Combining Spectra

No single spectroscopic technique is sufficient for complete structure determination. The power of modern organic chemistry comes from combining MS, IR, and NMR data systematically. This section gives you the strategy.

Step-by-Step Strategy

Step 1: Mass spectrometry.

• Find the M⁺ peak → molecular weight.
• Check isotope pattern → identify Cl, Br, or elemental composition.
• Apply the nitrogen rule → odd M⁺ = odd N count.
• Note base peak and prominent fragment losses → partial structural information.

Step 2: Determine molecular formula.

• Use accurate MW + constraints (nitrogen rule, heteroatom hints from MS) to deduce a molecular formula.
• Calculate degree of unsaturation: DoU = (2C + 2 + N - H - X) / 2, where C, H, N, X are the counts of carbon, hydrogen, nitrogen, halogen.
• DoU tells you the total number of rings + pi bonds.

Step 3: Infrared spectroscopy.

• Is there O-H or N-H? Check 3200-3500 broad/sharp.
• Is there a carbonyl (C=O) at 1700-1800? Note exact position to identify derivative type.
• Is there a triple bond (2100-2300)?
• Is the compound aromatic (peaks above 3000 for sp² C-H + multiple peaks at 1450-1600)?

Step 4: ¹H NMR.

• Count distinct environments.
• Note chemical shifts: what is near each H?
• Determine integration ratios: how many H’s of each type?
• Interpret splitting: how are the H’s connected?

Step 5: ¹³C NMR (if available).

• Count distinct carbons.
• Note carbonyl peaks (150-220 ppm).
• Distinguish alkyl (0-50), heteroatom-adjacent (50-90), sp²/aromatic (100-150) regions.

Step 6: Piece it together.

• Build a structural hypothesis consistent with all data.
• Check: does every piece of data match?
• If not, revise the hypothesis.

Degree of Unsaturation

A critical concept: the degree of unsaturation (DoU, or index of hydrogen deficiency) is calculated as:

DoU = (2C + 2 + N - H - X) / 2

where C, H, N, X are carbon, hydrogen, nitrogen, halogen counts. Oxygen is ignored (does not change the count).

Each DoU counts:

• 1 ring = 1 DoU.
• 1 double bond = 1 DoU.
• 1 triple bond = 2 DoU.
• Aromatic ring = 4 DoU (3 double bonds + 1 ring).

Examples:

• C₄H₁₀ (butane): DoU = (8+2-10)/2 = 0. No rings, no double bonds.
• C₄H₈ (butene OR cyclobutane): DoU = 1. One double bond OR one ring.
• C₆H₆ (benzene): DoU = (12+2-6)/2 = 4. Benzene.

Practice Example: Identify the Compound

Given:

• MS: M⁺ at m/z 88. No isotope pattern (probably no Cl/Br).
• IR: Strong peak at 1740 cm⁻¹. No O-H broad peak.
• ¹H NMR: Quartet (2H) at 4.1 ppm, singlet (3H) at 2.0 ppm, triplet (3H) at 1.2 ppm.

Analysis:

1. MW 88. Even → no odd number of N. Try molecular formula C₄H₈O₂ (MW = 4×12 + 8×1 + 2×16 = 88). Fits.
2. DoU = (8+2-8)/2 = 1. One ring or one double bond.
3. IR: C=O at 1740 is ester region. Consistent with an ester (which has the 1 DoU from the C=O).
4. NMR: ethyl group pattern (quartet at 4.1 + triplet at 1.2). The 3H singlet at 2.0 is an isolated methyl. Acetate type (CH₃-CO-O-).
5. Putting it together: methyl acetate is C₃H₆O₂ (MW 74) - doesn’t match. Ethyl acetate: CH₃-CO-O-CH₂-CH₃, C₄H₈O₂, MW 88. ✓

Answer: ethyl acetate.

Common Pitfalls

• Forgetting the nitrogen rule: odd M⁺ peak usually means odd N count.
• Ignoring DoU: always calculate it; it tells you how many rings + double bonds to expect.
• Confusing aldehyde C-H with other C-H: aldehyde gives a characteristic doublet at 2720+2820 in IR AND a 9-10 ppm peak in NMR.
• Misreading C=O position in IR: exact wavenumber identifies the derivative (ester 1735, ketone 1715, amide 1680, acyl halide 1800).