Answer:
(a) 3-bromo-1-chlorobutane
(b) 2-bromo-1-chlorobutane
(c) 3,3-dimethylcyclopentan-1-ol
(d) 1,1,2,2-tetrachloropropane
(e) 1-ethylcyclopentan-1-ol
Explanation:
We can analyze each reaction:
(a) 4-Chlorobut-1-ene + HBr
In this reaction, we have an alkene. So, we will have an addition reaction, the nucleophile "[tex]Br^-[/tex]" will be added to the most substituted carbon. In this case carbon 3 to produce 3-bromo-1-chlorobutane.
(b) 1-Chlorobut-1-ene + HBr
In this reaction, we also have an alkene. So, we will have an addition reaction, the nucleophile "[tex]Br^-[/tex]" will be added to the most substituted carbon. In this case carbon 2 to produce 2-bromo-1-chlorobutane.
(c) 4,4-Dimethylcyclopentene + H2O, H+
In this reaction, we have an alkene. So, we will have an addition reaction, the nucleophile "[tex]H_2O[/tex]" will be added to the most substituted carbon. In this case carbon 3 to produce 3,3-dimethylcyclopentan-1-ol.
(d) Propyne + 2HCl
In this reaction, we have an alkyne. So, we will have an addition reaction, the nucleophile "[tex]Cl^-[/tex]" will be added to the most substituted carbon, but we have 2 moles of the nucleophile, so would be added 2 times and we will have as product 1,1,2,2-tetrachloropropane.
(e) Cyclopentylethene + H3O+
In this reaction, we have an alkene. So, we will have an addition reaction, the nucleophile "[tex]H_3O^+[/tex]" will be added to the most substituted carbon. But in this case, the carbon cation would be produced in carbon 1 of the ethene. So, we can have a hydride shift to produce a tertiary carbocation. With this in mind the nucleohile will be addde to this tertiary carbocation and we will have 1-ethylcyclopentan-1-ol.
See figure 1 to further explanations
