Elimination of HBr from 2-bromobutane results in the formation of:
Predominantly 2-butene
Understanding Elimination of HBr from 2-Bromobutane
The question asks about the product distribution when hydrogen bromide (HBr) is eliminated from 2-bromobutane. This type of reaction is known as dehydrohalogenation, which is a type of elimination reaction (specifically E1 or E2, depending on conditions, but typically follows similar product distribution rules).
Identifying the Reactant: 2-Bromobutane
2-bromobutane is a secondary alkyl halide with the structure CH3-CHBr-CH2-CH3. In a dehydrohalogenation reaction, a halogen atom (Br in this case) is removed from one carbon atom, and a hydrogen atom is removed from an adjacent carbon atom.
Possible Elimination Pathways and Products
In 2-bromobutane, the bromine atom is attached to the second carbon. There are two adjacent carbon atoms from which a hydrogen atom can be removed:
- The methyl group (CH3-) at carbon 1.
- The methylene group (-CH2-) at carbon 3.
Removing HBr can thus lead to two different alkene products:
- Removal of H from carbon 1 and Br from carbon 2 leads to a double bond between C1 and C2.
CH3-CHBr-CH2-CH3 → CH2=CH-CH2-CH3 (1-butene) - Removal of H from carbon 3 and Br from carbon 2 leads to a double bond between C2 and C3.
CH3-CHBr-CH2-CH3 → CH3-CH=CH-CH3 (2-butene)
Applying Zaitsev's Rule
When an elimination reaction can produce more than one alkene product, Zaitsev's rule (also known as Saytzeff's rule) helps predict the major product. Zaitsev's rule states that the most substituted alkene is the most stable and is typically the predominant product.
- 1-butene (CH2=CH-CH2-CH3): This is a monosubstituted alkene (the double bond carbons are attached to one alkyl group, -CH2CH3).
- 2-butene (CH3-CH=CH-CH3): This is a disubstituted alkene (the double bond carbons are attached to two alkyl groups, two -CH3 groups).
According to Zaitsev's rule, the disubstituted alkene (2-butene) is more stable than the monosubstituted alkene (1-butene) due to greater hyperconjugation. Therefore, 2-butene is expected to be the predominant product of the elimination of HBr from 2-bromobutane.
Conclusion
The elimination of HBr from 2-bromobutane yields a mixture of 1-butene and 2-butene. Based on Zaitsev's rule, the more stable, more substituted alkene, 2-butene, is formed predominantly over 1-butene.
Important Questions from Chemical Reaction
- Which one of the following is not an example of a redox reaction?
- Which one of the following is not a pigment?
- Which one of the following reactions is an example of decomposition reaction?
Match List I with List II and select the correct answer using the code given below the Lists:
LIST I
(Chemical process)
LIST II
(Reaction)
A.
Electrolysis of water
1.
Double displacement
reactionB.
Burning of coal
2.
Combination reaction
C.
Iron nail immersed in copper sulphate solution
3.
Decomposition reaction
D.
Addition of barium chloride solution to aluminium sulphate solution
4.
Displacement reaction
Consider the following reaction:
Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s)
Which of the following statements about the given reaction is NOT correct?
