Answer:
The neutral organic product of the reaction is the 2-methoxy-2-methylpropane, shown in the attached figure.
Explanation:
By reacting an alkene (2-methylprop-1-ene) with an alcohol (methanol) in the presence of an acid it forms an ester (2-methoxy-2-methylpropane) through Markovnikov’s rule. First, a protonation of the alkene occurs where a tertiary carbocation is formed, then the nucleophile (methanol) attacks the previously formed carbocation, and finally ocurr the deprotonation of the hydrogen bound to the oxygen, thus forming the ester, in our case the neutral organic product 2-methoxy-2-methylpropane.