
The major product of the following reaction is:

(A)
(B)
(C)

(D)

Answer
161.7k+ views
Hint: Given compound name is \[\left( 2-bromo-2-methyl\text{ }butyl \right)\] benzene and sodium ethoxide (\[{{C}_{2}}{{H}_{5}}ONa\]) which acts as a strong nucleophile reagent and also a strong base and thus, it tend to donate electron or seek cation. The reaction between both the compounds is followed by a substitution reaction (one nucleophile will substitute for another nucleophile).
Complete Step by Step Solution:
In a given compound \[\left( 2-bromo-2-methyl\text{ }butyl \right)\], bromine’s electronegativity is quite high as compared to carbon to which it is bonded. Due to its high electronegativity, it tends to attract a bond pair of electrons towards itself and become negatively charged. This results in creation of three-degree carbonation such as

The resultant compound is unstable because it is a charge compound and not neutral compound and it tends to get neutral by the attack of another nucleophile. Sodium ethoxide (\[{{C}_{2}}{{H}_{5}}ONa\]) and ethanol (\[{{C}_{2}}{{H}_{5}}OH\]) are strong nucleophilic reagents (electron-rich), attack on it and side product will be\[NaBr\](attacked by sodium ethoxide) or\[HBr\] (attack by ethanol) such as

The major product is obtained by nucleophile substitution in this reaction and products obtained by elimination reaction are minor.
Thus, the correct option is A.
Note: Bromine (leaving group) is a good nucleophile due to which substitution reaction is preferred. If in the place of bromine, a strong and bulky base is present then an elimination reaction takes place and the major product will be obtained by elimination reaction and not by a substitution reaction. Also, neutral species are more stable than charged species.
Complete Step by Step Solution:
In a given compound \[\left( 2-bromo-2-methyl\text{ }butyl \right)\], bromine’s electronegativity is quite high as compared to carbon to which it is bonded. Due to its high electronegativity, it tends to attract a bond pair of electrons towards itself and become negatively charged. This results in creation of three-degree carbonation such as

The resultant compound is unstable because it is a charge compound and not neutral compound and it tends to get neutral by the attack of another nucleophile. Sodium ethoxide (\[{{C}_{2}}{{H}_{5}}ONa\]) and ethanol (\[{{C}_{2}}{{H}_{5}}OH\]) are strong nucleophilic reagents (electron-rich), attack on it and side product will be\[NaBr\](attacked by sodium ethoxide) or\[HBr\] (attack by ethanol) such as

The major product is obtained by nucleophile substitution in this reaction and products obtained by elimination reaction are minor.
Thus, the correct option is A.
Note: Bromine (leaving group) is a good nucleophile due to which substitution reaction is preferred. If in the place of bromine, a strong and bulky base is present then an elimination reaction takes place and the major product will be obtained by elimination reaction and not by a substitution reaction. Also, neutral species are more stable than charged species.
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