Question

Which of the following compounds shows a sharp band at 2150 cm$^{-1}$ and 3300 cm$^{-1}$ in the IR spectrum?

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

In IR spectra, alkyne C≡C stretches appear around 2150 cm$^{-1}$, and N-H stretches typically appear around 3300 cm$^{-1}$.

Answer 1

The Correct Option is A

Step 1: Identify IR Absorption Bands.
- A sharp band at 2150 cm$^{-1}$ indicates a C≡C stretch (alkyne).
- A broad band at 3300 cm$^{-1}$ indicates an N-H stretch (amine).

Step 2: Evaluate Options Based on Bands.
- (1) Alkyne (\( \text{C} \equiv \text{H} \)): Exhibits a sharp band at 2150 cm$^{-1}$ (C≡C stretch) and a stretch at 3300 cm$^{-1}$ (C-H stretch). This option matches both observed bands.
- (2) Alkene (\( \text{C} = \text{C} \)): Would show different stretching frequencies and does not align with both given bands.
- (3) Amines (\( \text{N-H} \)): Shows a broad stretch near 3300 cm$^{-1}$ but lacks the 2150 cm$^{-1}$ band.
- (4) Amine (\( \text{H}_2\text{N} \)): Displays a broad N-H stretch but not the characteristic C≡C stretch at 2150 cm$^{-1}$.

Step 3: Determine the Correct Identification.
Based on the analysis, the correct identification is (1) Alkyne (\( \text{C} \equiv \text{H} \)).

Final Answer:
\[ \boxed{\text{(1) Alkyne (C≡H)}} \]