Isotopic distribution in CHBr3 mass spectrum

Steps:

• Bromine isotopes are ^{79}Br (44 neutrons) and ^{81}Br (46 neutrons), differing by 2 mass units.
• CHBr3 has three Br atoms, each independently ^{79}Br or ^{81}Br.
• Possible combinations: 0, 1, 2, or 3 ^{81}Br atoms, giving total Br mass excesses of 0, 2, 4, or 6 units over all ^{79}Br.
• These yield four distinct molecular ion masses, producing four peaks.

Why C is correct:

• The binomial distribution of two isotopes in a tri-substituted molecule (n=3) gives n+1=4 unique mass peaks, per the rule for isotopomer patterns in mass spectrometry.

Why the others are wrong:

• A: Ignores intermediate combinations, assuming only pure isotopes.
• B: Misses one of the four possible isotopomer ratios.
• D: Overcounts by possibly including fragment ions, despite the "ignoring fragments" condition.

Final answer: C