Organic Analysis & Spectroscopy

In the mass spectrum of an organic compound, what does the molecular ion peak (M⁺) represent?

• AA fragment formed when the molecule breaks apart
• BThe whole molecule that has lost one electron to form a positive ion✓
• CThe most abundant ion in the spectrum
• DA molecule that has gained an electron

Answer: The molecular ion M⁺ is formed when the intact molecule loses a single electron during ionisation. Its mass-to-charge ratio (for a 1+ ion) equals the relative molecular mass of the compound, so it is used to determine Mr.

The mass spectrum of a ketone shows its molecular ion at m/z = 58 and a strong fragment peak at m/z = 43. Which fragment ion is most likely responsible for the m/z = 43 peak?

• A[CH₃]⁺
• B[C₂H₅]⁺
• C[CH₃CO]⁺ (acylium ion)✓
• D[OH]⁺

Answer: The compound (Mr 58) is propanone, CH₃COCH₃. A common fragmentation is loss of a methyl group (15) to give the acylium ion [CH₃CO]⁺, which has m/z = 43 (58 − 15 = 43). The complementary [CH₃]⁺ fragment appears at m/z = 15.

A small peak is often seen at one mass unit above the molecular ion peak (the 'M+1' peak). What is the main cause of this M+1 peak?

• AMolecules that have gained an extra proton
• BThe presence of ³⁵Cl atoms
• CTwo molecules sticking together
• DThe presence of a small percentage of the ¹³C isotope in the molecule✓

Answer: Carbon exists mostly as ¹²C but about 1.1% is ¹³C. Molecules containing one ¹³C atom have a mass one unit higher, producing the small M+1 peak. The relative height of M+1 can be used to estimate the number of carbon atoms.

In infrared spectroscopy, a strong, sharp absorption at around 1700 cm⁻¹ is characteristic of which bond?

• AC=O (carbonyl)✓
• BO–H in a carboxylic acid
• CC–H
• DC≡N

Answer: The carbonyl group (C=O) found in aldehydes, ketones, carboxylic acids and esters gives a strong absorption in the region ~1680–1750 cm⁻¹ (often quoted as 'around 1700 cm⁻¹'), making it a key diagnostic peak.

Two isomers, propan-1-ol and propanoic acid, are to be distinguished by infrared spectroscopy. Which feature most clearly identifies the carboxylic acid?

• AA sharp C–H absorption near 3000 cm⁻¹
• BA very broad O–H absorption from about 2500 to 3300 cm⁻¹ together with a strong C=O peak near 1700 cm⁻¹✓
• CA single sharp O–H absorption near 3600 cm⁻¹ only
• DThe complete absence of any absorption above 1500 cm⁻¹

Answer: A carboxylic acid shows a very broad O–H stretch (~2500–3300 cm⁻¹, due to hydrogen bonding) AND a strong C=O stretch near 1700 cm⁻¹. The alcohol has a C=O–free spectrum with a broad O–H around 3200–3550 cm⁻¹ but no carbonyl peak, so the C=O is the decisive difference.

In ¹H NMR spectroscopy, what is the purpose of adding tetramethylsilane, Si(CH₃)₄ (TMS), to the sample?

• AIt speeds up the relaxation of the nuclei
• BIt dissolves polar samples
• CIt acts as the standard reference, defined as a chemical shift (δ) of 0 ppm✓
• DIt removes oxygen from the sample

Answer: TMS is used as the reference standard; its single peak is defined as δ = 0 ppm and all other chemical shifts are measured relative to it. TMS is chosen because it is inert, volatile, non-toxic, gives one strong signal, and its protons are more shielded than almost all others.

In the high-resolution ¹H NMR spectrum of ethanol (CH₃CH₂OH, ignoring coupling to the OH proton), the CH₃ protons appear as which splitting pattern?

• AA singlet
• BA doublet
• CA quartet
• DA triplet✓

Answer: By the n+1 rule, the CH₃ protons are split by the 2 equivalent protons on the adjacent CH₂ group: n = 2, so 2 + 1 = 3 peaks, i.e. a triplet. (Reciprocally, the CH₂ protons are split into a quartet by the 3 CH₃ protons.)

A compound with molecular formula C₃H₆O₂ gives a ¹H NMR spectrum with three peaks in the integration ratio 3 : 2 : 1, the peak at δ ≈ 11.5 ppm (ratio 1) being a broad singlet. Which compound is consistent with this data?

• APropanoic acid, CH₃CH₂COOH✓
• BMethyl methanoate, HCOOCH₃
• CPropanone, CH₃COCH₃
• DHydroxypropanal, HOCH₂CH₂CHO

Answer: Propanoic acid CH₃CH₂COOH has three proton environments: CH₃ (3H), CH₂ (2H) and the acidic COOH (1H), giving a 3:2:1 ratio. The COOH proton appears far downfield (δ ≈ 11–12 ppm) as a broad singlet, matching the data. Propanone has only one environment (one peak).

How many peaks (carbon environments) would you expect in the ¹³C NMR spectrum of propan-2-ol, (CH₃)₂CHOH?

• A1
• B2✓
• C3
• D4

Answer: Propan-2-ol has two carbon environments: the two CH₃ groups are equivalent (related by symmetry) and count as one environment, and the central CH carbon is the second. So the ¹³C NMR spectrum shows 2 peaks.

In the ¹³C NMR spectrum of an organic compound, why is spin–spin splitting (coupling between adjacent carbons) normally not observed, so that each environment usually gives a single peak?

• A¹³C nuclei have no nuclear spin
• BCarbon atoms never have neighbouring atoms
• CThe low natural abundance of ¹³C (~1.1%) makes adjacent ¹³C–¹³C neighbours very rare, and the spectra are routinely proton-decoupled✓
• DAll carbon environments resonate at exactly the same frequency

Answer: Because ¹³C is only ~1.1% abundant, the chance of two ¹³C atoms being adjacent is very small, so C–C coupling is negligible. Spectra are also recorded with proton (broadband) decoupling, removing ¹H–¹³C coupling. Each distinct carbon environment therefore appears as a single line.

In thin-layer chromatography (TLC), a spot travels 3.6 cm while the solvent front travels 4.8 cm from the baseline. What is the Rf value of this component?

• A0.13
• B1.33
• C0.48
• D0.75✓

Answer: Rf = distance moved by spot ÷ distance moved by solvent front = 3.6 ÷ 4.8 = 0.75. Rf values lie between 0 and 1 and, under fixed conditions, are characteristic of a particular substance.

In gas chromatography (GC), what does the retention time of a component depend on, and how is the amount of each component usually determined?

• AIt depends on the balance between the component's solubility in the stationary phase and its volatility (tendency to stay in the mobile gas phase); the amount is proportional to the area under its peak✓
• BIt depends on the molecule's colour; the amount is found from the spot intensity
• CIt depends only on the molecular mass; the amount is read directly off the retention time
• DIt depends on the boiling point of the carrier gas; the amount is found from the baseline width

Answer: In GC the retention time reflects how strongly a component is retained by (dissolves in) the liquid stationary phase versus how readily it stays in the gaseous mobile phase — i.e. its relative solubility and volatility. The area under each peak is proportional to the amount of that component present.