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In some nucleophilic substitutions under
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- When (R)-6-bromo-2,6-dimethylnonane is dissolved in CH3OH, nucleophilic substitution yields an optically inactive solution. When the isomeric halide (R)-2-bromo-2,5-dimethylnonane is dissolved in CH3OH under the same conditions, nucleophilic substitution forms an optically active solution. Draw the products formed in each reaction, and explain why the difference in optical activity is observed.arrow_forwardWhen treated with NaOH, the bromide below gives an alkene by the E2 mechanism, by eliminationof the circled atoms: (a) Draw the Newman projection from which elimination takes place. (b) Draw the mechanism. (c) Draw the product with the proper stereochemistry. (d) Assign the proper stereochemical descriptor to the product. (Z, E? Trans,CiS?) (e)What would happen to the stereochemistry of the product if the enantiomer of the starting material were used inthe reaction?arrow_forwardWhen (ft)-6-bromo-2,6-dimethylnonane is dissolved in CH3OH, nucleophilic substitution yields an optically inactive solution. When the isomeric halide (fl)-2-bromo-2,5- dimethylnonane is dissolved in CH3OH under the same conditions, nucleophilic substitution forms an optically active solution. Draw the products formed in each reaction, and explain why the difference in optical activity is observed.arrow_forward
- When tert-butyl bromide is heated with an equal amount of ethanol in an inert solvent, one of the products is ethyl tert-butyl ether. (a) What happens to the reaction rate if the concentration of ethanol is doubled?arrow_forwardThis is not A or C. The second one is not A or Darrow_forwardConsider the tetracyclic compound with rings labeled A–D. (a) Which ring is the most reactive in electrophilic aromatic substitution? (b) Which ring is the least reactive in electrophilic aromatic substitution?arrow_forward
- Consider the following reaction scheme: (i) Provide the reagent(s) required for this transformation. (ii) Draw a curved arrow mechanism and comment on the stereoselectivity.arrow_forward(a) Draw all products formed by treatment of each dibromide (A and B) with one equivalent of NaNH2. (b) Label pairs of diastereomers and constitutional isomers. H Br H Br CH3 CH3 Br H H Br Aarrow_forwardConsider the tetracyclic aromatic compound drawn below, with rings labeled as A, B, C, and D. (a) Which of the four rings is most reactive in electrophilic aromatic substitution? (b) Which of the four rings is least reactive in electrophilic aromatic substitution? (c) What are the major product(s) formed when this compound is treated with one equivalent of Br2?arrow_forward
- For following substituted benzenes: [1] C6H5Br; [2] C6H5CN; [3] C6H5OCOCH3: Does the substituent activate or deactivate the benzene ring inelectrophilic aromatic substitution?arrow_forwardShow how you would convert hex-1-ene to each of the following compounds. You may use any additional reagents andsolvents you need.( (b) 1-methoxyhexanearrow_forwardExplain 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.arrow_forward
- Organic ChemistryChemistryISBN:9781305580350Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. FootePublisher:Cengage Learning