(a)
Interpretation:
The lactone group in each structure that merits the classification as macrolides is to be stated.
Concept introduction:
Erythromycin A is an antibiotic drug used in the treatment of many bacterial infections especially which causes infections on the skin and respiratory tract. Azithromycin is another antibiotic drug useful in the treatment of bacterial infections like ear infections, pneumonia and intestinal infections.
(b)
Interpretation:
The
Concept introduction:
Erythromycin A is an antibiotic drug used in the treatment of many bacterial infections especially which causes infections on the skin and respiratory tract. Azithromycin is another antibiotic drug useful in the treatment of bacterial infections like ear infections, pneumonia and intestinal infections.
(c)
Interpretation:
The
Concept introduction:
Erythromycin A is an antibiotic drug used in the treatment of many bacterial infections especially which causes infections on the skin and respiratory tract. Azithromycin is another antibiotic drug useful in the treatment of bacterial infections like ear infections, pneumonia and intestinal infections.
(d)
Interpretation:
The
Concept introduction:
Erythromycin A is an antibiotic drug used in the treatment of many bacterial infections especially which causes infections on the skin and respiratory tract. Azithromycin is another antibiotic drug useful in the treatment of bacterial infections like ear infections, pneumonia and intestinal infections.
(e)
Interpretation:
The reason as to why an amine is considered as a good choice to be the “chemical opposite of a ketone” is to be stated.
Concept introduction:
Erythromycin A is an antibiotic drug used in the treatment of many bacterial infections especially which causes infections on the skin and respiratory tract. Azithromycin is another antibiotic drug useful in the treatment of bacterial infections like ear infections, pneumonia and intestinal infections.
Want to see the full answer?
Check out a sample textbook solutionChapter 19 Solutions
Organic Chemistry Plus Masteringchemistry With Pearson Etext, Global Edition
- Shown below is the major resonance structure for a molecule. Draw the second best resonance structure of the molecule. Include all non-zero formal charges. H. H. +N=C H H H Cl: Click and drag to start drawing a structure. : ? g B S olo Ar B Karrow_forwardDon't used hand raitingarrow_forwardS Shown below is the major resonance structure for a molecule. Draw the second best resonance structure of the molecule. Include all non-zero formal charges. H H = HIN: H C. :0 H /\ H H Click and drag to start drawing a structure. ×arrow_forward
- Part II. two unbranched ketone have molecular formulla (C8H100). El-ms showed that both of them have a molecular ion peak at m/2 =128. However ketone (A) has a fragment peak at m/2 = 99 and 72 while ketone (B) snowed a fragment peak at m/2 = 113 and 58. 9) Propose the most plausible structures for both ketones b) Explain how you arrived at your conclusion by drawing the Structures of the distinguishing fragments for each ketone, including their fragmentation mechanisms.arrow_forwardPart V. Draw the structure of compound tecla using the IR spectrum Cobtained from the compound in KBr pellet) and the mass spectrum as shown below. The mass spectrum of compound Tesla showed strong mt peak at 71. TRANSMITTANCE LOD Relative Intensity 100 MS-NW-1539 40 20 80 T 44 55 10 15 20 25 30 35 40 45 50 55 60 65 70 75 m/z D 4000 3000 2000 1500 1000 500 HAVENUMBERI-11arrow_forwardTechnetium is the first element in the periodic chart that does not have any stable isotopes. Technetium-99m is an especially interesting and valuable isotope as it emits a gamma ray with a half life ideally suited for medical tests. It would seem that the decay of technetium should fit the treatment above with the result In(c/c) = -kt. The table below includes data from the two sites: http://dailymed.nlm.nih.gov/dailymed/druginfo.cfm?id=7130 http://wiki.medpedia.com/Clinical: Neutrospec_(Technetium_(99m Tc)_fanolesomab). a. b. C. Graph the fraction (c/c.) on the vertical axis versus the time on the horizontal axis. Also graph In(c/c.) on the vertical axis versus time on the horizontal axis. When half of the original amount of starting material has hours fraction remaining disappeared, c/c = ½ and the equation In(c/c.) = -kt becomes In(0.5) = -kt1/2 where t₁₂ is the half life (the time for half of the material to decay away). Determine the slope of your In(c/c.) vs t graph and…arrow_forward
- Organic ChemistryChemistryISBN:9781305580350Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. FootePublisher:Cengage LearningIntroduction to General, Organic and BiochemistryChemistryISBN:9781285869759Author:Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar TorresPublisher:Cengage Learning
- Organic Chemistry: A Guided InquiryChemistryISBN:9780618974122Author:Andrei StraumanisPublisher:Cengage LearningChemistry for Today: General, Organic, and Bioche...ChemistryISBN:9781305960060Author:Spencer L. Seager, Michael R. Slabaugh, Maren S. HansenPublisher:Cengage Learning