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Question
How is propanoic acid is prepared to start from an alkene?
Solution
\[\ce{\underset{(Ethene)}{CH2 = CH2} ->[HCl] \underset{(Ethyl chloride)}{CH3 - CH2 - Cl} ->[NaCN][-NaCl] \underset{(Ethyl cyanide)}{CH3 - CH2 - CN} ->[H^+/H2O][Excess] \underset{(Propanoic acid)}{CH3CH2COOH}}\]
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