# Major product formed when 1,1,1-trichloropropane is treated with aqueous KOH

What are (is) the product formed when 1,1,1-trichloropropane is treated with aqueous potassium hydroxide?

I think the major product should be propionic acid, but I'm not sure.

Figure: 1,1,1-trichloropropane

${\color{green}{\checkmark}}$ The above compound involves $\color{red}{\ce{C-H, C-C, C-Cl}}$ bonds, whose bond energies are $413, 348, 328 kj/mol$ respectively. So, initially $\color{orange}{\ce{C-Cl}}$ bond must break.

${\color{green}{\checkmark}}$ It may be noted that generally alcoholic $\color{blue}{\ce{KOH}}$ causes elimination (dehydrohalogenation reaction) in the molecule of haloalkane, while aqueous $\color{blue}{\ce{KOH}}$ leads to substitution. As in your reaction, aqueous $\color{blue}{\ce{KOH}}$ is used, substitution reaction is most preferable.

${\color{green}{\checkmark}}$ As primary carbon and strong $\color{orange}{\ce{C-Cl}}$ (compared to other $\ce{C-Br, C-I}$ bond) is involved in 1,1,1-trichloropropane, $S_{N2}$ mechanism is preferable.

From the above thought process, the most preferable reaction would be:
$$\ce{Cl_3C-CH_2-CH_3 + 3KOH(aq) ->[\text{heat}] (OH)3C-CH_2-CH_3 + 3KCl}$$

${\color{green}{\checkmark}}$ It may be noted that more than one $\color{blue}{OH}$ group cannot be present on the same carbon atom. In such a case, the compound will be extremely unstable and will readily lose a water molecule to form more stable aldehyde, ketone or carboxylic acid.

$$\ce{(OH)_3C-CH_2-CH_3 -> (OH)CO-CH_2-CH_3 + H_2O}$$

You are right, propionic acid along with water molecule is going to be the product.

The techinfo sheet for 1,1,1-trichloroethane says that major product of hydrolysis is 2,2-dichloroethylene, so I'd say that for 1,1,1-trichloropropane it would be 1,1-dichloropropene-1.

$\ce{OH^-}$ could displace $\ce{Cl3C^-}$ to give ethanol plus chloroform plus tetrachloroethylene (via cage effect on dichorocarbene).
$\ce{OH^-}$ could eliminate HCl to give 1,1-dichloro-1-propene, and on to 1-chloropropyne, and on to....
$\ce{OH^-}$ could displace $\ce{Cl^-}$ to give the acid chloride hemicarbonyl chloride, then on to propionic acid.

Solvent, concentration, and temperature are important for biasing reaction paths.