Question

The major product formed when 1, 1, 1-trichloro-propane is treated with aqueous potassium hydroxide is:
A. propyne
B. 1-propanol
C. 2-propanol
D. propionic acid

Answer 1

Hint: For this problem, we have to study the complete mechanism of the reaction in which we know that the aqueous potassium hydroxide acts to eliminate the chlorine atom and form the most stable carboxylic or aldehyde or ketone group.

Complete step by step answer:
- In the given question, we have to explain the product which will be formed when 1, 1, 1-trichloro-propane reacts with aqueous potassium hydroxide.
- As we know that the potassium hydroxide is responsible for the oxidation of the reactant because it replaces the electronegative element with the oxygen.
- Also, we know that the reaction of replacement of one element from the other element is known as elimination reaction.
- So, we know that the molecular formula of the compound 1, 1, 1-trichloro-propane is
\[\text{C}{{\text{l}}_{3}}\text{CC}{{\text{H}}_{2}}\text{C}{{\text{H}}_{3}}\text{+ KOH }\to \text{ (OH}{{\text{)}}_{3}}\text{CC}{{\text{H}}_{2}}\text{C}{{\text{H}}_{3}}\text{ + 3KCl}\]

- Now, as we can see that three molecules of hydroxide ion from potassium hydroxide replaces the chlorine atom and eliminates a by-product that is potassium chloride.
- But the presence of three hydroxyl groups causes the steric hindrance and also makes the molecule unstable.
- Due to the instability of the three hydroxyl groups, the product formed will release the water molecule due to which the formation of more stable carboxylic acid takes place.

- The balanced chemical will be as shown below:
\[{{(\text{OH})}_{3}}\text{CC}{{\text{H}}_{2}}\text{C}{{\text{H}}_{3}}\text{+ KOH }\to \text{ (OH)COC}{{\text{H}}_{2}}\text{C}{{\text{H}}_{3}}\text{ + 3}{{\text{H}}_{2}}\text{O}\]
- So, the final stable product which will be formed is propionic acid.
So, the correct answer is “Option D”.

Note: The action of the alcoholic potassium hydroxide and aqueous potassium hydroxide is different because alcoholic potassium hydroxide favours the substitution reaction whereas aqueous potassium hydroxide favours elimination reaction as shown in the answer.