Unit 2: CH02 Syllabus Node

Organic Analysis: Infrared Spectroscopy & Mass Spectrometry

Infrared spectroscopy absorption ranges, fingerprint region verification, mass spectrometry molecular ion and [M+1] peaks, and chemical testing.

1. Infrared (IR) Spectroscopy

Covalent bonds vibrate naturally by stretching and bending. Each bond absorbs infrared radiation matching its natural vibrational frequency. Absorption peaks appear as dips on the transmittance scale:

Annotated Infrared Spectrum Wavenumber (cm^-1) % Transmittance 4000 3000 2000 1500 500 O-H (Alcohol) C=O (Carbonyl) Fingerprint Region (<1500 cm^-1)
Bond Type Functional Group Wavenumber Range (\(\text{cm}^{-1}\)) Peak Appearance & Diagnostics
O-H Alcohols 3230 - 3550 Broad, smooth trough due to hydrogen bonding.
O-H Carboxylic acids 2500 - 3000 Very broad, jagged trough overlapping C-H stretching.
C=O Aldehydes, Ketones, Carboxylic acids, Esters 1680 - 1750 Sharp, intense, deep spike (unmistakable carbonyl peak).
C=C Alkenes 1620 - 1680 Moderate sharp absorption.
C-H Alkanes, Alkenes, Arenes 2850 - 3100 Sharp, multi-pointed absorption.

2. The Fingerprint Region

Fingerprint Region

The complex region of an infrared spectrum below \(1500\text{ cm}^{-1}\), caused by complex bending vibrations of the whole molecule.

While the functional group region (\(1500-4000\text{ cm}^{-1}\)) identifies specific bonds, the fingerprint region is unique to an individual compound. Unknown molecules are positively confirmed by computer comparison of their fingerprint spectrum against a library of known reference spectra.

3. Mass Spectrometry in Organic Analysis

In electron impact mass spectrometry, an intact organic molecule loses one electron to form a radical cation known as the molecular ion (\(M^{\bullet+}\)):

\[ \text{M} + e^- \rightarrow M^{\bullet+} + 2e^- \]

The \(m/z\) ratio of the molecular ion peak gives the relative molecular mass (\(M_r\)) of the intact molecule.

High-Resolution Mass Spectrometry

Standard low-resolution mass spectrometry measures \(m/z\) to the nearest integer. High-resolution spectrometers measure masses to four or five decimal places. This differentiates compounds that have identical integer \(M_r\) values:

  • Propanal (\(\text{C}_3\text{H}_6\text{O}\)): Accurate mass = \(3(12.00000) + 6(1.00782) + 15.99491 = \mathbf{58.04183}\).
  • Butane (\(\text{C}_4\text{H}_{10}\)): Accurate mass = \(4(12.00000) + 10(1.00782) = \mathbf{58.07820}\).

4. The [M+1] Peak and Carbon Counting

Examiner Trap: The [M+1] Peak is NOT an Impurity

Directly to the right of the molecular ion peak \(M\), a tiny peak is observed at \([M+1]\). This peak arises from the natural \(1.1\%\) abundance of the carbon-13 isotope (\(^{13}\text{C}\)) in the molecule.

The number of carbon atoms (\(n\)) in the molecule can be deduced by comparing the peak heights:

\[ n = \frac{\text{Height of }[M+1]\text{ peak}}{\text{Height of }M\text{ peak}} \times \frac{100}{1.1} \]

5. Chemical Identification Tests Summary

Functional Group Test Reagent Positive Observation
Alkene (\(\text{C}=\text{C}\)) Bromine water (\(\text{Br}_2(aq)\)) Orange to colourless.
Halogenoalkane (\(\text{R}-\text{X}\)) Warm with \(\text{NaOH}(aq)\), acidify with \(\text{HNO}_3\), add \(\text{AgNO}_3(aq)\) \(\text{AgCl}\) white ppt; \(\text{AgBr}\) cream ppt; \(\text{AgI}\) yellow ppt.
Alcohol (\(-\text{OH}\)) Acidified potassium dichromate(VI) Orange to green (\(1^\circ\) and \(2^\circ\) alcohols).
Aldehyde (\(-\text{CHO}\)) Tollens' reagent (warm) Silver mirror formed on glass.
Carboxylic acid (\(-\text{COOH}\)) Sodium hydrogencarbonate (\(\text{NaHCO}_3(aq)\)) Effervescence; gas turns limewater cloudy (\(\text{CO}_2\)).

6. Practice Questions

Practice Problem (4 Marks)
An organic compound X has the molecular formula C3H6O. Its infrared spectrum shows a sharp, strong absorption band at 1715 cm^-1 and no broad absorption between 3200 and 3600 cm^-1. When X is warmed with Tollens' reagent, no silver mirror forms. Deduce the structure of X, name the compound, and justify your answer using the data provided.

Deduction:

  • Absorption at \(1715\text{ cm}^{-1}\) indicates the presence of a carbonyl group (\(\text{C}=\text{O}\)). (1 mark)
  • Absence of absorption between \(3200-3600\text{ cm}^{-1}\) confirms compound X is not an alcohol. (1 mark)
  • Negative Tollens' test confirms X is a ketone, not an aldehyde. (1 mark)
  • With formula \(\text{C}_3\text{H}_6\text{O}\), the only possible ketone is propanone (\(\text{CH}_3\text{COCH}_3\)). (1 mark)