Q66E - Organic Chemistry (Mcmurry)

Question

Compound A, ${\mathbf{C}}_5{\mathbf{H}}_{10}\mathbf{O}$, is one of the basic building blocks of nature. All steriods and many other naturally occurring compounds are built from compound A. Spectroscopic analysis of A yields the following information:

IR : $\mathbf{3400}{\mathbf{cm}}^{-1}$; $\mathbf{1640}{\mathbf{cm}}^{-1}$

${}^1\mathbf{H}\mathbf{NMR}$: 1.63$\mathbf{\delta }$ (3H, singlet); 1.70 $\mathbf{\delta }$ (3H, singlet); 3.83 $\mathbf{\delta }$ (1H, broad singlet); 4.15 $\mathbf{\delta }$ (2H, doublet, J=7 Hz); 5.70$\mathbf{\delta }$ (1H, triplet, J = 7 Hz)

1. How many double bonds and/or rings does A have?
2. From the IR spectrum, what is the identity of the oxygen-containing functional group?
3. What kinds of protons are responsible for the NMR absorptions listed?
4. Propose a structure for A.

Step-by-Step Solution

Answer

1. Compound A has one double bond or ring.
2. The Presence of alcohol.
3. Unsplit methyl protons, alcohol protons and two protons bonded to a carbon atom bearing an electronegative atom.
4. The structure of compound A is,

Step 1: IR and NMR spectrum

The IR is a spectroscopic technique used to determine the functional groups present in a compound. The NMR is the information of radiofrequency radiation by nuclei present in a magnetic field. Both techniques are forms of absorption spectroscopy.

Step 2: The number of double bonds or ring in compound A

a.

Compound A has the chemical formula ${\text{C}}_{\text{5}}{\text{H}}_{\text{10}}\text{O}$ . From this, we can calculate the Double Bond Equivalent (DBE) to know how many double bonds or rings are present in a compound. The DBE is calculated by using the formula,

 $\text{DBE}=\text{C}+1-\frac{\text{H}}{\text{2}}-\frac{\text{X}}{\text{2}}+\frac{\text{N}}{\text{2}}$

Where,

H is the number of hydrogen.

C is the number of carbon.

N is the number of nitrogen.

X is the number of halogens.

$\begin{array}{c}\text{DBE}=\text{5}+1-\frac{\text{10}}{\text{2}}\\ =6-\frac{\text{10}}{\text{2}}\\ =\frac{2}{2}\\ =1\end{array}$

DBE comes out as 1.

Thus, compound A has one double bond or ring.

Step 3: The identity of the oxygen-containing functional group from the IR spectrum

b. From the given IR values, the IR absorption at ${\text{3400\hspace{0.17em}cm}}^{\text{-1}}$ indicates the presence of alcohol in the compound. The second IR absorption at ${\text{1640\hspace{0.17em}cm}}^{\text{-1}}$ is weak and is due to a C=C stretch.

Step 4: The kinds of protons which are responsible for the NMR absorptions

c.

From the given NMR values, the absorptions at 1.63 $\mathrm{\delta }$ (3H, singlet) and 1.70$\delta$ (3H, singlet) are due to unsplit methyl protons because of shifting slightly downfield. These protons are adjacent to an unsaturated centre.

The NMR value at 3.83$\delta$ (1H, broad singlet) is due to the presence of an alcohol proton.

The value at 4.15$\mathrm{\delta }$ (2H, doublet, J=7 Hz) is due to the bonding of two protons to a carbon-bearing an electronegative atom, and in this case, oxygen is an electronegative atom

The triplet at 5.70$\delta$ (1H, J = 7 Hz) is due to a vinylic proton.

Step 5: The structure for A

d. 

From the above parts, the structure of A should be.

The name of the compound is 3-Methyl-2-buten-1-ol.