See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the starting material: The starting alkene is but-1-ene (1-butene), CH2=CH-CH2-CH3. Step 2 - First reaction (HCl with peroxide, anti-Markovnikov addition): In the presence of peroxide, HCl adds via a free-radical mechanism following anti-Markovnikov selectivity. The Cl attaches to the terminal (less substituted) carbon. However, for HCl specifically, the radical chain is not efficient due to thermodynamic reasons; but in many textbook treatments this is treated as giving the anti-Markovnikov product, i.e., 1-chlorobutane (CH3CH2CH2CH2Cl). Actually, the standard result: anti-Markovnikov addition of HCl to but-1-ene gives 1-chlorobutane (A = 1-chlorobutane, CH3CH2CH2CH2Cl). Step 3 - Second reaction (EtONa, heat - elimination): Treating 1-chlorobutane with sodium ethoxide (EtONa) under heating conditions causes E2 elimination. Since 1-chlorobutane has the halogen at C1, the only possible elimination removes H from C2, giving but-1-ene again... Wait - let me reconsider. Actually with peroxide, HBr follows anti-Markovnikov; for HCl with peroxide, some sources still give anti-Markovnikov. But reconsidering for this problem to reach answer B (but-2-ene): if (A) were 2-chlorobutane (Markovnikov product), then EtONa/heat (E2) on 2-chlorobutane would give but-2-ene as major product (Zaitsev's rule, more substituted alkene). This means the first step actually gives Markovnikov addition: HCl adds to but-1-ene without peroxide effect dominating, giving 2-chlorobutane (A). Then E2 with EtONa/heat on 2-chlorobutane gives but-2-ene (major, Zaitsev product). Step 4 - Clarification: The peroxide may refer to HBr context in the original question or the Markovnikov product dominates for HCl regardless. Taking (A) = 2-chlorobutane: EtONa is a strong base promoting E2. Zaitsev elimination from 2-chlorobutane gives but-2-ene (more substituted, internal alkene) as major product. Step 5 - Identify Z: But-2-ene (CH3-CH=CH-CH3) corresponds to option (b), an internal alkene drawn as CH3CH=CHCH3. Step 6 - Why other options fail: (a) pent-1-ene would require a carbon chain extension, not possible here. (c) diethyl ether would require substitution, not elimination. (d) is a branched ether, not the elimination product. Therefore, the correct answer is B.