The frequency of O−H stretch occurs at ∼3600 cm−1 . The O−D stretch frequency (in cm−1 ) would be closest to
Frequency Calculation for O-D Stretch
The vibrational frequency of a bond, like the O-H or O-D stretch, can be approximated using the harmonic oscillator model. According to this model, the frequency (\(\nu\)) is proportional to the square root of the force constant (\(k\)) divided by the reduced mass (\(\mu\)) of the vibrating system.
\[ \nu \propto \sqrt{\frac{k}{\mu}} \]Where:
- \(k\) is the force constant of the bond.
- \(\mu\) is the reduced mass of the system.
For a diatomic molecule (or a bond treated as such) composed of two atoms with masses \(m_1\) and \(m_2\), the reduced mass is given by:
\[ \mu = \frac{m_1 m_2}{m_1 + m_2} \]In the case of O-H and O-D bonds, the force constant \(k\) is essentially the same because the electronic structure of the bond is not significantly affected by isotopic substitution (replacing H with D). The difference in frequency arises primarily from the difference in reduced mass.
Reduced Mass Calculation
Let's calculate the reduced mass for the O-H and O-D bonds. We use approximate atomic masses:
- Mass of Oxygen (\(m_O\)) ∼ 16 amu
- Mass of Hydrogen (\(m_H\)) ∼ 1 amu
- Mass of Deuterium (\(m_D\)) ∼ 2 amu
Reduced mass for O-H (\(\mu_{OH}\)):
\[ \mu_{OH} = \frac{m_O m_H}{m_O + m_H} = \frac{16 \times 1}{16 + 1} = \frac{16}{17} \text{ amu} \]Reduced mass for O-D (\(\mu_{OD}\)):
\[ \mu_{OD} = \frac{m_O m_D}{m_O + m_D} = \frac{16 \times 2}{16 + 2} = \frac{32}{18} = \frac{16}{9} \text{ amu} \]Frequency Relationship
Since the frequency is inversely proportional to the square root of the reduced mass (\(\nu \propto 1/\sqrt{\mu}\)) and the force constant \(k\) is the same for O-H and O-D:
\[ \frac{\nu_{OD}}{\nu_{OH}} = \sqrt{\frac{\mu_{OH}}{\mu_{OD}}} \]Now, we calculate the ratio of the reduced masses:
\[ \frac{\mu_{OH}}{\mu_{OD}} = \frac{16/17}{16/9} = \frac{16}{17} \times \frac{9}{16} = \frac{9}{17} \]Substitute this ratio into the frequency relationship:
\[ \frac{\nu_{OD}}{\nu_{OH}} = \sqrt{\frac{9}{17}} \]We are given that the frequency of O-H stretch is \(\nu_{OH} \approx 3600 \text{ cm}^{-1}\). We can now calculate \(\nu_{OD}\):
\[ \nu_{OD} = \nu_{OH} \times \sqrt{\frac{9}{17}} \] \[ \nu_{OD} \approx 3600 \text{ cm}^{-1} \times \sqrt{\frac{9}{17}} \]Calculating the square root:
\[ \sqrt{\frac{9}{17}} \approx \sqrt{0.5294} \approx 0.7276 \]Now, multiply by the O-H frequency:
\[ \nu_{OD} \approx 3600 \text{ cm}^{-1} \times 0.7276 \] \[ \nu_{OD} \approx 2619.36 \text{ cm}^{-1} \]Comparing with Options
The calculated O-D stretch frequency is approximately \(2619.36 \text{ cm}^{-1}\). We look for the option that is closest to this value.
- Option 1: 3000 cm−1
- Option 2: 2600 cm−1
- Option 3: 1800 cm−1
- Option 4: 900 cm−1
The value \(2619.36 \text{ cm}^{-1}\) is closest to \(2600 \text{ cm}^{-1}\).
Important Questions from IR Spectroscopy
In an IR spectra of 1-octyne and 4-octyne, the IR-spectra of 4-octyne does not have C ≡ C stretch absorption peak. The reason for this observation is that _________.
An IR-spectra is found to have a medium adsorption peak near 3400cm-1. This corresponds to which organic compound?
The number of CO bands for isomers from sets (i) and (ii) in their IR spectra
Set (i): Trigonal bipyramidal isomers, axial‐Fe(CO)4L (A)and equatorial‐Fe(CO) 4 L(B)
Set (ii): Octahedral isomers, fac‐Mo(CO)3L3 (C) and mer‐Mo(CO)3L3(D)
are
Match List I with List II
List I
List II
functional groups
respective approximate symmetric and asymmetric stretching frequencies
A.
N - H bonds of R - NH2
I.
1790 and 1810
B.
N - O bonds of R - NO2
II.
3300 and 3400
C.
C = O bonds of anhydride
III.
1350 and 1550
Choose the correct answer from the options given below:
The molecule that can absorb in the infra-red among the following is
