Concept explainers
Interpretation:
The distinguishing test for the given compounds is to be determined.
Concept Introduction:
▸ The pyranoses are six-membered monosaccharide ring structures. The name signifies heterocycle
▸ Pyranoses are the hemiacetals having ring structures of the
▸ Benedict’s test is a typical chemical method used to oxidize aldoses and ketoses. This particular test can be used as a colorimetric analysis tool and a positive or negative test can help in the distinction of different types of carbohydrates. Tollen’s reagent, which is composed of silver nitrate and ammonia can also selectively oxidize
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Organic Chemistry
- Identify A, B, C, and D in the preceding problem if D is oxidized to an optically inactive aldaric acid; if A, B, and C are oxidized to optically active aldaric acids; and if interchanging the aldehyde and alcohol groups of A leads to a different sugar.arrow_forward(1) what reagent is used to convert p-amino phenol to quinone (2) Draw two different kekule's structures for pyrene and under line the more stablearrow_forwardThe cyclic hemiacetal is more stable than the open-chain form, so very little of the open-chain form is present atequilibrium. Will an aqueous solution of glucose reduce Tollens reagent and give a positive Tollens test? Explain.arrow_forward
- 19 & 20. 19. 20. Consider the following synthetic estrogen shown below O= H Im Compound X The maximum number of stereoisomers in X is Illllllll CH3 a OH CH3 Draw compound X below and prioritize the groups present in the stereocenter labeled as a.arrow_forwardAnswer ALL parts of this question. (a) Provide the structure and name of the aldehyde with the formula, C2H4O. (b) Draw the two steps of the Strecker synthesis used for the conversion of the aldehyde in part (a) into racemic alanine. (c) Draw the (R) and (S) enantiomers of alanine. Which is the naturally occurring enantiomer? (d) Draw the structure of the dipeptide, Ala-Gly.arrow_forward22.47 Tertiary amines with three different alkyl groups are chiral but cannot be resolved because pyramidal inversion causes racemization at room temperature. Nevertheless, chiral aziridines can be resolved and stored at room temperature. Aziridine is a three-membered heterocycle containing a nitrogen atom. The following is an example of a chiral aziridine. In this compound, the nitrogen atom is a chiral center. Suggest a reason why chiral aziridines do not undergo racemization at room temperature.arrow_forward
- Synthesize octadecyl stearate using linoleic acid (18:2) (Given, stearic acid: 18:0) as the only starting material.arrow_forwardA chiral amine A having the R conguration undergoes Hofmann elimination to form an alkene B as the major product. B is oxidatively cleaved with ozone, followed by CH3SCH3, to form CH2 = O and CH3CH2CH2CHO. What are the structures of A and B?arrow_forward19.62 Naltrexone is used to help recovering narcotic addicts stay drug free. HO O OH naltrexone LO1,6,7 (a) Label and name all the functional groups. Where relevant, indicate whether the group is primary, secondary or tertiary. (b) Upon addition of dilute hydrochloric acid solution, naltrexone forms a water-soluble salt. Draw the structure of this salt.arrow_forward
- Arrange each group of compounds in order of increasing acidity.(b) p-toluenesulfonic acid, acetic acid, chloroacetic acidarrow_forwardAn aldose A, is reduced by sodium borohydride to an optically inactive alditol B. The Ruff degredation of A forms C. Oxidation of C by nitric acid generates the optically inactive diacid D. The ruff degreation of C forms D-glyceraldehyde. draw the structures for compounds A through D and discuss a mechanism for the reduction step using sodium borohydridearrow_forward24.82 When the compound shown here is heated, ethene gas is evolved and a product with the formula C14H3O2 is formed. The 'H NMR and 13C NMR spectra of C14H8O2 are shown below. (There are two signals >150 ppm in the 13C NMR spectrum. Recall that the 13C NMR signal at 77 ppm is from the CDCI3 solvent.) (a) Draw the structure of C,14H3O2. (b) Draw the mechanism that accounts for its formation. (c) What is the main driving force that favors the products of this reaction? ? C14H3O2 C14H8O2 9. 8 7 6 4 3 2 1 250 200 150 100 50 Chemical shift (ppm) Chemical shift (ppm)arrow_forward
- Chemistry for Today: General, Organic, and Bioche...ChemistryISBN:9781305960060Author:Spencer L. Seager, Michael R. Slabaugh, Maren S. HansenPublisher:Cengage LearningOrganic ChemistryChemistryISBN:9781305580350Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. FootePublisher:Cengage Learning