See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: Acid-catalyzed alkylation of isobutane with ethylene using HF as catalyst (industrial alkylation reaction). Step 1 - Initiation: HF protonates ethylene (H2C=CH2) to generate a carbocation. By Markovnikov's rule, the proton adds to give the more stable secondary carbocation: CH3-CH2(+) ... but actually ethylene gives a primary carbocation initially, which is not stable. In the HF-catalyzed alkylation mechanism, HF first generates a small amount of tert-butyl cation from isobutane via hydride abstraction or alternatively the ethylene is protonated to give ethyl cation which quickly reacts. Step 2 - Generation of tert-butyl cation: Under HF conditions, isobutane (2-methylpropane, CH3CH(CH3)2) loses a hydride to HF to generate the tert-butyl carbocation: (CH3)3C(+). Step 3 - Addition to ethylene: The tert-butyl cation (CH3)3C(+) adds to ethylene H2C=CH2 to give: (CH3)3C-CH2-CH2(+), a primary carbocation, which rearranges via hydride or methyl shift, or alternatively the tert-butyl cation adds and then the resulting carbocation rearranges. Step 4 - Rearrangement: The primary carbocation (CH3)3C-CH2-CH2(+) undergoes a 1,2-hydride shift to give (CH3)3C-CH(+)-CH3, a tertiary carbocation. This tertiary carbocation then loses a proton (to regenerate HF catalyst) to give the most substituted alkene by Zaitsev's rule: (CH3)2C=C(CH3)2, which is 2,3-dimethylbut-2-ene. Step 5 - Product identification: (CH3)2C=C(CH3)2 is 2,3-dimethylbut-2-ene, which is a tetrasubstituted alkene. This corresponds to option (b), which shows a double bond with two methyl groups on each carbon. Why other options fail: - Option (a): 3,3-dimethylbut-1-ene would result from simple addition without rearrangement; the primary carbocation intermediate is too unstable and rearranges. - Option (c): Shows a trisubstituted alkene with an H on the double bond carbon, which does not match the expected tetrasubstituted product. - Option (d): CH3CH(CH3)CH2CH=CH2 would be a simple addition product without carbocation rearrangement and is not the thermodynamic product under HF conditions. Therefore, the correct answer is B.