Unit 29: Spectroscopy-1 — Long Questions

Empirical formula moles C = 88.89/12 = 7.41; moles H = 11.1/1 = 11.1. Ratio C:H = 7.41:11.1 = 1:1.5 = 2:3, giving empirical formula C₂H₃ (empirical mass = 27). Molecular formula: n = M/empirical mass = 54/27 = 2, so molecular formula = C₄H₆ (M = 54, matching the molecular ion peak). IHD of C₄H₆ = (2×4+2−6)/2 = 2, consistent with reacting with H₂ in a 1:2 molar ratio (2 mol H₂ add across 2 C=C double bonds, confirming two reducible double bonds rather than a ring). A C₄H₆ compound with two conjugated double bonds that is a well-known vehicle-exhaust pollutant and recognised human carcinogen is buta-1,3-diene, CH₂=CH–CH=CH₂. It reacts completely with 2 mol H₂ (Ni catalyst) to give butane, CH₃CH₂CH₂CH₃, consistent with the 1:2 reaction ratio given.

A molecule can contain three kinds of electrons: σ electrons (in σ bonds), π electrons (in π bonds of unsaturated compounds) and n electrons (non-bonding lone pairs). In order of increasing energy, the molecular orbitals are: σ (bonding, lowest energy) < π (bonding) < n (non-bonding, lone pairs) < π* (antibonding) < σ* (antibonding, highest energy, normally empty). When a molecule absorbs UV/visible radiation of the correct energy, an electron is promoted from a filled bonding or non-bonding orbital to an empty antibonding orbital. The important transitions possible are: (1) σ→σ*: a σ electron is excited to the σ* orbital — needs the highest energy, occurs in the (far) UV region; (2) π→π*: a π electron is excited to the π* orbital — occurs in compounds with double/triple bonds, and its λmax increases with conjugation; (3) n→π*: a non-bonding electron is excited to the π* orbital — generally needs the least energy among the common transitions, so occurs at the longest wavelength; (4) n→σ*: a non-bonding electron is excited to the σ* orbital. Which transitions are possible depends on what bonds/lone pairs the compound has: a saturated compound with lone pairs (e.g. an alcohol or amine) can show n→σ*; an unsaturated compound (e.g. an alkene) can show π→π*; and an unsaturated compound with lone pairs (e.g. a carbonyl compound) can show both π→π* and n→π*.

Atomic spectra are of two types: (1) Emission spectrum: produced by heating a low-pressure gas, which supplies energy to excite electrons from lower to higher atomic energy levels. When the excited electrons fall back to lower levels, each transition releases a photon of a specific wavelength, seen as a set of distinct coloured lines on a black (dark) background. (2) Absorption spectrum: produced by passing white light through a cool, low-pressure gas. Electrons in the ground state absorb only photons whose energy exactly matches the energy gap to a higher level; the resulting spectrum shows dark lines, at the same wavelengths as the element's emission lines, superimposed on an otherwise continuous coloured/bright background. An absorption spectrum is essentially the photographic negative of the corresponding emission spectrum, and since each element's set of spectral lines is unique, both types can be used to identify elements (e.g. in stars).

In a mass spectrometer, atoms of the sample are bombarded with high-energy electrons, knocking an electron out of each atom and forming positively charged ions. These ions are separated and detected according to their mass-to-charge ratio (m/e) and recorded as a spectrum of % abundance versus m/e. Since each isotope of an element has a different mass, it produces a separate peak; the m/e value of each peak directly gives the isotopic mass of that isotope, and the height (or % abundance) of each peak gives its relative natural abundance. The (weighted-average) relative atomic mass of the element is then calculated using Ar = Σ(isotopic mass × isotopic abundance)/100. Example: the mass spectrum of boron shows two peaks — at m/e = 10 with 19.9% abundance (¹⁰B) and at m/e = 11 with 80.1% abundance (¹¹B). Ar(B) = [(19.9×10) + (80.1×11)]/100 = 1080.1/100 = 10.80, which is the relative atomic mass of boron listed on the periodic table.