krit.club logo

Measurement and Data Processing - Spectroscopic identification of organic compounds

Grade 12IBChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

The Index of Hydrogen Deficiency (IHD) is used to determine the number of rings or π\pi bonds in a molecule. Each IHD unit represents one ring or one double bond, while an IHD of 22 represents a triple bond.

•

Infrared (IR) Spectroscopy identifies functional groups by measuring the absorption of infrared radiation, which causes characteristic bond vibrations. Key regions include 1630−1820 cm−11630-1820\text{ cm}^{-1} for C=OC=O and 3200−3600 cm−13200-3600\text{ cm}^{-1} for O−HO-H (hydrogen bonded).

•

Mass Spectrometry (MS) determines the relative molecular mass (MrM_r) from the molecular ion peak (M+M^+). Fragmentation patterns, such as the loss of 1515 (CH3+CH_3^+) or 1717 (OH+OH^+), help deduce the structure.

•

Proton Nuclear Magnetic Resonance (1H^{1}H NMR) provides information on the environment of hydrogen atoms. The number of signals indicates the number of unique proton environments, the integration ratio shows the relative number of protons in each environment, and the chemical shift (δ\delta) indicates the chemical environment relative to Tetramethylsilane (TMS, δ=0\delta = 0).

•

Spin-spin coupling in 1H^{1}H NMR results in the splitting of signals into multiplets. The n+1n+1 rule states that a signal for a proton with nn neighboring protons on adjacent carbon atoms will be split into n+1n+1 peaks (e.g., a triplet indicates 22 neighboring protons).

📐Formulae

IHD=12(2c+2−h−x+n)IHD = \frac{1}{2}(2c + 2 - h - x + n)

E=hν=hcλE = h\nu = \frac{hc}{\lambda}

δ=νsample−νrefνref×106\delta = \frac{\nu_{sample} - \nu_{ref}}{\nu_{ref}} \times 10^6

M→M++e−M \to M^+ + e^-

💡Examples

Problem 1:

Determine the Index of Hydrogen Deficiency (IHD) for a compound with the molecular formula C6H5NO2C_6H_5NO_2.

Solution:

Using the formula: IHD=12(2(6)+2−5+1)=12(12+2−5+1)=102=5IHD = \frac{1}{2}(2(6) + 2 - 5 + 1) = \frac{1}{2}(12 + 2 - 5 + 1) = \frac{10}{2} = 5.

Explanation:

In the formula, c=6,h=5,n=1,x=0c=6, h=5, n=1, x=0. Oxygen is ignored in the IHD calculation. An IHD of 55 suggests a benzene ring (IHD=4IHD=4) plus one additional double bond or ring.

Problem 2:

An unknown compound with formula C3H6OC_3H_6O shows a strong IR absorption at 1715 cm−11715\text{ cm}^{-1} and a single sharp singlet in its 1H^{1}H NMR spectrum. Identify the compound.

Solution:

The compound is Propanone (CH3COCH3CH_3COCH_3).

Explanation:

The absorption at 1715 cm−11715\text{ cm}^{-1} indicates a carbonyl group (C=OC=O). With the formula C3H6OC_3H_6O, it could be propanal or propanone. A single singlet in 1H^{1}H NMR means all 66 protons are in the same environment. In propanone, the two methyl groups are equivalent and have no neighboring protons, resulting in one singlet. Propanal would show three different signals with splitting.

Spectroscopic identification of organic compounds Revision - Grade 12 Chemistry IB