Mechanism and reaction products

In summary, the nucleophile binds to the electrophilic carbon and the intermediate product is formed.
  • #1
charlie05
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6
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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  • #2
No. What does a nucleophile do in the presence of a carbonyl?
 
  • #3
nucleophilic addition...C is δ+ and O is δ-
 
  • #4
So what's the first step of the given reaction?
 
  • #5
the nucleophile binds to an electrophilic carbon...nucleophile is CH3O ?
 
  • #6
-
 
  • #7
sorry, I do not understand...
 
  • #8
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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  • #9
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...
 

What is a mechanism?

A mechanism is a detailed step-by-step explanation of how a chemical reaction occurs. It describes the movement of electrons and the breaking and forming of bonds between atoms.

How are reaction mechanisms determined?

Reaction mechanisms are determined through experiments, computer simulations, and theoretical calculations. By analyzing the reactants, products, and intermediates, scientists can propose a mechanism that is consistent with the observed data.

What are reaction products?

Reaction products are the substances that are formed as a result of a chemical reaction. They are the new compounds that are produced when atoms rearrange and bonds break and form during the reaction.

Why is understanding reaction mechanisms important?

Understanding reaction mechanisms is important because it allows scientists to predict and control the outcomes of chemical reactions. It also provides insight into the underlying principles of chemical reactions and can lead to the development of more efficient and sustainable processes.

Can reaction mechanisms change?

Yes, reaction mechanisms can change as new evidence or information is discovered. Scientists are constantly refining and updating their understanding of reaction mechanisms as they conduct more experiments and use more advanced techniques to study chemical reactions.

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