Respuesta :

An ester , propyl methanoate ( HCOOC₃H₇) when reacts with sodium hydroxide( NaOH) forms sodium methanoate (HCOONa) as the main product and propanol (C₃H₇OH).

The reaction is as follows:

HCOOC₃H₇+NaOH ⇒HCOONa + C₃H₇OH

So when propyl methanoate is hydrolyzed in water and in NaOH then sodium methanoate (HCOONa) as the main product and propanol (C₃H₇OH) forms

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