Choose the correct compound for the given IR spectrum. % Transmittance 100 80 60 40 20 0 4000 3500 3000 2500 2000 A) NH2 B) C) D) OH Wavenumber (cm
) 1500 1000
The Correct Answer and Explanation is:
The correct compound is D) 4-Methyl-1-pentanol.
Explanation
To identify the correct compound, we must analyze the key absorption bands in the provided Infrared (IR) spectrum and match them to the functional groups present in the given options.
- O-H Stretch (Alcohol): The most prominent feature in the spectrum is the very broad, strong absorption band centered at approximately 3300 cm⁻¹. This is the characteristic signal for the O-H stretching vibration of an alcohol. The significant broadening of this peak is due to intermolecular hydrogen bonding between alcohol molecules.
- C-H Stretch (Alkane): The sharp, strong peaks observed just below 3000 cm⁻¹ (specifically in the 2850-2960 cm⁻¹ range) are characteristic of C-H stretching vibrations from sp³-hybridized carbons, indicating the presence of an alkane-like (saturated) carbon skeleton.
- C-O Stretch: In the fingerprint region, there is a distinct peak around 1050 cm⁻¹. This absorption is consistent with the C-O stretching vibration of a primary alcohol, which typically appears between 1000-1250 cm⁻¹.
Now, let’s evaluate the options based on this analysis:
- A) Butylamine: This is a primary amine. A primary amine (-NH₂) would show two medium, sharp N-H stretching peaks in the 3300-3500 cm⁻¹ region, not a single, very broad peak like the one observed. Therefore, this option is incorrect.
- B) 1-Pentene: This is an alkene. It would exhibit a C=C stretching peak around 1640 cm⁻¹ and, importantly, a C-H stretching peak from the sp² carbons of the double bond above 3000 cm⁻¹. Neither of these key signals is present in the spectrum. Thus, this option is incorrect.
- C) Pentane: This is an alkane. Its spectrum would only show sp³ C-H stretching peaks below 3000 cm⁻¹ and C-H bending peaks in the fingerprint region. It would not have the strong, broad O-H absorption at 3300 cm⁻¹. Therefore, this option is incorrect.
- D) 4-Methyl-1-pentanol: This compound is a primary alcohol with a saturated alkyl chain. It contains an O-H group, which accounts for the broad peak at ~3300 cm⁻¹. It has several sp³ C-H bonds, matching the strong peaks below 3000 cm⁻¹. Finally, as a primary alcohol, it has a C-O bond, which corresponds to the peak observed at ~1050 cm⁻¹. This structure is a perfect match for all the major features of the given IR spectrum.
