How do we make #"2-butanol"# from #"ethyl chloride"# using inorganic reagents?

1 Answer
Oct 3, 2017

Well, #"butanone"# we can make from #"2-butanol"#....

Explanation:

And #"2-butanol"# we can make from #"ethyl magnesium chloride"# and #"acetaldehyde"#, and certainly we can make #"ethyl magnesium chloride"# and #"acetaldehyde"# from #"ethyl chloride"#.

And so....

#H_3C-CH_2Cl+H^(+)"/"H_2O rarr H_3C-CH_2OH#

#H_3C-CH_2OHstackrel(C_6H_5N, CrO_3)rarrH_3C-C(=O)H#

Meanwhile.....

#H_3C-CH_2Cl+Mg stackrel("dry "Et_2O)rarr H_3C-CH_2MgCl#

And so.........

#H_3C-CH_2MgCl+H_3C-C(=O)Hrarr#
#H_3C-CH_2C(O^(-))HCH_3stackrel(H^+)rarr"2-butanol"#

And finally.....

#"2-butanol "stackrel(H_3O^(+),K_2Cr_2O_7)rarr" butanone"# as required....