Question

2418 triglicerida cu formula procentuala 75,93%C,12,158%H si 11,91% O se supune hidrolizei totale cu 1,4 l solutie de KOH 10M.Excesul de baza se neutralizeaza cu solutie de acid sulfuric 1M. Se cere: 1. determina formula structurala a trigliceridei si denumirea daca contine un singur tip de acid gras
2. calculeaza indicele de saponificare al gliceridei
3. calculeaza volumul solutiei de acid sulfuric 1M necesar neutralizarii excesului de KOH.
4. propune o schema de reactie prin care sa se obtina 1,2,3-propantriolul, pornind de la metan ca sursa principala de substante organice

Answer 1

                                    :0,744375
C: 75,93/12=6,3275        8,5
H: 12,158/1=12,158        16,333
O: 11,91/16=0,744375    1
(C8,5H16,333O)n
insa o triglicerida are 6 atomi de O => n=6
f.m: C51H98O6
stim ca triglicerida e formata din 3 molecule de acid si 1 molecule de glicerol in urma caruia se elimina 3 molecule de apa
Mtriglicerida+3Mapa=3Macid+Mglicerol
Mtriglicerida=612+98+96=806g/mol
Mglicerol=MC3H8O3=36+8+48=92g/mol
Macid=(Mtriglicerida+3Mapa-Mglicerol)/3=(806+54-92)/3=256g/mol
acid are formula generala: CnH2nO2 daca e saturat si CnH2n-2O2 daca are o dubla legatura
presupunem ca e saturat: Macid=14n+32
14n+32=256 => n=16
formula structurala: CH3-(CH2)14-COOH => triglicerida este tripalmitina
b)
indicele de saponificare - cantitatea de KOH in mg necesara neutralizarii unui 1g de grasime
saponificarea se face cu 3 molecule de KOH
MKOH=39+16+1=56g/mol
806g grasime............168000mg KOH
1g grasime.................indicele
indicele=168000mg KOH/806g grasime=208,43mg KOH/g grasime
c)
806g..............3moli
C51H98O6 + 3KOH -> C3H8O3 + 3C16H31O2K
2418g............9moli
n=Cm*Vs => n=1,4*10=14moli
exces: 14-9=5 moli
2moli....1mol
2KOH + H2SO4 -> K2SO4 + 2H2O
5moli....2,5moli
Vs=n/Cm=2,5moli/1mol/l=2,5l H2SO4
d)
2CH4 -> C2H2 + 3H2
2C2H2 -> C4H4
C4H4 + 3H2 -> C4H10
C4H10 -> C3H6 + CH4
CH2=CH-CH3 + Cl2 -(500°C)-> CH2=CH-CH2-Cl + HCl
CH2=CH-CH2-Cl + Cl2 -> ClCH2-CH(Cl)-CH2-Cl
CH2(Cl)-CH(Cl)-CH2 + 3H2O -(HO-)-> CH2(OH)-CH(OH)-CH2(OH) + 3HCl