Identifying a Bromoalkane from Elemental, Mass and NMR Data
A student brominates an alcohol and distils the product, a colourless liquid, b.p. 71 °C. Elemental analysis gives C 29.2 %, H 5.7 %, Br 65.0 % (by difference). The mass spectrum shows two molecular-ion peaks of equal intensity at m/z 122 and 124, and no other significant high-mass peaks. The ¹H NMR spectrum (CDCl₃) shows three signals: a triplet at δ 1.05 (3H), a multiplet (sextet) at δ 1.85 (2H), and a triplet at δ 3.40 (2H). There is no absorption above 1600 cm⁻¹ in the IR other than C–H stretching. The student now plans two further reactions with this compound: (i) treatment with NaCN in dry DMSO, and (ii) conversion to a Grignard reagent followed by CO₂ and then dilute acid, using diethyl ether that has been standing open to the air and a flask rinsed with tap water. Ar(H) = 1.0, C = 12.0, Br = 79.9.
1. Determine the molecular formula and the structure of the compound, justifying each step from the data given.
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Formula. Moles: C 29.2/12.0 = 2.43; H 5.7/1.0 = 5.70; Br 65.0/79.9 = 0.814. Divide by 0.814 → C 2.99 : H 7.00 : Br 1.00, so the empirical formula is C₃H₇Br, M = 123.0 (for ⁷⁹Br).
One bromine. The 1:1 doublet at m/z 122 and 124 is the signature of a single Br atom, since ⁷⁹Br and ⁸¹Br are almost equally abundant.
No rings or π bonds. Degrees of unsaturation = (2×3 + 2 − 7 − 1)/2 = 0, and the IR shows no C=O or C=C. The compound is an acyclic saturated bromoalkane.
Regiochemistry. Three signals integrating 3:2:2 account for all seven hydrogens. Two triplets mean two CH₂/CH₃ groups each coupled to a CH₂; the intermediate multiplet (sextet) is a CH₂ coupled to five hydrogens. That is only possible for a straight chain, CH₃–CH₂–CH₂–Br (1-bromopropane). The 2-isomer would show a 6H doublet and a 1H septet, and the CH₂Br signal at δ 3.40 is characteristic of a carbon bearing both Br and hydrogens.
2. Predict and explain the outcome of reaction (i) with NaCN in dry DMSO for this compound, and compare its rate with that of 2-bromo-2-methylpropane under the same conditions.
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With 1-bromopropane: CN⁻ is a good nucleophile and DMSO is a polar aprotic solvent that leaves the anion bare and weakly solvates the leaving group. The substrate is primary, so the reaction is a clean Sₙ2: backside attack on C-1 displaces Br⁻ and gives butanenitrile, CH₃CH₂CH₂CN (hydrolysis would then give butanoic acid).
Comparison with 2-bromo-2-methylpropane: this is tertiary. The Sₙ2 pathway is sterically blocked - the nucleophile cannot reach the carbon through three alkyl groups - and in DMSO (poorly ionizing, no proton donor to solvate Br⁻) the Sₙ1 alternative is also slow. Under these conditions the primary bromide reacts far faster, and the tertiary one is largely recovered or slowly gives isobutene by E2/E1.
The judgement point is that tertiary ⇒ fast is only true for Sₙ1: switch the medium to aqueous ethanol or add Ag⁺ and the order inverts, because then the rate depends on carbocation stability rather than on steric access. Reaction conditions, not just the substrate class, decide the mechanism.
3. Predict the product of reaction (ii) as designed and explain what will actually go wrong. Correct the student's plan.
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Intended product is wrong by one carbon. CH₃CH₂CH₂Br + Mg → CH₃CH₂CH₂MgBr; then CO₂ inserts into the C–Mg bond and acid work-up gives CH₃CH₂CH₂CO₂H, butanoic acid (C₄H₈O₂) - a one-carbon chain extension, not propanoic acid.
Two failures from the reagents. (1) Diethyl ether standing open to the air accumulates water and peroxides. Water destroys the Grignard as fast as it forms: RmgBr + H₂O → RH + Mg(OH)Br, so the student would recover propane and see no acid. Peroxides are also a serious hazard when a Grignard is concentrated. (2) The flask rinsed with tap water leaves a film of water on the glass - enough to quench a small-scale preparation.
Correction. Use freshly dried ether (distilled from sodium/benzophenone or over molecular sieves), oven-dry glassware, a calcium chloride guard tube, and an inert atmosphere. Add a small crystal of iodine to activate the magnesium, initiate gently, then add the bromide at a rate that keeps reflux under control before introducing dry CO₂.