See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Concept: When an allylic or homoallylic alkyl halide is treated with water in the presence of a mild base (CaCO3, which neutralizes HCl as it forms), the reaction can proceed via neighboring group participation or carbocation rearrangement. Here, Me2C=CH-CH2-CH2-Cl is a delta,epsilon-unsaturated alkyl chloride (the double bond is separated from the leaving group by two carbons). Step 1 – Identify the substrate: Me2C=CH-CH2-CH2-Cl. The chloride is on carbon-1 (counting from Cl), and the double bond is between C3 and C4 (i.e., C3=C4 where C4 bears two methyl groups). This is a 4-pentenyl-type system (homoallylic chloride with one extra CH2). Step 2 – Ionization and neighboring group participation: Under aqueous conditions with CaCO3 (mild base, SN1-favoring), the C-Cl bond ionizes. The developing carbocation at C1 (primary) is unstable. The remote pi bond (C3=C4) can participate through intramolecular electrophilic cyclization. The electrons of the double bond attack the carbocation center (C1), forming a cyclopropylcarbinyl-type intermediate or directly a cyclopropane ring via pi participation. Step 3 – Cyclopropane formation: The pi electrons of Me2C=CH- attack the primary carbocation at C1, forming a three-membered ring (cyclopropane). This places the positive charge on the tertiary carbon (CMe2+), which is then captured by water (or OH-) to give a tertiary alcohol: cyclopropane ring with a -CMe2OH group attached. This is the product shown in option (d): 1-(2-hydroxypropan-2-yl)cyclopropane with an OH on the CMe2 group. Step 4 – Why other options fail: - (b) Me2C=CH-CH2-CH2-OH is the simple SN2 substitution product (direct displacement). This is disfavored because (i) the carbon bearing Cl is primary but the reaction uses aqueous/SN1 conditions, and (ii) intramolecular pi participation is much faster giving the cyclopropane. - (a) involves a rearranged tertiary alcohol without ring formation; the mechanism does not support a simple 1,2-shift to give that product. - (c) involves hydroxylation at an internal carbon, which is not consistent with either SN1 or SN2 on the terminal carbon. Step 5 – The major product is the cyclopropane ring bearing a -CMe2OH group (option d), formed via intramolecular pi-participation of the alkene onto the primary carbocation, followed by trapping of the resulting tertiary carbocation by water. Therefore, the correct answer is D.