Question

n-Butanol ОН Part 2: Draw the NMR spectrum Draw the structure of your molecule below, indicating each unique hydrogen atom with letters, starting with A. After complete the below table, summarizing the integration, approximate chemical shift number of neighbors, and splitting pattern of each hydrogen in your given molecule. Note, you do not need to fill every row in the below table. Structure: Hydrogen Integration Letter Chemical Number of Splitting Pattern Shift Neighbors Now, draw the NMR in the space provided. Note that peak locations should reflect the approximate chemical shift of each peak and each peak should be drawn with the correct splitting pattern. Finally, clearly indicate the integration of each peak.

Answer 1

4 На на н, ң, Ref н? “н нь о— на CH, Н. Н. на * ус л B. 4

1.

Hydrogen letter	Integration	Chemical shift	Number of Neighbours	Splitting pattern
a	1	4.7 ppm	2	Singlet (s)
b	2	3.64 ppm	2	Triplet (t)
c	2	1.56 ppm	4 (2 H on each side)	Quintet (q)
d	2	1.39 ppm	5 (2 H on one side, 3H on the other)	Sextet (s)
e	3	0.94 ppm	2	Triplet (t)

1. n-Butanol 1.39 3.64 0.94 0.94 VO OH 1.56 PPM

Note: the -OH proton does not show any splitting and is a singlet in the 1H-NMR. This is due to rapid interchange that occurs between the hydrogen atoms on the -OH group and either water molecules (alcohol samples almost always has water in them, even in very little amount) or other alcohol molecules and hence there is no time for resonance with the neigbouring H's of the -CH2 group.

2. The molecule, 3-bromopentane, is symmetrical form the middle (a plane of symmetry through the C-3) and hence  shows only 3 peaks in the 1H-NMR.

Hydrogen letter	Integration	Chemical shift	Number of Neighbours	Splitting pattern
a	1	3.60 ppm	4 (2 H on each side)	Quintet (q)
b	4	1.77 ppm	4 (1H of the central carbon and 3H of the methyl)	Quintet (q)
c	6	1.05 ppm	2	Triplet (t)

2. 3-bromopentane: 1.77 1.77 1,05 1.05 PPM