In substitution reactions, the leaving group is displaced by a nucleophile, leading to a new product. The order and nature of the nucleophile and leaving group affect the reaction outcome.
The given problem involves a substitution reaction, where an atom or group in a molecule is replaced by another atom or group. To solve the problem, we need to identify the reaction mechanism and the reagents involved to determine the correct product.
In substitution reactions, particularly in organic chemistry, two common types are:
SN1 Reactions: Unimolecular nucleophilic substitution, where a carbocation intermediate forms. These reactions are typically seen with tertiary substrates.
SN2 Reactions: Bimolecular nucleophilic substitution, where a direct displacement occurs without intermediates. These reactions occur with primary and secondary substrates.
A benzene ring with two bromine atoms and a nitro group ($NO_2$). Sodium ethoxide ($C_2H_5ONa$) in ethanol ($C_2H_5OH$) is a strong nucleophile and a base. The nitro group is an electron-withdrawing group, which activates the benzene ring towards nucleophilic aromatic substitution, particularly at the positions ortho and para to it.
In this case, one of the bromine atoms will be replaced by the ethoxide group ($OC_2H_5$). The bromine atom that is more activated by the nitro group will be substituted preferentially. Since both bromine atoms are ortho to the nitro group, they are equally activated. Therefore, the product will be the substitution of one of the bromine atoms with the ethoxide group.
<p>The product 'P' formed is: </p>
Was this answer helpful?
0
0
Show Solution
Verified By Collegedunia
Approach Solution -2
Step 1: Understand the reaction.
The given compound is 1,2-dibromo-4-nitrobenzene, and it reacts with sodium ethoxide (C₂H₅ONa) in ethanol (C₂H₅OH). The reaction is a nucleophilic aromatic substitution (SNAr) reaction facilitated by the presence of an electron-withdrawing group (NO₂) on the aromatic ring.
Step 2: Role of the nitro group.
The –NO₂ group at the para position relative to one of the bromine atoms increases the susceptibility of the ortho and para positions to nucleophilic attack. Hence, the bromine atom that is ortho or para to the –NO₂ group can be replaced more easily by the nucleophile (–OC₂H₅).
Step 3: Mechanism (Nucleophilic aromatic substitution).
In this case, sodium ethoxide provides the nucleophile (ethoxide ion, –OC₂H₅), which attacks the carbon attached to the bromine atom that is ortho to the –NO₂ group. The reaction proceeds via the formation of a Meisenheimer complex (an intermediate where the nucleophile adds to the ring before bromide leaves).
The bromine atom directly ortho to the –NO₂ group is substituted by –OC₂H₅, leading to the product formation.
Step 4: Product formed.
The product formed is: 2-Bromo-4-nitrophenetole (C₆H₄BrNO₂OC₂H₅)
which can be represented as: Final Answer:
The product P formed is:
\[
\boxed{\text{2-Bromo-4-nitrophenetole}}
\]
</p>
