Concept explainers
Interpretation:
To complete each of the hydroboration steps by adding the curved arrows, when borane add to propene to produce two different products, and to indicate the pertinent steric repulsion that is present in each reaction, as well as the partial charges that develop in the transition state, similar to Equations 12-36 and 12-37, and to determine which reaction is favored.
Concept introduction:
Hydroboration is stereospecific with the H and
In the transition states a partial positive charge appears on the C atom that has a partial bond to H. This is because the step is driven by the formation of a bond between the electron-rich alkene and the electron-poor B atom, so the C-B bond is formed to a greater extent in the transition state than the C-H bond. The transition state is more stable when the
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Organic Chemistry: Principles And Mechanisms
- Complete the mechanism of hydration of alkene below. Please follow curved arrows to predict the resultingstructures in each step.arrow_forwardCurved arrows are used to illustrate the flow of electrons. Using the provided starting and product structures, draw the curved electron-pushing arrows for the following reaction or mechanistic step(s). Be sure to account for all bond-breaking and bond-making steps. Na → 0:0-CHa NaOCH3, CH3OH Select to Add Arrows неон Q H3O+ Na → a Select to Add Arrows H Select to Add Arrows Harrow_forwardCurved arrows are used to illustrate the flow of electrons. Using the provided starting and product structures, draw the curved electron-pushing arrows for the following reaction or mechanistic steps. Be sure to account for all bond- breaking and bond-making steps. HH H H H H :0 Select to Add Arrows H H H H I I H3O+ heat H3O+ heat I HH H H3O+ Select to Add Arrows heat H HH H H H :0: HH H Select to Add Arrows H I I H HIarrow_forward
- Curved arrows are used to illustrate the flow of electrons. Using the provided starting and product structures, draw the curved electron-pushing arrows for the following reaction or mechanistic step(s). Be sure to account for all bond-breaking and bond-making steps. Drawing Arrows N :Br: NaⒸ N: C:O NaⒸ Br.arrow_forwardPlease show reaekson and don't use hend raitingarrow_forwardCurved arrows are used to illustrate the flow of electrons. Using the provided starting and product structures, draw the curved electron-pushing arrows for the following reaction or mechanistic step(s). Be sure to account for all bond-breaking and bond-making steps.arrow_forward
- Curved arrows are used to illustrate the flow of electrons. Using the provided starting and product structures, draw the curved electron-pushing arrows for the following reaction or mechanistic step(s). Be sure to account for all bond-breaking and bond-making steps. LL :0: F :0: H H 0. Select to Add Arrows :0: < H H H MacBooarrow_forwardSee image belowarrow_forwardCurved arrows are used to illustrate the flow of electrons. Using the provided starting and product structures, draw the curved electron-pushing arrows for the following reaction or mechanistic step(s). Be sure to account for all bond-breaking and bond-making steps. Then draw the organic product of the reaction. Include all lone pairs in the structures. Ignore inorganic byproducts and + counterions. Drawing Arrows NaNH2 Select to Draw Product H H I I O:Z: + S :O: H Undo H Na+ EP Resetarrow_forward
- Curved arrows are used to illustrate the flow of electrons. Using the provided starting and product structures, draw the curved electron-pushing arrows for the following reaction or mechanistic steps. Be sure to account for all bond-breaking and bond-making steps.arrow_forwardCurved arrows are used to illustrate the flow of electrons. Using the provided starting and product structures, draw the curved electron-pushing arrows for the following reaction or mechanistic step(s). Be sure to account for all bond-breaking and bond-making steps.arrow_forwardMy question is about part C of the question shown. The answer is shown and then I drew an alternative with a red arrow. The question asks to draw a halogen that will ONLY form the alkene shown, not most of the time or sometimes, as many of our questions ask us to draw all the possible alkenese, even unlikely ones. This one has to be ONLY that one. My question is why the structure on the far right does NOT lead to a double bond between the two rings. There is a hydrogen on the opposite side of the Cl, isn't there? If there's not a hydrogen there, why not? What am I not seeing? Or if there is a hydrogen but that bond can't be made, why not?arrow_forward
- Organic Chemistry: A Guided InquiryChemistryISBN:9780618974122Author:Andrei StraumanisPublisher:Cengage Learning