See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Step 1: Identify the substrate. The compound is 2-bromo-3-methylbutane: CH3-CH(Br)-CH(CH3)-CH3. The bromine is on C2, and C3 bears a methyl group. Step 2: Determine the mechanism. In EtOH (a polar protic solvent of moderate nucleophilicity) with a secondary substrate that can ionize, both SN1 and E1 pathways occur. The carbocation intermediate is a secondary carbocation at C2, but it can undergo a 1,2-hydride shift to form a more stable tertiary carbocation at C3. Step 3: Carbocation intermediates. - Initial 2° carbocation at C2: CH3-CH(+)-CH(CH3)-CH3 - After 1,2-hydride shift: tertiary carbocation at C3: CH3-CH2-C(+)(CH3)-CH3 Step 4: SN1 products (nucleophilic attack by EtOH acting as solvent/nucleophile, then deprotonation). - From 2° carbocation at C2: CH3-CH(OEt)-CH(CH3)-CH3. C2 is a stereocenter → R and S enantiomers → 2 stereoisomeric products. - From 3° carbocation at C3: CH3-CH2-C(OEt)(CH3)-CH3. C3 with two methyl groups and ethyl + OEt — C3 has CH3, CH3, CH2CH3, OEt: actually C3 becomes (CH3)2C(OEt)(CH2CH3) — this carbon has two identical methyl groups, so it is NOT a stereocenter → 1 product. SN1 products total: 2 + 1 = 3. Step 5: E1 products (elimination, loss of proton from adjacent carbon to carbocation). - From 2° carbocation at C2: remove H from C1 → CH2=CH-CH(CH3)-CH3 (but-1-ene, no stereoisomer) = 1 product; remove H from C3 → CH3-CH=C(CH3)-CH3 (2-methylbut-2-ene) = 1 product (no E/Z since same groups on one carbon). That gives 2 E1 products from 2° carbocation. - From 3° carbocation at C3: remove H from C2 → CH3-CH=C(CH3)-CH3 (2-methylbut-2-ene, same as above, already counted); remove H from C4 → CH3-CH2-C(CH3)=CH2 (2-methylbut-1-ene) = 1 new product. E1 products: but-1-ene (1) + 2-methylbut-2-ene (1) + 2-methylbut-1-ene (1) = 3 products. Step 6: Total products = SN1 products (3) + E1 products (3) = 6. Therefore, the correct answer is 6.