Draw all of the monochlorination products that you might obtain from the free-radical chlorination of ethylcyclopropane. you do not have to consider stereochemistry. draw one structure per sketcher. add additional sketchers using the dropdown menu in the bottom right corner. separate multiple products using the + sign from the dropdown menu.

-) (2-chloroethyl)cyclopropane

-) (1-chloroethyl)cyclopropane

-) 1-chloro-2-ethylcyclopropane

-) 1-chloro-1-ethylcyclopropane

Explanation:

The monochlorination are products in which Cl is added only 1 time. Now, in the ethylcyclopropane we will have 5 carbons, so we will have 5 options for monochlorination (see figure 1). But, for carbons 2 and 3 we will have equivalent structures, therefore at the end, we will have only 4 possible structures. In carbon 5, carbon 4, carbon 3 and carbon 1 (see figure 2).

i believe you are right for all except the 3rd ( look at sqdancefan his answer is much better)