Problem 118
Question
Alkyl halides and alcohols easily undergo nucleophilic substitution either through \(\mathrm{S}_{\mathrm{N}} 1\) or \(\mathrm{S}_{\mathrm{N}} 2\) mechanism. The relative case of these two processes depends upon the nature of the substrate (alkyl group as well as leaving group), nature of nucleophile and also upon the nature of solvent. \(\mathrm{S}_{\mathrm{N}} 1\) mechanism involves the formation of carbocation as intermediate while \(\mathrm{S}_{\mathrm{N}} 2\) mechanism involves the formation of a transition pentavalent state. \(\mathrm{S}_{\mathrm{N}} 1\) is the main mechanism in \(3^{\circ}\) alkyl halides and alcohols, while \(\mathrm{S}_{\mathrm{N}} 2\) mechanism is the path adopted by most of the \(1^{\circ}\) alkyl halides and \(2^{\circ}\) alkyl halides may follow \(\mathrm{S}_{\mathrm{N}} 1\) as well as \(\mathrm{S}_{\mathrm{N}} 2\). Neopentyl alcohol, \(\mathrm{Me}_{3} \mathrm{CCH}_{2} \mathrm{OH}\), reacts with HX according to (a) \(\mathrm{S}_{\mathrm{N}} 1\) mechanism (b) \(\mathrm{S}_{\mathrm{N}} 2\) mechanism (c) Both (a) and (b) (d) None
Step-by-Step Solution
VerifiedKey Concepts
SN1 mechanism
- Step 1: Leaving group detaches forming a carbocation.
- Step 2: Nucleophile attacks the carbocation to form the final product.
SN2 mechanism
- One-step mechanism involving simultaneous bond-forming and bond-breaking.
- Good for primary substrates due to less steric hindrance.
Steric hindrance
- Restricts nucleophilic approach to the electrophilic center.
- Commonly reduces the likelihood of SN2 reactions.
Primary alkyl groups
- Easily undergo SN2 reactions due to less steric hindrance.
- Carbocation formation in SN1 is usually less stable.