ochem Q, please help

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What is the product?

ICH2CH2I + Mg(ether) --> ???

the answer is CH2CH2

can someone explain please?
 
Cliff DAT 🙁

I got that question wrong too!

Only chem q that I got wrong on that practice test.

Cliff DAT prep book is too easy in difficulty, right?
 
yea i thought it was too good to be true too. i wonder if the real test is this way.

can you explain the answer?>
 
Strong base (Mg and ether = grignard reagent) so perform a double E2 reaction i believe.
 
with ether, 2 moles of grignard react with the reactant, thus Mg+ react with halogen (I-) and form a good leaving group. eventually it leaves with CH2CH2.
That's what I think, Correct me plz if I'm wrong.
 
with ether, 2 moles of grignard react with the reactant, thus Mg+ react with halogen (I-) and form a good leaving group. eventually it leaves with CH2CH2.
That's what I think, Correct me plz if I'm wrong.

OK I just spoke to my awesome professor from undergrad...it is a reaction is an internal grignard. You make the grignard reagent on one end and attack the other end to form a double bond.

hope that helps
 
Last edited:
to clarify the answer,

so ICH2CH2I gets attacked by Mg-ether to form Grignard Reagent, which is ICH2CH2-MgEther, right?

Then what attacks the other end and how does the reactions happen?

Could you please help me understand what becomes what and how it attacks?

Thank you very much.
 
2 Moles of Mg react with 1 Mole of ICH2CH2I..
2 to 1 ratio.. should answer ur qs.
 
i've always had trouble understanding Grignard stuff..

Please clarify for me, O chem masters!
 
the grignard is only formed on one end of the compound, then attacks the other.
 
okay, so the ICH2CH2-Mg grignard is formed, and another ICH2CH2-Mg attacks the I of the other, and as they are forming the double bond, the Mg on the other side of the attacked side gets kicked off??

could you please explain more detail than just throwing fragments?

i lack a lot of ochem knowledge to understand shortened versions 🙁

Thank you so much for helping!
 
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