The neutral necessary product the the reaction is the 2-methoxy-2-methylpropane, displayed in the fastened figure.
You are watching: Predict the neutral organic product of the reaction.
By reaction an alkene (2-methylprop-1-ene) with an alcohol (methanol) in the presence of an acid it forms an ester (2-methoxy-2-methylpropane) v Markovnikov’s rule. First, a protonation of the alkene occurs wherein a tertiary carbocation is formed, climate the nucleophile (methanol) attacks the previously created carbocation, and finally ocurr the deprotonation the the hydrogen bound to the oxygen, thus developing the ester, in our instance the neutral essential product 2-methoxy-2-methylpropane.
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