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Topic: Nuclephilic Substitution Questions  (Read 2297 times)

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Offline bshinz

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Nuclephilic Substitution Questions
« on: May 09, 2018, 07:59:21 PM »
Hi,
I'm doing a first year chemistry course and I'm having a lot of trouble when it comes to all the substitution questions. Can anyone direct me to how you know the answers to these? Thanks!!

Offline wildfyr

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Re: Nuclephilic Substitution Questions
« Reply #1 on: May 09, 2018, 11:00:12 PM »
You are asking us to do your homework. A. This is a lot of questions to answer B. according to the forum rules (see red link at the top) you must show your own efforts before we will help you. A hint here: focus on a few that give you the most trouble, maybe we can get you on the right path.

Offline bshinz

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Re: Nuclephilic Substitution Questions
« Reply #2 on: May 10, 2018, 02:41:53 AM »
Sorry, my friend directed me to this website and I'm not familiar with the rules/where I should be posting. My bad!

In regards to the first question I uploaded:
It's my understanding that alkyl halides are really reactive due to the high polarity of the C-halogen bond. I understand that the process in this question is a nucleophilic substitution, where the nucleophile will substitute with the I. I'm unsure as to what the nucleophile is in this example. Can the neg charge on the O of -OCH3 act as the nucleophile here? How does the O only have a -1 charge if it is in a single bond with the carbon?

Offline Babcock_Hall

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Re: Nuclephilic Substitution Questions
« Reply #3 on: May 10, 2018, 09:33:47 AM »
Any atom with a lone pair of electrons can be a nucleophile.  The formal charge on oxygen is -1, as you can verify from a simple calculation.  I realize that sometimes curved arrows are drawn from the negative charge on an atom, but doing so misleads the unwary IMO.

To identify a good leaving group, look for something that would be a weak base, not a strong base, if it departed with the pair of electrons.  IMO that is a better way of looking at leaving groups in substitution reactions than bond polarity.

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