73 - Organic Chemistry

Question

Treatment of compound D with $LiAl{H}_4$  followed by $H_{2}O$ forms compound E. D shows a molecular ion in its mass spectrum at m/z = 71 and IR absorptions at $3600-3200 c{m}^{-1}$  and . E shows a molecular ion in its mass spectrum at m/z = 75 and IR absorptions at  $3636 c{m}^{-1}$ and $3600-3200 c{m}^{-1}$ . Propose structures for D and E from these data and the given 1H NMR spectra.

Step-by-Step Solution

Answer

Structure of D:

Structure of E:

Rule of thirteen in mass spectroscopy

In mass spectroscopy, the molecular ion peak value is the molecular mass of the compound. 

The rule of thirteen states that the molecular mass of the compound divided by the mass of one unit of carbon and hydrogen (CH) gives the molecular formula of the compound.

The relation between the molecular mass(M) and subscripts n and r is given as:

  $\frac{M}{13}=n+\frac{r}{13}$

The molecular formula is expressed as:  $C_n{H}_{n+r}$

IR spectroscopy

IR spectroscopy is employed to get a better understanding of the functional groups present in a system. 

 

Every functional group absorbs at a specific frequency of infrared light, corresponding to the vibrations of the molecules in the compound.

NMR spectroscopy

Nuclear molecular resonance is an analytical technique employed to analyze the C-H framework of a compound.

 

There are two kinds of NMR spectroscopy: ${}^{1}H$NMR, where only peaks of hydrogen are shown.

 ${}^{13}C$NMR where the peaks of carbon s given.

 

Determination of structures of D and E

It is given that D reacts with $LiAl{H}_4$  to form E.

The reaction is given below:

                                        Reaction

Compound D shows molecular ion at m/z = 71. From the rule of thirteen, the molecular formula with respect to carbon and hydrogen is:

$\frac{71}{13}=5+\frac{6}{13}$

The molecular formula is expressed as: $C_5{H}_{11}$ 

The HNMR spectra of D is given below:

                                                  ${}^{1}HNMR$ of E

The spectra show that there are 5 hydrogen atoms present in the compound.

So, the rest 6 places of hydrogen are occupied possibly by nitrogen and oxygen.

 

The IR absorption at  $3600-3200 c{m}^{-1}$and the 1H singlet peak suggests that there is an alcoholic group present in the compound.

 

By substituting 1 oxygen in place of 1 carbon and 4 hydrogen atoms, the molecular formula becomes:

 $C_5{H}_{11}\underset{-C{H}_4}{\overset{+O}{\to }}{C}_4{H}_{7}O$

There is another IR absorption at $2263 c{m}^{-1}$  which suggests that there is a cyanide group present in the compound.

 

By replacing one carbon and 2 hydrogen with one nitrogen, the molecular formula becomes:

 $C_4{H}_{7}O\underset{-C{H}_2}{\overset{+N}{\to }}{C}_3{H}_{5}NO$

Thus, the possible structure of the compound having one -OH group, one -CH group, and the molecular formula $C_3{H}_{5}NO$ is:

                            Possible structure of D

The HNMR data given is compared with the structure of D, as shown below:

                            HNMR analysis of D

Now, when D is subjected to reduction, the product is shown below:

                                       Reduction of D

This structure of the product formed has 4 more hydrogen atoms. Thus the molecular mass becomes 75, the same as the molecular ion m/z value obtained for structure E.

The H-NMR peak of E is shown below:

                                               ${}^{1}HNMR$ of E

From the NMR data, it is observed that there are additional peaks of 2H and 3H, which suggests that there is an increase in the number of hydrogens.

 

The IR absorption at  $3600-3200 c{m}^{-1}$remains in structure E as well, suggesting the alcohol group is present.

The IR absorption at $3636 c{m}^{-1}$ suggests the presence of an amine functionality.

Since the given data matches with the product formed above, the possible structure of E is:

                                   Structure of E