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Consult your online resources for additional practice. IR spectroscopy is most useful for distinguishing: (A) double and triple bonds. (B) C–H bonds. (C) chirality of molecules. (D) relative percentage of enantiomers in mixtures.

Short Answer

Expert verified
Option (A) because IR spectroscopy effectively distinguishes double and triple bonds due to their unique vibrational frequencies.

Step by step solution

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01

Identify the Purpose of IR Spectroscopy

IR (Infrared) spectroscopy is a technique used to determine the presence of specific functional groups in a molecule by measuring the vibrational modes of molecular bonds.
02

Eliminate Option C

IR spectroscopy measures bond vibrations and is generally not used to analyze the chirality of molecules. Therefore, option (C) can be eliminated.
03

Eliminate Option D

IR spectroscopy does not provide information on the relative percentage of enantiomers in mixtures, which is better analyzed using techniques like polarimetry or chiral chromatography. Thus, option (D) can be eliminated.
04

Consider Option A

Double and triple bonds have distinct vibrational frequencies that can be readily detected using IR spectroscopy. This makes IR useful in distinguishing compounds with these types of bonds.
05

Consider Option B

C–H bonds are common in most organic compounds and their stretching vibrations are always present in IR spectra. However, the technique is not specialized for selectively distinguishing C–H bonds compared to other types of bonds.
06

Choose the Best Option

From the analysis, option (A) is the best choice because IR spectroscopy is particularly effective in distinguishing between double and triple bonds due to their unique vibrational frequencies.

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Functional Groups Identification
IR spectroscopy is a powerful tool in organic chemistry for identifying functional groups within a molecule. It works by measuring the absorption of infrared light by different chemical bonds. Different bonds absorb light at different frequencies, which correspond to the natural vibrational frequencies of the bonds.
This technique helps to determine which functional groups are present because each type of bond vibrates at a unique frequency.
For instance:
  • O-H bonds in alcohols show a strong, broad absorption around 3200-3550 cm-1
  • C=O bonds in carbonyl compounds exhibit a sharp absorption near 1700 cm-1
  • N-H bonds in amines can be detected around 3300-3500 cm-1
These distinctive absorption bands make it possible to identify various functional groups in a compound quickly and efficiently. Understanding these patterns is fundamental for interpreting IR spectra.
Double Bonds Detection
IR spectroscopy is particularly useful for detecting double bonds in organic molecules. Double bonds, such as those found in alkenes, have unique vibrational characteristics that appear as distinct peaks in the IR spectrum.
For example, a C=C double bond typically shows an absorption between 1620-1680 cm-1, which is not observed in single-bonded carbons (C–C).
This makes it straightforward to determine the presence of double bonds in a sample.
Observing the absorption peaks in this range can inform chemists about the structure of a molecule and confirm the presence of double bonds.
This application of IR spectroscopy is indispensable in organic synthesis and molecular characterization.
Triple Bonds Detection
Triple bonds, such as those found in alkynes, can also be readily detected using IR spectroscopy. The C≡C triple bond has a vibrational frequency that shows up as a sharp absorption peak between 2100-2260 cm-1.
This distinct range helps distinguish triple bonds from other types of carbon-carbon bonds (single or double bonds).
Additionally, the presence of sharp and significant peaks within this absorption range can also help differentiate between internal and terminal alkynes.
The strength and sharpness of these absorption bands make IR spectroscopy a reliable method for identifying molecules with triple bonds.
Chiral Analysis Techniques
While IR spectroscopy is excellent for identifying various bonds, it is not suitable for analyzing chirality. Chirality refers to the property of a molecule that makes it non-superimposable on its mirror image.
Other techniques, such as polarimetry, are used to determine the chirality of molecules by measuring the rotation of plane-polarized light.
Alternatively, chiral chromatography can separate enantiomers based on their interaction with a chiral stationary phase.
These techniques provide detailed information about the stereochemistry of molecules and the relative percentage of enantiomers in mixtures. Therefore, although IR spectroscopy is great for many structural analyses, it is not the go-to technique for chirality studies.

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Most popular questions from this chapter

The coupling constant, \(J,\) is: A. the value of \(n+1\) when determining splitting in \(\mathrm{NMR}\) spectra. B. measured in parts per million (ppm). C. corrected for by calibration with tetramethylsilane. D. a measure of the degree of splitting caused by other atoms in the molecule.

Oxygen \(\left(\mathrm{O}_{2}\right)\) does not exhibit an IR spectrum because: (A) it has no molecular motions. (B) it is not possible to record IR spectra of a gaseous molecule. (C) molecular vibrations do not result in a change in the dipole moment. (D) molecular oxygen contains four lone pairs overall.

In a molecule containing a carboxylic acid group, what would be expected in a \(^{1} \mathrm{H}-\mathrm{NMR}\) spectrum? A. A deshielded hydrogen peak for the hydroxyl hydrogen, shifted left B. A deshielded hydrogen peak for the hydroxyl hydrogen, shifted right C. A shielded hydrogen peak for the hydroxyl hydrogen, shifted left D. A shielded hydrogen peak for the hydroxyl hydrogen, shifted right

In an IR spectrum, how does extended conjugation of double bonds affect the absorbance band of carbonyl (C=0) stretches compared with normal absorption? (A) The absorbance band will occur at a lower wavenumber. (B) The absorbance band will occur at a higher wavenumber. (C) The absorbance band will occur at the same wavenumber. (D) The absorbance band will disappear.

The IR spectrum of a fully protonated amino acid would likely contain which of the following peaks A sharp peak at \(1750 \mathrm{cm}^{-1}\) A sharp peak at \(3300 \mathrm{cm}^{-1}\) A broad peak at \(3300 \mathrm{cm}^{-1}\) (A) I only (B) I and II only (C) II and III only (D) I, II, and III

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