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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EBK ORGANIC CHEMISTRY: PRINCIPLES AND M
- 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.arrow_forwardWhat is important about this slide? what is reactant and what is product?what reactant to give what?what principle is important here?what does it means that acetal can not be reduced?which one is acetal? explain the mechanism please? Lable out which is reactant and which is product please. I am totally confusedarrow_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. LL :0: F :0: H H 0. Select to Add Arrows :0: < H H H MacBooarrow_forward
- My question is asked on the picture that I am uploading, I sincerely hope Bartleby can help me more than Chegg ever did! Thank you! I need help finding the major products and minor products of the written reaction!arrow_forwardShown below is a carbocation intermediate in an electrophilic addition reaction of HCl with an alkene. For the conformation shown, select each hydrogen whose bond to carbon is aligned for effective hyperconjugation with the vacant p orbital on the positively charged carbon. Each adjacent carbon will have only one C-H bond so aligned. I need to click on the Hydrogens which then highlights the clicked hydrogens in blue. Much help needed. There are hydrogens and carbons shown.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 steps. Be sure to account for all bond-breaking and bond-making steps.arrow_forward
- Solve correctly please.arrow_forwardTROVIOW Topicaj (References) Draw the major organic product of the reaction shown below. HO, NANH2 You do not have to consider stereochemistry. • You do not have to explicitly draw H atoms. • Include cationic counter-ions, e.g., Na" in your answer, but draw them in their own sketcher. (Review Top Draw the major organic product of the reaction shown below. OH K2Cr207 H2SO4, H20 • You do not have to consider stereochemistry. • You do not have to explicitly draw H atoms. • In cases where there is more than one answer, just draw one.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
- When piperidine undergoes the series of reactions shown here, 1,4-pentadiene is obtained as the product. When the four different methyl- substituted piperidines undergo the same series of reactions, each forms a different diene: 1,5-hexadiene; 1,4-pentadiene; 2-methyl-1,4-pentadiene; and 3-methyl-1,4-pentadiene. Which methyl-substituted piperidine forms which diene?arrow_forward[Review Topics] [References) Draw a structural formula for the major product of the acid-base reaction shown. H2N. HCI (1 mole) (1 mole) You do not have to consider stereochemistry. • Do not include counter-ions, e.g., Na", I, in your answer. • In those cases in which there are two reactants, draw only the product from the compound that reacts. C P. opy eatearrow_forwardThe electrophilic addition reaction shown below involves a carbocation rearrangement. For the mechanism step below, draw curved arrows to show electron reorganization. Use the arrow tool to specify the origin and the destination of the reorganizing electrons. Consult the arrow-pushing instructions for the convention on regiospecific electrophilic attack on a double bond. I understand how it's supposed to be laid out. I'm just unsure of where the first arrow comes from as there are a lot of positions. I have attached examples on what positions I mean. Thank you.arrow_forward
- Organic Chemistry: A Guided InquiryChemistryISBN:9780618974122Author:Andrei StraumanisPublisher:Cengage Learning