The major product formed when 1, 1, 1-trichloropropane is treated with aqueous potassium hydroxide is:

The compound given is 1,1,1-trichloropropane, which has the structure $$CH_{3}-CH_{2}-CCl_{3}$$. Here, the carbon atom at position 1 (the terminal carbon) is bonded to three chlorine atoms.

When treated with aqueous potassium hydroxide (KOH), this compound undergoes hydrolysis. Aqueous KOH provides hydroxide ions ($$OH-$$) that act as a nucleophile. The reaction involves the substitution of chlorine atoms by hydroxyl groups, but due to the presence of three chlorine atoms on the same carbon, the process leads to the formation of a carboxylic acid.

The mechanism proceeds as follows:

First, one hydroxide ion attacks the carbon atom bearing the chlorine atoms, replacing one chlorine atom to form an intermediate:

$$CH_{3}-CH_{2}-CCl_{3} + OH- -\gt CH_{3}-CH_{2}-CCl_{2}OH + Cl-$$

This intermediate, $$CH_{3}-CH_{2}-CCl_{2}OH$$, is unstable because it has two chlorine atoms and a hydroxyl group on the same carbon. It readily loses a chloride ion and rearranges to form a dichlorocarbene-like species, but in aqueous conditions, it undergoes further hydrolysis.

The intermediate reacts with another hydroxide ion:

$$CH_{3}-CH_{2}-CCl_{2}OH + OH- -\gt CH_{3}-CH_{2}-CCl(OH)2 + Cl-$$

Now, we have $$CH_{3}-CH_{2}-CCl(OH)2$$, which is a geminal diol with one chlorine. This compound is highly unstable and undergoes rapid hydrolysis. The chlorine atom is replaced by another hydroxyl group:

$$CH_{3}-CH_{2}-CCl(OH)2 + OH- -\gt CH_{3}-CH_{2}-C(OH)3 + Cl-$$

The resulting compound, $$CH_{3}-CH_{2}-C(OH)3$$, is a geminal triol. Geminal triols are unstable and spontaneously lose a water molecule to form the corresponding carboxylic acid:

$$CH_{3}-CH_{2}-C(OH)3 -\gt CH_{3}-CH_{2}-COOH + H_{2}O$$

Thus, the overall reaction is:

$$CH_{3}-CH_{2}-CCl_{3} + 2KOH -\gt CH_{3}-CH_{2}-COOH + 2KCl + H_{2}O$$

The product is propionic acid, which has the formula $$CH_{3}CH_{2}COOH$$.

Now, comparing with the options:

• Option A: 2-Propanol is $$CH_{3}CH(OH)CH_{3}$$
• Option B: Propyne is $$CH_{3}C \equiv CH$$
• Option C: Propionic acid is $$CH_{3}CH_{2}COOH$$
• Option D: 1-Propanol is $$CH_{3}CH_{2}CH_{2}OH$$

The major product is propionic acid, corresponding to option C.

Hence, the correct answer is Option C.