Question

A). 3-Phenylpropanoyl chloride (C6H5CH2CH2COCl) reacts with AlCl3 to give a single product A with the formula C9H8O and an 1H NMR spectrum with signals at δ = 2.53 (t, J = 8 Hz, 2 H), 3.02 (t, J = 8 Hz, 2H) and 7.2-7.7 (m, 4H) ppm. Propose a structure for this compound.

B).

Answer 1

Concepts and reason

¹H NMR Spectroscopy:

Nuclear magnetic resonance $\left( {{\rm{NMR}}} \right)$ spectroscopy is one of the techniques in the analytical chemistry that is widely used to determine the purity of sample and also to predict the structure of the organic compounds.

The ${\rm{NMR}}$ spectroscopy is used to observe the phenomenon when the frequency of nuclei of atoms (sample) resonates with frequency of rotating magnetic field.

$^{\rm{1}}{\rm{HNMR}}$ spectroscopy determines the different types of hydrogens (chemically non-equivalent hydrogens) present in a molecule.

Fundamentals

¹H NMR spin-spin coupling patterns:

Equivalent protons: The protons with the same chemical environment will have the same chemical shift in NMR, so they are accounted as equivalent protons.

Non-equivalent: These protons have different environments with respect to other groups. Therefore, they are non-equivalent with different chemical shifts in NMR.

The chemical shift is the ratio of the radio frequency absorbed by a given nucleus to the operating frequency of the NMR instrument. It is expressed in terms of parts per million (ppm).

A typical NMR spectrum involves the plot of intensity associated with the given absorption against the ‘chemical shift’.

(A)

The reaction of $3 - Phenyl\,propanoyl\,chloride$ ( ${C_6}{H_5}C{H_2}C{H_2}COCl$ ) with $AlC{l_3}$ can be shown as follows:

The formula of the given compound is ${C_9}{H_8}O$ .

The compound gives the proton-NMR signals at the following values:

$\begin{array}{l}\\\delta = 2.53\,(t,\,J = 8\,Hz,\,2H)\\\\\delta = 3.02\,(t,\,J = 8\,Hz,\,2H)\\\\\delta = 7.2 - 7.7\,(m,\,4H)\\\end{array}$

Therefore, the proposed structure of the compound is shown below:

(B)

The product of the first reaction is shown below:

The reaction of the product involves the following sequence of reagents: $\begin{array}{l}\\NaB{H_4},\,\\\\{H_2}S{O_4}({100^0}C)\,and\,\\\\{H_2},\,Pd/C\\\end{array}$

This can be depicted as follows:

It is given that the resulting molecule gives five resonance lines in $^{13}C$ NMR spectrum. This is because for the compound obtained, there are five different carbon atoms.

Thus, the structure of the compound after each of the steps in the sequence is shown below:

Ans: Part A

The proposed structure for this compound is shown below:

Part B

The structure of the substance formed after each of the steps in the sequence is shown below: