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Elimination of bromine from 2- bromobutane results in the formation of
  • a)
    predominantly 2-butyne
  • b)
    predominantly 2-butene
  • c)
    predominantly 1-butene
  • d)
    equimolar mixture of 1 and 2-butene
Correct answer is option 'B'. Can you explain this answer?
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Elimination of bromine from 2- bromobutane results in the formation of...
Formation of 2-Butene from 2-Bromobutane by Elimination of Bromine

The elimination reaction of bromine from 2-bromobutane results in the formation of predominantly 2-butene. This can be explained by considering the reaction mechanism and the stability of the resulting products.

Reaction Mechanism:
The elimination reaction is a type of organic reaction in which a small molecule (such as a halide) is removed from a larger molecule to form a new compound. In this case, bromine is eliminated from 2-bromobutane to form a double bond.

The reaction proceeds through an E2 (bimolecular elimination) mechanism. In the E2 mechanism, the elimination of the halide and the formation of the double bond occur simultaneously in a concerted manner. The reaction is bimolecular because it involves the nucleophile (base) and the substrate (2-bromobutane) in the rate-determining step.

Stability of the Products:
The stability of the resulting products plays a crucial role in determining the predominant product formed. In this case, the elimination of bromine from 2-bromobutane can lead to the formation of two possible products: 1-butene and 2-butene.

1-Butene:
1-Butene is an alkene with the double bond located at the first carbon atom of the butene chain. However, the stability of this product is relatively low due to the presence of a primary carbon-carbon double bond. Primary carbons have only one alkyl group attached to them, making them less stable compared to secondary or tertiary carbons.

2-Butene:
2-Butene, on the other hand, is an alkene with the double bond located at the second carbon atom of the butene chain. This product is more stable than 1-butene because it has a secondary carbon-carbon double bond. Secondary carbons have two alkyl groups attached to them, providing more electron density and stability to the double bond.

Explanation for Predominant Formation of 2-Butene:
The stability of the products determines the relative rates of their formation. In the case of the elimination of bromine from 2-bromobutane, the stability of 2-butene makes it the predominant product. The greater stability of the secondary carbon-carbon double bond in 2-butene compared to the primary carbon-carbon double bond in 1-butene leads to a faster rate of formation for 2-butene.

Additionally, the E2 mechanism involves a concerted reaction, where the base abstracts the proton and the bromine is eliminated simultaneously. The regioselectivity of the reaction is determined by the stability of the transition state. The transition state leading to the formation of 2-butene is more stable than that for 1-butene, further favoring the formation of 2-butene as the major product.

In conclusion, the elimination of bromine from 2-bromobutane predominantly results in the formation of 2-butene due to its higher stability compared to 1-butene. The reaction follows an E2 mechanism, and the regioselectivity is determined by the stability of the transition state.
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Elimination of bromine from 2- bromobutane results in the formation ofa)predominantly 2-butyneb)predominantly 2-butenec)predominantly 1-butened)equimolar mixture of 1 and 2-buteneCorrect answer is option 'B'. Can you explain this answer?
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