X-ray crystallography is a method of determining crystal structures. X-rays are fired at a crystal sample and diffract to create three dimensional patterns and pictures that crystallographers can read and decipher to give information about the arrangements of particles and electron density in substances:

Bond lengths and bond angles are determined with x-ray crystallography.

The other techniques give a lot of information about the structure of organic compounds, but do not tend to give direct information about bond lengths and angles.

X-ray crystallography is the correct answer.

Infrared (IR) spectroscopy is used to identify the bonds that are present in a molecule. This allows the identification of potential functional groups.

The presence of a carbonyl group is therefore the correct answer.

The number of hydrogen environments could be determined by ¹H-NMR; The molecular mass by Mass Spectrometry; Bond lengths and angles by X-ray Crystallography.

¹H-NMR spectroscopy is the correct answer, as this would be the most effective technique: 1-chloropropane would exhibit three hydrogen environments (3:2:2) and 2-chloropropane would exhibit two hydrogen environments (6:1) as it is symmetrical (also the splitting patterns would be very different).

Infrared (IR) spectroscopy is used to identify the bonds that are present in a molecule. This allows the identification of potential functional groups, but would not easily distinguish between 2-chloropropane and 1-chloropropane.

Mass spectrometry, in the hands of skilled practitioner could be used, as the fragmentation patterns of the molecules would be different, but it would not be the most effective technique.

Reaction with silver nitrate solution, AgNO₃(aq) and ethanol, would generate a white precipitate (AgCl(s)) but would not distinguish between 2-chloropropane and 1-chloropropane.

There are two peaks in the spectrum, so the compound has two hydrogen environments.

The peak at approximately 1.2ppm is a triplet because it is split by two nearest neighbours (according to the n+1 rule); presumably a CH₂ group on the next carbon.

The peak at approximately 2.4ppm is a quartet because it is split by three nearest neighbours (according to the n+1 rule); presumably a CH₃ group on the next carbon.

Propanenitrile (CH₃CH₂CN) has three hydrogens in the CH₃ group that will split the neighbouring hydrogen environment (CH₂ group) at 2.4ppm into a quartet, and vice versa to give the triplet at 1.2ppm.

Thus CH₃CH₂CN is the correct answer.

CH₃CH₂OH and CH₃CHCH₂ both have three hydrogen environments. CH₃COCH₂Cl has two hydrogen environments, but they are separated by an C=O group, are therefore not neighbouring, so would not split one-another.

The NMR spectrum shows three hydrogen environments, with hydrogen ratio of 2:2:3. The splitting pattern suggests that the CH₂ and CH₃ are neighbouring (split into a quartet and triplet respectively), but the two protons at ppm 6.2 are not split by neighbouring protons.

Therefore CH₃CH₂CONH₂ is the correct answer.

CH₃CH₂CH₂OH has four hydrogen environments; CH₃CH₂CH₂Cl would have a different splitting pattern (all environments would be split as they are all neighbouring); CH₃CO₂CH₂CH₃ would show peak integrals of 3:2:3 (rather than 2:2:3).

¹H-NMR spectroscopy summary:

Peaks: The number of peaks tells us the number of different hydrogen environments.

Ratio: The integration trace or whole number written adjacent to each peak tells us the ratio of the number of hydrogens in each environment.

Splitting: The spin-spin splitting pattern of each peak tells us the number of hydrogens in neighbouring environments (on adjacent atoms).

Shift: The chemical shift table (data book) can help us with the types of hydrogens (functional groups etc.) in the molecule.

Therefore 2 and 3 only is the correct answer. Since ¹H-NMR will not always tell us the total number of hydrogen atoms in a molecule (as sometimes we aere just given the ratio).

Tetramethylsilane, (CH₃)₄Si, is used as a reference for ¹H-NMR spectroscopy because:

• All 12 hydrogen atoms are in the same environment so it gives one large sharp peak.
• The chemical shift is low (set as zero) relative other proton environments.
• It is unreactive, so does not intefere with the sample being tested.
• It is volatile (evaporates easily) so can easily be removed from a sample.

Therefore 1 and 2 only is the correct answer.

The 3D image of propan-2-ol in shown below.

The hydrogens in the two CH₃ groups on the ends of the molecule are all in the same environment (the molecule is symmetrical). So there are three hydrogen environments in total (not four).

The ratio of hydrogens in the environments is 1:1:6 (CH, OH and the two CH₃ groups) so the integration trace will show a 1:1:6 ratio.

The H on the middle carbon will split the two CH₃ (six hydrogen) peak into two (doublet), according to the n+1 rule. The two CH₃ groups (6 hydrogens) will spilt the single C-H into a multiplet (theoretically into 7).

Remember that OH hydrogens tend not to split or be split. So the hydrogen on the OH will be a singlet peak.

Therefore 2 only is the correct answer.