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My teacher did like this forming a really unstable cyclopropyl group but this is the correct answer instead. My question is why the product will be even formed? In the mechanism that my teacher had drawn he formed a really unstable carbo anion there whereas he could have formed a primary carbanion and next may it have formed a four membered ring.

Why did not the neucleophile (acetic acid) did not approach to the carbon atom with which the chlorine is attached directly? Forming

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I tried to justify the answer then it came to me that if the acetic acid does approach from right hand side may it be forming a stable carbocation by resonance enter image description here

And this next form a very stable cyclo propyl carbocation. (Which is one of the most stable carbocations) So which explaination should I rely on? Thank you.

  • $\begingroup$ See: resonance forms of cyclopropylmethyl carbocation $\endgroup$ – Eashaan Godbole Jul 13 '18 at 11:33
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    $\begingroup$ Acetic acid (not acetate anion) attacks an isolated double bond? Seems........unlikely. Anyone got reference for this reaction? $\endgroup$ – Waylander Jul 13 '18 at 11:45
  • $\begingroup$ To get a nucelophile to attack a alkene double bond you need to bind the alkene to a transition metal such as mercury or palladium $\endgroup$ – Nuclear Chemist Jul 13 '18 at 16:58
  • $\begingroup$ OK, unless someone produces a reference showing it, I am of the opinion that this reaction does not occur. $\endgroup$ – Waylander Jul 13 '18 at 18:12
  • $\begingroup$ So will it perform Sn1forming that stable cyclomethyl carbocation? Where as it should have easily performed Sn2 $\endgroup$ – Sarah jane Jul 18 '18 at 19:17

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