Destroyer O chem57

[utdent20]

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can anyone explain that to me, no. 57 on o chem destroyer.. the answer is 3 bromocycloheptane why is not going to be 1 or 2 becaue ROOR makes it go to more subsitttued whcih is where the double bond is.. I don't know i think i might have asked this question before but obviously did not get a good grasp

 

[xaznxcountx]

For this rxn you need to focus on NBS and not worry too much about the ROOR, I'm not sure the specifics of the mechanism, but when one utilizes NBS to add a Br to an alkene structure, the double bond does not get broken. Instead the Br gets added to the allylic position so the compound retains its double bond character as well as have a Bromine added to its structure. I hope that helps.

 

[Electrons]

yea, ditto. allylic carbocation is more stable if i remember correctly.

 

[utdent20]

thanks xanzx you've explained very well..and easy. unlike most ppl on here who get techinical and then i can't understand.. is ther any other trick with any other reactions?? I did nto know this about NBS. now ROOR with HBrwould take it the the least substituted? and waht about C6H5P+Ch2/DMSO? what happens there.. basically what is the function of DMSO?

 

[xaznxcountx]

thanks xanzx you've explained very well..and easy. unlike most ppl on here who get techinical and then i can't understand.. is ther any other trick with any other reactions?? I did nto know this about NBS. now ROOR with HBrwould take it the the least substituted? and waht about C6H5P+Ch2/DMSO? what happens there.. basically what is the function of DMSO?


Hey glad I could help.

For HBr/ROOR you're going to get the Bromine added to the least substituted carbon. So for example:

CH3CH2CH2=CH2, with HBr/ROOR the bromine will be added to the least substituted carbon. Personally I think knowing that is sufficient enough to answer the question if it were presented on the DAT.

So the compound becomes: CH3CH2CH3CH2BR

-Now if the same compound (CH3CH2CH2=CH2) were treated with HBr

then the compound becomes CH3CH2CH2BRCH3 (now the bromine ends up on the more substituted carbon of the double bond)


As for (C6H5)3 P+ CH2- this is the Wittig Reaction. Look on Road Map 1 of destroyer (pg. 101 for my edition), and look what happens.

Basically the the alkane that I highlighted in red will substitute the O on the carbonyl carbon. Keep in mind that the double bond DOES NOT get altered in anyway.


Example: CH3CH2C=O use (C6H5)3 P+ CH2 ----> CH3CH2C=CH2

Hope that helps, good luck studying 😀

 

[utdent20]

ok i get. it. thanks!

 

[xaznxcountx]

O yeah as for DMSO, it is a polar aprotic solvent meaning it will favor SN2