7-Dimethyloctane For The E2 Reaction

The E2 reaction performed began by dehydrohalogenating the 3-chloro-3, 7-dimethyloctane with potassium hydroxide in aqueous ethanol. The 3-chloro-3, 7-dimethyloctane will lose a hydrogen from three different carbons that cause the formation of three constitutional isomers. If the hydroxide ion attacks a hydrogen on the 3-methyl carbon, then 2-ethyl-6-methyl-1-heptene will be the product. If a hydrogen on the carbon-2 was attacked, then the product would be 3, 7-dimethyl-2-octene. If a hydrogen was attacked on the carbon-4, then 3, 7-dimethyl-3-octene will result (Figure 1). As a side note, the potassium hydroxide acts as a polarizable nucleophile that will cause the overall reaction to favor an elimination reaction over a substitution reaction.