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:
A - II, B - III, C - I
Functional Group Stretching Frequencies Match
Infrared (IR) spectroscopy is a valuable technique used to identify functional groups present in a molecule. This identification is based on the absorption of infrared radiation by molecular vibrations, specifically stretching and bending modes. Different functional groups absorb IR radiation at characteristic frequencies, often expressed in wavenumbers ($\text{cm}^{-1}$). Stretching vibrations involve changes in bond length, while bending vibrations involve changes in bond angle. Functional groups often exhibit both symmetric and asymmetric stretching modes, leading to different absorption frequencies.
Let's analyze the given functional groups and their typical IR stretching frequency ranges:
| List I (Functional Groups) | List II (Approximate Symmetric and Asymmetric Stretching Frequencies ($\text{cm}^{-1}$)) |
|---|---|
| A. N - H bonds of R - NH2 (Primary Amine) | I. 1790 and 1810 |
| B. N - O bonds of R - NO2 (Nitro Compound) | II. 3300 and 3400 |
| C. C = O bonds of anhydride | III. 1350 and 1550 |
Stretching Frequencies Analysis
- N - H bonds of R - NH2 (Primary Amine): Primary amines (R-NH2) have two N-H bonds. These bonds can vibrate symmetrically and asymmetrically. These stretching vibrations typically appear in the IR spectrum as two sharp bands in the region of 3300 to 3500 $\text{cm}^{-1}$. The given frequencies in option II (3300 and 3400 $\text{cm}^{-1}$) fall perfectly within this expected range for N-H stretches of a primary amine. Therefore, A matches with II.
- N - O bonds of R - NO2 (Nitro Compound): Nitro compounds (R-NO2) have N=O bonds which exhibit strong stretching vibrations. There are two characteristic bands due to asymmetric and symmetric stretching. The asymmetric stretch typically appears around 1500-1600 $\text{cm}^{-1}$, and the symmetric stretch is typically around 1300-1390 $\text{cm}^{-1}$. The given frequencies in option III (1350 and 1550 $\text{cm}^{-1}$) correspond well to the symmetric (1350 $\text{cm}^{-1}$) and asymmetric (1550 $\text{cm}^{-1}$) stretching frequencies of the N-O bonds in a nitro group. Therefore, B matches with III.
- C = O bonds of anhydride: Anhydrides contain two C=O double bonds linked by an oxygen atom. Due to the coupling of the two carbonyl groups, anhydrides typically show two very strong absorption bands in the carbonyl region (around 1700-1850 $\text{cm}^{-1}$). The higher frequency band (around 1780-1830 $\text{cm}^{-1}$) is due to the asymmetric stretch, and the lower frequency band (around 1740-1770 $\text{cm}^{-1}$) is due to the symmetric stretch. The given frequencies in option I (1790 and 1810 $\text{cm}^{-1}$) are characteristic of the two strong C=O stretching bands observed for anhydrides. Therefore, C matches with I.
Based on this analysis:
- A (N-H bonds) matches II (3300 and 3400 $\text{cm}^{-1}$).
- B (N-O bonds) matches III (1350 and 1550 $\text{cm}^{-1}$).
- C (C=O bonds of anhydride) matches I (1790 and 1810 $\text{cm}^{-1}$).
This corresponds to the match A - II, B - III, C - I.
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
The molecule that can absorb in the infra-red among the following is
Consider the following statements about Infrared (IR) spectroscopy.
A. It is used to determine the band gap, the band structure, and the charge carrier concentration of a compound.
B. It is used to identify the functional group(s) of a compound.
C. It is used to characterize different stretching and bending modes of vibration in molecules.
D. Heteronuclear diatomic molecules are IR-active.
The correct statements are
