Chemical Detectives

Question 1

mass spectroscopy (electron impact ionisation (EI))

Answer 1

• determines the molecular weight of a compound
• sample is ionised by an electron beam
• charged molecules are accelerated and passed through a magnetic region where they are deflected - degree of deflection (m/z or mass:charge ratio) allows mass to be determined

Question 2

resolution and isotope distributions

Answer 2

high or low - important for molecules of similar mass (sometimes can only tell the difference at high res)

isotope distribution charts show the relative intensity if different isotopes of an atom and you can determine the ratio

Question 3

microanalysis and molecular formula determination

Answer 3

given percentages from mass spectrometry and can work out the molecular formula and then the empirical formula

Question 4

index of hydrogen deficiency (double bond equivalents)

Answer 4

presence or rings or unsaturation (double or triple bonds) can be obtained from molecular formula

index = (Hreference - H molecule)/2

if there are molecules from group 15, add 1H to reference

group 16 require no correction

group 17, subtract 1H from reference

reference is CnH2n+2

Question 5

fundamentals of IR spectroscopy

Answer 5

absorption of IR provides info on functional groups

• absorbed directly by bonds with a dipole moment (IR active bonds)
• vibrations (stretching, scissoring, rocking, twisting and wagging) give rise to absorption
• dips in graphs indicate absorption
• hard to read beyond 1400cm-1

Question 6

aldehydes and ketones

Answer 6

aldehydes end in al, ketones in one

both have a C=O carbonyl group

aldehydes have an R group and an H (CHO), ketones have Cs on either side (CO).

• trigonal planar around a central carbon
• carbon has sp2 hybridisation - unhybridised orbital forms a pi bond
• CO is polar
• alcohols are oxidised to form ketones and aldehydes

Question 7

rules of IR

Answer 7

• bending usually occurs at a lower wavelength than stretching
• stronger the bond, the higher the wavenumber
• triple bonds have higher wavenumber than double and single
• wavenumber decreases with an increase of mass in Y (C-Y)
• hybridisation causes stretching to decrease sp to sp3 (higher wavenumber for triple bond)

Question 8

carboxylic acids

Answer 8

COOH
weak acids
form when an alcohol becomes an aldehyde and then an acid
eg. ethanoic acid

Question 9

ester

Answer 9

odours
COO
formed when a carboxylic acid reacts with an alcohol
eg. methyl ethanoate

Question 10

NMR

Answer 10

• functional groups
• number of different groups
• connectivity of different groups

looks at chemical shifts
environments appear as different peaks
size of peaks indicates number of hydrogens

Question 11

why do we not observe splitting in C13 NMR?

Answer 11

low abundance of C13

proton decoupling in the NMR - one sharp peak instead of many (for all the H)