Question

4) Chemical Formula: CH100 IR: strong broad peak at 3300cm- 244)+2 = 10 PPM 5) Chemical Formula: C8H11N IR: Two weak peaks at 3350cm! Doublet, IH Triplet, IH Triplet, 1H Doublet, IH PPM

Answer 1

Question 4

Strong broad peak at 3300 in IR shows the presence of OH group in the molecule.

1HNMR gives us following information

The split in the NMR peaks occurs due to coupling with the neighbouring hydrogens. If there is 1 neighbouring hydrogen then a doublet will be observed. If 2 neighbouring hydrogens are there then a triplet will be observed and so on.

a. There are 2 CH3 groups that have a 1 neighbouring hydrogen. This is inferred by the doublet signal at around 0.9.

b. There is 1 CH group that has multiple neighbouring hydrogens. This signal is appearing as a multiple split peaks at 1.4

c. There is a CH2 group that has one neighbouring hydrogen. This signal appears as a doublet at 3.5

d. Now the remaining signal is a broad short peak with 1 Hydrogen. There is no split observed and this will be due to OH group.

So the structure of this compound can be written as

Do notice the number of neighbouring hydrogens next to each marked peak

ta С la&1 0.9 doublet 2 – 1.4 multiplet 3 3.4 doublet 4 4.2 singlet

Question 5

IR shows 2 weak peaks at around 3350. The weak peaks at this region are due to NH bonds. As there are 2 such peaks, the compound has 2 NH bonds. As molecular formula shows only 1 N, so the group here is NH2.

1HNMR gives following information

a. It shows a CH3 group at 1.1 ppm that has 2 neighbouring hydrogens as it gives a triplet split.

b. There is a CH2 group with at 2.5 ppm that has 3 neighbouring hydrogens as it gives a quartet split.

Note: this type of pattern is typical of ethyl groups (CH3-CH2)

c. There are 4 peaks in the aromatic region between 6.5 to 8 ppm. As aromatic Carbons can have only 1 Hydrogen bonded to them, the splits here will mean hydrogens present on multiple carbons. For example, if a doublet is observed, then there is one neighbouring Carbon that has a hydrogen bonded to it. The second neighbouring carbon should not have a hydrogen atom (C=CH=CH, middle carbon will give a doublet). If the split is a triplet, then both neighbouring Carbons have a hydrogen atom (CH=CH=CH, middle carbon will give a triplet split).

d. There are 2 carbons that have only 1 neighbouring CH as they are split into a doublet

e. There are 2 carbons that have 2 neighbouring CH as they are split into a triplet.

f. None of these are equivalent positions. There fore, we can straight away rule out the para substituted benzene ring. In para substituted benzene ring, there will be 2 sets of equivalent protons.

g. All hydrogens are showing a split, so all of these have a neighbouring proton. So we can also rule out meta substituted benzene ring. In meta substituted ring, there will be 1 hydrogen that will not have any neighbouring protons. So 1 singlet should be seen

h. This leaves the possibility of ortho substituted benzene ring.

i. We have already determined CH3 CH2 group as one of the substituents. The second substituent will be NH2 group. Apart from the IR information, the broad peak at 7.5 is also an indicator of NH2 peak.

The following structure is possible

6 NHQ 43 CH₂ - CH₂ 2 3 4 1.1 triplet 2.6 do quartet 6.9 doublet 6.5 triplet 5 6.8 triplet 6.7 doublet 7 7.5 singlet 6