Question

Chemistry 8A Combined Spectroscopy Assignment - 1 Determine the structure of each unknown compound based on its IR and NMR spectra. Compound A: Compound F: Compound B: Compound G: Compound C: Compound H: Compound D: Compound I: Compound E: Compound J:

Answer 1

For proper understanding, you should know the IR frequency values and NMR chemical shift value for various functional groups.

In NMR, remember that the splitting of peaks occur following (n+1) rule. ie. If there is 'n' number of hydrogen atoms in the neighbourhood of a hydrogen atom(maximum to 3 bond distance) , then the peak due to this H atom splits in (n+1) numbers.

Compound A:

6H peak at 1.2ppm corresponds to two methyl groups(-CH₃) having same chemical environment.This peak appears as a doublet ,means the spin of these methyl hydrogen atoms couple with the spin of ONE hydrogen in its neighbourhood. ie. in simple terms, the methyl groups are attached to a carbon having one hydrogen atom (-CH- group).

The peak of this -CH- hydrogen is split into septet (7 peaks, due to two methyl groups or 6H atoms in neighbourhood ,splitting as per (n+1) rule - n=6, n+1=7, hence 7 peaks), appearing at about 2.6 ppm.

Further, there is a singlet appearing at 12ppm, which corresponds to carboxylic acid proton (-COOH).

carboxylic acids exist as dimer via hydrogen bonding.Hence electron density is reduced from the H atom and its peak appear at downfield about 10-12 ppm.

From the data available, we have a COOH group, CH group appearing as a septet and two -CH₃ groups appearing as doublet.

Hence the compound is (CH₃)₂CHCOOH

This is further confirmed by IR data and C¹³ data

From C¹³ spectra , it is clear that there is 3 different type of carbon atoms and from the IR data the characteristic peaks at 3000cm-1 and 1600-1750cm-1 is of acid O-H and C=O stretching respectively.

Compound B:

Analysing the C¹³ spectra, there are five different types of Carbon atoms ( confirmed by 5 peaks).

The peak above 200 ppm corresponds to carbonyl carbon. It could be of a keto group or an aldehyde group. But from Proton NMR data , there is no peak in the range 8-9 ppm. Aldehyde peak appears at this range. Hence we can confirm a keto group in our compound.

Observe the proton Nmr peak. The 3H peaks at about 1ppm correspond to a -CH₃ group. It appear as a triplet , means there is two H atoms adjacent to it. ie. this methyl group is attached to a -CH₂ group. ( n=2 , n+1 = 3 peaks or triplet).

The 2H multiplet (6 peaks) at 1.5 ppm corresponds to -CH₂ group in between a -CH₃ and -CH₂ groups (here n= 2+3 =5 & n+1 = 6)

2H triplet peak at 2.5ppm corresponds to a CH₂ group adjacent to a CH₂ group and also to a C=O group ( reason for this peak appearing downfield is that it is attached directly to carbonyl group,which is electron withdrawing and hence provides deshielding where as the previous mentioned CH₂ group is attached to two alkyl groups and appear at upfield).

Also a 3H singlet appear at 2.0ppm. This corresponds to a -CH3 group attached directly to -C=O group.

From the data we have, a -CH₃ group attached to a -CH₂ group. A -CH₂ group attached to both -CH₃ and -CH₂ groups a -CH2 group attached to another -CH₂ group and carbonyl group & a -CH₃ group attached to carbonyl group.

Hence the compound is : CH₃CH₂CH₂COCH₃

Compound C

The number of protons corresponding to the peaks at 1 ppm is not clear in the image. I assume it is 6H.

But from IR data, the broad peak in the region above 3500cm-1 corresponds to O-H stretching.

C-O stretching frequency at 1200cm-1 is also present.

This confirms the presence of an -OH group, which is further confirmed by proton nmr singlet peak at about 4 ppm.

6H doublet at 1ppm corresponds to two -CH₃ groups attached to a -CH group(n=1,n+1 =2 ,hence doublet). This -CH peak appears as multiplet (9 peaks), hence it is adjacent to 8 H atoms ie. 6 H atoms corresponding to the two -CH₃ groups and two H atoms from a -CH₂ group. This -CH₂ peak hence appear at about 3.4 ppm as a doublet , means this -CH₂ is attached to the -CH group.

Hence the compound is (CH₃)₂CHCH₂OH

C¹³ spectra show three different types of carbon.

Compound D:

It is said in the question that D is isomer of C.

Only Two peaks are obtained in its proton nmr.

The compound is (CH₃)₃C-OH.

9H singlet corresponds to three equivalent -CH₃ groups and the 1H singlet corresponds to -OH.