Analytical Techniques (OCR A Level Chemistry A): Flashcards

Exam code: H432

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  • Define fingerprint region

Cards in this collection (18)

  • Define fingerprint region

    A fingerprint region is the region of an IR spectrum below approximately 1500 cm−1 that is unique to each molecule, used to identify a compound by comparison with a spectral database.

  • What bond causes the broad absorption between 2500 and 3300 cm−1 in an IR spectrum?

    The O–H bond of a carboxylic acid, which produces a characteristically broad absorption in this region.

  • True or False?

    A molecule absorbs infrared radiation only if it has a permanent dipole that changes as it vibrates.

    True.

    Molecules such as H2 and O2 are IR inactive because they are symmetrical and have no changing dipole during vibration.

  • In IR spectroscopy, a sharp absorption between 1630 and 1820 cm−1 indicates a .......... bond, while a broad absorption between 3200 and 3600 cm−1 indicates an .......... group in an alcohol.

    In IR spectroscopy, a sharp absorption between 1630 and 1820 cm−1 indicates a C=O bond, while a broad absorption between 3200 and 3600 cm−1 indicates an O–H group in an alcohol.

  • How is infrared spectroscopy used in roadside breathalysers?

    IR radiation is passed through the exhaled breath; the characteristic bonds of ethanol absorb IR at specific wavenumbers, and the degree of absorption indicates the concentration of ethanol present.

  • True or False?

    The fingerprint region is useful for distinguishing between different members of the same homologous series.

    True.

    All members of a homologous series show similar bond absorptions, but each has a unique fingerprint region that can be matched to a database.

  • What is wavenumber and what are its units?

    Wavenumber is the reciprocal of wavelength and is measured in cm−1. It is used as the x-axis unit in IR spectra.

  • Define molecular ion (M+)

    A molecular ion (M+) is the ion formed when a molecule loses one electron in the mass spectrometer; the m/z value of the molecular ion peak equals the relative molecular mass of the compound.

  • What causes the [M+1] peak in a mass spectrum?

    The natural abundance of carbon-13 isotopes; the more carbon atoms in the molecule, the larger the [M+1] peak relative to the molecular ion peak.

  • True or False?

    The base peak in a mass spectrum corresponds to the fragment ion with the highest m/z value.

    False.

    The base peak is the most abundant (tallest) fragment ion peak; the highest m/z peak is the molecular ion peak.

  • In the mass spectrum of an alcohol, a peak at 18 below the molecular ion peak is due to the loss of a .......... molecule, and a peak at m/z = 31 corresponds to the .......... fragment.

    In the mass spectrum of an alcohol, a peak at 18 below the molecular ion peak is due to the loss of a water molecule, and a peak at m/z = 31 corresponds to the CH2OH+ fragment.

  • Why is fragmentation analysis needed in addition to the molecular ion peak?

    Different compounds may share the same molecular mass; fragmentation patterns reveal structural differences that identify the specific compound.

  • True or False?

    A peak at m/z = 29 in a mass spectrum can correspond to the C2H5+ fragment ion.

    True.

    m/z = 29 is a characteristic fragment for C2H5+ (mass = 29), commonly seen in the fragmentation of straight-chain carbon compounds.

  • Define empirical formula

    An empirical formula is the simplest whole-number ratio of atoms of each element in a compound, determined from elemental analysis data.

  • How do you determine the molecular formula of a compound from its empirical formula?

    Calculate the empirical mass and divide the relative molecular mass (from mass spectrometry) by it to find the multiplier, then scale up the empirical formula.

  • True or False?

    A compound with molecular formula C3H6O and a sharp IR absorption around 1750 cm−1 is likely to be an aldehyde or ketone.

    True.

    The sharp absorption at ~1750 cm−1 indicates a C=O (carbonyl) group, which is present in both aldehydes and ketones.

  • To distinguish between propanal and propanone using mass spectrometry, propanone would show a characteristic peak at m/z = .........., whereas propanal would show a peak at m/z = .......... corresponding to the CHO+ fragment.

    To distinguish between propanal and propanone using mass spectrometry, propanone would show a characteristic peak at m/z = 15 (CH3+), whereas propanal would show a peak at m/z = 29 corresponding to the CHO+ fragment.

  • Why should IR spectroscopy and mass spectrometry be used together when identifying an unknown organic compound?

    IR identifies functional groups present while MS confirms the relative molecular mass and fragmentation pattern; used together they provide complementary evidence for the full structure.

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