Question

Which of the following on heating gives an ether as major product?
P : C\(_6\)H\(_5\)CH\(_2\)Br + CH\(_3\)ONa
Q : C\(_6\)H\(_5\)ONa + CH\(_3\)Br
R : (CH\(_3\))\(_3\)C–Cl + CH\(_3\)ONa
S : C\(_6\)H\(_5\)CH = CHCl + CH\(_3\)ONa

• A. Both P and Q
• B. Both R and S
• C. Both P and R
• D. Both Q and S

Answer 1

The Correct Option is A

Let's evaluate each reaction to see if it results in the formation of an ether:

1. For P: C\(_6\)H\(_5\)CH\(_2\)Br + CH\(_3\)ONa:
- This is a nucleophilic substitution reaction. The CH\(_3\)O\(^-\) (methoxide ion) attacks the electrophilic carbon attached to the bromine (C\(_6\)H\(_5\)CH\(_2\)Br), resulting in the formation of an ether (C\(_6\)H\(_5\)CH\(_2\)OCH\(_3\)). - Therefore, this reaction gives an ether as the major product.

2. For Q: C\(_6\)H\(_5\)ONa + CH\(_3\)Br:
- This is also a nucleophilic substitution. The phenoxide ion (C\(_6\)H\(_5\)O\(^-\)) attacks the methyl group (CH\(_3\)) attached to bromine, forming phenyl methyl ether (C\(_6\)H\(_5\)OCH\(_3\)). - Therefore, this reaction also gives an ether as the major product.

3. For R: (CH\(_3\))\(_3\)C–Cl + CH\(_3\)ONa:
- In this case, the methoxide ion (CH\(_3\)O\(^-\)) attacks the tertiary carbon of the alkyl chloride (C\(_3\)Cl). However, because this is a tertiary carbocation intermediate, the reaction is more likely to result in a rearrangement or a different product rather than an ether. - Therefore, this reaction does not give an ether as the major product.

4. For S: C\(_6\)H\(_5\)CH = CHCl + CH\(_3\)ONa:
- The CH\(_3\)O\(^-\) will attack the vinyl chloride (C\(_6\)H\(_5\)CH = CHCl), but this is not likely to form an ether under normal conditions due to the nature of the vinyl group.
Therefore, this reaction does not give an ether as the major product. Thus, the reactions P and Q both lead to the formation of an ether as the major product.

So, the correct answer is (A): Both P and Q.

Answer 2

P: C\(_6\)H\(_5\)CH\(_2\)Br + CH\(_3\)ONa: This is a nucleophilic substitution reaction (S\(_N\)2), where the \(CH_3\)O\(^-\) ion (from sodium methoxide) attacks the electrophilic carbon of the C\(_6\)H\(_5\)CH\(_2\)Br molecule, replacing the bromine atom and forming an ether as the product: C\(_6\)H\(_5\)CH\(_2\)OCH\(_3\).

Q: C\(_6\)H\(_5\)ONa + CH\(_3\)Br: This reaction also leads to the formation of an ether by nucleophilic substitution (S\(_N\)2), where methoxide ion (CH\(_3\)O\(^-\)) from sodium methoxide attacks the electrophilic methyl group from CH\(_3\)Br, forming the ether product C\(_6\)H\(_5\)OCH\(_3\).

R: (CH\(_3\))\(_3\)C–Cl + CH\(_3\)ONa: This reaction involves a tertiary alkyl chloride. The nucleophilic substitution does not occur readily due to the steric hindrance of the tertiary carbon. Therefore, this reaction is unlikely to form an ether as the major product.

S: C\(_6\)H\(_5\)CH = CHCl + CH\(_3\)ONa: This is an elimination reaction, where CH\(_3\)ONa would likely cause the elimination of HCl to form a double bond, resulting in an alkene, not an ether.

Thus, the correct answer is (A), as both P and Q will give an ether as the major product.