Question

How are the following conversions carried out?
2- chloropropane to 1- propanol

Answer 1

To answer this question, you must recall the reactions and reagents involved in the formation of various organic compounds. The conversion in the question involves the formation of an alcohol from an alkyl halide.

Complete answer:
The chlorine molecule is attached to the second carbon atom in the carbon chain and the hydroxyl group is attached to the first carbon in the chain. So we cannot carry out a substitution reaction directly for the formation of 1- propanol from 2- chloropropane.
First we convert the 2- chloropropane into a haloalkane with the halogen molecule attached at the terminal carbon atom. To do this, react the given compound with an alcoholic solution of ${\text{KOH}}$ . The alcoholic solution of ${\text{KOH}}$ contains alkoxide ion, which is a strong base. It extracts a proton from the $\beta$ - carbon atom of the alkyl chloride and the chlorine substituent leaves as chloride ion.
In total, a molecule of hydrogen chloride is lost resulting in the formation of prop-1- ene. Now we need a functional group to be attached to the terminal carbon atom and thus, we react it with $HBr$ in presence of peroxide and form 1- bromopropane.
Finally we react to bromopropane with aqueous ${\text{KOH}}$ . The hydroxide group in aqueous medium is a strong nucleophile and thus, when it reacts with bromopropane, substitution reaction takes place. Thus, the bromide ion is lost and hydroxide ion replaces it resulting in the formation of 1- propanol.
$C{H_3} - CH\left( {Cl} \right) - C{H_3}\xrightarrow[{^\Delta }]{{alc.KOH}}C{H_3} - CH = C{H_2}\xrightarrow[{Peroxide}]{{HBr}}C{H_3} - C{H_2}C{H_2} - Br$
$C{H_3} - C{H_2}C{H_2} - Br\xrightarrow[\Delta ]{{Aq.KOH}}C{H_3}C{H_2}C{H_2} - OH$

Note:
When aqueous ${\text{KOH}}$ is used, ${\text{O}}{{\text{H}}^ - }$ ion is both a strong base as well as a strong nucleophile. But in an aqueous solution, the hydroxide ions are highly hydrated. This reduces the basic character of ${\text{O}}{{\text{H}}^ - }$ ions and thus it fails to abstract a hydrogen from the $\beta$ - carbon of the alkyl chloride to form a double bond.