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How could you use infrared spectroscopy to distinguish between the following pairs of isomers? (a) \(\mathrm{HC} \equiv \mathrm{CCH}_{2} \mathrm{NH}_{2}\) and \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{C} \equiv \mathrm{N}\) (b) \(\mathrm{CH}_{3} \mathrm{COCH}_{3}\) and \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHO}\)

Short Answer

Expert verified
Use distinct IR bands for alkynes vs. nitriles and ketones vs. aldehydes to differentiate isomers.

Step by step solution

01

Understand the Structure of Isomers

Examine the molecular structures of each pair of isomers to identify functional groups that might display distinct absorbance bands in an infrared (IR) spectrum. (a) For \( \mathrm{HC} \equiv \mathrm{CCH}_{2} \mathrm{NH}_{2} \) and \( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{C} \equiv \mathrm{N} \), note the presence of the terminal alkyne group (C-H) in the first and the nitrile group (C鈮) in the second compound.(b) For \( \mathrm{CH}_{3} \mathrm{COCH}_{3} \) (acetone) and \( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHO} \) (propanal), note the differences between the ketone (C=O in the middle of the chain) and aldehyde (C=O at the end of the chain) groups.
02

Identify Key IR Absorption Bands

Recall typical IR absorption bands for functional groups. (a) A terminal alkyne has a sharp C鈮-H stretch around 3300 cm鈦宦, while a nitrile group has a broad and strong C鈮 stretch near 2200-2250 cm鈦宦. (b) A ketone displays a strong C=O stretch around 1715 cm鈦宦, whereas an aldehyde shows two additional peaks: one for the C=O stretch around 1720-1740 cm鈦宦 and another weak peak near 2720 cm鈦宦 due to the C-H bond adjacent to the carbonyl.
03

Analyze the IR Spectra to Differentiate Isomers

To distinguish between each pair:(a) Check the IR spectrum of \( \mathrm{HC} \equiv \mathrm{CCH}_{2} \mathrm{NH}_{2} \) for a terminal alkyne C鈮-H stretch near 3300 cm鈦宦, absent in \( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{C} \equiv \mathrm{N} \), which will show a characteristic C鈮 peak around 2200-2250 cm鈦宦.(b) Examine the IR spectrum for acetone and propanal; acetone will have a single strong peak for the C=O stretch near 1715 cm鈦宦. For propanal, identify the C=O stretch near 1720-1740 cm鈦宦 and check for the unique aldehyde peak around 2720 cm鈦宦.

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Key Concepts

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

Isomers
Isomers are fascinating because they have the same molecular formula but different structural arrangements. This means they can have very different chemical and physical properties despite having the same types and numbers of atoms. Understanding isomers is crucial because they illuminate how the arrangement of atoms affects a compound's behavior.

There are several types of isomers, but the focus here is on structural isomers. Structural isomers differ in the connectivity of atoms. For example, in the given pairs of isomers, one might contain a nitrile group, while the other might have a terminal alkyne. Each type of isomer will show different characteristics in infrared spectroscopy based on these structural differences.
Functional Groups
Functional groups are specific groups of atoms within molecules that have characteristic properties and predictable chemical behaviors. They are like the identifying badges that give the molecule its functionality and are key when analyzing chemical behavior.

In infrared spectroscopy, different functional groups display distinct absorption patterns, which makes it easier to identify and differentiate substances. For instance, a ketone will show a particular IR absorption due to its C=O group, whereas an aldehyde will display an additional peak. Recognizing these groups in molecular structures can help predict where these absorption bands will appear in an IR spectrum.
IR Absorption Bands
Infrared (IR) absorption bands are critical in identifying different compounds based on the vibrations of their bonds. These bands appear at specific frequencies, correlating with the energy required to stretch or bend a particular bond within a molecule.

The IR absorption spectrum is like a molecular fingerprint. Important absorption bands arise from different functional groups, such as the sharp peak near 3300 cm鈦宦 for a terminal alkyne or the strong stretch around 2200-2250 cm鈦宦 for a nitrile group. By examining the position and intensity of these absorption bands in the IR spectrum, chemists can deduce the presence of certain functional groups in a compound.
Terminal Alkyne
A terminal alkyne is an organic compound with a carbon-carbon triple bond at the end of a carbon chain. This feature is notable due to the unique C鈮-H bond.

In IR spectroscopy, terminal alkynes are easily recognized by their distinct absorption band. They show a sharp, strong C鈮-H stretch near 3300 cm鈦宦. This sharpness and position make it easier to differentiate terminal alkynes from other types of triple bonds. Knowing this helps identify whether a molecule contains a terminal alkyne, which can significantly affect its reactivity and interactions.
Nitrile Group
The nitrile group, characterized by a carbon-nitrogen triple bond (C鈮), is a functional group that can significantly influence a molecule's properties. This group is often found in organic compounds like acrylonitrile, which is used in many industrial applications.

In an IR spectrum, the nitrile group's presence is signaled by a strong and broad absorption band near 2200-2250 cm鈦宦. This IR absorption is particularly broad and distinct, making it relatively easy to spot compared to similar compounds without a nitrile group. By identifying this band, chemists can confirm the presence of nitriles and further investigate the compound鈥檚 potential uses and reactions.

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Most popular questions from this chapter

It's useful to develop a feeling for the amounts of energy that correspond to different parts of the electromagnetic spectrum. Calculate the energies in \(\mathrm{kJ} / \mathrm{mol}\) of each of the following kinds of radiation: (a) A gamma ray with \(\lambda=5.0 \times 10^{-11} \mathrm{~m}\) (b) An X ray with \(\lambda=3.0 \times 10^{-9} \mathrm{~m}\) (c) Ultraviolet light with \(\nu=6.0 \times 10^{15} \mathrm{~Hz}\) (d) Visible light with \(\nu=7.0 \times 10^{14} \mathrm{~Hz}\) (e) Infrared radiation with \(\lambda=2.0 \times 10^{-5} \mathrm{~m}\) (f) Microwave radiation with \(\nu=1.0 \times 10^{11} \mathrm{~Hz}\)

The hormone cortisone contains \(\mathrm{C}, \mathrm{H},\) and \(\mathrm{O},\) and shows a molecular ion at \(\mathrm{M}^{+}=360.1937\) by high-resolution mass spectrometry. What is the molecular formula of cortisone? (The degree of unsaturation for cortisone is 8.2

Halogenated compounds are particularly easy to identify by their mass spectra because both chlorine and bromine occur naturally as mixtures of two abundant isotopes. Recall that chlorine occurs as \({ }^{35} \mathrm{Cl}(75.8 \%)\) and \({ }^{37} \mathrm{Cl}(24.2 \%) ;\) and bromine occurs as \({ }^{79} \mathrm{Br}(50.7 \%)\) and \({ }^{81} \mathrm{Br}(49.3 \%)\) At what masses do the molecular ions occur for the following formulas? What are the relative percentages of each molecular ion? (a) Bromomethane, \(\mathrm{CH}_{3} \mathrm{Br}\) (b) 1 -Chlorohexane, \(\mathrm{C}_{6} \mathrm{H}_{13} \mathrm{Cl}\)

Would you expect two diastereomers such as meso-2,3-dibromobutane and \((2 R, 3 R)\) -dibromobutane to have identical or different IR spectra? Explain.

Which is stronger, the \(\mathrm{C}=\mathrm{O}\) bond in an ester \(\left(1735 \mathrm{~cm}^{-1}\right)\) or the \(\mathrm{C}=\mathrm{O}\) bond in a saturated ketone \(\left(1715 \mathrm{~cm}^{-1}\right) ?\) Explain.

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