In the reaction below, which of the following sets a \(-\mathrm{d}\) is the correct sequence for I, II, and III
A. (i) \(\mathrm{SOCl}_2\) (ii) propanol (iii) Mg/dry ether B. (i) \(\mathrm{Mg} /\) dry ether (ii) propanol (iii) \(\mathrm{PCl}\). C. (i) propanol (ii) \(\mathrm{Mg} /\) dry ether (ii) \(\mathrm{SOC} / 2\) D. (i) \(\mathrm{SOCl}_2\) (ii) \(\mathrm{Mg} /\) dry ether(iii) propanol
Correct Answer: D
Explanation
The sequence represents the scheme for the production of monocarboxylic acid. Alkanol first reacts with a chlorinating agent such as \(\mathrm{SOCl}_2\) or \(\mathrm{PCl} / 5\). \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}+\mathrm{SOCl}_2 \rightarrow \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{Cl}+\mathrm{HCl}+\mathrm{SO}_2\) The chloroalkane formed is then passed into magnesium with dry ether to form Grignard reagent \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{Cl}+\mathrm{Mg} /\) dry ether \(\rightarrow \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{MgCl}\) Alkanoic acid is thus obtained using the scheme below \(\mathrm{COOH}+\mathrm{Mg}^{2+}+\mathrm{Cl}\)
