See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the substrate: 3,3-dimethyl-1-butene is (CH3)3C-CH=CH2. Its carbon skeleton has a tert-butyl group attached to a vinyl group. Step 2 - Markovnikov addition of HCl: The proton adds to the terminal carbon (C1) of the double bond, placing the positive charge on C2, giving the secondary carbocation (CH3)3C-CH(+)-CH3. This is carbocation 2: (CH3)3C(+)CHCH3. Step 3 - Rearrangement: The secondary carbocation at C2 can undergo a 1,2-hydride or 1,2-methyl shift from the adjacent quaternary C3. A 1,2-methyl shift from the tert-carbon converts the secondary carbocation into a tertiary carbocation: (CH3)2C(+)-CH(CH3)2, which is carbocation 4: (CH3)2C(+)CH(CH3)2. This rearranged carbocation has a different (rearranged) carbon skeleton. Step 4 - Products: Carbocation 2 captures Cl- to give 2-chloro-3,3-dimethylbutane (normal skeleton product). Carbocation 4 captures Cl- to give 2-chloro-2-methylpentane... actually 2-chloro-2,3-dimethylbutane (rearranged skeleton product). Step 5 - Evaluate all options: - Carbocation 1: (CH3)3C(+)CHCH2Cl - this would require anti-Markovnikov addition or a different mechanism; it is not an intermediate in simple HCl addition to this alkene. - Carbocation 2: (CH3)3C(+)CHCH3 - YES, this is the direct Markovnikov carbocation intermediate. - Carbocation 3: (CH3)2C(+)C(CH3)2Cl - has a Cl already on the carbon skeleton at a position inconsistent with being a carbocation intermediate before Cl- capture; not a valid intermediate. - Carbocation 4: (CH3)2C(+)CH(CH3)2 - YES, this is the rearranged tertiary carbocation formed by methyl shift from carbocation 2. Step 6 - The two possible carbocation intermediates are 2 and 4, corresponding to option (e). Therefore, the correct answer is E.