Organic chemistry - Carboxylic acids (3)
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1.
Explain why acidified potassium manganate(VII) is used as an oxidizing agent in the reaction of ethanol to ethanoic acid. What observations would you make during the reaction and what would these observations indicate?
Acidified potassium manganate(VII) (KMnO4) is used as an oxidizing agent in the reaction of ethanol to ethanoic acid because it has a high oxidizing potential. This means it readily accepts electrons from other substances, causing them to be oxidized. In this case, it oxidizes ethanol to ethanoic acid.
The reaction is carried out in an acidic solution (typically dilute sulfuric acid). The acid is essential for the following reasons:
- Provides H+ ions: The acid provides hydrogen ions (H+) which are required for the oxidation process.
- Maintains a suitable pH: The acidic conditions help to maintain a pH that is optimal for the reaction to proceed.
During the reaction, several observations can be made:
- Colour Change: The purple colour of the potassium manganate(VII) solution will fade as it is reduced to manganese(II) ions (MnO2). This fading indicates that the oxidation reaction is occurring.
- Formation of a Black Precipitate: A black precipitate of manganese(II) oxide (MnO2) will form. This precipitate is a characteristic sign of the reduction of manganate(VII) to manganate(II).
- Release of Heat: The reaction is exothermic, meaning it releases heat. This can be observed by a slight increase in the temperature of the solution.
These observations indicate that the oxidation of ethanol to ethanoic acid is taking place. The fading of the purple colour and the formation of the black precipitate are direct evidence of the reduction of manganate(VII) to manganate(II), confirming that the reaction is proceeding as expected.
CH3CH2OH (ethanol) + KMnO4 (acidified) → CH3COOH (ethanoic acid) + MnO2 + KOH + H2O
2.
Ethanoic acid reacts with metals to produce a salt and hydrogen gas. Describe the reaction of ethanoic acid with magnesium, including the chemical equation and the name of the salt produced. Also, briefly explain why the reaction rate might vary depending on the metal used.
Ethanoic acid reacts with metals, generally those higher in the reactivity series than hydrogen, to produce a salt and hydrogen gas. The general equation for the reaction is:
- CH3COOH(aq) + M(s) → CH3CO2M(aq) + H2(g)
Where M represents a metal. For example, with magnesium:
CH3COOH(aq) + Mg(s) → Mg(CH3COO)2(aq) + H2(g)
The salt produced is magnesium ethanoate (Mg(CH3COO)2). The reaction rate depends on the metal's position in the reactivity series. More reactive metals (e.g., sodium, potassium) will react more vigorously and rapidly with ethanoic acid than less reactive metals (e.g., copper, zinc). This is because the energy released during the reaction is greater with more reactive metals, leading to a faster rate of electron transfer.
3.
Question 2
Explain why an acid catalyst is needed for the esterification reaction. Include a description of the mechanism by which the acid catalyst speeds up the reaction.
An acid catalyst is needed for the esterification reaction because it increases the rate of the reaction by making the carbonyl carbon of the carboxylic acid more electrophilic. Without a catalyst, the reaction is very slow and requires high temperatures to proceed at a reasonable rate.
The mechanism involves the following steps:
- The acid catalyst (e.g., H2SO4) protonates the carbonyl oxygen of the carboxylic acid. This makes the carbonyl carbon more positive and therefore more susceptible to nucleophilic attack.
- The alcohol (R'-OH) acts as a nucleophile and attacks the protonated carbonyl carbon, forming a tetrahedral intermediate.
- The tetrahedral intermediate collapses, eliminating water and forming the ester (R-COOR').
- The acid catalyst is regenerated in the process.