Answer
Verified
97.5k+ views
Hint: When 2-Chloro-2-methylbutane reacts with sodium methoxide in methanol, it undergoes \[{S_N}1\] and \[{E_1}\]. \[{S_N}1\]means unimolecular nucleophilic substitution reaction and \[{E_1}\]means unimolecular elimination reaction. On knowing what type of reaction can take place in this \[{S_N}1\]and \[{E_1}\]mechanism we can determine the product.
Complete step by step answer:
2-Chloro-2-methylbutane when reacted with sodium methoxide in methanol, it will form a carbocation by the removal of \[NaCl\]. The formed carbocation will undergo \[{S_N}1\] and \[{E_1}\].
In \[{S_N}1\]mechanism, nucleophile will come and attack the same position of the leaving group, that means that the nucleophile, methoxy group \[C{H_3}{O^ - }\] will be attached in the position of the leaving chlorine \[C{l^ - }\] group. Thus, leading to the formation of 2-Methoxy-2-methylbutane i.e. compound (a) in the figure.
In the\[{E_1}\] mechanism, the deprotonation will occur in the carbon atom which is nearer to the carbocation and hence leading to the formation of alkene. In 2-methylbutane carbocation, there are two carbon atoms which are nearer to the carbocation. One carbon atom is at the terminal position and the other carbon atom is at the third position.
In the terminal carbon atom, a proton is removed, thus leading to the formation of 2-methyl-1-butene i.e. compound (b) in the figure.
In the third carbon atom, a proton will be removed, thus leading to the formation of 2-methyl-2-butene i.e. compound (c) in the figure.
Therefore, all the three compounds(a), (b) and (c) during the reaction of 2-Chloro-2-methylbutane with sodium methoxide in methanol.
Hence, the correct answer is option (A) all the above.
Additional information:
Comparison of \[{S_N}1\] and \[{S_N}2\]
Note: In the reaction both elimination and substitution reaction take place. In the elimination reaction the deprotonation of the carbocation takes places to form alkene, while in the substitution reaction, substitution of leaving group by a nucleophile takes place. Here substitution takes place at only one position while elimination is taking place at two positions.
Complete step by step answer:
2-Chloro-2-methylbutane when reacted with sodium methoxide in methanol, it will form a carbocation by the removal of \[NaCl\]. The formed carbocation will undergo \[{S_N}1\] and \[{E_1}\].
In \[{S_N}1\]mechanism, nucleophile will come and attack the same position of the leaving group, that means that the nucleophile, methoxy group \[C{H_3}{O^ - }\] will be attached in the position of the leaving chlorine \[C{l^ - }\] group. Thus, leading to the formation of 2-Methoxy-2-methylbutane i.e. compound (a) in the figure.
In the\[{E_1}\] mechanism, the deprotonation will occur in the carbon atom which is nearer to the carbocation and hence leading to the formation of alkene. In 2-methylbutane carbocation, there are two carbon atoms which are nearer to the carbocation. One carbon atom is at the terminal position and the other carbon atom is at the third position.
In the terminal carbon atom, a proton is removed, thus leading to the formation of 2-methyl-1-butene i.e. compound (b) in the figure.
In the third carbon atom, a proton will be removed, thus leading to the formation of 2-methyl-2-butene i.e. compound (c) in the figure.
Therefore, all the three compounds(a), (b) and (c) during the reaction of 2-Chloro-2-methylbutane with sodium methoxide in methanol.
Hence, the correct answer is option (A) all the above.
Additional information:
Comparison of \[{S_N}1\] and \[{S_N}2\]
\[{S_N}1\] | \[{S_N}2\] |
Unimolecular | Bimolecular |
Rate depends on two species | Rate depends only on one species |
Second order | First order |
Solvent used is polar protic | Solvent used is Polar aprotic |
Alkyl halide reacts in the order\[3^\circ > 2^\circ > 1^\circ \] | Alkyl halide reacts in the order\[1^\circ > 2^\circ > 3^\circ \] |
Note: In the reaction both elimination and substitution reaction take place. In the elimination reaction the deprotonation of the carbocation takes places to form alkene, while in the substitution reaction, substitution of leaving group by a nucleophile takes place. Here substitution takes place at only one position while elimination is taking place at two positions.
Recently Updated Pages
Write a composition in approximately 450 500 words class 10 english JEE_Main
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
Write an article on the need and importance of sports class 10 english JEE_Main
Name the scale on which the destructive energy of an class 11 physics JEE_Main
Choose the exact meaning of the given idiomphrase The class 9 english JEE_Main
Choose the one which best expresses the meaning of class 9 english JEE_Main
Other Pages
The capacity of a pure capacitor is 1 farad In dc circuits class 11 physics JEE_Main
Formula for number of images formed by two plane mirrors class 12 physics JEE_Main
If a wire of resistance R is stretched to double of class 12 physics JEE_Main
Electric field due to uniformly charged sphere class 12 physics JEE_Main
An object is moving with speed v0 towards a spherical class 12 physics JEE_Main