What would the skeletal structure of this be? Molecular formula consistent with mass spec: C3H6Br2 1H
The Correct Answer and Explanation is:
Correct Skeletal Structure:
The molecular formula C₃H₆Br₂ corresponds to 1,2-dibromopropane (or alternatively 1,3-dibromopropane, but the NMR data can distinguish them — see below).
A likely skeletal structure for C₃H₆Br₂ is:
Br Br
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CH3–CH–CH2 or Br–CH2–CH2–CH2–Br
However, based on ¹H NMR (which was mentioned), the correct structure is:
1,2-dibromopropane:
CH₃–CHBr–CH₂Br
Explanation
The molecular formula C₃H₆Br₂ tells us the molecule contains three carbon atoms, six hydrogens, and two bromine atoms. First, calculate the degree of unsaturation (index of hydrogen deficiency, or IHD):
- Saturated hydrocarbon formula: CₙH₂ₙ₊₂ = C₃H₈
- Actual hydrogen count = 6 → missing 2 hydrogens
- Each degree of unsaturation = 2 hydrogens → IHD = 1
But each halogen (Br) is counted like a hydrogen for IHD purposes. So:
Adjusted hydrogen count = 6 + 2 (for 2 Br) = 8
Now compare to C₃H₈ → No hydrogen deficiency → IHD = 0
✅ Therefore, no rings or double bonds are present.
We now consider isomers with this formula:
- 1,2-dibromopropane: CH₃–CHBr–CH₂Br
- 1,3-dibromopropane: BrCH₂–CH₂–CH₂Br
- 2,2-dibromopropane: (CH₃)₂CBr₂ → Not possible with only 6 H
¹H NMR Evidence (mentioned in your prompt):
- In 1,2-dibromopropane, we expect:
- A doublet for CH₃ (next to CH)
- A multiplet for CH (next to CH₃ and CH₂Br)
- A triplet or complex pattern for CH₂Br
→ This gives three types of protons → matches 1,2-dibromopropane
In contrast, 1,3-dibromopropane would give only two types of protons (CH₂Br and central CH₂), due to symmetry.
So the correct structure consistent with C₃H₆Br₂ and ¹H NMR is:
CH₃–CHBr–CH₂Br (1,2-dibromopropane)
This molecule has three unique proton environments, no unsaturation, and matches both molecular formula and NMR.
