each of the following spectra (IR and C-NMR)arrow_forward1. For an unknown compound with a molecular formula of C8H100: a. What is the DU? (show your work) b. Solve the structure and assign each of the following spectra. 8 6 2 ō (ppm) 4 2 0 200 150 100 50 ō (ppm) LOD D 4000 3000 2000 1500 1000 500 HAVENUMBERI -11arrow_forward16. The proton NMR spectral information shown in this problem is for a compound with formula CioH,N. Expansions are shown for the region from 8.7 to 7.0 ppm. The normal carbon-13 spec- tral results, including DEPT-135 and DEPT-90 results, are tabulated: 7 J Normal Carbon DEPT-135 DEPT-90 19 ppm Positive No peak 122 Positive Positive cus и 124 Positive Positive 126 Positive Positive 128 No peak No peak 4° 129 Positive Positive 130 Positive Positive (144 No peak No peak 148 No peak No peak 150 Positive Positive してしarrow_forward
- 3. Propose a synthesis for the following transformation. Do not draw an arrow-pushing mechanism below, but make sure to draw the product of each proposed step (3 points). + En CN CNarrow_forwardShow work..don't give Ai generated solution...arrow_forwardLabel the spectrum with spectroscopyarrow_forward
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