This is in regards to a fischer esterification lab. 1. How would you change the

Question

This is in regards to a fischer esterification lab. 1. How would you change the reaction conditions of the acid catalyzed esterification to favor ester hydrolysis rather than ester formation? 2. Propanoic acid and methyl acetate are isomers. Which has the higher boiling point and why? Which would have the greater Rf value on TLC? 3. If 1g of isopentyl alcohol is reacted with an excess of acetic acid and 1 mL of isopentyl acetate (density=0.876 g/mL) is obtained what is the percent yield? Also, in my case I used ~1.5 mL of hexanol and reacted it with 3 mL of acetic acid and 5 drops of sulfuric acid could you please explain what role sodium bicarbonate played in the extraction of the product as well as what a reaction equation supporting this would look like? I'm thinking that the sodium bicarbonate was used to neutralize the acetic acid in the reaction mixture. Thank you.

Explanation / Answer

Q1.

ANSWER: Fischer esterfication is shown below

RCOOH + R'OH <--------> RCOOR' + H2O

Farward reaction is esterfication and bacckward is ester hydrolysis. Ester hydrolysis can be favoured by adding water to the reaction mixture. As per Le Chtliers Law bacward reaction will be favoured by the addition of water.

Q2. Out of propionic acid (CH3CH2COOH) and methyl acetate (CH3COOCH3), propionic acid will have hogher boiling point because it has more polar bonds and are capable of H-bond formation, while as in methyl acetate does not possess such tendencies.

Methyl acetate is less polar hence will travel greater length on TLC compared to propionic acid. Hence methyl acetate will have higher Rf value.

NOTE: Please send the other questions separately, as one question only is to be solved one time.

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