Question

An infra-red spectrum shows a broad peak at $3000c{m^{ - 1}}$ and a strong peak at $1710c{m^{ - 1}}$. Which substance could have produced this spectrum?
A. Methyl propanoate
B. Propan-2-ol
C. Propanoic acid
D. Propanone

Answer 1

Hint:Almost all the compound having covalent bonds, whether organic or inorganic, absorbs various frequencies of electromagnetic radiation in the infrared region of the electromagnetic spectrum. This region lies at a wavelength longer than those of visible light, but lies shorter than those of microwaves.

Complete answer:
In the case of IR spectrum, we are mainly interested in the vibrational portion of the infrared region and it includes radiation with wavelength lies at $2.5 \times {10^{ - 6}} - 25 \times {10^{ - 6}}m$.
In the absorption process in this region, those frequencies of infrared radiation that match the natural vibrational frequencies of the molecule are absorbed. Here is the main thing, as all the bonds in a molecule are capable of absorbing the infrared energy, even if the frequency of the radiation exactly matches that of the bond motion. Only those bonds that have a dipole moment that changes as a function of time are capable of absorbing infrared radiation. That means, that bond must present an electric dipole that is changing at the same frequency as the incoming radiation for energy to be transferred.
The important use of the infrared spectrum is to determine structural information about the molecule. The absorption of each type of bond is regularly found in this region. Here when we talk about frequency, then we are just talking about fundamental frequency, that arises from excitation from the ground state to the lowest excited state.
We already have the correlation chart with us for IR. We just have to check the frequency that is given corresponds to which bond. In the given question, $3000c{m^{ - 1}}$ broad peak that corresponds to –OH bond. The second peak given is $1700c{m^{ - 1}}$ that corresponds to C=O bond. Hence from the given option carboxylic acid has both the bonds present.

Therefore the correct option will be C. Propanoic acid

Note:
Symmetric bonds such as ${H_2},C{l_2}$ do not absorb in the IR region, because there is no dipole moment in symmetric bonds and stronger bonds which have large force constant value vibrate at higher frequency and also bonds between atoms of higher masses vibrate at lower frequency than bonds between lighter atoms