Answer
64.8k+ views
Hint: Here we have an alkyl halide reacting with methoxide ion (\[C{H_3}{O^ - }\] ) in the presence of methanol (\[C{H_3}OH\]). The alkyl halide undergoes an elimination reaction called dehydrohalogenation.
Complete Step by Step Solution:
In this question, we have to predict the product that will be formed when 1-bromo-2-methylpropane reacts with a methoxide ion (\[C{H_3}{O^ - }\] ) in the presence of methanol (\[C{H_3}OH\]).
The methoxide ion is an electron-rich species having lone pairs on the oxygen atom. The carbon-bromine (\[C - Br\] ) bond in 1-bromo-2-methylpropane is polarised towards bromine because of its higher electronegativity. This makes the carbon atom a bit deficient in electrons. Therefore, one might think that the methoxide ion would substitute the bromine in a nucleophilic substitution reaction to form an ether as the final product.
However, this does not happen because the methoxide ion does not act as a nucleophile. Instead, it prefers to attack a hydrogen atom on 1-bromo-2-methylpropane and abstracts it as a proton (\[{H^ + }\]). Thus, the methoxide ion acts as a base.
When the methoxide ion acts as a base, it initiates an elimination reaction instead of a substitution reaction.
Although elimination reactions can occur via multiple mechanisms, in this case, the mechanism followed is a “bimolecular” elimination mechanism, also known as the E2 mechanism.
This is a single-step mechanism involving the abstraction of a beta-hydrogen as a proton and the elimination of bromine as a bromide ion (\[B{r^ - }\]) to form an alkene as the final, major product.
![](https://www.vedantu.com/question-sets/d2ae2900-4872-40d5-8726-ce69f5970fc35225064408196397343.png)
Image: Elimination reaction of 1-bromo-2-methylpropane to form 2-methylprop-1-ene
Thus, option C is correct.
Note: Nucleophilicity and basicity are not the same. Nucleophilicity refers to the ability of an electron-rich species to donate its lone pairs of electrons to electron-deficient species. Therefore, all nucleophiles are Lewis bases. But basicity refers to the ability of an electron-rich species to accept a proton. When an electron-rich species acts as an electron donor (i.e., a nucleophile), it causes substitution. When an electron-rich species acts as a proton acceptor (i.e., a base), it causes elimination. It is highly advised that students learn about the extent of nucleophilicities and basicities of different electron-rich species.
Complete Step by Step Solution:
In this question, we have to predict the product that will be formed when 1-bromo-2-methylpropane reacts with a methoxide ion (\[C{H_3}{O^ - }\] ) in the presence of methanol (\[C{H_3}OH\]).
The methoxide ion is an electron-rich species having lone pairs on the oxygen atom. The carbon-bromine (\[C - Br\] ) bond in 1-bromo-2-methylpropane is polarised towards bromine because of its higher electronegativity. This makes the carbon atom a bit deficient in electrons. Therefore, one might think that the methoxide ion would substitute the bromine in a nucleophilic substitution reaction to form an ether as the final product.
However, this does not happen because the methoxide ion does not act as a nucleophile. Instead, it prefers to attack a hydrogen atom on 1-bromo-2-methylpropane and abstracts it as a proton (\[{H^ + }\]). Thus, the methoxide ion acts as a base.
When the methoxide ion acts as a base, it initiates an elimination reaction instead of a substitution reaction.
Although elimination reactions can occur via multiple mechanisms, in this case, the mechanism followed is a “bimolecular” elimination mechanism, also known as the E2 mechanism.
This is a single-step mechanism involving the abstraction of a beta-hydrogen as a proton and the elimination of bromine as a bromide ion (\[B{r^ - }\]) to form an alkene as the final, major product.
![](https://www.vedantu.com/question-sets/d2ae2900-4872-40d5-8726-ce69f5970fc35225064408196397343.png)
Image: Elimination reaction of 1-bromo-2-methylpropane to form 2-methylprop-1-ene
Thus, option C is correct.
Note: Nucleophilicity and basicity are not the same. Nucleophilicity refers to the ability of an electron-rich species to donate its lone pairs of electrons to electron-deficient species. Therefore, all nucleophiles are Lewis bases. But basicity refers to the ability of an electron-rich species to accept a proton. When an electron-rich species acts as an electron donor (i.e., a nucleophile), it causes substitution. When an electron-rich species acts as a proton acceptor (i.e., a base), it causes elimination. It is highly advised that students learn about the extent of nucleophilicities and basicities of different electron-rich species.
Recently Updated Pages
Write a composition in approximately 450 500 words class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What is the common property of the oxides CONO and class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What happens when dilute hydrochloric acid is added class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
If four points A63B 35C4 2 and Dx3x are given in such class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
The area of square inscribed in a circle of diameter class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)