Organic Chemistry - Product please

1

Debosmit Majumder ·2012-01-18 23:56:24

will this be the product?

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71

Vivek @ Born this Way ·2012-01-19 05:26:11

It depends highly on where the C14 atom is ?

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rishabh ·2012-01-19 07:52:36

i gues debosmit is right.
first eto^- will attack c¹⁴ and c-o bond will break rendering a negative charge on o.
now internal nucleophilic rxn. => o^- will attack the carbon and cl^- will leave.

EDIT: ^ c¹⁴ is just for marking the carbon.doesn't affect the reaction.

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71

Vivek @ Born this Way ·2012-01-20 04:02:04

Your mechanism is right.

But even if OEt attacks dirrectly at cl- attached carbon in an SN2 fasion, the product will be same ?

So why it isn't so?

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rishabh ·2012-01-20 07:26:56

beacause the epoxide ring is more reactive.

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262

Aditya Bhutra ·2012-01-20 08:42:18

because OEt will attack from least hindered site . (ie the C14 atom)
hence in the product given by debosmit, the rightmost carbon is C14

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71

Vivek @ Born this Way ·2012-01-20 22:30:54

Thanks..

My doubt was cleared regarding steric hindrance with 3D viewer..

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