Carry out the transformation: starch – glucose – ethanol – ethyl formate – formic acid – silver

Let’s solve the chain of transformations:
1. We write down the formulas of substances:
(С6Н10О5) n – C6H12O6 – C2H5OH – HCOO – C2H5 – HCOOH – Ag.
2. We carry out transformations:
(C6H10O5) n + nH2O = n C6H12O6 – hydrolysis of starch is carried out by heating in the presence of sulfuric acid, glucose is obtained;
С6Н12О6 = 2С2Н5ОН + 2СО2 – fermentation, ethanol is released;
С2Н5ОН + НСООН = НСОО – С2Н5 + Н2О – esterification, ethyl formate was formed;
НСОО – С2Н5 + Н2О = НСООН + С2Н5ОН – ether hydrolysis, formic acid was obtained;
НСООН + Н2 = СН3ОН – addition, methanol is released;
CH3OH + CuO = HCOH + Cu + H2O – alcohol oxidation, methanal was formed;
НСОН + Ag2O = HCOOH + 2Ag – qualitative reaction to aldehyde, silver and methanoic acid are released.

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