μCnH2n=12n+2n=14n alchena
14n.....12n grame carbon
100.....x x=1200n/14n=85,71
a=alchena A+H2O=CnH2n+2O=14n+18
14n+18.......12n grame carbon
100.........y y=1200n/14n+18
Aplicam diferenta de procent de care ni se spune in problema
procent carbon A-procent carbon a=25,71
85,71-1200n/(14n+18)=25,71
dupa ce amplicam cu 14n+18 pentru a avea acelasi numitor obtinem:
85,71(14n+18)-1200n=25,71(14n+18)
85,71-25,71(14n+18)=1200n
60(14n+18)=1200n
14n+18=20n ⇒6n=18⇒n=3⇒alchena cautata este propena C3H6
C3H6+ H2O⇒alcool izopropilic/2 propanol (a)
C3H6+HCl⇒2 cloropropan (b)
cloropropan+H2O(NaOH)⇒alcool izopropilic (a)
C3H6+Cl2(500°C)⇒3 cloropropena (c)
3 cloropropena+H2O(NaOH)⇒CH2=CH-CH2-OH (d)
CH2=CH-CH2-OH + H2⇒1 propanol (e)
Afirmatiile adevarate sunt: 1 si 3
