Isopropylamine and n-pentylamine are to be synthesized. In the laboratory synthesis, outline all the steps from alcohols of four carbons or less, using any needed inorganic reagents.

The synthesis of isopropylamine initiates from isopropanol and that of n-pentylamine starts from n-butanol. Firstly, these alcohols are converted into alkyl halides, specifically isopropyl chloride and n-butyl chloride, through a reaction with hydrochloric acid. Subsequently, these alkyl halides are converted to azides; isopropyl azide and n-butyl azide, via a nucleophilic substitution reaction with sodium azide. Lastly, reduction of these azides with a reducing agent like lithium aluminium hydride, yields the desired amines, isopropylamine, and n-pentylamine. The relevant chemical reactions can be represented as follows: For Isopropylamine: Conversion to alkyl halide: \(CH_{3}CH(OH)CH_{3} + HCl \rightarrow CH_{3}CH(Cl)CH_{3} + H_{2}O\) Conversion to azide: \(CH_{3}CH(Cl)CH_{3} + NaN_{3} \rightarrow CH_{3}CH(N_{3})CH_{3} + NaCl\) Reduction to amine: \(CH_{3}CH(N_{3})CH_{3} + LiAlH_{4} \rightarrow CH_{3}CH(NH_{2})CH_{3} + LiAlH_{4}O\) For n-pentylamine: Conversion to alkyl halide: \(CH_{3}(CH_{2})_{2}CH_{2}OH + HCl \rightarrow CH_{3}(CH_{2})_{2}CH_{2}Cl + H_{2}O\) Conversion to azide: \(CH_{3}(CH_{2})_{2}CH_{2}Cl + NaN_{3} \rightarrow CH_{3}(CH_{2})_{2}CH_{2}N_{3} + NaCl\) Reduction to amine: \(CH_{3}(CH_{2})_{2}CH_{2}N_{3} + LiAlH_{4} \rightarrow CH_{3}(CH_{2})_{2}CH_{2}NH_{2} + LiAlH_{4}O\)