Tutorial 9 - Inorganic NMR Answers PDF

Preview

Summary

Paul Elliott 16/17...

Description

SHC4001 Inorganic Chemistry 3 – Tutorial 9

Inorganic Spectroscopy and Instrumental Techniques 1 - Inorganic NMR spectroscopy 1. Below is shown the 1H NMR spectrum of potassium borohydride (KBH4). Explain the appearance of the spectrum.

Both

10

B (I = 3, nat. ab. 20 %) &

11

B (I = 3/2, nat. ab. 80 %) NMR active and

1

couple of H. Hence 80 % of the sample contains 1

11

BH4 which gives rise to a

1:1:1:1 H NMR signal, whilst the remaining sample contains gives rise to a 1:1:1:1:1:1:1 septet signal. overlapping signals should be 4:1

10

BH4 which

Overall integration of the two

SHC4001 Inorganic Chemistry 3 – Tutorial 9

2.

1

H NMR (250 MHz) spectra of the complex [(5-C5H5)W(=CH2)(CO)2(PEt3)]+

were recorded between -110 and -20 C.

The region for the alkylidene

protons are shown below

a) Draw the structure of the complex (the carbonyl ligands in this complex are equivalent).

b) Explain why a doublet is observed at -20 C. 1H-31P coupling (31P I = ½) c) Explain why two resonances are observed at -110 C and suggest a dynamic process to explain the observed changes of the spectra with temperature. Rotation of the CH2 unit about the W=C bond is slow at low temp. One proton points up towards Cp ring and the other away making them inequivalent.

The resonance at  15.5 shows the large 1H-31P

coupling and the smaller 1H-1H geminal coupling.

SHC4001 Inorganic Chemistry 3 – Tutorial 9

d) What is the coalescence temperature? -70 C e) What is the rate constant for exchange at the coalescence temperature? 𝑘=

𝜋∆𝜈

√2

Spectrum recorded at 250 MHz, therefore 1 ppm = 250 Hz. Signals at 15.5 and 13.9 ppm, therefore  = 1.6 ppm = 1.6*250 = 400 Hz. Therefore k = 889 s-1

f) If the rate constant calculated in d) is given by

𝑘= (

𝑘

)

∆

⁄

what is the approximate activation energy for the observed exchange process? (Boltzmann constant = 1.381 x 10-23 J K-1, Planck’s constant = 6.626 x 10-34 J s, gas constant = 8.314)

Rearranging gives: = ( (

𝑘

)

T = -70 + 273 = 203 K ( (2

)

𝑘

2 =

)=2

𝑘)

SHC4001 Inorganic Chemistry 3 – Tutorial 9

3. The cyclooctatetraene complex [Ru(CO)3(C8H8)] exhibits a room temperature 1

H NMR spectrum that contains a sharp singlet resonance. When recorded at

low temperature, this signal broadens and then separates into 4 distinct resonances. Thinking carefully, draw the structure of this complex (hint: what is the electron count of the Ru(CO)3 fragment) and then account for the observed NMR data. Ru(0) complex therefore requires 10 e- to be 18 e- complex. 6 provided by CO ligands hence 4 from C8H8 ligand. Therefore only two of the four C=C bonds coordinated to metal. Must coordinate to metal like 1,3-butadiene as this gives four proton environments. Coordinating to C=C bonds on opposite sides of ring would give two environments. Fluctionality arises from the Ru centre jumping between C=C bonds on the C 8H8 possibly by 16 e- intermediate coordinated to only 1 C=C bond, which is fast at room temperature giving a single resonance whereas two signals for coordinated and two signals for non-coordinated alkene protons observed at low temperature, slow exchange

4. a) The following fluxional trinuclear complex is

13

CO labelled. Bridging CO

ligands give rise to signals at approximately 200 ppm whereas terminal CO ligands give signals at around 180 ppm. temperature (slow exchange) ligands (think carefully!).

13

Sketch the expected low

C NMR spectrum arising from the

13

CO

SHC4001 Inorganic Chemistry 3 – Tutorial 9

Rh3 is trigonal planar. For Rh with terminal CO, this gives doublet due to coupling to Rh. Because this CO is one side of Rh3 plane and Cp is the other side, bridging CO ligands are inequivalent and appear as triplets due to coupling to two Rh centres.

b) Based on the mechanism for the fluxional process outlined below, sketch the expected high temperature (fast exchange)

13

C NMR spectrum and

explain its appearance.

In the closed bis-bridging forms above a CO – terminal  bridging  terminal  bridging … b CO – bridging  terminal  bridging  terminal … c CO – bridging  bridging  bridging  bridging … therefore a and b become equivalent as they alternative between bridging and terminal sites whereas c is always bridging.

SHC4001 Inorganic Chemistry 3 – Tutorial 9

Hence high temp spectrum has two signals 2:1 in size, both are quartets through coupling to three Rh centres. The signal for a and b will appear near 190 ppm, the average for the two sites....