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
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.