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Measurement and Data Processing - Spectroscopic identification of organic compounds

Grade 11IBChemistry

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

🔑Concepts

The Index of Hydrogen Deficiency (IHDIHD) provides the number of rings and/or π\pi bonds in a molecule. For a compound with the general formula CcHhNnOoXxC_c H_h N_n O_o X_x, the IHDIHD is calculated to determine degrees of unsaturation.

Mass Spectrometry (MSMS) is used to determine the relative atomic and molecular masses. The molecular ion peak (M+M^+) gives the molar mass of the compound, while fragmentation patterns (e.g., m/z=15m/z = 15 for CH3+CH_3^+ or m/z=29m/z = 29 for C2H5+C_2H_5^+) help deduce the structure.

Infrared (IRIR) Spectroscopy identifies functional groups based on the absorption of infrared radiation, which causes specific bonds to vibrate. Key absorptions include OHO-H (broad peak at 32003600 cm13200-3600 \text{ cm}^{-1}) and C=OC=O (strong peak at 17001750 cm11700-1750 \text{ cm}^{-1}).

Proton Nuclear Magnetic Resonance (1H^1H NMR) provides information on the chemical environment of hydrogen atoms. The chemical shift (δ\delta) is measured in parts per million (ppmppm) relative to the standard Tetramethylsilane (TMSTMS), which is assigned δ=0\delta = 0.

In high-resolution 1H^1H NMR, the splitting pattern of a signal (multiplicity) follows the n+1n+1 rule, where nn is the number of protons on adjacent carbon atoms.

The fingerprint region in an IRIR spectrum is the area below 1500 cm11500 \text{ cm}^{-1} which is unique to every compound and is used for specific identification by comparison with a database.

📐Formulae

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

E=hνE = h \nu

c=νλc = \nu \lambda

ν~=1λ\tilde{\nu} = \frac{1}{\lambda}

💡Examples

Problem 1:

A compound has a molecular formula of C4H8OC_4H_8O. Calculate its IHDIHD and predict the IRIR absorption if it contains a carbonyl group.

Solution:

IHD=12(2(4)+28)=1IHD = \frac{1}{2}(2(4) + 2 - 8) = 1. The IRIR spectrum would show a strong absorption peak between 17001750 cm11700-1750 \text{ cm}^{-1}.

Explanation:

An IHDIHD of 11 indicates either one double bond or one ring. Since the problem specifies a carbonyl group (C=OC=O), this accounts for the double bond. Carbonyl groups have a characteristic strong stretch in the 17001750 cm11700-1750 \text{ cm}^{-1} region.

Problem 2:

In the 1H^1H NMR spectrum of ethanol (CH3CH2OHCH_3CH_2OH), describe the splitting pattern for the CH2CH_2 group.

Solution:

The CH2CH_2 group will appear as a quartet.

Explanation:

According to the n+1n+1 rule, the number of peaks is determined by the neighbors. The CH2CH_2 group is adjacent to a CH3CH_3 group (n=3n=3 protons). Therefore, the splitting is 3+1=43+1 = 4 (a quartet). Note: In standard IB conditions, the OHOH proton does not usually cause splitting of adjacent protons due to rapid exchange.

Problem 3:

Identify the fragment responsible for a peak at m/z=45m/z = 45 in the mass spectrum of a carboxylic acid.

Solution:

COOH+COOH^+

Explanation:

The molar mass of a carboxyl fragment is calculated as 12.01(C)+(16.00×2)(O)+1.01(H)4512.01 (C) + (16.00 \times 2) (O) + 1.01 (H) \approx 45. In mass spectrometry, the lost fragment or the ion detected must account for this mass-to-charge ratio.

Spectroscopic identification of organic compounds Revision - Grade 11 Chemistry IB