Untitled

Quiz 2 Study Notes

Section A

Question 1: Fill in the Blanks

1. Write the word or phrase that best completes the statement or answers the question. [4 marks]
   • (i) In an IR spectrometer, the monochromator uses prisms or diffraction gratings to allow only one frequency of light to enter the detector at a time.
   • (ii) The wavelength and frequency of a given wave of electromagnetic radiation are inversely proportional.
   • (iii) The region of the IR spectrum which contains the most complex vibrations (1400–600 cm⁻¹) is called the fingerprint region.
   • (iv) Absorption spectrum is the measurement of the amount of light absorbed by a compound as a function of the wavelength of light.

Question 2: IR Stretch Identification [1 mark]

2. Which of the following bonds has an IR stretch around 1600 cm⁻¹?
   • A. C–N
   • B. C=N
   • C. C≡N
   • Correct Answer: B (C=N)

Question 3: Wavenumber Order

3. List the following compounds in order of increasing wavenumber of the absorption band:
   • A. Amide
   • B. Ester
   • C. Ketone
   • Order: A < B < C
   • Explanation: Typically, ketones absorb at higher wavenumbers than esters, which in turn absorb at higher wavenumbers than amides due to differences in bond strength and atomic masses.

Question 4: Calculate Vibrational Degrees of Freedom [4 marks]

4. Calculate the vibrational degrees of freedom for the following molecules:
   • (i) For CO:
     • CO has 2 atoms; thus, the formula for degrees of freedom is:
       $ext{Degrees of freedom} = 3N - 5 = 3(2) - 5 = 1$
   • (ii) For NH₃:
     • NH₃ has 4 atoms; thus, using the same formula:
       $ext{Degrees of freedom} = 3N - 6 = 3(4) - 6 = 6$

Question 5: Distinction of Compounds [6 marks]

5. How could IR spectroscopy be used to distinguish between the following pairs of compounds?
   • (i) H₃C–CH₂–C≡C–H and H₃C–C≡C–CH₃
     • Analysis: The terminal alkyne (H₃C–CH₂–C≡C–H) shows O–H stretch absorption characteristics that differ from that of a symmetrical alkyne (H₃C–C≡C–CH₃).
   • (ii) Further examples could explore other functional groups, e.g., alcohols vs. ketones, to use their distinguishing absorption peaks for identification.

Section B

Question 1: Causes of Vibration

1. Which of the following causes the vibration of atoms?
   • A. The number of protons contained in a nucleus.
   • B. Electron movement to higher energy levels.
   • C. The molecule’s total molecular weight.
   • D. Dipole moments between atoms. (Correct Answer)

Question 2: Order of Increasing Wavenumber

2. Which is the correct order of increasing wave number of the stretching vibrations of:
   • (1) C–H (alkane)
   • (2) O–H (alcohol)
   • (3) C=O (ketone)
   • (4) C≡C (alkyne)
   • A. (4) < (3) < (2) < (1)
   • B. (3) < (4) < (2) < (1)
   • C. (3) < (4) < (1) < (2)
   • D. (4) < (3) < (1) < (2) (Correct Answer)

Question 3: Dependence of Frequency

3. The frequency of the stretching vibration of a bond in IR spectroscopy depends on what two quantities?
   • A. The nuclear charges of the atoms and the atomic radii
   • B. The electronegativity of the atoms and the nuclear charges of the atoms
   • C. The masses of the atoms and the stiffness of the bond (Correct Answer)
   • D. The stiffness of the bond and the electronegativity of the atoms

Question 4: Frequency Comparison of C–O and C–N

4. In IR spectroscopy, the C–O bond has frequency than the C–N bond because .
   • A. higher, an O atom has an even number of neutrons
   • B. lower, an O atom has more mass than an N atom
   • C. higher, an O atom has more electronegativity than an N atom (Correct Answer)
   • D. higher, an O atom has more mass than an N atom

Question 5: True or False Statements [4 marks]

5. Indicate whether each of the following statements is True or False:
   • (i) The O–H stretch of a concentrated solution of an alcohol occurs at a higher frequency than the O–H stretch of a dilute solution. True
   • (ii) Propyne will not have an absorption band at 3100 cm⁻¹ because there is no change in dipole moment. True
   • (iii) Light of 2 μm is of higher energy than light of 3 μm. False
   • (iv) IR spectroscopy can be used to identify functional groups quickly and easily. True

Question 6: Distinction of Amines

6. How could IR spectroscopy be used to distinguish between the following pairs of compounds?
   • (i) A) A tertiary amine
   • B) A primary amine
   • Analysis: Primary amines exhibit N–H stretching frequencies, while tertiary amines do not, allowing for distinction.

Question 7: Deduction of Compound Structure

7. Deduce the possible structure of the compound with the molecular formula C₄H₈O given the IR absorptions: 2950, 2820, 2715, 1720 cm⁻¹.
   • Analysis: The absorption at 1720 cm⁻¹ suggests the presence of a carbonyl group (C=O), while the remaining absorptions suggest the presence of aliphatic CH groups. Possible structures include ketones or aldehydes.

Question 8: Calculate Vibrational Degrees of Freedom

8. Calculate the vibrational degrees of freedom for the molecule: CH₃–CH₃.
   • Analysis: For ethane (CH₃–CH₃):
     • It has 6 atoms; thus, the degrees of freedom are calculated as:
       $ext{Degrees of freedom} = 3N - 6 = 3(6) - 6 = 12$