Determine the number of chiral carbon atoms present in the major product formed from the electrophil — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

The reaction proceeds as follows: 1) 3-methyl-1-butene ($CH_3-CH(CH_3)-CH=CH_2$) undergoes electrophilic attack by $H^+$ to form the most stable initial carbocation, which is a secondary carbocation: $CH_3-CH(CH_3)-C^+H-CH_3$. 2) A 1,2-hydride shift occurs because it converts the $2^\circ$ carbocation into a much more stable $3^\circ$ carbocation: $CH_3-C^+(CH_3)-CH_2-CH_3$. 3) The bromide nucleophile ($Br^-$) attacks this $3^\circ$ carbocation to yield the major product: 2-bromo-2-methylbutane ($CH_3-C(Br)(CH_3)-CH_2-CH_3$). Analyzing the structure of 2-bromo-2-methylbutane, the C2 carbon is bonded to a $-Br$, an ethyl group ($-CH_2CH_3$), and two identical methyl groups ($-CH_3$). Since it does not have four different groups attached, it is achiral. Thus, there are 0 chiral carbons.