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

Step 1: What controls IR absorption position. 
In IR spectroscopy, the position of an absorption peak (wavenumber) depends mainly on two factors:
– how strong the bond is, and
– how heavy the atoms forming the bond are.

Step 2: Relevant relation for bond vibrations.
Bond vibrations approximately follow Hooke’s law, which gives the wavenumber as:

\[ \bar{\nu} = \frac{1}{2\pi c}\sqrt{\frac{k}{\mu}} \]

Here:
\( k \) represents the force constant (a measure of bond strength).
\( \mu \) is the reduced mass of the two bonded atoms.

Step 3: Effect of bond strength.
Stronger bonds vibrate at higher frequencies.
Therefore, triple bonds absorb at higher wavenumbers than double bonds, and double bonds absorb higher than single bonds:

\[ C \equiv C \;>\; C = C \;>\; C - C \]

Step 4: Effect of atomic mass.
The reduced mass has a very strong influence on the wavenumber.
When one of the atoms is very light, the reduced mass becomes small, which significantly increases the wavenumber.

Hydrogen is the lightest atom, so bonds involving hydrogen have very small reduced mass.
As a result, X–H bonds absorb at much higher wavenumbers than most other bonds.

Step 5: Comparing common IR absorptions.
Typical stretching frequencies are:
– O–H: \( 3200\!-\!3600 \, \text{cm}^{-1} \)
– C≡C: \( 2100\!-\!2250 \, \text{cm}^{-1} \)
– C=C: \( 1600\!-\!1680 \, \text{cm}^{-1} \)
– C–C: below \( 1300 \, \text{cm}^{-1} \)

Even though the C≡C bond is very strong, the extremely low mass of hydrogen makes the O–H bond vibrate at a higher wavenumber.

Step 6: Final conclusion.
Among the given bonds, the O–H bond shows absorption at the highest wavenumber in IR spectroscopy.