Assume these 2 reactions. I must write the Lewis structures of the products.

  1. $$\ce{C6H5COOH + CH3CH2OH ->[\ce{H+}] ?}$$ (comment: A benzene ring + $\ce{CO_2H}$ on one end, I don't know its name)

  2. $$\ce{CH3CH2OH + (CH3CO)2O ->[\ce{H+}]?}$$ (comment: acetic anhydride)

My approach:

For the first reaction I will get acetic anhydride (source)

acetic anhydride

For the second reaction I will get aspirin and acetic acid.


  1. Is the above correct?
  2. Is there a connection between the two reactions? Is it an intermediate step of some sort? What is the importance between the two reactions?
  3. Why is $\ce{H+}$ added?
  4. What is the benzene ring and the $\ce{CO2H}$ called?

1 Answer 1

  1. No, I hesitate to give the correct products since this is homework, but maybe some of what's below will help

  2. Yes, the two reactions are both esterifications; converting an acid, an acid anhydride, an acid chloride, etc., to the corresponding ester

  3. The proton is a catalyst, it is not consumed in the reaction. It protonates the carbonyl oxygen, making the carbonyl carbon more susceptible to nucleophilic attack

  4. benzoic acid

  • $\begingroup$ Well even if it is homework how will I learn if I don't see the right answer at least once? $\endgroup$ Jun 3, 2014 at 0:46
  • $\begingroup$ Is this the only product of the first reaction? en.wikipedia.org/wiki/Ethyl_benzoate $\endgroup$ Jun 3, 2014 at 0:49
  • $\begingroup$ For the second reaction will I get CH3C(O)OC6H5 + CH3COO-? $\endgroup$ Jun 3, 2014 at 0:52
  • $\begingroup$ Yes, you got the first one and I think you got the second one, but your alpha-structure is a little ambiguous. You can find the second structure here sigmaaldrich.com/content/dam/sigma-aldrich/structure3/016/… $\endgroup$
    – ron
    Jun 3, 2014 at 0:56
  • $\begingroup$ @ron In an esterification of un-symmetrical anhydride, how to decide which part of anhydride gets added to form ester? Please reply. $\endgroup$ Aug 21, 2022 at 7:20

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