Consider the following organic halides: (I) Bromobenzene, (II) 2-Bromopropane, (III) 1-Bromopropane. The correct order of reactivity towards \(S_N2\) reaction is:
Show Hint
Steric hindrance is the primary factor limiting the rate of \(S_N2\) reactions.
Step 1: Recall what controls SN2 speed. The SN2 reaction happens by a backside attack of the nucleophile on the carbon bearing the halogen. So the less crowded that carbon is, the faster the reaction. Crowding (steric hindrance) slows SN2 down. Step 2: Classify each halide. Bromobenzene (I) is an aryl halide. 2-Bromopropane (II) is a secondary halide. 1-Bromopropane (III) is a primary halide. Step 3: Rank the alkyl ones. A primary carbon is less crowded than a secondary carbon, so primary reacts faster in SN2. Thus III (primary) is faster than II (secondary). Step 4: Place bromobenzene. In bromobenzene the C-Br bond has partial double bond character from resonance and the ring blocks backside attack, so aryl halides essentially do not undergo SN2. So I is the slowest by far. Step 5: Combine the order. Putting it together, primary beats secondary beats aryl: $III > II > I$. Step 6: Conclude. The correct reactivity order toward SN2 is $III > II > I$. \[ \boxed{III > II > I} \]
Was this answer helpful?
0
Top Questions on Nucleophilic and electrophilic substitution reactions (both aromatic and aliphatic)
