Compound 3 below, called mycoepoxydiene, has been isolated from a marine fungus and has been shown to possess anti-cancer andanti-inflammatory properties. It contains an unusual oxygen-bridged cyclooctadiene skeleton. Both enantiomers of this compoundwere made in the laboratory. (Org. Lett. 2002, 4, 2941-2943). Show the reagents you would use to convert compound 1 into compound2.(CCH31OTBDMS2A. conc. H₂SO4, heatD. 1) BH3-THF; 2) H₂O2, NaOHB. Br2OTBDMSE. HBrCH3C. H₂, PtManystepsF. t-BuOKH3COCO,,,The transformation above can be performed with some combination of the reagents listed below. Give the necessary reagents in thecorrect order, as a string of letters (without spaces or punctuation, such as "EBF"). If there is more than one correct solution, providejust one answer.3'HCH3CH3TDBMS = -Si-t-BuCH3

We have to carry out the below given transformation. The given reagents are Introduction: Reaction…

Compound 3 below, called mycoepoxydiene, has been isolated from a marine fungus and has been shown to possess anti-cancer and anti-inflammatory properties. It contains an unusual oxygen-bridged cyclooctadiene skeleton. Both enantiomers of this compound were made in the laboratory. (Org. Lett. 2002, 4, 2941-2943). Show the reagents you would use to convert compound 1 into compound 2. H3COCO,,, 1 OTBDMS CH3 2 A. conc. H₂SO4, heat D. 1) BH3-THF; 2) H₂O2, NaOH OTBDMS B. Br2 E. HBr CH3 C. H₂, Pt Many steps F. t-BuOK O 3 "O The transformation above can be performed with some combination of the reagents listed below. Give the necessary reagents in the correct order, as a string of letters (without spaces or punctuation, such as "EBF"). If there is more than one correct solution, provide just one answer. ""H CH3 CH3 TDBMS = -Si-t-Bu CH₂

Compound 3 below, called mycoepoxydiene, has been isolated from a marine fungus and has been shown to possess anti-cancer and anti-inflammatory properties. It contains an unusual oxygen-bridged cyclooctadiene skeleton. Both enantiomers of this compound were made in the laboratory. (Org. Lett. 2002, 4, 2941-2943). Show the reagents you would use to convert compound 1 into compound 2. H3COCO,,, 1 OTBDMS CH3 2 A. conc. H₂SO4, heat D. 1) BH3-THF; 2) H₂O2, NaOH OTBDMS B. Br2 E. HBr CH3 C. H₂, Pt Many steps F. t-BuOK O 3 "O The transformation above can be performed with some combination of the reagents listed below. Give the necessary reagents in the correct order, as a string of letters (without spaces or punctuation, such as "EBF"). If there is more than one correct solution, provide just one answer. ""H CH3 CH3 TDBMS = -Si-t-Bu CH₂

Organic Chemistry

8th Edition

ISBN:9781305580350

Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. Foote

Publisher:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. Foote

Chapter22: Reactions Of Benzene And Its Derivatives

Section: Chapter Questions

Problem 22.50P: Following is the structure of miconazole, the active antifungal agent in a number of...

See similar textbooks

Similar questions

An unknown compound A (molecular formula C7H14O) was treated with NaBH4 in CH3OH to form compound B (molecular formula C7H16O). Compound A has a strong absorption in its IR spectrum at 1716 cm−1. Compound B has a strong absorption in its IR spectrum at 3600–3200 cm−1. The 1H NMR spectra of A and B are given. What are the structures of A and B?
The reaction of 3,4-dimethyl-3-hexanol (3,4-dimethylhexan-3-ol) with HBr generates compound A as the major product. Treatment of compound A with a strong base gives two isomers of compound B as the major product, along with one isomer of compound C and one isomer of compound D as minor products, all of which have one double-bond equivalent. Identify compounds A, B, C, and D and give their names. By what mechanism does the reaction of 3,4-dimethyl-3-hexanol with HBr occur? By what mechanism does the reaction of A with strong base to form B occur? Propose reaction conditions for an alternative, one-step method for converting 3,4-dimethyl-3-hexanol directly to compound B.
Identify products A and B from the given 1H NMR data. Treatment of acetone [(CH3)2C=O] with dilute aqueous base forms B. Compound B exhibits four singlets in its 1H NMR spectrum at 1.3 (6 H), 2.2 (3 H), 2.5 (2 H), and 3.8 (1H) ppm. What is the structure of B?
An unknown compound A (molecular formula C7H14O) was treated withNaBH4 in CH3OH to form compound B (molecular formula C7H16O).Compound A has a strong absorption in its IR spectrum at 1716 cm−1.Compound B has a strong absorption in its IR spectrum at 3600−3200cm−1. The 1H NMR spectra of A and B are given. What are the structuresof A and B?
Compound A Br₂, H₂O Compound B (C8H15BrO) + enantiomer CH₂O O Compound C + enantiomer Draw the structure of Compound B (watch out for stereochemistry), and mechanisms for its formation from Compound A, and its conversion to Compound C.
Compound X is optically inactive and has the formula C 16H 16Br 2. On treatment with strong base, X gives hydrocarbon Y, C 16H 14. Compound Y absorbs 2 equivalents of hydrogen when reduced over a palladium catalyst and reacts with ozone to give two fragments. One fragment Z, is an aldehyde with formula C 7H 6O. The other fragment is glyoxal, (CHO)2. Which of the following answers is correct? Select all that are correct.
Compound A with molecular formula C6H10 has two peaks in its 1H NMR spectrum, both of which are singlets (with ratio 9 : 1). Compound A reacts with an acidic aqueous solution containing mercuric sulfate to form compound B, which gives a positive iodoform test and has an 1H NMR spectrum that shows two singlets (with ratio 3 : 1). Identify A and B.
Compound X (molecular formula C10H12O) was treated with NH2NH2, -OH to yield compound Y (molecular formula C10H14). Based on the 1H NMR spectra of X and Y given below, what are the structures of X and Y?
An unknown compound A of molecular formula C10H18O reacts with H2SO4 to form two compounds (B and C)of molecular formula C10H16. B and C both react with H2 in the presence of Pd-C to form decalin. Ozonolysis of B forms D, and ozonolysis of C forms a diketone E of molecular formula C10H16O2. Identify the structures of compounds A, B, C, and E.
Treatment of compound A (C8H17Br) with NaOCH2CH3 affords two constitutional isomers B and C. Ozonolysis of B affords CH2=O and (CH3CH2CH2)2C=O. Ozonolysis of C affords CH3CH2CH2COCH3 and CH3CH2CHO. What is the structure of A?
Piperazinone derivatives, such as compound 4, are being investigated for their potential use in the treatment of migraines as well as a variety of other diseases, including hypertension and sepsis. The following two-step synthetic procedure was developed as a general method for producing a variety of piperazinone derivatives (Tetrahedron Lett. 2009, 50, 3817-3819). Compound 1 is a readily available starting material that can be converted to compound 2 via an SN2 process. Reductive amination of compound 2 affords compound 3 which is short-lived and rapidly undergoes acid-catalyzed amide formation to the cyclic product 4a, piperazinone derivative. 1 H₂N. OMe Br NaHCO3 DMF 2 PhCH, NH NaBH3CN AcOH, heat 4 [3] Challenge Problem 22.100a Incorrect. OMe Draw the structure of compound 3. If it is a pair of diastereomers, draw both diastereomers. Do NOT use abbreviations such as Ph or OMe when drawing the structure(s). Draw Your Solution
Explain this observation: Ethyl 3-phenylpropanoate (C6H5CH2CH2CO2CH2CH3) reacts with electrophiles to afford ortho- and para-disubstituted arenes, but ethyl 3-phenylprop-2-enoate (C6H5CH=CHCO2CH2CH3) reacts with electrophiles to afford meta- disubstituted arenes.