​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:​

Correct option is B

Explanation-

Given:
Molecular ion peak [M]⁺ at m/z = 149, 100% abundance.
Another peak at m/z = 150 with 6.7% abundance → indicates ¹³C isotope peak.
[M + 2]⁺ peak with 5% abundance → suggests presence of Sulfur (S), especially ³⁴S.

Step 1: Check [M + 1]⁺ peak (m/z = 150, 6.7%)
This peak is due to ¹³C, which has a natural abundance of 1.1%.
Calculation of the number of carbon atoms:
                                                               $$\frac{\text{Relative abundance of } [M + 1]^+}{\text{Abundance of } ^{13}C} = \frac{6.7}{1.1} \approx 6.1$$

So, approximately 6 carbon atoms. Most likely 5–6 C atoms.
Step 2: Check [M + 2]⁺ peak (m/z = 151, 5%)
This is due to the ³⁴S isotope, which has 4.2% abundance.
If the [M+2] peak is around 5%, it's strongly indicative of one sulfur atom:
For one sulfur atom:     M+2 peak ≈ 4.2
Observed: 5% => confirms one sulfur atom.​​

Step 3: Molecular Weight Matching
Look for the compound whose molecular weight is 149.

Option b : C₅H₁₁NO₂S

C = 5 × 12 = 60
H = 11 × 1 = 11
N = 1 × 14 = 14
O = 2 × 16 = 32
S = 1 × 32 = 32
Total = 60 + 11 + 14 + 32 + 32 = 149 g/mol 
Also:
~5 carbon atoms → Matches [M+1] peak (6.7%)
1 sulfur atom → Matches [M+2] peak (5%)

Final Answer: Option b  – C₅H₁₁NO₂S
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