Question

The order of reactivity of the compounds C₆H₅CH₂Br, C₆H₅CH(C₆H₅)Br, C₆H₅CH(CH₃)Br and C₆H₅C(CH₃)(C₆H₅)Br in S_N² reaction is

• A.
• B.
• C.
• D.

Answer 1

The Correct Option is A

The reactivity of alkyl halides in S_N² reactions is primarily influenced by the steric hindrance around the carbon atom to which the halogen is attached. The order of reactivity follows from the fact that less sterically hindered compounds react more readily in S_N² reactions. Hence, the reactivity decreases as the size of the substituents (such as phenyl groups or alkyl groups) increases.

• The compound with a simple alkyl group (C₆H₅CH₂Br) will react the fastest due to minimal steric hindrance.
• Adding an additional alkyl group (C₆H₅CH₂CH₂Br) increases the steric hindrance slightly, reducing the reactivity compared to (A).
• When substituting a phenyl group (C₆H₅) onto the carbon attached to the bromine, as in C₆H₅CH(C₆H₅)Br, the reactivity decreases further because of steric hindrance from the bulky phenyl group.

The correct order of reactivity is (A) C₆H₅CH₂Br < C₆H₅CH₂CH₂Br < C₆H₅CH(C₆H₅)Br < C₆H₅CH(C₆H₅)Br < C₆H₅CH₂CH₃Br.

The correct answer is (A) : .

Answer 2

To solve this problem, we need to determine the order of reactivity of the given compounds in an \(S_N2\) reaction. The reactivity in \(S_N2\) reactions is influenced by the steric hindrance and the ability of the leaving group (Br⁻ in this case) to leave.

1. Understanding \(S_N2\) Mechanism:
The \(S_N2\) reaction proceeds through a backside attack by the nucleophile (in this case, Br⁻), resulting in the inversion of configuration. The rate of the reaction depends on the steric hindrance at the electrophilic carbon (where the leaving group is attached). Less steric hindrance means a faster reaction.

2. Analyzing the Structure of Each Compound:
The compounds are different in the way alkyl groups are attached to the benzene ring and the position of the leaving group (Br):

• The first compound is a benzyl bromide (C₆H₅CH₂Br). The carbon attached to the bromine is a primary carbon with minimal steric hindrance, making it highly reactive in \(S_N2\) reactions.
• The second compound is a phenyl ethyl bromide (C₆H₅CH₂CH₂Br). It has a primary carbon attached to the bromine, but with slightly more steric hindrance compared to the benzyl bromide.
• The third compound is a para-methyl phenyl methyl bromide (C₆H₅CH(CH₃)Br). The carbon attached to the bromine is a secondary carbon, meaning it has more steric hindrance than the primary carbon compounds, reducing the reactivity in \(S_N2\) reactions.
• The fourth compound is a para-methyl benzyl bromide (C₆H₅CH₂CH₃Br). This compound has a primary carbon attached to the bromine but with a methyl group in the para position, which can cause additional steric hindrance compared to benzyl bromide, making it less reactive in \(S_N2\) reactions.

3. Comparing Reactivity:
The reactivity in \(S_N2\) reactions decreases with increasing steric hindrance. Thus, the order of reactivity is as follows: - Benzyl bromide (C₆H₅CH₂Br) is the most reactive due to minimal steric hindrance. - Phenyl ethyl bromide (C₆H₅CH₂CH₂Br) is slightly less reactive because of the longer alkyl chain. - Para-methyl phenyl methyl bromide (C₆H₅CH(CH₃)Br) is even less reactive due to the secondary carbon. - Para-methyl benzyl bromide (C₆H₅CH₂CH₃Br) is the least reactive due to additional steric hindrance from the methyl group in the para position.

Final Answer:
The correct order of reactivity in \(S_N2\) reactions is (A):
C₆H₅CH₂Br < C₆H₅CH₂CH₂Br < C₆H₅CH(CH₃)Br < C₆H₅CH₂CH₃Br.