Q55E - Organic Chemistry (Mcmurry)

Question

A compound of unknown structure gave the following spectroscopic data:

Mass spectrum: ${\mathbf{M}}^{+}\mathbf{=}\mathbf{88}\mathbf{.}\mathbf{1}$

IR:  $\mathbf{3600}{\mathbf{cm}}^{-1}$

 ${}^1\mathbf{H}$NMR: 1.4$\mathbf{\delta }$  (2H, quartet, J=7Hz); 1.2$\mathbf{\delta }$ (6H, singlet); 1.0$\mathbf{\delta }$ (1H, singlet); 0.9$\mathbf{\delta }$ (3H, triplet, J=7Hz)

 ${{}^1}^3\mathbf{C}$NMR: 74, 35, 27, 25$\mathbf{\delta }$

(a) Assuming that the compound contains C and H but may or may not contain O, give three possible molecular formulas.

(b) How many protons (H) does the compound contain? 

(c) What functional group(s) does the compound contain?

(d) How many carbons does the compound contain?

(e) What is the molecular formula of the compound? 

(f) What is the structure of the compound?

(g) Assign peaks in the molecule’s NMR spectrum corresponding to specific protons.

Step-by-Step Solution

Answer

The answer is 

(a) The possible molecular formula of the compound are ${\text{C}}_{\text{5}}{\text{H}}_{\text{12}}{\text{O,\hspace{0.17em}C}}_{\text{4}}{\text{H}}_{\text{8}}{\text{O}}_{\text{2}}{\text{,\hspace{0.17em}C}}_{\text{3}}{\text{H}}_{\text{4}}{\text{O}}_{\text{3}}$ .

(b) The ${}^1\mathrm{H}$NMR data of the compound show that the given compound has twelve protons in it.

(c) The IR absorption spectrum peak at ${\text{3600\hspace{0.17em}cm}}^{\text{-1}}$ shows that the compound is an alcohol.

(d) The given compound contains five carbons, out of which two are identical.

(e) The molecular formula of the compound is ${\text{C}}_{\text{5}}{\text{H}}_{\text{12}}\text{O}$.

(f) The structure of the compound is given below.

2-Methyl-2-butanol

(g) 

Where, a = 0.9 $\text{δ}$

             b = 2.24$\text{δ}$

             c = 5.00$\text{δ}$

             d = 1.4 $\text{δ}$

Step-by-step Solution

The spectroscopic techniques can be used to identify and distinguish between any compounds. Every compound will have unique NMR spectrums. The structural isomers are distinguished based on the coupling of the protons in different compounds.

Step 1: Subpart (a)

(a) The compound contains C, H and O. Hence, the possible molecular formula of the compound is ${\text{C}}_{\text{5}}{\text{H}}_{\text{12}}{\text{O,\hspace{0.17em}C}}_{\text{4}}{\text{H}}_{\text{8}}{\text{O}}_{\text{2}}{\text{,\hspace{0.17em}C}}_{\text{3}}{\text{H}}_{\text{4}}{\text{O}}_{\text{3}}$.

Step 2: Subpart (b)

(b) The ${}^{\text{1}}\text{H}$NMR data of the compound show that the given compound has a total of twelve protons in it. (2 + 6 + 1+ 3 = 12)

Step 3: Subpart (c)

(c) The data indicates the IR absorption spectrum peak at ${\text{3600\hspace{0.17em}cm}}^{\text{-1}}$ , hence the compound is an alcohol.

Step 4: Subpart (d)

(d) The number of protons is 12 and can be concluded that the given compound contains five carbons, out of which two are identical.

Step 5: Subpart (e)

(e) From the number of protons and the number of carbon, we can conclude that the molecular formula of the compound is ${\text{C}}_{\text{5}}{\text{H}}_{\text{12}}\text{O}$.

Step 6: Subpart (f)

(f)The structure of the compound is given below.

2-Methyl-2-butanol

Step 7: Subpart (g)

(g) The shift value of the protons on specific carbon atoms is labelled as shown below.

Where, a = 0.9$\text{δ}$ 

            b = 2.24$\text{δ}$

            c = 5.00$\text{δ}$

             d = 1.4 $\text{δ}$