Compound L is a colourless, fruity-smelling liquid used as a flavouring in the food industry. In its mass spectrum, compound L shows a molecular ion peak at m/z = 88 and a prominent fragment ion peak at m/z = 57. Its infrared spectrum shows a strong, sharp absorption at 1735 cm^…

Answer: ((a)) Mr = 88 ((b)) Ester: the C=O absorption at 1735 cm^-1 falls in the ester range, and the C–O absorption at 1200 cm^-1 confirms a C–O single bond is also present (which a ketone would not show); the absence of any absorption at 2500–3000 or 3200–3600 rules out a carboxylic acid and an alcohol r… ((c)) Mass lost = 88 - 57 = 31, corresponding to loss of 3 (methoxy radical), leaving the acylium fragment CH3CH2CO^+ at m/z = 57. ((d)) CH3CH2COOCH3, methyl propanoate.. Part (a): Relative molecular mass from the molecular ion The molecular ion, M^+, is the whole molecule having lost just one electron, so its m/z value gives the relative molecular mass directly: Mr(L) = 88 Part (b): Identifying the functio…

AS Chemistry 9701: Combining Mass Spectrometry and IR to Identify an Ester

Bond	Functional group	Wavenumber / cm⁻¹
C–O	hydroxy, ester	1040–1300
C=O	ester	1710–1750
O–H	carboxylic acid	2500–3000
O–H	alcohol	3200–3600

Bond

Functional group

Wavenumber / cm⁻¹

C–O

hydroxy, ester

1040–1300

C=O

ester

1710–1750

O–H

carboxylic acid

2500–3000

O–H

alcohol

3200–3600

Part (a): Relative molecular mass from the molecular ion

The molecular ion, $\text{M}^+$ , is the whole molecule having lost just one electron, so its $m/z$ value gives the relative molecular mass directly:

$M_r(\text{L}) = 88$

Part (b): Identifying the functional group from the infrared spectrum

Two absorptions are present, and two are conspicuously absent:

$1735\ \text{cm}^{-1}$ lies in the C=O ester range ( $1710$ – $1750\ \text{cm}^{-1}$ ).
$1200\ \text{cm}^{-1}$ lies in the C–O range ( $1040$ – $1300\ \text{cm}^{-1}$ ), confirming a carbon–oxygen single bond is also present.
There is no absorption in $2500$ – $3000\ \text{cm}^{-1}$ , ruling out a carboxylic acid (which would show a strong, broad O–H absorption there).
There is no absorption in $3200$ – $3600\ \text{cm}^{-1}$ , ruling out an alcohol (which would show an O–H absorption there).

A ketone or aldehyde would show the C=O absorption but not the C–O absorption (it has no C–O single bond), so the combination of both the C=O and the C–O bands, together with the absence of any O–H band, identifies compound L as an ester.

Part (c): Interpreting the fragment ion at m/z = 57

The neutral fragment lost has mass:

$88 - 57 = 31$

A mass of $31$ matches the methoxy group, $\text{OCH}_3$ ( $16 + 12 + 3 = 31$ ), lost as the radical $\bullet\text{OCH}_3$ . This leaves behind the acylium cation:

$\text{CH}_3\text{CH}_2\text{CO}^+ \quad \left(m/z = 3(12) + 5(1) + 16 = 57\right)$

Part (d): Deducing the structure of compound L

Putting the evidence together:

$M_r = 88$ together with an ester functional group (parts (a) and (b)) fits the molecular formula $\text{C}_4\text{H}_8\text{O}_2$ .
Loss of $\bullet\text{OCH}_3$ (mass 31) to leave the propanoyl cation, $\text{CH}_3\text{CH}_2\text{CO}^+$ (mass 57), shows that the $\text{OCH}_3$ group is attached directly to a propanoyl unit, $\text{CH}_3\text{CH}_2\text{CO}-$ .

This is consistent with:

$\text{CH}_3\text{CH}_2\text{COOCH}_3$

Methyl propanoate.

Final answers

(a) $M_r = 88$
(b) Ester, the C=O and C–O absorptions together, with no O–H absorption anywhere, rule out both a carboxylic acid and an alcohol.
(c) Fragment lost $= 31$ ( $\bullet\text{OCH}_3$ ); ion formed is $\text{CH}_3\text{CH}_2\text{CO}^+$ .
(d) $\text{CH}_3\text{CH}_2\text{COOCH}_3$ , methyl propanoate.