Compound 1 has molecular formula C7H15Cl. It shows two signals in the 1H-NMR spectrum, one at 1.08 ppm and one at 1.59 ppm. The relative integrals of these two signals are 3 and 2, respectively. Compound 2 has molecular formula C6H12. It shows three signals in the 1H-NMR spectrum, one at 0.96 ppm, one at 2.03 ppm, and one at 5.33 ppm. The relative integrals of these three signals are 3, 2, and 1, respectively. Propose structures for compounds 1 and 2, explaining how you reach your conclusion.

2-chloro-2,3,3-trimethylbutane;  hex-3-ene

Explanation:

1. Compound 1  

(a) Calculate the unsaturation value

C₇H₁₅Cl. The corresponding alkane is C₇H₁₆.

U = (16 - 16)/2 = 0/2 = 0. There are no rings or double bonds.

(b) Determine the structure

With 15 protons and only two peaks, the compound must be quite symmetrical.

The area ratio is 3:2, so there are nine protons of one type and six of another type (multiples of 3 = methyl groups?).

There may be three methyl groups in one environment and two in another.

One possibility is 2-chloro-2,3,3-tetramethylbutane, which I shall write as (CH₃)₃C-C(CH₃)₂-Cl .  

The two methyl groups are in one environment, and the three methyl groups are in a different environment.

c) Confirmatory evidence

An ordinary CH₃ group normally appears at 0.9 ppm. An electronegative Cl on an adjacent C should pull it downfield by about 0.6 ppm to 1.5 ppm. We see a 6H peak at 1.59 ppm. This confirms the pair of CH₃ groups.

The three CH₃ groups are further from the Cl, so they should feel a smaller effect. They would be pulled downfield by about 0.2 ppm to 1.1 ppm. We see a 9H peak at 1.09 ppm. This confirms the three CH₃ groups.

Compound 1 is 2-chloro-2,3,3-tetramethylbutane.

2. Compound B

(a) Calculate the unsaturation value.

C₆H₁₂. The formula of an alkane is C₆H₁₄.

U = (14 - 12)/2 = 2/2 = 1. The compound contains a ring or a double bond.

(b) Determine the structure

With 12 protons and only three peaks, the compound must be symmetrical.

The area ratio is 3:2:1, so there are six protons of one type, four of another type, and two of a third type.

If the compound has twofold symmetry, there may be two CH₃, two CH₂, and two CH groups.

The peak at 5.33 ppm suggests a double bond: -CH=CH-

One possibility is hex-3-ene — CH₃-CH₂-CH=CH-CH₂-CH₃.  

(c) Confirmatory evidence

An alkene H normally appears at about 5.3 ppm, and we see a 2H peak at 5.32 ppm.

An ordinary CH₂ group normally appears around 1.1 ppm. The adjacent C=C bond should pull it downfield by about 0.9 ppm to 2.0 ppm. We see a 4H peak at 2.03 ppm. This confirms the CH₂ groups.

An ordinary CH₃ group appears at 0.9 ppm. The CH₃ groups in Compound 2 are two C-C bonds away from the alkene, so the should be affected only slightly, perhaps by about 0.1 ppm. We see a 6H peak at 0.96 ppm. This confirms the CH₃ groups.  

Compound 2 is hex-3-ene.