See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the starting material and product. The starting material is 3,3-dimethyl-1-butene: (CH3)3C-CH=CH2. The observed product has bromine on a tertiary carbon adjacent to another carbon bearing two methyl groups, consistent with 2-bromo-2,3-dimethylbutane: (CH3)2CBr-CH(CH3)2. Step 2 - Apply Markovnikov's rule for initial protonation. HBr adds H+ to the terminal alkene carbon (C1), placing the positive charge on C2 (the internal carbon). This gives a secondary carbocation: (CH3)3C-CH+(CH3) ... wait, let us re-examine. The alkene is at C1-C2 with C3 being the neopentyl-type carbon. Protonation of C1 gives a secondary carbocation at C2: (CH3)3C-C+H-CH3 ... Actually, re-examining: 3,3-dimethyl-1-butene is (CH3)3C-CH=CH2. Protonation of CH2 (C1) by Markovnikov gives carbocation at C2 (secondary): (CH3)3C-C+H2... no. Protonation at C1 gives + at C2: structure is (CH3)3C-C+(H)-... this is a secondary carbocation at C2. Step 3 - Carbocation rearrangement via methyl shift. The secondary carbocation at C2 is adjacent to the quaternary C3 bearing three methyl groups. A 1,2-methyl shift from C3 to C2 converts the secondary carbocation into a tertiary carbocation at C3: the result is (CH3)2C+-CH(CH3)2, a tertiary carbocation. This is more stable. Step 4 - Bromide attacks the tertiary carbocation. Br- attacks the tertiary carbocation at C3 to give (CH3)2CBr-CH(CH3)2, which is 2-bromo-2,3-dimethylbutane. This matches the observed product. Step 5 - Evaluate why other options fail. (a) A hydride shift from C3 to C2 would give a different tertiary carbocation but the carbon framework would not match the observed product as cleanly as a methyl shift does; moreover the product shown is better explained by methyl shift. (b) A double bond shift before protonation is not a standard ionic mechanism step. (c) Bromide does not add to alkenes without protonation in HBr addition; this is not the correct mechanism. (d) Correctly describes: protonation at C1 (Markovnikov, secondary carbocation at C2), then 1,2-methyl shift to give tertiary carbocation at C3, then Br- addition to give the observed product. Therefore, the correct answer is D.