Carbon-13 NMR Spectroscopy (CIE A Level Chemistry)

During the production of an NMR spectrum, tetramethylsilane (TMS) is mixed with the sample.

State the number of peaks in the C-13 NMR spectrum of 1,3-dichlorobenzene.

Compound A contains the elements carbon and hydrogen forming an aromatic ring along with the elements oxygen and nitrogen.

Part of the mass spectrum of A is shown in Fig. 2.2.

Give the identity of the molecular ion that gives rise to the peak at m / e = 76 in Fig. 2.2.

Suggest the structures of the three possible dinitrobenzene isomers of A that contain a benzene ring.

The C-13 NMR spectrum of compound A has four peaks.

Identify the structure of A. Explain your reasoning by labelling the different carbon environments in all the structures drawn in part (ii).

An unknown alcohol was analysed and found to contain 64.9% carbon, 13.5% hydogen and the rest oxygen. 

The results of mass spectrometry found the mass of the alcohol to be 74.12 g mol^-1.

Determine the molecular formula of the unknown alcohol. Show your working.

The unknown alcohol can exist as four possible isomers.

The ¹³C NMR spectra of one of the isomers is shown in Fig. 3.1.

Fig. 3.1

Deduce which isomer produced the spectrum shown in Fig. 3.1. Explain your answer with reference to Table 3.1.

Table 3.1

Hybridisation of the carbon atom	Environment of carbon atom	Example	Chemical shift range δ/ppm
sp3	alkyl	CH3–, CH2–, –CH<, >C<	0 – 50
sp3	next to alkene / arene	–C–C=C, –C–Ar	25 – 50
sp3	next to carbonyl / carboxyl	C–COR, C–O2R	30 – 65
sp3	next to halogen	C–X	30 – 60
sp3	next to oxygen	C–O	50 – 70
sp2	alkene or arene	>C=C<,	110 – 160
sp2	carboxyl	R−COOH, R−COOR	160 – 185
sp2	carbonyl	R−CHO, R−CO−R	190 – 220
sp	nitrile	R−C≡N	100 – 125

Three hydrocarbons, L, M and N, have the molecular formula C₈H₁₀. Information about the number of peaks seen in the carbon-13 (¹³C) NMR spectrum of the three isomers is shown in Table 1.1.

Table 1.1

Complete Table 1.1 to give details of the proton NMR spectra for isomers L, M and N.

Table 1.1

Explain which of the three isomers, L, M or N has the highest melting point.

A chemist analyses a naturally occurring compound, K.

The percentage composition by mass is carbon 70.58%; hydrogen 5.92% and oxygen 23.50%.

The mass spectrum of the compound is shown in Fig. 5.1.

Fig. 5.1

Determine the molecular formula of the compound K. Show your working.

The results of qualitative tests performed on compound K are shown in Table 5.1.

Table 5.1

Identify the functional groups present in compound K. Explain your answer.

The carbon-13 (¹³C) NMR of compound K is shown in Fig. 5.2.

Fig. 5.2

Table 5.1 

Hybridisation of the carbon atom	Environment of carbon atom	Example	Chemical shift range δ / ppm
sp3	alkyl	CH3–, CH2–, –CH<, >C<	0 – 50
sp3	next to alkene / arene	–C–C=C, –C–Ar	25 – 50
sp3	next to carbonyl / carboxyl	C–COR, C–O2R	30 – 65
sp3	next to halogen	C–X	30 – 60
sp3	next to oxygen	C–O	50 – 70
sp2	alkene or arene	>C=C<,	110 – 160
sp2	carboxyl	R−COOH, R−COOR	160 – 185
sp2	carbonyl	R−CHO, R−CO−R	190 – 220
sp	nitrile	R−C≡N	100 – 125

Suggest the structure of compound K using your answers to (a) and (b) along with information from Fig. 5.2 and Table 5.1. Explain your answer.