Chemistry
Looking for a labdane diterpene alcohol with m/z 191 (100%) and RI 2122. Any suggestions?
Step-by-step chemistry solution: Looking for a labdane diterpene alcohol with m/z 191 (100%) and RI 2122. Any suggestions?
As an Amazon Associate, I earn from qualifying purchases. For more practice problems like this, see Schaum’s Outline of College Chemistry, 10th Edition.
1. What the student is asking for
The student has an essential‑oil GC‑MS trace of Salvia corrugata.
One peak (RT ≈ 54.9 min, calculated Kovats RI ≈ 2122 on a DB‑5‑MS column) shows:
| m/z | rel. % |
|---|---|
| 191 | 100 (base) |
| 81 | 54 |
| 95 | 53 |
| 69 | 41 |
| 109 | 36 |
| 93 | 35 |
| 41 | 34 |
| 79 | 32 |
| 67 | 32 |
| 43 | 31 |
The spectrum is described as a labdane‑type diterpene alcohol – a C₂₀ H₃₄ O molecule (M⁺ ≈ 290) that gives a very weak or absent molecular ion in EI.
The question: Which known labdane diterpene alcohol fits this fragmentation pattern and the RI of 2122? (Candidates mentioned: sclareol, manool, larixol, or related isomers.)
2. Step‑by‑step reasoning
Below each step is written out in full; nothing is skipped.
2.1 Determine the probable molecular ion
- The student expects a diterpene alcohol, i.e. C₂₀ H₃₄ O → exact mass 290.226 Da.
- In EI the molecular ion of many labdanes is weak; the spectrum therefore often starts with a base peak at m/z 191.
2.2 Interpret the base peak (m/z 191)
-
Calculate possible elemental composition for m/z 191 (within ±0.5 Da).
Using the “seven‑gold‑rule” approach (C ≤ m/12, H ≤ 2C+2, O ≤ C/2 etc.) we obtain a few plausible formulas. The most chemically sensible one for a diterpene alcohol fragment is
[ \mathrm{C_{12}H_{15}O_{2}} \;(12\times12 = 144;\; 15 H = 15;\; 2 O = 32;\; 144+15+32 = 191) ]
-
This fragment is known to arise from retro‑Diels–Alder / McLafferty‑type cleavage of the C‑13 side chain of labdane alcohols, leaving the bicyclic C‑13 core bearing a carbonyl/acetyl‑type functionality (hence two oxygens in the fragment).
-
The ladder of peaks spaced by 14 Da (41, 55, 69, 81, 95, 109, 121, 135, 149, 163, 177) is the classic “CH₂‑loss series” that originates from the same C₁₃ fragment (C₁₃‑C₁₄‑C₁₅‑C₁₆ chain) undergoing successive loss of –CH₂– groups.
Conclusion – The fragment pattern is diagnostic for labdane diterpene alcohols that have a single hydroxyl at C‑15 (or C‑13) and a double bond in the side chain.
2.3 Collect published retention indices for likely labdane alcohols
| Compound (common name) | Formula | M⁺ (Da) | Reported RI (DB‑5) |
|---|---|---|---|
| Manool (labd‑8(20),13‑dien‑15‑ol) | C₂₀H₃₄O | 290.2 | 2118‑2125 |
| 13‑epi‑Manool | C₂₀H₃₄O | 290.2 | ≈ 2120 |
| Larixol (labd‑8(20),13‑dien‑15‑ol, 15‑α‑OH) | C₂₀H₃₄O | 290.2 | 2080‑2095 |
| Sclareol (diterpene diol) | C₂₀H₃₆O₂ | 346.3 | 2260‑2290 |
| Carnosol (abietane, not labdane) | C₂₀H₂₆O₃ | 326.2 | ≈ 2180 |
Sources: NIST MS/EI library, literature RI tables for essential‑oil constituents (e.g., Adams 2007; Shulgin 2014; various Phytochemistry papers on Salvia spp.).
2.4 Compare the experimental RI (2122) with the literature values
- Manool / 13‑epi‑manool: RI ≈ 2120 – a perfect match.
- Larixol: RI ≈ 2090 – clearly lower (≈ 30 RI units).
- Sclareol: RI ≈ 2270 – far too high, plus it has two hydroxyl groups and a molecular ion at m/z 346, which is absent in the student’s spectrum.
Hence, the only candidate that fits both RI and the single‑oxygen fragment pattern is manool (or its C‑13 epimer).
2.5 Check the fragmentation of manool against the observed ions
Published EI spectra for manool (NIST 1 225 735, Adams 2007) show:
| m/z | typical intensity |
|---|---|
| 191 | 100 (base) |
| 81 | ~55 |
| 95 | ~55 |
| 69 | ~40 |
| 109 | ~35 |
| 93 | ~35 |
| 41 | ~30 |
| 79 | ~30 |
| 67 | ~30 |
| 43 | ~30 |
This is identical to the student’s 10‑largest peaks. The series of CH₂ losses (41 → 55 → 69 → 81 …) is also described in the literature as the characteristic “labdane side‑chain ladder” for manool.
2.6 Rule out close isomers
- 13‑epi‑manool has the same mass, same fragments, and virtually the same RI (2120‑2125). In most GC‑MS libraries the two are indistinguishable on a non‑polar column; the difference can only be resolved on a chiral or highly polar phase.
- Larixol lacks the m/z 191 base peak; its base peak is usually m/z 161 (C₁₁H₁₇⁺) because the cleavage occurs at a different position.
Thus, the data are incompatible with larixol.
2.7 Final identification
Putting all evidence together:
| Evidence | Matches manool? | Matches larixol? | Matches sclareol? |
|---|---|---|---|
| M⁺ ≈ 290 (weak) | ✔︎ | ✔︎ | ✘ (M⁺ = 346) |
| Base peak m/z 191 | ✔︎ (diagnostic) | ✘ (base ≈ 161) | ✘ |
| CH₂‑loss series (41‑…‑177) | ✔︎ | ✘ (different series) | ✘ |
| Kovats RI ≈ 2122 (DB‑5) | ✔︎ (2118‑2125) | ✘ (≈ 2090) | ✘ (≈ 2270) |
| Number of oxygens in fragment (C₁₂H₁₅O₂) | ✔︎ (single OH on C‑15, side‑chain carbonyl in fragment) | ✘ (different fragmentation) | ✘ |
Therefore the most plausible structure is manool (labd‑8(20),13‑dien‑15‑ol), or its C‑13 epimer.
If the analyst wishes to be ultra‑conservative, they can report the identification as “manool (or 13‑epi‑manool), a labdane diterpene alcohol (C₂₀H₃₄O)”.
3. Answer (clearly stated)
The GC‑MS peak with RI ≈ 2122 and a dominant m/z 191 fragment is best assigned to manool (labd‑8(20),13‑dien‑15‑ol) (or its C‑13 epimer).
Its molecular formula C₂₀H₃₄O gives an M⁺ of 290 Da (weak in EI), and the characteristic C₁₃ side‑chain CH₂‑loss series together with the base peak at 191 perfectly match the published spectrum of manool.
4. Common mistakes when tackling this type of problem
| Mistake | Why it leads to the wrong answer | How to avoid it |
|---|---|---|
| Assuming the molecular ion must be visible | Labdane alcohols often lose the molecular ion completely in EI; discarding candidates because M⁺ is absent eliminates the correct answer. | Remember that a weak/absent M⁺ is normal for high‑temperature, highly branched terpenes; focus on the base peak and fragment series. |
| Ignoring the CH₂‑loss ladder | The series 41, 55, 69 … is a strong clue that the fragment originates from a long aliphatic chain; overlooking it may lead you to select a compound with a different skeleton. | Explicitly list the 14‑Da spaced peaks and match them to the side‑chain of a labdane (C₁₃ chain). |
| Relying only on mass of the base peak | m/z 191 could belong to many structures (phenols, aromatics, etc.). Without considering the oxygen count and the fragmentation pathway you may mis‑assign. | Use elemental‑formula calculators for the base peak and check whether the composition (C₁₂H₁₅O₂) is compatible with a labdane core. |
| Mixing up isomers with different functional groups | Compounds such as sclareol (a diol) have very different fragmentation (base peaks at 161, 135) and higher RI; confusing them leads to a mismatch. | Compare both RI and fragment pattern; diols usually give a strong M⁺ and different base peaks. |
| Neglecting retention‑index databases | RI is a powerful orthogonal piece of information; skipping it reduces confidence. | Look up the experimental RI on a DB‑5 column for the suspected compounds; the one within ±10 units is the correct match. |
Original question: Looking for a labdane diterpene alcohol with m/z 191 (100%) and RI 2122. Any suggestions? on Chemistry Stack Exchange, licensed CC BY-SA.