Question

when a tertiary alkyl halide is dissolved in a polar solvent that is both a weak base and weak nucleophile, (such an ethanol, EtOH) we expect both ____ and ___ process to be disfavored because it is teriary and nuc AND BASE Are weak

Answer 1

Both nucleophilic substitution (SN1) and elimination (E1) processes are disfavored when a tertiary alkyl halide is dissolved in ethanol due to steric hindrance and ethanol's weak basicity and nucleophilicity.

Step-by-step explanation:

When a tertiary alkyl halide is dissolved in a polar solvent that is both a weak base and weak nucleophile, such as ethanol (EtOH), we expect both nucleophilic substitution (SN1) and elimination (E1) processes to be disfavored. This is because in the case of a tertiary alkyl halide, the steric hindrance at the carbon center is significant, making it difficult for a nucleophile to approach and carry out a substitution reaction. Moreover, since ethanol is a weak base, it will not efficiently promote elimination.

The tertiary halide is prone to dissociation in polar protic solvents, fitting the characteristic of SN1 and E1 mechanisms, where the rate-determining step involves the formation of a carbocation. However, due to the weakly basic and nucleophilic nature of ethanol, the following steps in both SN1 and E1 reactions are disfavored.