Predict the products of the following SN2 reactions. e.
+ CH3I —> f. (CH3)3C—CH2CH2Br + excess NH3 —>

Verified step by step guidance

1. Identify the nucleophile and electrophile in each reaction. For reaction (e), the nucleophile is pyridine (C5H5N) and the electrophile is methyl iodide (CH3I). For reaction (f), the nucleophile is ammonia (NH3) and the electrophile is 2-bromo-2-methylbutane ((CH3)3C—CH2CH2Br).
2. In an SN2 reaction, the nucleophile attacks the electrophilic carbon from the opposite side of the leaving group, leading to inversion of configuration at the carbon center.
3. For reaction (e), the lone pair on the nitrogen of pyridine attacks the methyl group of CH3I, displacing the iodide ion and forming N-methylpyridinium iodide.
4. For reaction (f), the lone pair on the nitrogen of NH3 attacks the carbon bonded to the bromine in (CH3)3C—CH2CH2Br, displacing the bromide ion and forming a primary amine. Since excess NH3 is present, further substitution can occur, leading to the formation of a secondary amine.
5. Consider the stereochemistry and regiochemistry of the products, ensuring that the inversion of configuration is accounted for in the SN2 mechanism.