Complete the folling reactions by writing the structures of the expected products & by naming the reactants & products
HCl
a) CH3COOCH3 + H20 ---> ?
NaOH
b) CH3CH2CH2COOCH2CH2CH3 + H20 ---> ?
KOH
C) HCOOCH2CH3 + H20 ---> ?
a) To determine the products of the reaction between CH3COOCH3 and HCl, we need to consider the reaction type. In this case, it's an acid-base reaction.
The reactant CH3COOCH3 (ethyl acetate) is an ester, while HCl is an acid. During the reaction, the HCl molecule will donate a proton (H+) to the ethyl acetate molecule.
The acid-base reaction can be represented as follows:
CH3COOCH3 + HCl → CH3COOH + CH3OH
The product is acetic acid (CH3COOH) and methanol (CH3OH).
b) This reaction involves the compound CH3CH2CH2COOCH2CH2CH3 (butyl propanoate) and NaOH. Again, this is an acid-base reaction.
The reactant butyl propanoate is an ester, while NaOH is a strong base. In this reaction, the NaOH will donate a hydroxide ion (OH-) to the butyl propanoate molecule.
The acid-base reaction can be represented as follows:
CH3CH2CH2COOCH2CH2CH3 + NaOH → CH3CH2CH2COOH + CH3CH2CH2CH2OH
The product is propanoic acid (CH3CH2CH2COOH) and butanol (CH3CH2CH2CH2OH).
c) In this reaction, the compound HCOOCH2CH3 (ethyl formate) reacts with KOH. Once again, this is an acid-base reaction.
The reactant ethyl formate is an ester, while KOH is a strong base. The KOH will donate a hydroxide ion (OH-) to the ethyl formate molecule.
The acid-base reaction can be represented as follows:
HCOOCH2CH3 + KOH → HCOOK + CH3CH2OH
The product is potassium formate (HCOOK) and ethanol (CH3CH2OH).