See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the starting material: The starting diene is 2-methylpenta-1,4-diene, i.e., CH2=C(CH3)-CH2-C(CH3)=CH2 (two terminal methylenated carbons separated by a CH2 group, each with a methyl substituent). Step 2 - Action of 2 eq. KNH2 in liquid NH3: KNH2 is a very strong base. With terminal allenes or dienes, KNH2 in NH3(l) isomerizes terminal alkenes to internal alkynes (or in this case, isomerizes the double bonds). For a diene like 2,4-dimethyl-1,4-pentadiene, two equivalents of KNH2 deprotonate allylic/propargylic positions and cause isomerization. The net result of treating 2-methylpenta-1,4-diene with 2 eq. KNH2 in NH3(l) is formation of a dianion (dienyl dianion or bis-allylic dianion). Specifically, KNH2 abstracts the allylic protons from both terminal methylene groups, generating a stabilized dianion at the two terminal carbons of the system: [-CH2-C(CH3)=CH-C(CH3)=CH2] type dianion. The most stable dianion places negative charges at the terminal carbons (C1 and C5 of the original skeleton, after full deprotonation and resonance). Step 3 - Reaction with n-C4H9Br: One equivalent of n-butyl bromide alkylates one terminal carbanion. Since 2 eq. KNH2 generates a dianion, alkylation occurs at the less hindered terminal carbon. The n-butyl group attaches to one terminus. Step 4 - Workup with H3O+: Protonation of remaining anion and hydrolysis of any enol forms. The enol tautomers of the allylic/vinylic system tautomerize to ketones upon aqueous acid workup. Step 5 - Determine the product structure: Starting from CH2=C(CH3)-CH2-C(CH3)=CH2, after dianion formation and monoalkylation at one terminal carbon with n-Bu, followed by acid workup (tautomerization of terminal =CH2 or =C- to C=O via enol), the product is a 1,5-diketone. The two terminal =CH2 groups each become CH3-C=O after tautomerization (the =C(CH3)- becomes -C(=O)-CH3 type). With one n-Bu group added at a terminal position, the product is: CH3-C(=O)-CH2-CH2-C(=O)-n-C4H9, i.e., 1-(n-butyl)-4-pentanedione type, which corresponds to a structure with two ketone groups and a butyl chain - matching option (d), which shows two ketone groups with a butyl substituent on one end. Step 6 - Why other options fail: Option (a) shows alkylation at the central carbon (C3) giving a 3-substituted-2,4-pentanedione with a pentyl chain - wrong regiochemistry and wrong chain length. Option (b) shows a longer chain suggesting different connectivity. Option (c) shows a propyl group instead of butyl. Option (d) correctly shows the product of terminal monoalkylation with n-butyl followed by tautomerization to give a diketone with the n-butyl chain at one terminal ketone carbon. Therefore, the correct answer is D.