Discussion
An organic compound with molecular formula \(\mathrm{C_{10}H_{13}Cl}\) gives the following NMR data.
| (i) | Singlet, δ 1.57, 6H |
| (ii) | Singlet, δ 3.07, 2H |
| (iii) | Singlet, δ 7.27, 5H |
Identify the structure.
Answer: C
The formula \(\mathrm{C_{10}H_{13}Cl}\) has an aromatic ring and the NMR data show three singlets, so none of the groups is split by neighbouring hydrogens.
The 5H signal at 7.27 is a monosubstituted benzene ring, \(\mathrm{C_6H_5}\). The 6H singlet at 1.57 is two equivalent methyl groups on a carbon that carries no hydrogen but bears the chlorine, \(\mathrm{C(CH_3)_2Cl}\). The 2H singlet at 3.07 is a \(\mathrm{CH_2}\) between the ring and the quaternary carbon, so there is no neighbour to split it. Together this is \(\mathrm{C_6H_5CH_2C(CH_3)_2Cl}\), option C.
Option A has the formula \(\mathrm{C_{10}H_{14}Cl}\) with no CH2, option B would show a doublet and multiplets, and option D would give several triplets, so they do not match.
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