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Elimination of bromine from 2-bromobutane results in the formation of
- a)Predominantly 2-butene
- b)Equimolar mixture of 1 and 2-butene
- c)Predominantly 1-butene
- d)Predominantly 2-butyne.
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
Elimination of bromine from 2-bromobutane results in the formation ofa...
2-bromobutane → CH3CH = CHCH3 + CH3CH2CH=CH2
The major product follows Saytzeff rule.
Hence option (A) is the correct answer.
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Elimination of bromine from 2-bromobutane results in the formation ofa...
Elimination of Bromine from 2-Bromobutane
The elimination reaction of bromine from 2-bromobutane is a type of E2 (bimolecular elimination) reaction. During this reaction, the bromine atom is removed from the carbon chain, resulting in the formation of a double bond.
E2 Reaction Mechanism
The E2 reaction occurs in a single step, involving the simultaneous removal of a proton and a leaving group (in this case, bromine) from adjacent carbon atoms. The leaving group takes a pair of electrons from the carbon-hydrogen bond, forming a double bond between the two carbon atoms.
The reaction mechanism can be summarized as follows:
1. The base (usually a strong base such as hydroxide ion, OH-) approaches the hydrogen atom on the carbon adjacent to the bromine atom.
2. The base abstracts the proton, creating a carbanion intermediate and leaving a bromide ion.
3. The carbanion intermediate then rearranges to form the most stable alkene through a transition state.
4. Finally, the bromide ion combines with a proton from the solvent to form HBr.
Stability of Alkenes
The stability of alkenes is determined by the degree of substitution at the double bond. The more substituted the double bond, the more stable the alkene. This stability is due to the increased electron density around the double bond, which is provided by the alkyl groups attached to the carbon atoms.
Explanation of the Correct Answer
In the case of 2-bromobutane, elimination of bromine results in the formation of 2-butene. This is because the double bond in 2-butene is more substituted than the double bond in 1-butene. Therefore, the stability of the alkene product, 2-butene, is higher compared to that of 1-butene.
The stability of alkenes follows the order: tertiary > secondary > primary. In 2-butene, the double bond is formed between two secondary carbons, making it more stable than 1-butene, which has a double bond between a primary and a secondary carbon.
Hence, the correct answer is option 'A' - Predominantly 2-butene.
It is important to note that while the formation of 2-butene is favored, a small amount of 1-butene may also be formed as a minor product due to the presence of a primary carbon adjacent to the double bond. However, the major product will be 2-butene.
The elimination reaction of bromine from 2-bromobutane is a type of E2 (bimolecular elimination) reaction. During this reaction, the bromine atom is removed from the carbon chain, resulting in the formation of a double bond.
E2 Reaction Mechanism
The E2 reaction occurs in a single step, involving the simultaneous removal of a proton and a leaving group (in this case, bromine) from adjacent carbon atoms. The leaving group takes a pair of electrons from the carbon-hydrogen bond, forming a double bond between the two carbon atoms.
The reaction mechanism can be summarized as follows:
1. The base (usually a strong base such as hydroxide ion, OH-) approaches the hydrogen atom on the carbon adjacent to the bromine atom.
2. The base abstracts the proton, creating a carbanion intermediate and leaving a bromide ion.
3. The carbanion intermediate then rearranges to form the most stable alkene through a transition state.
4. Finally, the bromide ion combines with a proton from the solvent to form HBr.
Stability of Alkenes
The stability of alkenes is determined by the degree of substitution at the double bond. The more substituted the double bond, the more stable the alkene. This stability is due to the increased electron density around the double bond, which is provided by the alkyl groups attached to the carbon atoms.
Explanation of the Correct Answer
In the case of 2-bromobutane, elimination of bromine results in the formation of 2-butene. This is because the double bond in 2-butene is more substituted than the double bond in 1-butene. Therefore, the stability of the alkene product, 2-butene, is higher compared to that of 1-butene.
The stability of alkenes follows the order: tertiary > secondary > primary. In 2-butene, the double bond is formed between two secondary carbons, making it more stable than 1-butene, which has a double bond between a primary and a secondary carbon.
Hence, the correct answer is option 'A' - Predominantly 2-butene.
It is important to note that while the formation of 2-butene is favored, a small amount of 1-butene may also be formed as a minor product due to the presence of a primary carbon adjacent to the double bond. However, the major product will be 2-butene.
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