What isotope pattern is produced when a molecule contains two bromine atoms and one chlorine atom? With the usual rounded natural-abundance assumptions—⁷⁹Br:⁸¹Br ≈ 1:1 and ³⁵Cl:³⁷Cl ≈ 3:1—the principal molecular-ion groups occur at M, M+2, M+4 and M+6 with approximate intensity ratio 3:7:5:1. The result follows by combining the two-bromine pattern with the one-chlorine pattern: (1+x)²(3+x) = 3+7x+5x²+x³.
A concrete molecular example is dibromochloromethane, CHBr₂Cl, also called chlorodibromomethane. Its light-isotope nominal molecular mass, using ¹²C, ¹H, two ⁷⁹Br and one ³⁵Cl, is 206. If a suitable intact singly charged molecular ion is detected, the other main isotope groups are expected near nominal m/z 208, 210 and 212. The strongest group in the idealised four-member cluster is M+2, not M.
For families and students searching for an H3 Chemistry tutor in Hougang, this is a useful mathematical extension of the official 2027 H3 Chemistry 9813 mass-spectrometry objectives. SEAB explicitly names halogen M+2 and M+4 isotope patterns. The three-heavy-halogen M+6 calculation is not a separately named compulsory topic; it uses the same physical reasoning to tackle an unfamiliar but manageable stimulus. The skill is more valuable than another memorised number sequence.
- Formula: CHBr₂Cl, one carbon and H, two bromines and one chlorine.
- Rounded isotope weights: bromine light:heavy 1:1; chlorine light:heavy 3:1.
- Main isotope ratio: M:M+2:M+4:M+6 ≈ 3:7:5:1.
- Nominal masses: 206, 208, 210 and 212 for an intact singly charged CHBr₂Cl ion.
- Dominant group: M+2, with coefficient 7.
- Average molecular weight: approximately 208.28 g mol⁻¹ from natural isotope abundances, which differs from the light-isotope nominal mass 206.
- Reason for the difference: average mass, monoisotopic mass and nominal isotope-group m/z answer distinct questions.
- Context: treat CHBr₂Cl as a spectral case study, not a casual laboratory preparation or handling instruction.
Official SEAB H3 Chemistry syllabus 9813 (2027) · NIST Chemistry WebBook: dibromochloromethane CHBr₂Cl · PubChem formula, mass and safety context · Hougang JC Chemistry tuition hub.
Contents: The isotope principle · Start with one chlorine · Two bromines · Derive 3:7:5:1 · Masses of CHBr₂Cl · Every isotope combination · Convert to percentages · Why 208.28 differs from 206 · Compare other halogen counts · Misinterpretations · Worked spectra · Original practice · Teaching and career links · FAQs.
Step 1: Understand Why Isotope Groups Are Two Nominal Units Apart
A molecular ion contains nuclei with definite isotope masses. Bromine is dominated by ⁷⁹Br and ⁸¹Br, and chlorine by ³⁵Cl and ³⁷Cl. Each heavy isotope contains two additional neutrons compared with the corresponding light isotope. Replacing a light atom with a heavy one therefore increases the nominal mass of the ion by about two mass units.
In the simplest singly charged mass-spectral interpretation, that change increases the nominal mass-to-charge ratio, m/z, by two. A molecule with several such halogens can have combinations with no heavy isotopes, one heavy isotope, two heavy isotopes or more. Each count gives a main group separated by successive two-unit shifts.
This is why an M+2 group is not necessarily a different chemical molecule in the sense of different connectivity. It may represent the same bonding structure containing one heavy isotope. Isotopologues share elemental identities and connections but differ in isotope composition.
Actual isotope mass differences are close to, rather than exactly, 2.0000 atomic mass units, so high-resolution measurement can show small deviations from the simplified nominal values. But for common 2027 H3 examination-style isotope-counting problems, the nominal separation is a useful first interpretation.
Step 2: Why One Chlorine Gives a 3:1 Pair
Natural ³⁵Cl occurs at about 75.8% and ³⁷Cl at about 24.2%. A school-level simplification is 75% to 25%, equivalent to relative weights 3:1. One chlorine-containing molecular ion therefore commonly shows principal light and heavy partners at M and M+2 with an approximate intensity ratio 3:1.
The larger M group represents molecules carrying ³⁵Cl, while M+2 represents molecules carrying ³⁷Cl. This is a statistical property of a very large collection of molecular ions, not an event where an ion changes isotopic identity during flight through the instrument.
The simplified one-chlorine generating expression is (3+x), where 3 represents the light isotope and x represents the heavy-isotope choice with its +2 nominal shift. We are not claiming that x is a chemical atom or an instrument reading. It is a bookkeeping symbol for isotope substitution.
Step 3: Why Two Bromines Give a 1:2:1 Triplet
Natural bromine is dominated by ⁷⁹Br and ⁸¹Br with nearly equal abundances. For a single bromine atom, the principal isotope group is therefore close to 1:1 between light and heavy choices. For two bromines, the relative statistical weights multiply, giving (1+x)² = 1+2x+x².
- Both bromines ⁷⁹Br: relative weight 1, adding zero nominal mass units.
- Exactly one ⁸¹Br: there are two possible isotope assignments, giving weight 2 at +2.
- Both bromines ⁸¹Br: relative weight 1, giving +4.
- Thus the bromine-only distribution is 1:2:1 at the three principal groups.
This is the same physics behind a molecular-ion isotope pattern of Br₂ under the simple model. But we are not claiming the parent CHBr₂Cl molecular ion is Br₂. The bromine-only distribution is one component that must be combined with the additional chlorine atom.
A student who remembers 1:2:1 for two bromines may be tempted to stop there. That loses the extra chlorine and misses the additional M+6 group. The central challenge is to combine two independent isotope distributions correctly.
Step 4: Derive the Complete Two-Bromine–One-Chlorine Pattern
For two bromines, use (1+x)² = 1+2x+x². For one chlorine, use (3+x). Multiply them: (1+2x+x²)(3+x) = 3+7x+5x²+x³. The power of x tells us how many heavy halogen isotopes have been selected, and its coefficient gives their total rounded relative weight.
M: No heavy halogens
Both Br atoms are ⁷⁹Br and Cl is ³⁵Cl. Bromine weight is 1×1, chlorine weight is 3, so the total rounded weight is 3. This is the light-isotope group called M.
M+2: Exactly one heavy isotope
There are two possibilities of isotope type. One of the two Br atoms can be ⁸¹Br while the other Br is light and Cl is light; there are two assignments, each with chlorine weight 3, giving 6. Alternatively, both bromines are light and the chlorine is heavy ³⁷Cl, giving 1. The total coefficient is 6+1=7.
M+4: Exactly two heavy isotopes
Both bromines can be heavy ⁸¹Br while chlorine is light ³⁵Cl, giving weight 3. Alternatively, exactly one bromine is heavy and the chlorine is heavy; there are two bromine assignments, each weight 1, giving 2. The combined coefficient is 3+2=5.
M+6: All three halogens are heavy
Both bromines are ⁸¹Br and the chlorine is ³⁷Cl. There is one heavy-heavy-heavy combination with rounded weight 1. The nominal mass shift is +2+2+2=+6.
- M: coefficient 3.
- M+2: coefficient 7.
- M+4: coefficient 5.
- M+6: coefficient 1.
- Complete ratio: 3:7:5:1.
This derivation is more reliable than looking up an unfamiliar ratio. If a question changes two bromines into two chlorines, the student can change the generating expression and rebuild the answer. Understanding the physical meaning of each term is what makes the mathematics useful.
Step 5: Calculate the Nominal Masses of Dibromochloromethane
Dibromochloromethane contains one carbon, one hydrogen, two bromines and one chlorine. The light-isotope composition is ¹²C, ¹H, two ⁷⁹Br atoms and one ³⁵Cl atom. Its nominal mass is 12 + 1 + 2(79) + 35 = 206. For a singly charged intact molecular ion, the corresponding nominal m/z is 206.
Each heavy isotope substitution adds about two nominal mass units. The remaining principal groups therefore lie at 208, 210 and 212. The strongest expected group is 208 under the rounded distribution, with coefficient seven.
Do not confuse the molecular ion with every mass-spectral peak
Electron-impact and other ionisation processes may produce fragments. An observed peak at a lower m/z can come from a fragment retaining one bromine, both bromines or a different halogen combination. Its isotope pattern depends on the composition of that particular fragment, not necessarily on the full parent molecule.
The stated 206/208/210/212 sequence describes the principal isotope partners of an intact singly charged CHBr₂Cl molecular ion in the simple model. A real laboratory spectrum may have additional satellite peaks from carbon-13 or other isotope combinations, as well as many fragment ions.
Step 6: Count Every Heavy-Isotope Combination Without Algebra
The generating-function result can be checked by counting the eight nominal heavy/light configurations for two distinguishable isotope positions on bromine and one on chlorine. Although the two Br atoms are chemically equivalent in the molecular formula, exchanging their isotope identities represents two configurations of the naturally distributed sample, and those contributions count toward the same nominal peak group.
- Br light / Br light / Cl light → M; weight 3.
- Br heavy / Br light / Cl light → M+2; weight 3.
- Br light / Br heavy / Cl light → M+2; weight 3.
- Br light / Br light / Cl heavy → M+2; weight 1.
- Br heavy / Br heavy / Cl light → M+4; weight 3.
- Br heavy / Br light / Cl heavy → M+4; weight 1.
- Br light / Br heavy / Cl heavy → M+4; weight 1.
- Br heavy / Br heavy / Cl heavy → M+6; weight 1.
Add the weights by nominal shift and the coefficients are again 3 at M, 7 at M+2, 5 at M+4 and 1 at M+6. The algebra and enumeration agree. This double-check is useful in a pressured paper because it reveals a missed mixed-isotope placement.
Notice the non-intuitive result: the most intense nominal isotope group does not have the lowest mass. In a three-halogen mixture, several ways of choosing exactly one heavy isotope make the M+2 group more probable than the all-light group.
Step 7: Convert the Ratio to Percentages
The coefficients sum to 3+7+5+1=16. Under the rounded 50/50 bromine and 75/25 chlorine assumptions, divide each coefficient by 16. The ideal probabilities are 18.75% for M, 43.75% for M+2, 31.25% for M+4 and 6.25% for M+6. The four sum to 100%.
If a graph normalises its largest isotope peak to intensity 100 instead of making total intensity 100%, divide all coefficients by seven and multiply by 100. The relative peak heights become approximately 42.9 : 100 : 71.4 : 14.3. That is the same pattern expressed with a different normalisation.
A student should not confuse these percentages with the proportion of all ions in a real full mass spectrum. They apply to the chlorine/bromine isotope distribution within a correctly identified ion species under the simplified abundance model. Fragmentation can shift the actual total ion counts into other m/z values.
Step 8: Why NIST Gives Molecular Weight 208.28 but M Starts at 206
The NIST Chemistry WebBook reports a relative molecular weight of approximately 208.280 for CHBr₂Cl. The light-isotope nominal formula gives 206 instead. Both values can be correct, because they answer different questions.
The nominal mass 206 describes a particular isotopologue containing ¹²C, ¹H, two ⁷⁹Br atoms and one ³⁵Cl atom. The average molecular weight 208.28 accounts for the natural mixture of isotopes and their relative abundances, using more precise isotopic masses than rounded integers.
The intensity-weighted mean of the four nominal mass groups under the simplified 3:7:5:1 model is (3×206 + 7×208 + 5×210 + 1×212)/16. The numerator is 618+1456+1050+212 = 3336, giving 208.5 as a rough rounded-isotope nominal-mass average. It differs slightly from the real 208.28 because the simplified 3:1 and 1:1 abundances and integer isotope masses are approximate.
This is a useful quantitative check: the true average should sit among the isotope masses, closer to the more abundant middle groups. A student who calls 206 the natural-abundance average weight has overlooked the heavier isotope contributions. A student who calls 208.28 the exact m/z of the lightest isotopologue has confused average and isotope-specific quantities.
Step 9: Compare One Cl/Two Br with Two Cl/One Br and Three Cl
Suppose a related hypothetical formula contains two chlorine atoms and one bromine, rather than one chlorine and two bromines. The generating expression is (3+x)²(1+x), giving 9+15x+7x²+x³. Its principal intensity ratio becomes 9:15:7:1, not 3:7:5:1.
For a molecule with three chlorines and no bromine, the expression is (3+x)³ = 27+27x+9x²+x³, producing the well-known extended 27:27:9:1 example. These cases all have up to three heavy-halogen substitutions and therefore M+6, but their relative intensities differ because the natural isotope distributions are different.
For two bromines without chlorine, (1+x)² = 1+2x+x² gives only three nominal groups through M+4. The addition of one chlorine introduces the fourth group and changes all relative weights. Merely counting the presence of M+6 without examining the shape of the whole cluster would throw away useful chemical information.
The corresponding Hougang articles are three chlorines and the M+6 pattern and one chlorine and one bromine in an M+4 pattern. The present problem is a different mixed-halogen three-atom case.
Ten Frequent Mass Spectrometry Errors
- Using molecular weight 208.28 as the light-isotope m/z. Average and monoisotopic or nominal isotope masses differ.
- Using 3:1 for bromine. Bromine’s two major isotopes are nearly equal in abundance.
- Using 1:1 for chlorine. Chlorine’s light isotope is about three times as abundant.
- Writing (3+x)²(1+x) for two bromines plus one chlorine. That expression counts two chlorines and one bromine instead.
- Forgetting the mixed-isotope multiplicity. Different placements contribute to the same nominal group.
- Stopping at M+4. Three heavy halogen choices can generate M+6.
- Calling all peaks in the spectrum molecular-ion isotope partners. Many are fragments or different ions.
- Assuming the first or tallest group must be M. M+2 is largest in the ideal CHBr₂Cl pattern.
- Demanding exact whole-number ratios from natural samples. True isotope fractions differ from the rounded model.
- Inferring complete connectivity from isotope counts alone. Mass composition does not uniquely specify a molecule.
Three Original Worked Mass-Spectrum Cases
Case A: Four peaks beginning at m/z 206
A hypothetical mass spectrum of a correctly assigned singly charged intact molecular ion shows groups at m/z 206, 208, 210 and 212. After suitable correction and normalisation, their approximate relative intensities are 3:7:5:1. The independently established formula is CHBr₂Cl. Explain the pattern.
Solution: Two nearly 1:1 bromines contribute (1+x)² and one approximately 3:1 chlorine contributes (3+x). Multiplication gives 3:7:5:1, with each heavy substitution adding two nominal mass units to the light-isotope molecular ion at 206. This is consistent with CHBr₂Cl.
Case B: The same nominal spacing, different intensities
A second hypothetical ion shows four groups spaced by two units, but the rounded ratio is closer to 9:15:7:1. Is that the same two-Br/one-Cl composition?
Solution: Not under the same simplified natural-abundance assumptions. This ratio matches two chlorines and one bromine, (3+x)²(1+x). The full formula and mass still need to be checked, but the pattern is different.
Case C: A fragment with only one Br retained
An unknown parent has formula CHBr₂Cl, but a lower-mass fragment ion contains just one bromine and no chlorine. What simple halogen isotope signature is expected for that fragment?
Solution: Approximately a 1:1 M/M+2 pair from one Br. The fragment’s isotope pattern does not have to match the parent four-group pattern because its elemental composition has changed.
These three cases illustrate how an isotope pattern is best read as evidence about the ion actually detected. A full mass spectrum can contain several isotopic clusters, each associated with a different ion or fragment.
Twenty-Seven Original H3 Chemistry Questions With Answers
What is dibromochloromethane’s formula?
Answer: CHBr₂Cl: one carbon, one hydrogen, two bromine atoms and one chlorine atom.
Is dibromochloromethane the same as bromochloromethane?
Answer: No. Dibromochloromethane has two Br atoms and one Cl; bromochloromethane has only one of each.
Why does each heavy halogen add approximately two nominal mass units?
Answer: Replacing ⁷⁹Br with ⁸¹Br or ³⁵Cl with ³⁷Cl increases the isotope mass by roughly two u.
How many main nominal isotope groups arise from two bromines and one chlorine?
Answer: Four: M, M+2, M+4 and M+6.
What is the rounded abundance factor for chlorine?
Answer: ³⁵Cl:³⁷Cl approximately 3:1.
What is the rounded abundance factor for bromine?
Answer: ⁷⁹Br:⁸¹Br approximately 1:1.
Why do two bromines by themselves give 1:2:1?
Answer: The combinations are light-light, two ways of mixed light/heavy, and heavy-heavy.
What generating expression represents two Br plus one Cl?
Answer: (1+x)²(3+x), with x indicating the +2 nominal heavy-isotope choice.
Expand the expression.
Answer: 3+7x+5x²+x³, giving a coefficient ratio 3:7:5:1.
Calculate the light-isotope nominal molecular mass of CHBr₂Cl.
Answer: 12+1+2×79+35=206.
What are the other principal nominal mass groups?
Answer: 208, 210 and 212 for one, two and three heavy halogen choices.
Which isotopic group is expected to have the greatest intensity in the rounded model?
Answer: M+2, coefficient 7 out of a total 16.
Why is M+2 larger than M here?
Answer: There are multiple ways to introduce one ⁸¹Br or one ³⁷Cl, making the combined one-heavy probability larger than the all-light probability.
What is the M+4 coefficient?
Answer: 5, representing the combinations with two heavy halogen isotopes.
What is the M+6 coefficient?
Answer: 1, from both bromines being ⁸¹Br and the chlorine being ³⁷Cl.
Normalise the 3:7:5:1 pattern into percentages.
Answer: 18.75%, 43.75%, 31.25%, 6.25% for M, M+2, M+4, M+6 in the rounded 75/25 and 50/50 model.
Why does NIST report molecular weight about 208.28 while the light isotope M is 206?
Answer: 208.28 is an average molecular weight over natural isotope abundances; nominal 206 is the light-isotope composition of an individual molecular ion.
Do the four peak intensities have to be exactly 3:7:5:1 in a real spectrum?
Answer: No. Actual Br and Cl abundances differ slightly from the rounded fractions, and fragments and other isotope satellites can overlap.
What do carbon-13 M+1 satellites do?
Answer: A ¹³C substitution can produce weaker nominal +1 partner peaks beside the main isotope groups; this molecule has one carbon.
Is the whole mass spectrum made only of four peaks?
Answer: No. Ionisation and fragmentation can produce other ions; the four-group pattern concerns a correctly assigned intact molecular-ion isotope cluster.
How would one bromine plus two chlorines differ?
Answer: The generating expression is (1+x)(3+x)²=9+15x+7x²+x³, giving 9:15:7:1 rather than 3:7:5:1.
What does an M+6 signal alone prove?
Answer: By itself it is not unique evidence for CHBr₂Cl. The full isotope pattern, mass and other chemical information must be consistent.
Can the isotope pattern establish exactly which atom bonds to which?
Answer: No. It constrains elemental isotope composition, not complete connectivity or stereochemistry.
Can this compound’s formula support an ordinary degree-of-unsaturation calculation?
Answer: CHBr₂Cl has one carbon, one hydrogen and three monovalent halogens: DBE=(2+2−1−3)/2=0.
Is M+6 separately named in the 2027 H3 Chemistry 9813 outcomes?
Answer: The syllabus names M+2 and M+4 halogen patterns; M+6 is a justified extension of that isotope-counting method.
Which ions might carry different halogen patterns?
Answer: A fragment retaining only one Br might show an approximately 1:1 Br M/M+2 pair, whereas a fragment retaining both Br may show approximately 1:2:1.
What experimental safety assumption should be avoided?
Answer: Do not handle halogenated chemicals as casual student demonstration materials; use validated published spectra and supervised laboratory practice.
How I Would Teach This in an H3 Chemistry Small Group
I would give students three cards: two marked Br and one Cl. Each has a light and heavy isotope choice. Before any algebra, they enumerate the eight possible heavy/light assignments and group them by the number of heavy atoms. This reveals why some nominal masses receive contributions from several assignments.
Then we assign the rounded abundance weights, 1:1 for each bromine and 3:1 for chlorine, and sum them. One student independently expands the algebraic expression, another enumerates combinations, and a third calculates the nominal m/z values for the proposed CHBr₂Cl formula. All three must then reproduce the full reasoning alone.
The next exercise introduces an average molecular weight of 208.28 and asks why the lowest nominal isotope group begins at 206. This separates average, nominal and isotope-specific mass—concepts often confused when students merely read one periodic-table value.
Finally, I replace one bromine with chlorine and ask learners to predict the changed ratio without looking it up. If they can obtain 9:15:7:1 after deriving 3:7:5:1, the method has transferred. If they can also explain why a fragment might display a different halogen pattern, they are reading a real mass spectrum with appropriate caution.
For Hougang families, the eduKate JC2 H3 Chemistry small-group overview describes the tutorial model. Confirm specialist availability, the student’s H2 Chemistry foundation, actual venue near Sixth Avenue MRT and class timetable directly; the Hougang title does not imply a separate teaching room in the neighbourhood.
Why Mixed-Halogen Mass Patterns Matter in Applied Chemistry
Halogenated organic compounds appear in industrial chemistry and environmental analytical work. Multiple chlorine or bromine atoms can create distinctive isotope patterns that help narrow an unknown ion’s elemental composition. Researchers then combine isotope information with exact mass, fragmentation, NMR and chromatography to distinguish structural possibilities.
A well-designed laboratory investigation also considers chemical safety and environmental implications. Dibromochloromethane is a regulated hazardous halogenated chemical, and the purpose of this page is educational spectrum interpretation, not instructions for informal synthesis or handling.
The broader scientific method is simple and transferable: derive the expected pattern from isotope probabilities, compare it with measured ions, check molecular mass, and avoid claiming more about structure than the data justify.
Two Bromines and One Chlorine: FAQs
What is the ideal M:M+2:M+4:M+6 ratio?
3:7:5:1, using rounded natural-abundance weights.
Why is M+2 larger than M?
Several one-heavy-isotope arrangements contribute to the same nominal mass, giving a combined weight of seven.
What formula gives the pattern?
Any suitably identified ion with two Br and one Cl may show this halogen signature; CHBr₂Cl is one concrete example.
What is the light-isotope nominal m/z for CHBr₂Cl?
206 for an intact singly charged molecular ion in the rounded ¹²C/¹H/⁷⁹Br/³⁵Cl calculation.
Why is the average molar mass about 208.28?
Because it averages the natural isotopes using their actual abundances and masses.
Does the 2027 H3 Chemistry syllabus name M+6?
No. The official 9813 syllabus names M+2 and M+4 halogen patterns; M+6 is an application extending the same combinatorial reasoning.
A 3:7:5:1 pattern may look exotic when first encountered, but it is only the familiar chlorine and bromine isotope choices combined correctly. Students who can reconstruct it from individual atomic abundances are prepared to interpret unfamiliar but fair mass-spectral evidence.
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