Răspuns :
a)
notam compusul bromurat
CnH2n+2-xBrx, x = nr. at. Br din compus
100% ............................ 216 g
74,075% Br ................ 80x g Br
=> x = 2 => compus dibromurat
=> CnH2nBr2
=> 14n+160 = 216 => n = 4
=> C4H8Br2
b)
I
CH2Br I CH2Br
Br-C-H I H-C-Br
CH2 I CH2
CH3 I CH3
I
c)
I l I
CH3 I CH3 l CH3 I CH3
Br-C-H I H-C-Br l H-C-Br I Br-C-H
Br-C-H I H-C-Br l Br-C-H I H-C-Br
CH3 I CH3 l CH3 I CH3
I I I
1 2 3 4
=>
1 + 2 = mezoforme = 1 enantiomer
3 + 4 stereizomeri (diastereoizomeri)
=> teoretic = 4 stereoizomeri
=> practic = 3 stereizomeri