Question

Complete the following equation: \[ \text{Anisole} + CH_3Cl \xrightarrow{\text{Anhyd. } AlCl_3} \; ? \]

Answer 1

Concept:
This reaction represents a Friedel–Crafts alkylation, which involves the introduction of an alkyl group onto an aromatic ring using an alkyl halide and a Lewis acid catalyst. In this case, the –OCH\(_3\) (methoxy) group is present, which activates the benzene ring, making it more reactive towards electrophilic aromatic substitution.
The –OCH\(_3\) group also acts as a directing group, specifically activating the ortho and para positions on the aromatic ring for substitution.
Step 1: Understand the activating effect of the –OCH\(_3\) group.
The methoxy group (-OCH\(_3\)) is an electron-donating group through resonance and inductive effects. This electron donation makes the benzene ring more electron-rich and more susceptible to attack by an electrophile (in this case, the alkyl group). As a result, the methoxy group directs the electrophilic substitution to the ortho and para positions relative to itself.
Step 2: Mechanism of the Friedel–Crafts alkylation.
In the Friedel–Crafts alkylation, the alkyl halide reacts with a Lewis acid (such as AlCl\(_3\)) to generate an alkyl cation, which is the electrophile. This alkyl cation then attacks the electron-rich benzene ring at the ortho or para positions, leading to the formation of alkylated products.
Step 3: Specific substitution pattern due to the –OCH\(_3\) group.
Due to the activating nature of the –OCH\(_3\) group, the substitution occurs predominantly at the ortho and para positions of the benzene ring. The para substitution is generally favored due to less steric hindrance compared to the ortho positions.
Step 4: Final products.
The products of the reaction are:

• o-Methylanisole (ortho product)
• p-Methylanisole (para product, major product)

The para product (p-Methylanisole) is the major product due to its lower steric hindrance and more favorable formation.
Final Answer:
The reaction produces both o-Methylanisole and p-Methylanisole, with p-Methylanisole being the major product due to the para substitution preference.

Answer 2

Concept:
Nitration is a type of electrophilic aromatic substitution reaction, in which a nitro group (\(NO_2\)) is introduced onto an aromatic ring. In the case of a methoxy group (\(-OCH_3\)) attached to the ring, it activates the benzene ring and directs the nitration reaction to the ortho and para positions relative to itself. The methoxy group is an electron-donating group, which makes the ring more reactive towards electrophilic substitution.

Step 1: Understand the directing effect of the methoxy group.
The methoxy group (\(-OCH_3\)) is an electron-donating group through resonance, which increases the electron density on the aromatic ring. This makes the ring more susceptible to attack by electrophiles, such as the nitronium ion (\(NO_2^+\)) in nitration reactions. The electron-donating effect of the methoxy group directs the substitution to the ortho and para positions relative to its position on the ring.
Step 2: Mechanism of nitration.
In nitration, the nitronium ion (\(NO_2^+\)) is generated in the presence of a strong acid like \(HNO_3\) and \(H_2SO_4\). The nitronium ion acts as the electrophile, which attacks the electron-rich benzene ring. The methoxy group makes the ortho and para positions more reactive, so the nitration occurs predominantly at these positions.
Step 3: Substitution pattern due to the methoxy group.
Since the methoxy group activates the ring and directs substitution to the ortho and para positions, the resulting products are primarily the ortho and para nitro compounds. The para substitution is typically the major product because it experiences less steric hindrance compared to the ortho position.
Step 4: Final products.
The products of the nitration reaction are:

• o-Nitroanisole (ortho product)
• p-Nitroanisole (para product, major product)

The para-substituted product (p-Nitroanisole) is favored due to the steric and electronic factors that make it more stable.
Final Answer:
The nitration of anisole results in the formation of o-Nitroanisole and p-Nitroanisole, with p-Nitroanisole being the major product due to its preference for para substitution.

Answer 3

Concept:
The Williamson synthesis is an ether synthesis reaction that involves the reaction of an alkyl halide with an alkoxide ion. To minimize the possibility of elimination and ensure the formation of the desired ether, it is important to use a primary alkyl halide, as primary alkyl halides are less prone to elimination compared to secondary or tertiary ones.
To prepare tert-butyl ethyl ether, the reaction is as follows: \[ (C_2H_5)Cl + (CH_3)_3CONa \]
Step 1: Reacting alkyl halide with sodium alkoxide.
The alkyl halide used is ethyl chloride (\(C_2H_5Cl\)), which is a primary alkyl halide. The sodium alkoxide used is sodium tert-butoxide (\((CH_3)_3CONa\)), which is a strong nucleophile that will attack the ethyl chloride to form tert-butyl ethyl ether.
Step 2: Formation of the ether.
In the Williamson synthesis, the alkoxide ion attacks the primary alkyl halide via an SN2 mechanism, resulting in the formation of the desired ether (tert-butyl ethyl ether) and the elimination of sodium chloride (NaCl).
Final Answer:

• Alkyl halide: Ethyl chloride
• Sodium alkoxide: Sodium tert-butoxide

Answer 4

Concept:
In the cleavage of alkyl aryl ethers, the bond between the alkyl group and oxygen is broken, rather than the aryl–oxygen bond. This is because:

• The aryl–O bond has partial double bond character.
• This bond is stronger and less likely to undergo cleavage.

The reaction for the cleavage of an alkyl aryl ether is as follows: \[ C_6H_5OCH_3 + HI \rightarrow C_6H_5OH + CH_3I \]
Step 1: Bond cleavage mechanism.
In this reaction, the alkyl group (in this case, methyl) is cleaved from the oxygen atom, forming methyl iodide (\(CH_3I\)) and phenol (\(C_6H_5OH\)). The bond between the alkyl group and the oxygen atom is broken due to the attack by the iodide ion (I⁻), which displaces the methyl group.
Step 2: Why methanol and iodobenzene are not formed.
The cleavage does not occur at the aryl–oxygen bond because:

• The aryl–O bond is much stronger due to its partial double bond character.
• This bond does not break easily under normal conditions, making iodobenzene and methanol unlikely products.

Final Answer:
The products formed in the cleavage of methyl phenyl ether (anisole) are:

• Phenol (\(C_6H_5OH\))
• Methyl iodide (\(CH_3I\))

Methanol and iodobenzene are not formed due to the stronger aryl–oxygen bond.

Answer 5

Concept:
In a tetrahedral geometry, the bond angle is typically 109.5°. In ethers, two bulky alkyl groups are attached to the oxygen atom. The presence of these large groups leads to mutual repulsion, which slightly increases the bond angle from the typical tetrahedral angle.

• The two bulky alkyl groups attached to oxygen experience repulsion.
• This repulsion causes a slight increase in the bond angle.

Therefore, the bond angle in an ether is approximately: \[ \text{C–O–C bond angle} \approx 111^\circ \]
Step 1: Understanding the influence of alkyl groups.
The bulky alkyl groups attached to the oxygen atom in ethers cause the electron clouds to push against each other. This repulsion increases the bond angle slightly above the regular 109.5° seen in a perfect tetrahedral structure.
Step 2: Final Conclusion.
The bond angle in ethers is slightly greater than the ideal tetrahedral angle due to the repulsion between the bulky alkyl groups, which causes the C–O–C bond angle to be around 111°.