Draw the products of the following hydrolysis

Draw the products of the following hydrolysis.

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Answer

General guidance

Concepts and reason
Hydrolysis:

It is the chemical reaction between a water molecule and another compound, in which the water molecule ionizes and splits the other compound forming two or more products.

Fundamentals

Hydro means water and lysis means unbind. The reaction involves the cleavage of chemical bonds using water. Hydrolysis of an ester gives both an acid and an alcohol as products.

The mechanism of the hydrolysis of an ester in the presence of an acid or proton source is as follows:

1.The double bonded oxygen of an ester abstracts a proton and forms a hydroxonium ion.

н
н*
R—С—О—R
R—с.
-о—R

2.The attack of the water molecule on an electrophilic carbon.

н
Н
R—С—О—R
R—с
-о—R
Н
Н
т

3.A proton transfer followed by the cleaving of the alcohol group, forming a carbocation.

Proton transfer:
H
Н
Proton transfer
R C
R C
-R
-R
Н
н
Н
Departure of alcohol:
н
н
R
—с
-R
R CO
НО—R
Alcohol
Н
Н
: O
I
:O
I
.

4.Deprotonation to yield the final product.

н
-н*
R C
Ht
C OH
R
Caroboxylic acid
H

Step-by-step

Step 1 of 3

-н
Н2
н*
Н2
CHз
Нас
CНз
Hас
Н
:O
—н
Н2
С.
CHз
Hас
H
н
O
т
O:
O:

The double bonded oxygen of the ester abstracts the proton and forms a hydroxonium ion, making the adjacent carbon atom more electrophilic in nature. The oxygen from the water molecule attacks the electrophilic carbon and forms a hydronium ion.

Step 2 of 3

:O
-н
—Н
О
Н2
с.
На
с
CHз
CHз
Нас
Hас
Н
Н
: O
-н
Н2
с.
Со
CHз
Нас
н
O
-т
O
с

The oxygen attached to the alkyl group abstracts the proton from the hydronium ion and bears a positive charge on it. The carbon-oxygen bond with a positive charge cleaves, forming an ethanol and a carbocation.

Step 3 of 3

-н
-н*
С
HзС
H*
Hас
: ОН
Н

Н2
Н2
Н*
CH3
НО
Hас
н
Hас
Н
CHз
ОН
о—о


The proton of the hydroxyl group from the molecule that has carbocation (carbon with positive charge) deprotonates to form a neutralized stable product called acetic acid.

Answer

Н2
Н2
Н*
CH3
НО
Hас
н
Hас
Н
CHз
ОН
о—о

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