Skip to main content
Ch. 12 - Infrared Spectroscopy and Mass Spectrometry
Wade - Organic Chemistry 9th Edition
Wade9th EditionOrganic ChemistryISBN: 9780135213728Not the one you use?Change textbook
All textbooksWade 9th EditionCh. 12 - Infrared Spectroscopy and Mass SpectrometryProblem 21a,b
Chapter 12, Problem 21a,b

A C-D (carbon–deuterium) bond is electronically much like a C-H bond, and it has a similar stiffness, measured by the spring constant, k. The deuterium atom has twice the mass (m) of a hydrogen atom, however.
(a) The infrared absorption frequency is approximately proportional to km\(\sqrt{\frac{k}{m}\)} , when one of the bonded atoms is much heavier than the other, and m is the lighter of the two atoms (H or D in this case). Use this relationship to calculate the IR absorption frequency of a typical C-D bond. Use 3000 cm–1 as a typical C-H absorption frequency.
(b) A chemist dissolves a sample in deuterochloroform (CDCl3) and then decides to take the IR spectrum and simply evaporates most of the CDCl3. What functional group will appear to be present in this IR spectrum as a result of the CDCl3 impurity?

Verified step by step guidance
1
Step 1: Understand the relationship between the IR absorption frequency and the reduced mass of the bond. The frequency (ν) is proportional to the square root of the spring constant (k) divided by the reduced mass (μ). For this problem, the reduced mass is approximately equal to the mass of the lighter atom (m) since the other atom (carbon) is much heavier. The relationship can be expressed as: ν ∝ √(k/m).
Step 2: Compare the C-H and C-D bonds. Since the spring constant (k) is the same for both bonds, the ratio of their frequencies will depend on the square root of the ratio of their masses. For the C-H bond, the lighter atom is hydrogen (m_H), and for the C-D bond, the lighter atom is deuterium (m_D). The ratio of the frequencies can be written as: ν(C-D)/ν(C-H) = √(m_H/m_D).
Step 3: Substitute the given values. The mass of deuterium (m_D) is twice the mass of hydrogen (m_H), so m_H/m_D = 1/2. Using this ratio, calculate the proportional frequency of the C-D bond relative to the C-H bond. The typical C-H absorption frequency is given as 3000 cm⁻¹.
Step 4: For part (b), consider the chemical nature of deuterochloroform (CDCl3). When CDCl3 is evaporated, residual CDCl3 may remain in the sample. CDCl3 contains a C-D bond, which will appear in the IR spectrum. The C-D bond absorbs at a lower frequency than the C-H bond due to the heavier mass of deuterium. This absorption typically appears in the range of 2100–2200 cm⁻¹.
Step 5: Conclude that the functional group appearing in the IR spectrum due to the CDCl3 impurity is the C-D bond. This is because the residual deuterochloroform contributes its characteristic C-D stretching absorption in the IR spectrum.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
14m
Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Spring Constant (k)

The spring constant (k) is a measure of the stiffness of a bond, indicating how much force is needed to stretch or compress it. In the context of molecular vibrations, a higher spring constant corresponds to a stronger bond, resulting in higher vibrational frequencies. This concept is crucial for understanding how bond strength influences infrared absorption frequencies.
Recommended video:
Guided course
02:19
The relationship between equilibrium constant and pKa.

Mass of Atoms (m)

In the context of vibrational spectroscopy, the mass of the atoms involved in a bond significantly affects the frequency of the bond's vibrations. The relationship between mass and frequency is inversely proportional; as the mass increases, the vibrational frequency decreases. This principle is essential for calculating the infrared absorption frequency of C-D bonds compared to C-H bonds.
Recommended video:
Guided course
01:44
The difference between atomic numbers and atomic mass.

Infrared (IR) Spectroscopy

Infrared spectroscopy is a technique used to identify molecular structures based on the absorption of infrared light by molecular vibrations. Different functional groups absorb IR radiation at characteristic frequencies, allowing chemists to deduce the presence of specific groups in a compound. Understanding how solvents like deuterochloroform (CDCl3) can introduce additional absorption peaks is vital for interpreting IR spectra.
Recommended video:
Guided course
16:04
General Features of IR Spect
Related Practice
Textbook Question

Predict the masses and the structures of the most abundant fragments observed in the mass spectra of the following compounds.

(c) 4-methylpentan-2-ol

1
views
Textbook Question

Predict the masses and the structures of the most abundant fragments observed in the mass spectra of the following compounds.

(b) 3-methylhex-2-ene

1
views
Textbook Question

A common lab experiment is the dehydration of cyclohexanol to cyclohexene.

(a) Explain how you could tell from the IR spectrum whether your product was pure cyclohexene, pure cyclohexanol, or a mixture of cyclohexene and cyclohexanol. Give approximate frequencies for distinctive peaks.

(b) Explain why mass spectrometry might not be a good way to distinguish cyclohexene from cyclohexanol.

2
views
Textbook Question

A laboratory student added 1-bromobutane to a flask containing dry ether and magnesium turnings. An exothermic reaction resulted, and the ether boiled vigorously for several minutes. Then she added acetone to the reaction mixture and the ether boiled even more vigorously. She added dilute acid to the mixture and separated the layers. She evaporated the ether layer, and distilled a liquid that boiled at 143 °C. GC–MS analysis of the distillate showed one major product with a few minor impurities. The mass spectrum of the major product is shown here.

(a) Draw out the reactions that took place and show the product that was formed.

<IMAGE>

1
views
Textbook Question

Three common lab experiments are shown. In each case, describe how the IR spectrum of the product would differ from that of the reactant. Give approximate frequencies for distinctive peaks in the IR spectrum of the reactant and also that of the product.

1
views
Textbook Question

A laboratory student added 1-bromobutane to a flask containing dry ether and magnesium turnings. An exothermic reaction resulted, and the ether boiled vigorously for several minutes. Then she added acetone to the reaction mixture and the ether boiled even more vigorously. She added dilute acid to the mixture and separated the layers. She evaporated the ether layer, and distilled a liquid that boiled at 143 °C. GC–MS analysis of the distillate showed one major product with a few minor impurities. The mass spectrum of the major product is shown here.

(b) Explain why the molecular ion is or is not visible in the mass spectrum, and show what ions are likely to be responsible for the strong peaks at m/z 59 and 101.

<IMAGE>

2
views