Which of the patterns (A, B, C or D) fits best with the ¹³C NMR spectrum of TiCl₃(CDH₂) [Given: ¹.J(C‐H) > ¹J(C‐D)]

Concept:

• C¹³NMR detects the ¹³C isotopes present in the compound.
• the peaks get split by the active nuclei in the surrounding when magnetic field is applied. The numbers of splitting lines is given by  2nI+1. where, n is the number of neighbouring nuclei and   I is the nuclear spin value.
• The peaks are more downfield compared to H-NMR and lies in the range of 30 - 200 ppm.
• Coupling constant (J) gives a measure of  interaction by neighbouring nuclei. It is independent of applied field strength.

Explanation:

Since coupling constant is higher between C and H, the signal will first split because of H nuclei. The number of splitting lines will be

= 2nI_H + 1

 = \(2\times2\times\frac{1}{2} + 1\)     \( (I_H = \frac{1}{2} )\)

= 3

Intensity ratio of lines will be = 1:2:1

Further, the lines will split because of coupling between nuclei of C and D. Number of splitting lines = 2nI_D + 1

= \(2\times 1\times1+1\)    \((I_D=1) \)

= 3

Intensity ration of split lines will be = 1:1:1

final split diagram for given compound in ¹³CNMR will be:

Conclusion:

Therefore, the pattern for splitting in 13C NMR spectrum of TiCl3(CDH2) will be: