# 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

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 $S_N1$: