Question

The ratio of number of oxygen atoms to bromine atoms in the product $Q$ is ___ $\times 10^{-1}$.

Answer 1

Correct Answer: 15

The initial compound with \( \text{OC}_2\text{H}_5 \) undergoes nitration to introduce a \( \text{NO}_2 \) group, yielding intermediate \( P \).

\[ \text{OC}_2\text{H}_5 \xrightarrow{\text{HNO}_3, \text{H}_2\text{SO}_4} \text{OC}_2\text{H}_5\text{NO}_2 \]

Bromination of \( P \) in the presence of Fe introduces two bromine atoms on the aromatic ring, resulting in product \( Q \).

\[ \text{OC}_2\text{H}_5\text{NO}_2 \xrightarrow{\text{2 Br}_2, \text{Fe}} \text{Br-OC}_2\text{H}_5\text{NO}_2\text{Br} \]

Count the number of oxygen and bromine atoms in product \( Q \):

• Oxygen atoms: 3 (one from \( \text{OC}_2\text{H}_5 \) and two from \( \text{NO}_2 \))
• Bromine atoms: 2

The ratio of oxygen to bromine atoms in \( Q \) is:

\[ \frac{3}{2} \times 10^{-1} = 15 \times 10^{-1} \]

Answer: 15

Answer 2

The problem asks for the ratio of the number of oxygen atoms to bromine atoms in the final major product Q, formed from a two-step reaction starting with ethoxybenzene. The result should be expressed in the form of \( \text{__} \times 10^{-1} \).

Concept Used:

1. Electrophilic Aromatic Substitution (EAS): This is the core mechanism for both reactions. The reactivity and orientation of substitution on a substituted benzene ring are governed by the nature of the substituent already present.

2. Directing Effects of Substituents:

• Activating Groups: Groups that donate electron density to the ring, making it more reactive towards electrophiles. They are typically ortho, para-directing. The ethoxy group (\(-OC_2H_5\)) is a strong activating group due to the +R (resonance) effect of the oxygen atom.
• Deactivating Groups: Groups that withdraw electron density from the ring, making it less reactive. They are typically meta-directing. The nitro group (\(-NO_2\)) is a strong deactivating group due to its -R and -I effects.

3. Nitration of Benzene Derivatives: The reaction with a mixture of concentrated nitric acid (\(HNO_3\)) and sulfuric acid (\(H_2SO_4\)) introduces a nitro group (\(-NO_2\)) onto the aromatic ring.

4. Bromination of Benzene Derivatives: The reaction with bromine (\(Br_2\)) in the presence of a Lewis acid catalyst (like Fe or \(FeBr_3\)) introduces a bromine atom (\(-Br\)) onto the aromatic ring.

When multiple substituents are present, the directing effect is determined by the more powerful activating group.

Step-by-Step Solution:

Step 1: Identify the major product P from the nitration of ethoxybenzene.

The starting material is ethoxybenzene. The ethoxy group (\(-OC_2H_5\)) is an ortho, para-directing and activating group. During nitration, the nitro group (\(-NO_2\)) will be directed to the ortho and para positions. Due to steric hindrance from the bulky ethoxy group, the para-substituted product is the major product.

Reaction:

\[ \text{Ethoxybenzene} \xrightarrow{HNO_3, H_2SO_4} \text{1-Ethoxy-4-nitrobenzene (major product P)} \]

So, P is 1-ethoxy-4-nitrobenzene.

Step 2: Identify the major product Q from the bromination of product P.

The reactant for the second step is P (1-ethoxy-4-nitrobenzene). This molecule has two substituents on the benzene ring:

• The ethoxy group (\(-OC_2H_5\)) at position 1, which is strongly activating and ortho, para-directing.
• The nitro group (\(-NO_2\)) at position 4, which is strongly deactivating and meta-directing.

The position of the incoming electrophile (bromine) is determined by the more powerful activating group, which is the ethoxy group. The positions ortho to the ethoxy group are positions 2 and 6. These are also the positions meta to the nitro group, so both groups direct the incoming bromine to the same positions.

The reaction uses \(2Br_2\), indicating that two bromine atoms will be substituted onto the ring. Since positions 2 and 6 are equivalent and activated, both will be brominated.

Reaction:

\[ \text{1-Ethoxy-4-nitrobenzene (P)} \xrightarrow{2Br_2, Fe} \text{2,6-Dibromo-1-ethoxy-4-nitrobenzene (major product Q)} \]

The final product Q is 2,6-Dibromo-1-ethoxy-4-nitrobenzene.

Final Computation & Result:

Step 3: Count the number of oxygen and bromine atoms in the final product Q.

The structure of Q is 2,6-Dibromo-1-ethoxy-4-nitrobenzene.

• Number of Oxygen atoms: There is one oxygen atom in the ethoxy group (\(-OC_2H_5\)) and two oxygen atoms in the nitro group (\(-NO_2\)). Total number of oxygen atoms = 1 + 2 = 3.
• Number of Bromine atoms: There are two bromine atoms substituted on the ring at positions 2 and 6. Total number of bromine atoms = 2.

Step 4: Calculate the ratio and express it in the required format.

The ratio of the number of oxygen atoms to bromine atoms is:

\[ \text{Ratio} = \frac{\text{Number of oxygen atoms}}{\text{Number of bromine atoms}} = \frac{3}{2} = 1.5 \]

We need to express this result in the form \( \text{__} \times 10^{-1} \).

\[ y \times 10^{-1} = 1.5 \] \[ y = \frac{1.5}{10^{-1}} = 1.5 \times 10 = 15 \]

The ratio of the number of oxygen atoms to bromine atoms in the product Q is 15 \( \times 10^{-1} \).