What is the major product of the reaction of 2-ethoxypentane with hydrogen iodide?

What is the major product formed by heating 2-ethoxypentane with $\ce{HI}$?

The options are:

According to me the correct answer should be option-1 because 2-methylbutyl carbocation is more stable than ethyl carbocation as there are more +I groups attached in 2-methylbutyl carbocation. But according to my book the second option is correct. Is my answer correct? If not please explain

1 Answer

Strictly speaking, none of those answers are correct, provided your stem is right. The reaction will start with protonation of the ether to give you the following species:

And the iodide nucleophile can attack either side via $S_{N}2$ mechanism. Since the left side is more accessible, it'll probably attack there first giving you:

Then, since we're still in the presence of HI and it is a strong acid, it protonates the alcohol and converts it into a corresponding alkyl iodide via either $S_N2$ like in the example above or $S_N1$:

A general rule of thumb here for the future reference is that in a hydrolysis of an ether with a strong hydrohalogen acid (HI, HBr), you get a mix of two corresponding alkyl halides with halogen where the oxygen used to be.

• can you please explain how the left side is more accessible ? Is it because of less steric hindrance? – TheNewGenGamer Jul 4 '16 at 15:01
• precisely so. Remember, that things constantly move, groups rotate around single bonds, so on the left you have a relatively small ethyl, while on the right you have a huge propeller of a group that will knock off anything that comes (or attempts to do so) close to attack that secondary carbon with the oxygen. – ChemistryHelpCenter Jul 4 '16 at 20:37