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Topic: GOC  (Read 3183 times)

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Offline AMEDIO

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GOC
« on: March 12, 2011, 03:37:59 AM »
Which of the following resonating structures of 1-methoxy-1,3-butadiene is least stable??
http://img861.imageshack.us/i/14920168.png/

i am confused C and D....how to decide between the 2??

Offline Mandeep Deka

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Re: GOC
« Reply #1 on: March 12, 2011, 08:32:33 AM »
So far as i can see it,
in (D), the carbon atom on which the negative charge resides has an oxygen atom attached to it which will obviously lead to repulsion between the lone pair on oxygen and the negative charge on the carbon atom, wheres it is not the case in (C). This shall be the reason!

Offline orgstudy

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Re: GOC
« Reply #2 on: March 17, 2011, 02:16:43 PM »
i disagree with Mandeep

in (d) the - charge is closest to O but i (a)&(b) its on O which is more instable

in (b), the - charge is 2 degree and maybe its stabilized by 2 allylic groups but it has a very good chance of bonding to right allylic group and making a highly stable compound
Chemistry is a subject in which even the Professor can get confused at any time.

Offline orgopete

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Re: GOC
« Reply #3 on: March 18, 2011, 09:49:17 AM »
a), b), and c) are all resonance structures one would draw for 1-methoxybutadiene. d) however has the reverse polarization of a double bond next to the oxygen. This should have a destabilizing effect as noted by Mandeep.

If 1-methoxybutadiene were hydrolyzed with acid, the product would be 2-butenal (crotonaldehyde). Therefore protonation must have occurred at C-2 or C-4 and stabilized by the oxygen. This is shown in the resonance structures a) and b). d) would not predict this reaction to occur.
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