Question

Which product is obtained when 1-pentene reacts with HCl in the presence of benzoyl peroxide?

• A. 1-chloro pentane
• B. 2-chloro pentane
• C. 1-chloro, 3-methyl butane
• D. 2,2-dimethyl 1-chloropropane

Hint:

The peroxide (Kharasch) effect only works for one specific hydrogen halide.

Answer 1

The Correct Option is B

Step 1: A common exam trap is to assume that benzoyl peroxide always means anti-Markovnikov addition. This is only true for HBr; it is important to check which hydrogen halide is used before applying the peroxide rule. Step 2: The reason lies in bond energies. The propagation steps of the radical chain mechanism require that both the addition of the halogen radical to the alkene and the abstraction of hydrogen from HX be exothermic. Step 3: For $HBr$, both propagation steps are exothermic, so the chain mechanism runs efficiently, and bromine ends up on the terminal, less substituted carbon, giving the anti-Markovnikov product. Step 4: For $HCl$, the $H-Cl$ bond is too strong compared to $H-Br$, so the hydrogen abstraction step becomes endothermic and the chain cannot propagate. No radical pathway is available for HCl regardless of peroxide. Step 5: With the radical route shut down, 1-pentene, $CH_2=CHCH_2CH_2CH_3$, simply reacts with $HCl$ through ordinary electrophilic addition. Protonation occurs at C1 to generate the more stable secondary carbocation at C2, and then chloride ion attacks that carbocation. \[ CH_2=CHCH_2CH_2CH_3 + H^+ \rightarrow CH_3-\overset{+}{C}H-CH_2CH_2CH_3 \] \[ CH_3-\overset{+}{C}H-CH_2CH_2CH_3 + Cl^- \rightarrow CH_3CHClCH_2CH_2CH_3 \] This confirms the product is the Markovnikov product, 2-chloropentane, and benzoyl peroxide plays no real role. \[\boxed{\text{2-chloropentane}}\]