Otrzymać kwas n-masłowy mając do dyspozychi butan-1-ol i dowolne odczynniki nieorganiczne
CH3-CH2-CH2-CH2-OH + CuO ---> CH3-CH2-CH2-CHO + Cu + H2O
CH3-CH2-CH2-CHO + Ag2O ---> CH3-CH2-CH2-COOH + 2Ag
CH₃-CH₂-CH₂-CH₂OH + CuO --> CH₃-CH₂-CH₂-CHO + Cu + H₂O
CH₃-CH₂-CH₂-CHO + 2Cu(OH)₂ --> CH₃-CH₂-CH₂-COOH + Cu₂O + 2H₂O
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Litterarum radices amarae sunt, fructus iucundiores
Pozdrawiam :)
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CH3-CH2-CH2-CH2-OH + CuO ---> CH3-CH2-CH2-CHO + Cu + H2O
CH3-CH2-CH2-CHO + Ag2O ---> CH3-CH2-CH2-COOH + 2Ag
CH₃-CH₂-CH₂-CH₂OH + CuO --> CH₃-CH₂-CH₂-CHO + Cu + H₂O
CH₃-CH₂-CH₂-CHO + 2Cu(OH)₂ --> CH₃-CH₂-CH₂-COOH + Cu₂O + 2H₂O
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Litterarum radices amarae sunt, fructus iucundiores
Pozdrawiam :)