A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • one year ago

This question is about reactions for organic chemistry! Please help!

  • This Question is Open
  1. aaronq
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    where is the question?

  2. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Its an attachment

  3. aaronq
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    you didn't attach anything

  4. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Let me try attaching it here.

    1 Attachment
  5. Photon336
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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.

  6. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Thank you!

  7. Photon336
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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

  8. Photon336
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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.

  9. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.