Characteristic organic reactions — practice questions

1. Identify the reactants and structure: Reactant is 2-chloro-2-methylpropane, which is a tertiary halogenoalkane. The nucleophile is the hydroxide ion, OH⁻, from NaOH(aq).

2. Determine the mechanism type: Tertiary halogenoalkanes undergo nucleophilic substitution via the SN1 mechanism because they can form a stable tertiary carbocation intermediate.

3. Predict the product: The OH⁻ nucleophile will replace the Cl⁻ leaving group. The product is 2-methylpropan-2-ol.

4. Draw the mechanism: Step 1: Heterolytic fission of the C-Cl bond to form a tertiary carbocation and a chloride ion. This is the slow, rate-determining step. (CH₃)₃CCl → (CH₃)₃C⁺ + Cl⁻

Step 2: Rapid attack by the OH⁻ nucleophile on the planar carbocation. (CH₃)₃C⁺ + OH⁻ → (CH₃)₃COH

Mechanism Diagram: Draw the structure of 2-chloro-2-methylpropane. Show a curly arrow from the C-Cl bond to the Cl atom. Use a 'slow' label on the reaction arrow leading to the tertiary carbocation (CH₃)₃C⁺ and Cl⁻. Then, draw the planar carbocation. Show a curly arrow from the lone pair on the oxygen of an OH⁻ ion to the positive carbon atom. Use a 'fast' label on the reaction arrow leading to the final product, 2-methylpropan-2-ol.

1. Identify reactants and reaction type: Propene (CH₃CH=CH₂) is an unsymmetrical alkene. Hydrogen bromide (HBr) is a hydrogen halide. The reaction is electrophilic addition.

2. Apply Markovnikov's Rule: The major product will be formed via the more stable carbocation. HBr provides the electrophile H⁺. Pathway A (Major): H⁺ adds to the terminal CH₂ carbon, forming a secondary carbocation (CH₃C⁺HCH₃). This is more stable. Pathway B (Minor): H⁺ adds to the central CH carbon, forming a primary carbocation (CH₃CH₂C⁺H₂). This is less stable.

3. Predict the major product: The Br⁻ ion will attack the more stable secondary carbocation. The major product is 2-bromopropane.

4. Draw the mechanism for the major product: *Step 1: Draw propene and H-Br. Show the H-Br dipole (H$^{\delta+}$-Br$^{\delta-}$). Draw a curly arrow from the C=C double bond to the H$^{\delta+}$ of HBr. Draw another curly arrow from the H-Br bond to the Br atom. This forms the secondary carbocation (CH₃C⁺HCH₃) and a bromide ion (Br⁻).

Step 2: Draw the secondary carbocation. Draw a curly arrow from a lone pair on the bromide ion (Br⁻) to the positively charged carbon atom. This forms the final product, 2-bromopropane (CH₃CHBrCH₃).