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A $^{13}$C NMR spectrum is shown - HSC - SSCE Chemistry - Question 5 - 2020 - Paper 1
Question 5
A $^{13}$C NMR spectrum is shown. Which compound gives rise to this spectrum? A. chloroethane B. 1-chloropropane C. 1,2-dichloroethane D. 1,2-dichloropropane
Worked Solution & Example Answer:A $^{13}$C NMR spectrum is shown - HSC - SSCE Chemistry - Question 5 - 2020 - Paper 1
Step 1
Identify the Peaks in the NMR Spectrum
Answer
The given C NMR spectrum features two distinct peaks. The first peak appears around 20 ppm, and the second peak is at a higher chemical shift, approximately 35 ppm. The number of peaks indicates the number of different carbon environments present in the compound.
Step 2
Analyze the Given Compounds
Answer
To determine which compound corresponds to the spectrum, we need to analyze the structures of the options:
- A. Chloroethane: Contains only one type of carbon signal as there is only one carbon attached to chlorine. This would likely result in fewer peaks.
- B. 1-Chloropropane: Also leads to just one distinct carbon environment, likely producing only one peak in the NMR.
- C. 1,2-Dichloroethane: This compound would produce two distinct environments due to different chlorination which creates two different peaks.
- D. 1,2-Dichloropropane: Similar to the previous compounds, this may lead to multiple peaks as well but may differ in chemical shift. Based on the number of distinct peaks observed in the spectrum, the best candidate appears to be 1,2-dichloroethane.
Step 3
Determine the Correct Option
Answer
Considering the analysis of peaks and the structures of the compounds, the compound that gives rise to the NMR spectrum is C. 1,2-dichloroethane. However, upon detailed consideration, since the structure supports (Chloroethane assumption) may yield to less peak appearance, it’s strongly tied to general observation. Therefore, it's A. chloroethane.