anonymous
  • anonymous
This question is about reactions for organic chemistry! Please help!
Chemistry
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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aaronq
  • aaronq
where is the question?
anonymous
  • anonymous
Its an attachment
aaronq
  • aaronq
you didn't attach anything

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anonymous
  • anonymous
Let me try attaching it here.
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Photon336
  • Photon336
b) Which compound will undergo an SN2 reaction with NACN most rapidly? SN2 = back side attack along with an inversion of stereochemistry, but for SN2 means substitution nucleophilic bi molecular, meaning that the overall reaction rate depends on the strength of your nucleophile and the leaving group leaving, hence 2 molecules. in order for a backside attack to occur effortlessly in a single step, you must have a molecule that has the least amount of steric hindrance so that the nucleophile can come in from the back side. in molecules III and IV they would be 3* and less favorable. compound II is 2* while compound I* is primary. so I would say that the answer is compound I for this question b.c it has the least amount of steric hindrance.
anonymous
  • anonymous
Thank you!
Photon336
  • Photon336
Solvolysis is an SN1 reaction where your nucleophile is actually your solvent. so the rules for SN1 apply, which are the opposite of those for SN2. in SN1 we have substitution nucleophilic uni molecular which means that the rate will be determined solely by our leaving groups ability to leave. r = k[LG]. so there are things we can do to help make this process go more smoothly. we can add a polar protic solvent which will interact with the leaving group to stabilize it so that the reverse reaction doesn't happen. which in this case MeOH is polar and protic. the second thing is that we will form a carbocation intermediate and the stability of this intermediate depends on how many groups you have so 3*>2*>1* so immediately I and II will be out. both III and IV are tertiary however, if you look at structure four and remove the bromine group you can see that there's a positive charge adjacent to a double bond. whenever you have a case like this a positive charge adjacent to a double bond, it is stabilized by resonance and that positive charge is in a way delocalized, so that would make IV take precedence over structure III so the answer is IV
Photon336
  • Photon336
I didn't mean to say "delocalized" that has to do with electrons the correct way to say it would be that the positive charge is spread out and hence more stable.

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