Question

2-Chloro-2-methylbutane on reaction with sodium methoxide in methanol yields
a.
b.
c.
(A) All of these
(B) (a) and (c)
(C) (c)only
(D) (a) and (b)

Answer 1

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 ${S_N}2$

${S_N}1$	${S_N}2$
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.