Answer:
See explanation
Step-by-step explanation:
In this case, we have to check two variables:
1) The leaving group
2) The carbon bonded to the leaving group.
Let's check one by one:
2-chloro-3-methylbutane
In this molecule, the leaving group is "Cl", the carbon bonded to the leaving group has two neighbors. Therefore, we have a secondary substrate.
1-phenylpropan-1-ol
In this molecule, the leaving group is "OH", the carbon bonded to the hydroxyl group has two neighbors also. So, we have a secondary substrate.
(E)-pent-3-en-2-yl 4-methylbenzenesulfonate
In this case, the leaving group is "OTs" (Tosylate), the carbon bonded to the tosylate group has as a neighbor a double bond. Therefore, we have an allylic substrate.
3a-bromooctahydro-1H-indene
In this molecule, the leaving group is "Br", the carbon bonded to the bromine has three neighbors. So, we have a tertiary substrate.
1-iodo-3-methylbutane
In this molecule, the leaving group is "I", the carbon bonded to the iodide has only one neighbor. So, we have a primary substrate.
See figure 1
I hope it helps!