Answer 5


Iodoethane reacts with aqueous NaOH to give mostly ethanol (A) via nucleophilic substitution:

CH3CH2I + OH - CH3CH2OH + I -

and with ammonia in ethanolic solution in a sealed tube to give aminoethane (B), although further substitution also occurs:

CH3CH2I + 2NH3 CH3CH2NH2 + NH4I

Potassium cyanide in ethanol gives propanonitrile (ethyl cyanide) (C), which can be reduced by lithium aluminium hydride to 1-aminopropane, (D):

CH3CH2I + CN - CH3CH2CN + I -

CH3CH2CN + 4[H] CH3CH2CH2NH2

Propanonitrile is hydrolysed by aqueous alkali to the corresponding carboxylic acid anion, which gives the free acid, propanoic acid (E), with hydrochloric acid:

CH3CH2CN + H2O + OH - CH3CH2COO - + NH3

CH3CH2COO - + H3O + CH3CH2COOH + H2O

Propanoic acid gives propanoyl chloride (F) with phosphorus(V) chloride:

CH3CH2COOH + PCl5 CH3CH2COCl + POCl3 + HCl

 


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