Question

In IR spectroscopy, which bond absorbs at the highest wavenumber?

• A. C-C
• B. C=C
• C. C \(\equiv\) C
• D. O-H

Hint:

Bonds to Hydrogen (X-H) always appear in the high-frequency region (\( 2500-4000 \text{ cm}^{-1} \)) of the IR spectrum because of the low atomic mass of Hydrogen.

Answer 1

The Correct Option is D

To determine which bond absorbs at the highest wavenumber in IR spectroscopy, we need to understand the relationship between bond strength and wavenumber absorption.

Concept Explanation:

• In IR spectroscopy, the absorption of infrared radiation causes molecular vibrations. The wavenumber (in cm^-1) at which a bond absorbs is directly related to the bond strength and the masses of the atoms involved.
• Stronger bonds and lighter atoms typically absorb at higher wavenumbers because they vibrate at higher frequencies.

Now, let's analyze the given options:

1. C-C (Carbon-Carbon) Single Bond: Typically absorbs at a low wavenumber, around 1200 cm^-1.
2. C=C (Carbon-Carbon) Double Bond: Absorbs at a higher wavenumber than single bonds, typically around 1600-1680 cm^-1.
3. C ≡ C (Carbon-Carbon) Triple Bond: Even stronger and stiffer than double bonds, absorbing higher at around 2100-2260 cm^-1.
4. O-H (Oxygen-Hydrogen) Bond: This bond is stronger and involves a lighter hydrogen atom, leading to an absorption at a significantly higher wavenumber, around 3200-3550 cm^-1. This high absorption is characteristic of the O-H stretch, often seen in alcohols and phenols.

Based on this analysis, O-H bonds absorb at the highest wavenumber due to their bond strength and the light mass of the hydrogen atom.

Conclusion: The correct answer is O-H, which absorbs at the highest wavenumber in IR spectroscopy.

Answer 2

Step 1: Understanding the Concept:
The wavenumber \( (\bar{\nu}) \) of absorption in Infrared (IR) spectroscopy is determined by the strength of the bond and the mass of the atoms involved.
Step 2: Key Formula or Approach:
According to Hooke's Law for bond vibration:
\[ \bar{\nu} = \frac{1}{2\pi c} \sqrt{\frac{k}{\mu}} \]
Where:
\( k \) is the force constant (bond strength).
\( \mu \) is the reduced mass \( (\frac{m_1 m_2}{m_1 + m_2}) \).
Step 3: Detailed Explanation:
- Bond Strength: Higher bond order (triple \(>\) double \(>\) single) increases \( k \), which increases wavenumber. So \( C \equiv C>C = C>C-C \).
- Reduced Mass: As the mass of the atoms decreases, \( \mu \) decreases, which significantly increases the wavenumber.
- Comparing \( C \equiv C \) and \( O-H \): Although \( C \equiv C \) has a high force constant, the reduced mass of the \( O-H \) bond is extremely small because of the Hydrogen atom (\( m_H \approx 1 \)).
- Typically, \( O-H \) stretching occurs at \( 3200-3600 \text{ cm}^{-1} \), while \( C \equiv C \) occurs at \( 2100-2250 \text{ cm}^{-1} \), \( C=C \) at \( 1600-1680 \text{ cm}^{-1} \), and \( C-C \) in the fingerprint region (\(<1300 \text{ cm}^{-1} \)).
Step 4: Final Answer:
The O-H bond absorbs at the highest wavenumber due to the very low reduced mass of Hydrogen.