Question

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Answer 1

1)
a) CH3-CH(Cl)-CH2-CH3 si C4H9Cl
b) prezinta izomerie optica => se poate scrie in oglinda
.......CH3...........|...........H3C
……..|………………..|……………….|
Cl - C - H…….……|…..……H - C - Cl
.......|.................|……………….|
......C2H5..........|…………H5C2
........................(±)
si 1-clorobutan - izomer de pozitie
c)
2CH3-CH2-CH2-CH3 + Cl2/lumina -> CH3-CH2-CH2-CH2-Cl + CH3-CH(Cl)-CH2-CH3 + 2HCl
CH2=CH-CH2-CH3 + HCl -> CH3-CH(Cl)-CH2-CH3
CH3-C≡C-CH3 + H2/Pd/Pb²+ -> CH3-CH=CH-CH3 -(+ HCl)-> CH3-CH(Cl)-CH2-CH3
2)
CH2=CH-CH3 + Br2 -> CH2(Br)-CH(Br)-CH3
b) avem 1,1dibromopropan, 1,2dibromopropan si 1,3dibromopropan
c)
c/100=md/ms => md=c*ms/100=4*80/100=3,2g Br2
MC3H6Br2=24+4+160=188g/mol
.............160g.....188g
C3H6 + Br2 -> C3H6Br2
.............3,2g......x=?
x=3,2*188/160=3,76g
4)
notam formula: CxHyBr2
NE=[2+x(4-2)+y(1-2)+2(1-2)]/2=1
2=2+2x-y-2 => y=2x-2
formula: CxH2x-2Br2
Mcompus..........100%
160g................86,02%
Mcompus=160*100/86,02=186g/mol
Mcompus=12x+2x-2+160=14x+158
14n+158=186 => 14n=28 => n=2
F.m: C2H2Br2
CH2=CBr2 sau Br-CH=CH-Br