Predict the major product(s) for each of the following reactions. H3O (a) 4-Chlorobut-1-ene HBr (b) 1-Chlorobut-1-ene HBr (c) 4,4-Dimethylcyclopentene H2O, H (d) Propyne 2 HCl (e) Cyclopentylethene

(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 "" 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 "" 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 "" 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 "" 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 "" 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