$Ankadien:C_{n}H_{2n-2}$
$C_{n}H_{2n-2}+2Br_{2} \to C_{n}H_{2n-2}Br_{4}$
0,09 0,18
$mdd_{giảm}=mBr_{2}\text{ phản ứng}=600.60\%=360g$
$nBr_{2}=\frac{360.8\%}{160}=0,18$
$MAnkadien=\frac{4,86}{0,09}$
⇔$14n-2=54⇔n=4$
⇒$CTPT:C_{4}H_{6}$
CTCT:
+$CH_{2}=CH-CH=CH_{2}$: buta-1,3-dien
+$CH_{2}=C=CH-CH_{3}$: buta-1,2-dien