Question

In the reaction,

$NaNH_2$/liq.ammonia $\longrightarrow$ A

The major product A is

Answer 1

2-amino-4-methoxybenzene

Answer 2

The reaction involves 1-chloro-4-methoxybenzene reacting with sodium amide ($NaNH_2$) in liquid ammonia. This is a classic example of nucleophilic aromatic substitution via the benzyne mechanism.

Step 1: Benzyne Formation (Elimination)

The strong base, amide ion ($NH_2^-$), abstracts an acidic proton ortho to the leaving group (Cl). The carbon bearing the leaving group (C1) and an adjacent carbon (C2 or C6) must bear a hydrogen atom. Simultaneously, the chloride ion departs, forming a triple bond (benzyne).

Let's number the carbons:

• C1: bonded to -Cl
• C2: ortho to C1, meta to -OCH3
• C3: meta to C1, ortho to -OCH3
• C4: para to C1, bonded to -OCH3
• C5: meta to C1, ortho to -OCH3
• C6: ortho to C1, ortho to -OCH3

The two possible ortho hydrogens available for abstraction are at C2 and C6.

The -OCH3 group is an electron-donating group (EDG) by resonance (+M effect) and an electron-withdrawing group by induction (-I effect).

The +M effect of -OCH3 makes the ortho-hydrogens (at C6) less acidic. The -I effect of -OCH3 makes the meta-hydrogens (at C2) slightly more acidic. Therefore, the abstraction of the proton at C2 is kinetically favored.

Abstraction of H at C2 and elimination of Cl at C1 leads to the formation of 1,2-didehydro-4-methoxybenzene (benzyne intermediate).

In this benzyne, the triple bond is between C1 and C2. The -OCH3 group is at C4.

• C1 is para to -OCH3.
• C2 is ortho to -OCH3.

Structure of 1,2-didehydro-4-methoxybenzene:

      C1≡C2
     /    \
    C6-----C3
     \    /
      C5--C4-OCH3

Step 2: Nucleophilic Attack and Protonation

The nucleophile ($NH_2^-$) attacks one of the carbons of the benzyne triple bond, and the negative charge forms on the other carbon, creating a carbanion intermediate. This carbanion is then protonated by liquid ammonia.

There are two possible attack sites for the nucleophile: C1 or C2.

Let's analyze the stability of the resulting carbanions. The regioselectivity of nucleophilic attack on substituted benzynes is governed by the electronic effects of the substituents on the stability of the intermediate carbanion.

The -OCH3 group is an EDG by resonance (+M effect) and a weak EWG by induction (-I effect). The +M effect is generally dominant.

• Electron-donating groups (EDGs) destabilize carbanions at ortho and para positions due to resonance electron donation.
• Electron-withdrawing groups (EWGs) stabilize carbanions.

For EDGs like -OCH3, the nucleophile preferentially attacks the carbon of the benzyne triple bond that is ortho to the EDG. This leads to the carbanion forming at the carbon para to the EDG. This is because the inductive effect of -OCH3 (being electron-withdrawing) can stabilize the carbanion, and it's stronger at closer positions. Also, forming the carbanion at the para position minimizes the destabilizing resonance effect from the EDG compared to an ortho carbanion.

Applying this rule to 1,2-didehydro-4-methoxybenzene:

• C2 is ortho to -OCH3.
• C1 is para to -OCH3.

The nucleophile ($NH_2^-$) will preferentially attack C2 (ortho to -OCH3).

This attack leads to the formation of a carbanion at C1 (para to -OCH3).

Intermediate carbanion:

      C1(-)---C2-NH2
     /
    C6-----C3
     \    /
      C5--C4-OCH3

This carbanion is then protonated by liquid ammonia ($NH_3$).

The -NH2 group is at C2, and the -OCH3 group is at C4.

The major product A is 2-amino-4-methoxybenzene.

Final Product Structure:

SMILES: COC1=CC(N)=C(C=C1)

The product is 2-amino-4-methoxybenzene (also known as 4-methoxy-2-aminobenzene).

The final answer is 2-amino-4-methoxybenzene

Explanation of the solution:

The reaction proceeds via the benzyne mechanism. The strong base $NaNH_2$ abstracts an ortho-hydrogen to the chlorine (specifically, the hydrogen at C2, which is meta to the methoxy group, is more acidic than the one at C6, which is ortho to methoxy). This forms a benzyne intermediate with the triple bond between C1 and C2. The methoxy group is an electron-donating group. In benzyne reactions with electron-donating groups, the nucleophile preferentially attacks the carbon ortho to the EDG (C2 in this case), leading to the formation of a carbanion at the para position (C1). This carbanion is then protonated by liquid ammonia. Thus, the amino group attaches at C2, and the methoxy group remains at C4, yielding 2-amino-4-methoxybenzene.