(a) From the elemental analysis the empirical formula of Compound P is C₅H₉O₂Cl.

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(b) The M⁺ peak at m/z = 136 leads to the conclusion that the molar mass containing just ³⁵Cl is 136 g mol^-1 and with the  less abundant isotope of chlorine,  ³⁷Cl it is 138 g mol^-1 so the molecular mass is the same as the empirical mass and the molecular formula of Compound P is therefore C₅H₉O₂Cl. The fragment at m/z = 107 suggests loss of an ethyl group to leave C₃H₄O₂³⁵Cl⁺ and the fragment at m/z = 73 could be due to loss of CH₃CH³⁵Cl to leave C₃H₅O₂⁺. The lost CH₃CH³⁵Cl⁺ ion could then be responsible for the peak at m/z = 63.

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(c) The absorptions centred just below 3000 cm^-1 are due to the C–H bonds and the sharp absorption at about 1750 cm^-1  shows the presence of a carbonyl group, C=O. There are no broad absorptions due to an O–H bond but there is an absorption at about 1190 cm^-1 which could be due to a C–O bond making Compound P an ester.  C–Cl bonds vibrate lower than 1000 cm^-1 so cannot be seen on this spectrum.

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(d) The ¹H NMR spectrum data can be summarised as follows:

Normally it is possible to distinguish between isomers of esters from their ¹H  NMR spectra. However the presence of the chlorine atom makes it difficult in this case as chlorine, like oxygen, also affects the chemical shift. The splitting patterns and integration traces for CH₃CHClCOOCH₂CH₃ and CH₃CH₂COOCHClCH₃ will be the same so the compound could be an ester derived either from 2-chloropropanoic acid or from propanoic acid. The methyl group with the signal split into a triplet is bonded to a –CH₂− group and the methyl group split into a doublet is bonded to a –CH−group however the shifts for the signals for the two methyl groups are very close (1.31 ppm and 1.67 ppm) so difficult to assign. Similarly the two quartets with integration traces of one and two are equally close and both lie within the 3.7 – 4.7 ppm range of the shift for –CO-OCH₂− or –CO-OCHCl−.

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Therefore from the information given Compound P could be either ethyl 2-chloropropanoate or 1-chloroethyl propanoate. If you had the compound in the lab you could easily identify which isomer it is by hydrolysing it and seeing whether chloropropanoic acid or propanoic acid is obtained.

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