See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

Step 1: Determine the degree of unsaturation. C5H11Br has the formula CnH(2n+1)Br, which corresponds to a saturated alkyl bromide (monobromopentane). Degree of unsaturation = 0, so all isomers are open-chain (no rings or double bonds). Step 2: Enumerate all carbon skeletons for C5 (pentane isomers) and place Br on each unique carbon position. Skeleton 1: n-pentane (CH3-CH2-CH2-CH2-CH3) - Br on C1: 1-bromopentane - Br on C2: 2-bromopentane - Br on C3: 3-bromopentane This gives 3 isomers. Skeleton 2: isopentane (2-methylbutane, CH3-CH(CH3)-CH2-CH3) - Br on C1 (terminal CH3 of main chain, equivalent positions exist): 1-bromo-2-methylbutane - Br on C2 (the branching carbon): 2-bromo-2-methylbutane - Br on C3: 3-bromo-2-methylbutane (i.e., Br on the CH2 group) - Br on C4 (terminal CH3): 1-bromo-3-methylbutane (note: the two terminal methyls on C2 are equivalent, but C1 and C4 of main chain are different from the methyl branch) Let me label carefully: 2-methylbutane carbons: C1(CH3)-C2(CH)-C3(CH2)-C4(CH3) with a methyl on C2. Unique positions: C1/C4 terminal (but C1 and C4 are NOT equivalent due to branching), C2 (tertiary), C3 (secondary), and the methyl branch on C2. - Br on C1: 1-bromo-2-methylbutane - Br on C2: 2-bromo-2-methylbutane - Br on C3: 3-bromo-2-methylbutane - Br on C4: 1-bromo-3-methylbutane - Br on methyl branch of C2: same as 1-bromo-2-methylbutane? No, the methyl on C2 gives: CH3-C(CH2Br)(CH3)-CH2-CH3 = 1-bromo-2-methylbutane... Actually Br on the branch methyl of C2 gives (CH3)C(Br)(CH2CH3) attached to C1... Let me recount: the methyl substituent on C2 is a distinct position giving 1-bromo-2-methylbutane, while C1 of the main chain gives a different connectivity. Re-examining: 2-methylbutane has 4 distinct H-bearing carbons: C1(primary), C2(tertiary), C3(secondary), C4(primary, equivalent to C1? No, not equivalent). And the branch methyl. Unique carbons: main-C1, C2, C3, main-C4, branch-CH3. C1 and branch-CH3 are both primary on C2 but C1 connects further to C3 while branch-CH3 does not — they are non-equivalent. So 5 positions giving 5 isomers. Skeleton 3: neopentane (2,2-dimethylpropane, C(CH3)4) - Only one unique position (all 4 methyls equivalent): neopentyl bromide, 1-bromo-2,2-dimethylpropane This gives 1 isomer. Step 3: Total = 3 (n-pentane) + 4 (isopentane, correcting to 4 unique positions after careful analysis) + 1 (neopentane) = 8. Recounting isopentane: CH3-CH(CH3)-CH2-CH3. Label: Ca(CH3 on left of branch point), Cb(CH branch point), Cc(CH2), Cd(CH3 terminal), and Ce(CH3 branch). Ca and Ce are both methyls on Cb but Ca has a longer chain beyond Cb (Cc-Cd) while Ce is a dead-end — non-equivalent. So unique positions: Ca, Cb, Cc, Cd, Ce = 5? But Ca (1-bromo-2-methylbutane from C1) and Ce (also gives 1-bromo-2-methylbutane from branch) — placing Br on Ca gives BrCH2-CH(CH3)-CH2-CH3 and on Ce gives CH3-CH(CH2Br)-CH2-CH3 — these are the same compound (1-bromo-2-methylbutane). So Ca and Ce ARE equivalent. Thus isopentane gives 4 unique positions: C1/branch (equivalent), C2, C3, C4 → 4 isomers. Total = 3 + 4 + 1 = 8. Therefore, the correct answer is 8.