Hydrocarbon A, C5H8, reacts with 2 moles of Br2 to give 1,2,3,4-tetrabromo-2-methylbutane. What is the structure of hydrocarbon A?
2. Write line-angle formulas for all compounds with the molecular formula C4H8. Which are sets of constitutional isomers? Which are sets of cis-trans isomers?
The Correct Answer and Explanation is:
1. Structure of Hydrocarbon A (C₅H₈)
Hydrocarbon A reacts with 2 moles of Br₂ to form 1,2,3,4-tetrabromo-2-methylbutane, meaning four bromines add across double or triple bonds. This product suggests complete addition across two double bonds or a triple bond in the starting molecule. The name of the product gives a big clue: 2-methylbutane backbone with Br atoms on C-1, C-2, C-3, and C-4, implying a conjugated diene or alkyne in the reactant.
The structure of Hydrocarbon A is:
2-Methyl-1-butyne
Line-angle formula:
CH3
|
CH≡C–CH2–CH3
This terminal alkyne (triple bond between C-1 and C-2) undergoes electrophilic addition with 2 moles of Br₂ across the triple bond, resulting in the tetra-bromo derivative.
2. All Line-Angle Formulas for Compounds with Formula C₄H₈
C₄H₈ has two degrees of unsaturation, indicating either a ring or a double bond (alkenes or cycloalkanes). The possible structures include:
A. Alkenes
- 1-Butene
- cis-2-Butene
- trans-2-Butene
- 2-Methyl-1-propene (isobutene)
B. Cycloalkanes
- Cyclobutane
- Methylcyclopropane
- Ethylcyclopropane
- 1,1-Dimethylcyclopropane
- 1-Butyne (if you consider alkynes)
- 2-Butyne
Line-angle formulas:
(You can imagine a zigzag line for carbon chains)
- 1-Butene: CH₂=CH–CH₂–CH₃
- cis-2-Butene: CH₃–CH=CH–CH₃ (same side)
- trans-2-Butene: CH₃–CH=CH–CH₃ (opposite side)
- 2-Methylpropene: (CH₃)₂C=CH₂
- Cyclobutane: 4-carbon ring
- Methylcyclopropane: 3-carbon ring + CH₃
- 1-Butyne: CH≡C–CH₂–CH₃
- 2-Butyne: CH₃–C≡C–CH₃
Isomer Classification
Constitutional Isomers
These differ in connectivity of atoms:
- All the above compounds are constitutional isomers of each other (they share C₄H₈ formula but differ in structure and bonding).
Cis-Trans Isomers
- cis-2-Butene and trans-2-Butene are geometric (cis-trans) isomers, differing in spatial arrangement around the double bond.
- These occur only when each double-bonded carbon has two different groups.
Summary
The hydrocarbon C₅H₈ reacting with two moles of Br₂ to form 1,2,3,4-tetrabromo-2-methylbutane indicates the presence of a triple bond, since each Br₂ molecule adds across a π-bond. The final product suggests a 2-methylbutane backbone with bromines on the first four carbons, implying that the starting compound must be 2-methyl-1-butyne. This alkyne undergoes electrophilic addition of two Br₂ molecules across its triple bond, saturating the molecule and forming the observed tetrabromo product.
For the molecular formula C₄H₈, there are multiple possible isomers due to the presence of two degrees of unsaturation, which can result from either one double bond (alkenes) or one ring (cycloalkanes). These isomers include straight-chain alkenes like 1-butene and 2-butene (which has both cis and trans isomers), branched alkenes like 2-methylpropene, and ring compounds such as cyclobutane and methylcyclopropane. There are also alkynes like 1-butyne and 2-butyne that fulfill the molecular formula.
All of these compounds are constitutional isomers because they have the same molecular formula but different connectivity. Among them, cis-2-butene and trans-2-butene form a pair of geometric (cis-trans) isomers, which occur due to restricted rotation around the double bond and different spatial positioning of groups. Recognizing such isomer types is crucial in organic chemistry, as they significantly influence physical and chemical properties.
