55. The molecular ion peak [M]⁺ of an analyte as measured by Electron Ionization Mass Spectrometry has an m/z of 149 and a relative abundance of 100%. The [M]⁺ has a relative abundance of 6.7% and the [M + 2]⁺ peak has a relative abundance of 5%. The abundance of the major isotope of H, C, N, O, and S are ¹H–100%, ¹²C–98.9%, ¹³C–1.1%, ¹⁴N–99.6%, ¹⁵N–0.4%, ¹⁶O–99.8%, ¹⁸O–0.2%, ³²S–95.0%, ³³S–0.75% and ³⁴S–4.2%. The most probable molecular formula of the compound is:

(A) C₇H₂₁N₂O

(B) C₅H₁₁NO₂S

(C) C₆H₁₃O₂S

(D) C₆H₁₅NOS

Determining the Molecular Formula from Molecular Ion Isotopic Pattern in Electron Ionization Mass Spectrometry

Correct Answer

Option (2): C₅H₁₁NO₂S

Explanation

One of the most powerful applications of Electron Ionization (EI) Mass Spectrometry is the determination of molecular formula from the isotopic distribution of the molecular ion. The molecular ion peak (M) represents molecules containing only the most abundant isotopes, whereas the M+1 and M+2 peaks arise because naturally occurring heavier isotopes such as ¹³C, ¹⁵N, ¹⁸O, ³³S, and ³⁴S are incorporated into a small fraction of molecules.

Step 1: Interpretation of the M+1 Peak

The largest contributor to the M+1 peak is ¹³C, whose natural abundance is approximately 1.1%. Therefore, the percentage intensity of the M+1 peak provides an estimate of the number of carbon atoms.

The observed intensity is

M+1 = 6.7%

Number of carbon atoms ≈

6.7 ÷ 1.1 ≈ 6

Thus, the molecule should contain approximately six carbon atoms.

Among the options, only Options (3) and (4) contain six carbon atoms.

Step 2: Interpretation of the M+2 Peak

The M+2 peak is especially useful for identifying sulfur because ³⁴S has a relatively high natural abundance of approximately 4.2%. Molecules containing one sulfur atom therefore exhibit an M+2 peak close to 4–5% of the molecular ion intensity.

The observed value is

M+2 = 5.0%

This strongly indicates the presence of one sulfur atom.

Options (2), (3), and (4) each contain one sulfur atom.

Step 3: Verify the Molecular Mass

The molecular ion appears at m/z = 149. The molecular masses of the remaining candidate formulas are calculated using the major isotopes.

Option (2): C₅H₁₁NO₂S

= (5 × 12) + (11 × 1) + (14) + (2 × 16) + (32)

= 60 + 11 + 14 + 32 + 32

= 149 ✔

Option (3): C₆H₁₃O₂S

= 72 + 13 + 32 + 32

= 149 ✔

Option (4): C₆H₁₅NOS

= 72 + 15 + 14 + 16 + 32

= 149 ✔

All three formulas satisfy the molecular mass, so the isotopic distribution must be examined further.

Step 4: Compare the M+1 Contribution

The M+1 peak receives contributions from several isotopes.

For Option (2)

From 5 carbons:

5 × 1.1 = 5.5%

From one nitrogen:

0.4%

From one sulfur (³³S):

0.75%

Total ≈

5.5 + 0.4 + 0.75 = 6.65%

This agrees almost exactly with the observed 6.7%.

For Option (3)

Carbon contribution alone is

6 × 1.1 = 6.6%

Adding sulfur contribution gives

6.6 + 0.75 = 7.35%

This is significantly higher than the observed value.

For Option (4)

Carbon contribution:

6.6%

Nitrogen contribution:

0.4%

Sulfur contribution:

0.75%

Total ≈

7.75%

This is even further from the observed value.

Therefore, only Option (2) matches both the molecular mass and the observed isotopic abundances.

Why Option (1) is Incorrect

Although this formula has the correct molecular mass, it contains no sulfur atom. A molecule lacking sulfur would not produce an M+2 peak close to 5%. In addition, seven carbon atoms would be expected to generate an M+1 peak of approximately 7.7%, which is inconsistent with the observed spectrum.

Why Option (2) is Correct

This formula has the correct molecular mass of 149, contains one sulfur atom that explains the approximately 5% M+2 peak arising from ³⁴S, and predicts an M+1 intensity of approximately 6.65%, which matches the observed value of 6.7%. Therefore, it is the only formula fully consistent with the experimental isotopic pattern.

Why Option (3) is Incorrect

Although the molecular mass and sulfur content are correct, six carbon atoms together with sulfur predict an M+1 peak of approximately 7.35%, which is higher than the observed 6.7%. Therefore, the isotopic distribution does not support this molecular formula.

Why Option (4) is Incorrect

This formula also has the correct molecular mass, but the presence of six carbon atoms, one nitrogen, and one sulfur predicts an M+1 peak close to 7.75%, considerably larger than the experimental value. Hence, it is inconsistent with the measured isotopic abundances.

Role of M+1 and M+2 Peaks in Molecular Formula Determination

The M+1 peak primarily originates from naturally occurring ¹³C atoms and is therefore widely used to estimate the number of carbon atoms present in a molecule. Additional contributions from ¹⁵N, ²H, ¹⁷O, and ³³S are usually much smaller but become important when distinguishing between closely related molecular formulas.

The M+2 peak is particularly useful for identifying elements such as sulfur, chlorine, and bromine because their heavier isotopes occur at relatively high natural abundance. A single sulfur atom typically contributes an M+2 peak of approximately 4–5%, whereas chlorine and bromine produce much larger and highly characteristic isotope patterns.

Conclusion

The observed molecular ion at m/z 149, together with an M+1 peak of 6.7% and an M+2 peak of 5.0%, indicates a molecule containing one sulfur atom and an isotopic distribution that matches C₅H₁₁NO₂S. Therefore, the correct answer is Option (2).