Chemistry How is Methyl Propanoate Formed from Propanoyl Chloride and Sodium Methoxide?

AI Thread Summary
Methyl propanoate is formed from propanoyl chloride and sodium methoxide through a nucleophilic addition reaction. The nucleophile, methoxide ion (CH3O-), attacks the electrophilic carbon of the carbonyl group, leading to the formation of a tetrahedral intermediate. This intermediate then decomposes to yield methyl propanoate and sodium chloride (NaCl) as byproducts. Proper notation and charge conservation are crucial in representing the reaction mechanism accurately. Understanding these concepts is essential for clarity in chemical communication and academic evaluation.
charlie05
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Homework Statement
Describe in detail the mechanism of reaction between propanoyl chloride and sodium methoxide
Relevant Equations
CH3CH2COCl + CH3ONa
CH3CH2COCl + CH3ONa = CH3CH2COOCH3 + NaCl ?

from propanoyl chloride is formed acylium ion?
 
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No. What does a nucleophile do in the presence of a carbonyl?
 
nucleophilic addition...C is δ+ and O is δ-
 
So what's the first step of the given reaction?
 
the nucleophile binds to an electrophilic carbon...nucleophile is CH3O ?
 
-
 
sorry, I do not understand...
 
Sorry, in some places I have been accused of being long winded sometimes so I thought I would write the shortest post ever .
CH3O- Do you get it now?
 
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Yes thank you...an intermediate product is formed - anion CH3O is bound to the positive carbon, to the negative carbonyl oxygen is bounded Na +? or does sodium bind to chlorine?
 
  • #10
product is methylpropionate and NaCl?
 
  • #11
charlie05 said:
product is methylpropionate and NaCl?
Yes, you already said that in #1.
Look up the reaction for the more frequently given reaction of an acyl chloride with an alcohol. Which already goes readily enough. Both the -Cl and the =O withdraw electrons, leaving the C extra susceptible to nucleophilic attack. This is usually portrayed as from lone pair oxygen atom of the alcohol, sometimes with the alcohol's proton being transferred to a base at the same time. In the case of an alkoxide ion the proton has already been extracted. Alkoxide ions should be powerful nuclepohile. So your mechanism should be a small variation of what is more usually portrayed, and go through the tetrahedral intermediate which then gives product by Cl- leaving.

Second time you have written CH3O without the - charge, this is disturbing to chemists and teachers and you could lose marks for it so respect the conservation of charge principle.

You ought to write the mechanism out explicitly to be sure and for your Prof to be sure you have got it.
 
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  • #12
Yes, thank you very much...
Intermediate product...?
1570715696282.png
 
  • #13
On the right lines, but
(i) Don't put a + charge on the C and a - charge on the OCH3. The formation of the dative O→C bond neutralises these formal charges.
(ii) Only a single bond between the C and the O-.
 
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  • #14
1570722236976.png
 
  • #15
I still have a question - what is the driving force of decomposition of intermediates to the product ... is it the formation of NaCl?
 
  • #16
Yes I was going to say, certain things get imprinted when you deal with them a lot, it is quite like grammar, so your Prof might have a fit he saw that double bond and pentavalent carbon in your first attempt.
 
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  • #17
:-) i understand...
 
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