
Concept explainers
(a)
Interpretation: The product formed by the treatment of
Concept introduction: Monosaccharides are the small units of simple sugars. The hydroxyl groups of monosaccharides are converted into the ether groups in presence of base and

Answer to Problem 28.49P
The product formed by the treatment of
Figure 1
Explanation of Solution
The conversion of hydroxyl groups of
Figure 2
The product formed by the treatment of
(b)
Interpretation: The product formed by the treatment of
Concept introduction: Monosaccharides are the small units of simple sugars The hydroxyl groups of monosaccharides are converted into the ether groups in presence of base and alkyl halide.

Answer to Problem 28.49P
The products formed by the treatment of
Figure 3
Explanation of Solution
The two anomers are formed by the treatment of monosaccharides with methanol in the presence of hydrochloric acid. One product with methoxy group on the equatorial position and other product with methoxy group on the axial position. The corresponding chemical reaction is shown below.
Figure 4
The products formed by the treatment of
(c)
Interpretation: The product formed by the treatment of
Concept introduction: Monosaccharides are the small units of simple sugars. The hydroxyl groups of monosaccharides are converted into the ether groups in presence of base and alkyl halide.

Answer to Problem 28.49P
The product formed by the treatment of
Figure 5
Explanation of Solution
In the presence of acetic anhydride and pyridine, the conversion of alcoholic groups in
Figure 6
The product formed by the treatment of
(d)
Interpretation: The product formed by the treatment of
Concept introduction: Monosaccharides are the small units of simple sugars. The hydroxyl groups of monosaccharides are converted into the ether groups in presence of base and alkyl halide.

Answer to Problem 28.49P
The products formed by the treatment of
Figure 7
Explanation of Solution
The first step is the conversion of hydroxyl groups of gulose into the ether groups in presence of base and methyl iodide as shown in Figure 2.
The product
Figure 8
The products formed by the treatment of
(e)
Interpretation: The product formed by the treatment of
Concept introduction: Monosaccharides are the small units of simple sugars The hydroxyl groups of monosaccharides are converted into the ester groups in presence of base and acetic anhydride or acetyl chloride.

Answer to Problem 28.49P
The products formed by the treatment of
Figure 9
Explanation of Solution
The conversion of alcoholic groups of product
Figure 10
The products formed by the treatment of
(f)
Interpretation: The product formed by the treatment of
Concept introduction: Monosaccharides are the small units of simple sugars. The hydroxyl groups of monosaccharides are converted into the ether groups in presence of base and alkyl halide.

Answer to Problem 28.49P
The products formed by the treatment of
Figure 11
Explanation of Solution
The conversion of hydroxyl groups of product
Figure 12
The products formed by the treatment of
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Chapter 28 Solutions
ORGANIC CHEMISTRY
- ● Biological Macromolecules Identifying the parts of a disaccharide Take a look at this molecule, and then answer the questions in the table below it. CH2OH O H H H OH OH OH H H CH2OH H O OH H OH H H H H OH Is this a reducing sugar? Does this molecule contain a glycosidic bond? If you said this molecule does contain a glycosidic bond, write the symbol describing it. If you said this molecule does contain a glycosidic bond, write the common names (including anomer and enantiomer labels) of the molecules that would be released if that bond were hydrolyzed. If there's more than one molecule, separate each name with a comma. Explanation Check O yes X O no ○ yes O no Uarrow_forwardThe aim of the lab is to measure the sodium content from tomato sauce using the Mohr titration method. There are two groups being: Regular Tomato sauce & Salt Reduced tomato sauce QUESTION: State how you would prepare both Regular & Salt reduced tomato sauce samples for chemical analysis using the Mohr titration methodarrow_forwardUsing the conditions of spontaneity to deduce the signs of AH and AS Use the observations about each chemical reaction in the table below to decide the sign (positive or negative) of the reaction enthalpy AH and reaction entropy AS. Note: if you have not been given enough information to decide a sign, select the "unknown" option. reaction observations conclusions A The reverse of this reaction is always spontaneous but proceeds faster at temperatures above -48. °C. ΔΗ is (pick one) ✓ AS is (pick one) B This reaction is spontaneous except below 114. °C but proceeds at a slower rate below 135. °C. ΔΗ is (pick one) AS is (pick one) ΔΗ is C This reaction is exothermic and proceeds faster at temperatures above -43. °C. (pick one) AS is (pick one) v Х 5 ? 18 Ararrow_forward
- ion. A student proposes the following Lewis structure for the perchlorate (CIO) io : :0: : Cl : - - : :0: ك Assign a formal charge to each atom in the student's Lewis structure. atom central O formal charge ☐ top O ☐ right O ☐ bottom O ☐ Cl ☐arrow_forwardDecide whether these proposed Lewis structures are reasonable. proposed Lewis structure Yes. Is the proposed Lewis structure reasonable? Cl- : 2: :Z: :Z: N—N : 0: C C1: O CO No, it has the wrong number of valence electrons. The correct number is: ☐ No, it has the right number of valence electrons but doesn't satisfy the octet rule. The symbols of the problem atoms are:* ☐ Yes. No, it has the wrong number of valence electrons. The correct number is: ☐ No, it has the right number of valence electrons but doesn't satisfy the octet rule. The symbols of the problem atoms are:* | Yes. No, it has the wrong number of valence electrons. The correct number is: No, it has the right number of valence electrons but doesn't satisfy the octet rule. The symbols of the problem atoms are:* | If two or more atoms of the same element don't satisfy the octet rule, just enter the chemical symbol as many times as necessary. For example, if two oxygen atoms don't satisfy the octet rule, enter "0,0". ☑arrow_forwardUse the observations about each chemical reaction in the table below to decide the sign (positive or negative) of the reaction enthalpy AH and reaction entropy AS. Note: if you have not been given enough information to decide a sign, select the "unknown" option. reaction observations conclusions ΔΗ is (pick one) A This reaction is faster above 103. °C than below. AS is (pick one) ΔΗ is (pick one) B This reaction is spontaneous only above -9. °C. AS is (pick one) ΔΗ is (pick one) C The reverse of this reaction is always spontaneous. AS is (pick one) 18 Ararrow_forward
- Use the observations about each chemical reaction in the table below to decide the sign (positive or negative) of the reaction enthalpy AH and reaction entropy AS. Note: if you have not been given enough information to decide a sign, select the "unknown" option. reaction observations conclusions A The reverse of this reaction is always spontaneous but proceeds slower at temperatures below 41. °C. ΔΗ is (pick one) AS is (pick one) ΔΗ is (pick one) B This reaction is spontaneous except above 94. °C. AS is (pick one) This reaction is always spontaneous, but ΔΗ is (pick one) C proceeds slower at temperatures below −14. °C. AS is (pick one) Х 00. 18 Ar 무ㅎ B 1 1arrow_forwardDraw the product of the reaction shown below. Ignore inorganic byproducts. + H CH3CH2OH HCI Drawingarrow_forwardplease explain this in simple termsarrow_forward
- K Most Reactive Na (3 pts) Can the metal activity series (shown on the right) or a standard reduction potential table explain why potassium metal can be prepared from the reaction of molten KCI and Na metal but sodium metal is not prepared from the reaction of molten NaCl and K metal? Show how (not). Ca Mg Al с Zn Fe Sn Pb H Cu Ag Au Least Reactivearrow_forward(2 pts) Why is O2 more stable as a diatomic molecule than S2?arrow_forwardDraw the Lewis structure for the polyatomic phosphite (PO¾³¯) a anion. Be sure to include all resonance structures that satisfy the octet rule. C I A [ ]¯arrow_forward
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