Quick questions on Mass spectrometry and infrared spectroscopy: VCE Chemistry Unit 4

A mass spectrometer ionises the molecule (usually with electron impact), accelerates the ions in a vacuum, separates them by mass-to-charge ratio (m/z), and detects them. The output is a stick spectrum of relative abundance against m/z. Three things to look at in any organic mass spectrum:

The peak at the highest m/z (ignoring the small isotope peaks above it) corresponds to the whole molecule with a single electron removed. Its m/z equals the molecular mass (in u). For an unknown C3H8O, M+ appears at m/z = 60.

A small peak at one mass unit above the molecular ion (the M+1 peak) is due to natural ^13C (about 1.1% abundance). Its relative intensity is roughly 1.1% per carbon atom, which can be used to estimate the number of carbons.

The high-energy collisions break the molecular ion into smaller cations. Common diagnostic losses:

State the position (in $\text{cm}^{-1}$) of the IR absorption that would distinguish a carboxylic acid from an ester. [2 marks]

A compound has $M^{+} = 88$, $M+1 = 4.4\%$ of $M$, no $M+2$. IR shows strong absorption at $1730 \, \text{cm}^{-1}$ and $1200 \, \text{cm}^{-1}$, no broad O-H. (a) Calculate carbons.

A compound has $M^{+} = 122$, $M+2$ equal in height to $M$. IR: $1690 \, \text{cm}^{-1}$, $3050 \, \text{cm}^{-1}$, $1600 \, \text{cm}^{-1}$. (a) State what the $M+2$ pattern implies.