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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.

Short Answer

Expert verified
The C=O bond in the ester (1735 cm\(^{-1}\)) is stronger than in the saturated ketone (1715 cm\(^{-1}\)).

Step by step solution

01

Understand Bond Strength and Frequency

Bond strength is correlated with bond length. Shorter bonds are generally stronger, with a higher bond dissociation energy. The wavenumber in cm\(^{-1}\) refers to the frequency at which a bond vibration occurs in infrared spectroscopy. Higher frequencies usually imply stronger bonds.
02

Compare the Frequencies

We are given that the vibration frequency for the C=O bond in an ester is 1735 cm\(^{-1}\), while for a saturated ketone, it is 1715 cm\(^{-1}\).
03

Interpret the Frequency Values

Since 1735 cm\(^{-1}\) is higher than 1715 cm\(^{-1}\), it indicates that the C=O bond in the ester vibrates at a higher frequency, suggesting it is shorter and stronger compared to the C=O bond in the saturated ketone.

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Key Concepts

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

Bond Strength
Bond strength is a crucial concept in chemistry, as it determines how tightly atoms are held together in a molecule. It is related to the bond length and the energy required to break the bond. Shorter bonds are typically stronger because the atoms are closer together, resulting in a stronger attractive force. This makes them harder to break and hence require more energy, known as bond dissociation energy.

In infrared spectroscopy, bond strength can be inferred from the frequency of bond vibrations. The wavenumber, given in cm\(^{-1}\), indicates the frequency at which a bond vibrates. Higher wavenumbers usually mean stronger bonds, as the bonds require more energy to vibrate at higher frequencies. Thus, by analyzing the frequencies, one can deduce the relative bond strength in different molecules.
Esters
Esters are a class of organic compounds commonly used in a variety of applications, from perfumes to pharmaceuticals. They are characterized by the presence of a carbonyl group (\(\mathrm{C}=\mathrm{O}\)), which is responsible for many of their chemical properties. The carbonyl group in esters typically shows absorption in the infrared spectroscopy at higher frequencies compared to some other functional groups.

In the context of infrared spectroscopy, the ester bond’s absorption frequency at 1735 cm\(^{-1}\) indicates a relatively strong carbonyl bond. This is because the ester linkage often creates strain or angles that increase the electron demand of its carbonyl carbon, resulting in a shorter and stronger bond.
  • Used commonly in perfumes due to characteristic fragrances.
  • Serve as solvents and plasticizers in industry.
  • In biochemistry, esters form the backbone of lipid molecules.
Saturated Ketones
Saturated ketones are organic compounds characterized by a carbonyl group (\(\mathrm{C}=\mathrm{O}\)) bonded to two carbon atoms that are part of non-branching carbon chains. They differ from unsaturated ketones, where one or more double bonds exist in the carbon chain. Saturated ketones tend to be less reactive than their unsaturated counterparts due to the stability provided by the single carbon bonds.

In infrared spectroscopy, saturated ketones exhibit a C=O bond vibration frequency of 1715 cm\(^{-1}\). Although slightly lower in frequency compared to esters, this still indicates a significant bond strength. The slightly lower frequency signifies a marginally longer bond compared to esters, suggesting a weaker, yet still strong, bond.
  • Commonly used in flavorings and fragrances.
  • Play a role in various biological processes.
  • Used as solvents in industrial applications.

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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 \(\nu=1.0 \times 10^{11} \mathrm{~Hz}\)

How might you use IR spectroscopy to distinguish between the following pairs of isomers? (a) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) and \(\mathrm{CH}_{3} \mathrm{OCH}_{3}\) (b) Cyclohexane and hex-1-ene (c) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H}\) and \(\mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{CHO}\)

Camphor, a saturated monoketone from the Asian camphor tree, is used as a moth repellent and as a constituent of embalming fluid, among other things. If camphor has \(\mathrm{M}^{+}=152.1201\) by high-resolution mass spectrometry, what is its molecular formula?

Carvone is an unsaturated ketone responsible for the odor of spearmint. If carvone has \(\mathrm{M}^{+}=150\) in its mass spectrum and contains three double bonds and one ring, what is its molecular formula?

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 \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHO}\)

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