Assuming you mean...
#H_3color(blue)(C)-(Hcolor(blue)(C) CH_2OH)-color(blue)(C)H_2-(Hcolor(blue)(C) CH_3)-color(blue)(C)H_3#
where the blue carbons are in the main chain.
You can see that the carbon center with the ethoxy group (#R-CH_2OH#) has one connection to #CH_3#, one connection to #H#, one connection to #CH_2OH#, and one connection to everything to the right.
That would actually indicate a sterically-hindered primary alcohol. If it was just #OH# instead of #CH_2OH#, then it would be tertiary.
Since it's a primary alcohol, it oxidizes into an aldehyde (just change #CH_2OH# to #C-(C=O)H#), and if the oxidizing agent is further-reactive, the reaction would continue towards a carboxylic acid (just change #CH_2OH# to #C-(C=O)OH#). This oxidization implies it either loses a hydrogen or gains an oxygen.
