Answer:
52. The alcohol USED => methanol, CH3OH
The carboxylic acid USED => propanoic acid, CH3CH2COOH.
53. The alcohol USED => Ethanol, CH2CH3OH
The carboxylic acid USED => Formic acid, HCOOH.
Explanation:
52. To obtain Methyl propanoate, CH3CH2COOCH3, we simply react propanoic, CH3CH2COOH and methanol, CH3OH together as shown below:
CH3CH2COOH + CH3OH —> CH3CH2COOCH3 + H2O
The alcohol used: methanol, CH3OH
The carboxylic acid used: propanoic acid, CH3CH2COOH.
53. To obtain Ethyl methanoate, HCOOCH2CH3, we simply react
Formic acid, HCOOH and ethanol, CH3CH2OH together as show below:
HCOOH + CH3CH2OH —> HCOOCH2CH3 + H2O
The alcohol USED => Ethanol, CH2CH3OH
The carboxylic acid USED => Formic acid, HCOOH.
