2. Arrange the compounds of each set in order of reactivity toward an Sn2 reaction.

a)t-butyl chloride, 1-chloropropane, 1-chloro-2,2-dimethylpropane, 1-chloro-2-methylpropane
b)2-bromopropane, 2-chloropropane, 2-iodopropane

To determine the order of reactivity towards an Sn2 reaction for each set of compounds, you need to consider the factors that affect Sn2 reactivity. These factors include the nature of the leaving group and the steric hindrance around the carbon atom that undergoes the Sn2 reaction. Here's how to analyze each set:

a) t-butyl chloride, 1-chloropropane, 1-chloro-2,2-dimethylpropane, 1-chloro-2-methylpropane:

1. Evaluate the nature of the leaving group: In this case, all the compounds have the same leaving group, which is chloride (Cl-). Therefore, the leaving group does not affect the reactivity order in this set.

2. Consider steric hindrance: Steric hindrance refers to the presence of bulky groups around the carbon atom undergoing the Sn2 reaction. The more bulky substituents there are, the more difficult it is for the nucleophile to approach the carbon atom. As a result, steric hindrance decreases reactivity.

Based on steric hindrance, the order of reactivity towards an Sn2 reaction in this set is as follows (from most reactive to least reactive):
1-chloropropane > 1-chloro-2-methylpropane > t-butyl chloride > 1-chloro-2,2-dimethylpropane

b) 2-bromopropane, 2-chloropropane, 2-iodopropane:

1. Evaluate the nature of the leaving group: In this set, the leaving groups are bromide (Br-), chloride (Cl-), and iodide (I-). Generally, iodide is a better leaving group than bromide, which is better than chloride. Therefore, considering the nature of the leaving group alone, the reactivity order would be as follows:
2-iodopropane > 2-bromopropane > 2-chloropropane

2. Consider steric hindrance: Steric hindrance can also influence the reactivity order within this set. As before, the more bulky substituents, the lower the reactivity.

Considering both the nature of the leaving group and steric hindrance, the final order of reactivity towards an Sn2 reaction in this set is as follows (from most reactive to least reactive):
2-iodopropane > 2-bromopropane > 2-chloropropane

Remember that the reactivity order can change depending on specific reaction conditions, but this analysis should give you a good starting point.

a) To arrange the compounds in order of reactivity towards an Sn2 reaction, we need to consider the stability of the carbocations that would be formed during the reaction. The more stable the carbocation, the slower the reaction.

1. t-butyl chloride: In this compound, the carbocation formed after the chloride ion leaves is a tertiary carbocation. Tertiary carbocations are highly stable due to the presence of three alkyl groups, which provide electron-donating inductive effects. Therefore, t-butyl chloride is the least reactive towards an Sn2 reaction.

2. 1-chloropropane: This compound has a primary carbocation, which is less stable than a tertiary carbocation. The primary carbocation has only one alkyl group providing electron-donating inductive effects. Hence, 1-chloropropane is more reactive than t-butyl chloride.

3. 1-chloro-2,2-dimethylpropane: In this compound, the carbocation formed after the chloride ion leaves is a secondary carbocation. Secondary carbocations are more stable than primary carbocations but less stable than tertiary carbocations. The two methyl groups in 1-chloro-2,2-dimethylpropane provide some electron-donating inductive effects, increasing the stability of the carbocation. Therefore, it is more reactive than 1-chloropropane but less reactive than 1-chloro-2-methylpropane.

4. 1-chloro-2-methylpropane: This compound has a secondary carbocation similar to 1-chloro-2,2-dimethylpropane. However, it has only one methyl group, resulting in slightly lower electron-donating inductive effects. Therefore, it is more reactive than 1-chloro-2,2-dimethylpropane but less reactive than 1-chloropropane.

Arranging the compounds in order of reactivity towards an Sn2 reaction:
t-butyl chloride < 1-chloropropane < 1-chloro-2,2-dimethylpropane < 1-chloro-2-methylpropane

b) To arrange the compounds in order of reactivity towards an Sn2 reaction, we need to consider the leaving group ability and the strength of the bond between the carbon and the leaving group.

1. 2-iodopropane: Iodine is a better leaving group than bromine and chlorine. Therefore, 2-iodopropane is the least reactive towards an Sn2 reaction.

2. 2-bromopropane: Bromine is a better leaving group than chlorine but less reactive than iodine. Therefore, 2-bromopropane is more reactive than 2-iodopropane.

3. 2-chloropropane: Chlorine is the least reactive leaving group among the three compounds. Hence, 2-chloropropane is more reactive than both 2-bromopropane and 2-iodopropane.

Arranging the compounds in order of reactivity towards an Sn2 reaction:
2-iodopropane < 2-bromopropane < 2-chloropropane