Final answer:
In a 3º alkyl halide, a strong base will likely result in an elimination reaction, while a weak base will result in an S1 substitution reaction. A poor nucleophile may not react significantly or may react with the solvent instead.
Step-by-step explanation:
In a 3º alkyl halide, the carbon that is bonded to the halogen atom is connected to three other alkyl groups. When a 3º alkyl halide reacts with a strong base, the reaction is likely to undergo an elimination reaction instead of a substitution reaction. The base will remove a hydrogen atom from a neighboring carbon, resulting in the formation of an alkene. For example, if tert-butyl bromide reacts with a strong base like sodium ethoxide, tert-butyl bromide will undergo an E2 elimination reaction and produce 2-methylpropene.
When a weak base reacts with a 3º alkyl halide, the reaction is likely to undergo an SN1 substitution reaction. A weak base will not be able to remove a hydrogen atom from a neighboring carbon, so the halogen atom will be replaced by the weak base through a two-step mechanism. For example, if tert-butyl bromide reacts with a weak base like water, tert-butyl bromide will undergo an SN1 reaction and produce tert-butyl alcohol.
In the case of a poor nucleophile, the reaction may not occur at a significant rate because the nucleophile is not strong enough to displace the halogen atom. The poor nucleophile may react with the solvent instead, or the reaction may not proceed at all. For example, if tert-butyl bromide reacts with a poor nucleophile like sodium iodide, the reaction may not proceed because sodium iodide is not a strong enough nucleophile to replace the bromine atom.