See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: The rate of electrophilic addition of HBr to alkenes depends on the stability of the carbocation intermediate formed in the rate-determining step (Markovnikov addition). A more stable (higher substituted) carbocation intermediate leads to a faster reaction rate. Step 1 - Analyze each option: (a) Cyclohexane with a Br substituent reacting with HBr: Cyclohexane is an alkane (saturated), not an alkene. Alkanes do not readily undergo electrophilic addition with HBr under normal conditions. This reaction is very slow or does not occur readily. (b) 2-methylbut-1-ene (CH2=C(Me)(Et)) reacting with HBr: Protonation of the terminal CH2 gives a tertiary carbocation (tertiary carbon bearing methyl and ethyl groups). A tertiary carbocation is relatively stable, leading to a reasonably fast rate. (c) Benzene reacting with HBr to give 1-bromocyclohex-2-ene: Benzene is aromatic and highly stable due to resonance. Electrophilic addition to benzene disrupts aromaticity, requiring much more energy. This reaction is extremely slow and unfavorable compared to simple alkene addition. (d) 2-methylbut-1-ene (or the alkene shown, which upon protonation gives a tertiary carbocation): Looking at option (d), the alkene shown is 2-methylbut-1-ene (CH2=C(CH3)-CH(CH3)- or similar trisubstituted arrangement). The product shown is a tertiary bromide (Br on a tertiary carbon). The carbocation intermediate formed is tertiary, and the alkene in option (d) appears to be more substituted or forms a more stable tertiary carbocation more readily. Step 2 - Compare (b) and (d): Both form tertiary carbocations via Markovnikov addition. However, in option (d), the alkene structure (2-methylbut-1-ene with the double bond positioned to give a more substituted/stable tertiary carbocation) reacts faster. The key distinction is that option (d)'s alkene is more electron-rich or the resulting carbocation is more stable (tertiary with better hyperconjugation/inductive stabilization), making it the fastest among all options. Step 3 - Eliminate other options: (a) alkane - no addition reaction; (c) benzene - aromatic stability prevents addition; (b) forms tertiary carbocation but slightly less favorable than (d). Step 4 - Conclusion: Option (d) proceeds through the most stable tertiary carbocation intermediate, giving the fastest rate of HBr addition among the four options. Therefore, the correct answer is D.