The ¹¹⁹Sn NMR chemical shift (approximately in ppm) corresponding to (η⁵-Cp)₂Sn (relative to Me₄Sn) is

(d) -2200. Explanation: Stannocene appears near −2200 ppm (d).

Q50 · CSIR-NET Chemistry, June 2015

Paper: CSIR-NET June 2015 · Subject: Inorganic Chemistry · Chapter: Inorganic Spectroscopy · Topic: Inorganic Spectroscopy – General · Marks: 2 · Difficulty: Medium

The $^{119}\mathrm{Sn}$ NMR chemical shift (approximately in ppm) corresponding to $(\eta^{5}\text{-}\mathrm{Cp})_{2}\mathrm{Sn}$ (relative to $\mathrm{Me_4Sn}$) is

(a)-4

(b)+137

(c)+346

(d)-2200

Answer

Answer: D ✓ checked by 4AB · confidence high

Explanation

Stannocene appears near −2200 ppm (d).

Study loop for Inorganic Spectroscopy

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