See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Step 1: Identify the substrate. 1,2-dimethyl propyl p-toluene sulfonate is the tosylate of 1,2-dimethylpropan-1-ol (i.e., 2-methylbutan-1-yl tosylate). The structure is: CH3CH(CH3)CH(CH3)-OTs, more precisely the carbon skeleton is (CH3)2CHCH(CH3)-OTs — this is 1,2-dimethylpropyl tosylate, meaning C1 bears the OTs and a methyl group, C2 bears a methyl group and is also a CH, and C3 is a CH3. Actually let us be precise: '1,2-dimethylpropyl' refers to the 1,2-dimethylpropyl group = (CH3)(C2H5)CH- type. The parent is propyl (3 carbons), with methyl substituents at C1 and C2: CH3-C(CH3)(OTs)-CH2CH3? No — 1,2-dimethylpropyl = C1 has OTs and methyl, C2 has methyl, C3 is CH3: structure is CH3C(OTs)(H)-CH(CH3)-CH3, i.e., 2-methylbutan-1-yl tosylate with OTs on C1. Wait — '1,2-dimethylpropyl' as a group means the propyl chain has methyls at positions 1 and 2: C1(CH3)(OTs)(H)-C2(CH3)(H)-C3H3. This is 1,2-dimethylpropyl = same as 1,2-dimethyl-1-propyl meaning C1 is (CH3)(OTs)(H) and C2 is (CH3)(H) giving overall: CH3CH(OTs)CH(CH3)CH3 — i.e., 3-methylbutan-2-yl tosylate (a secondary tosylate with a beta methyl branch). Step 2: Ionization. Under solvolysis (SN1/E1 conditions) in acetic acid, the OTs leaves to form a secondary carbocation at C2 of 2-methyl butane skeleton: (CH3)CH(+)CH(CH3)CH3... Actually with the structure CH3-CH(OTs)-CH(CH3)-CH3, ionization gives carbocation at C2: CH3-CH(+)-CH(CH3)-CH3. This is a secondary carbocation adjacent to a tertiary carbon (C3 bears two methyls and one H... let us recount: CH3-C+(H)-CH(CH3)-CH3 — C3 = CH(CH3) meaning C3 has one methyl, one H, connected to C2 and C4(CH3). So C3 is secondary. The initial carbocation at C2 is secondary. Step 3: 1,2-hydride or methyl shift. The adjacent C3 is CH(CH3), so a 1,2-hydride shift from C3 to C2 gives a tertiary carbocation at C3: CH3-CH2-C+(CH3)-CH3 (2-methylbutan-3-yl cation = tertiary). Alternatively a 1,2-methyl shift could occur. Step 4: Products from original secondary carbocation (C2+): (a) Substitution with AcOH at C2 → CH3CH(OAc)CH(CH3)CH3 (one substitution product, retention/inversion — counted as one product assuming no stereochemistry distinction here, or two if stereoisomers counted). (b) E1 elimination: remove H from C1 → CH2=CHCH(CH3)CH3 (but-1-ene derivative: 3-methylbut-1-ene); remove H from C3 → CH3CH=C(CH3)CH3 (2-methylbut-2-ene). Step 5: Products from rearranged tertiary carbocation (C3+) = (CH3)2C+-CH2CH3: (c) Substitution → (CH3)2C(OAc)CH2CH3 (one substitution product). (d) E1 elimination: remove H from C2 → CH3CH=C(CH3)... wait, cation is at C3 in 2-methylbutane numbering after shift: the tertiary cation (CH3)2C+CHCH3 — elimination gives (CH3)2C=CHCH3 (2-methylbut-2-ene, same as above) or (CH3)C(=CH2)CH2CH3 (2-methylbut-1-ene). Step 6: Counting unique products: Substitution products: (1) CH3CH(OAc)CH(CH3)CH3 from secondary cation, (2) (CH3)2C(OAc)CH2CH3 from tertiary cation. Alkene products: (3) 3-methylbut-1-ene (CH2=CHCH(CH3)CH3), (4) 2-methylbut-2-ene ((CH3)2C=CHCH3) from either cation, (5) 2-methylbut-1-ene (CH2=C(CH3)CH2CH3) from tertiary cation. Total = 5 distinct products (2 substitution + 3 alkene). Step 7: Why other options fail: Options (a)2, (b)3, (c)4 undercount because they miss either the rearrangement pathway products or one of the elimination products. The full accounting of both cations and all possible elimination/substitution pathways gives 5 products. Therefore, the correct answer is D.