mass spectrum of phenol C6H6O C6H5OH fragmentation pattern of m/z m/e ions for analysis and identification of phenol image diagram doc brown's advanced organic chemistry revision notes

Advanced Organic Chemistry: The mass spectrum of phenol C₆H₅OH

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Interpreting and explaining the mass spectrum of phenol C₆H₅OH

[Author © Dr Phil Brown GRIC, PhD: Doc Brown's advanced level organic chemistry exam revision notes suitable for students of UK A level chemistry courses & US K12 grade 11, grade 12 and AP honors chemistry courses: Molecular spectroscopy - analysing the mass spectra of phenol [spectra page updated RE-EDIT]

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Practise exam questions on the mass spectrum of phenol with answers!

Introductory note on the mass spectrum of phenol

Students and teachers please note my explanation of the mass spectrum of phenol is designed for advanced, but pre-university, chemistry courses.

If M represents the phenol molecule, the initial ionisation to give the molecular ion is:

M_(g) + high KE e^- ==> [M•]⁺_(g) + 2e^- and fragmentation equations assume [M]⁺ is the start of the processes and all species are in a gaseous state.

I've not usually shown an unpaired electron on e.g. an ion or a non-ionised alkyl radical R e.g.

[M•]⁺ ==> [X]⁺ + R•, but you should be aware this is a more accurate depiction of some processes.

I've used simplified equations to show how some of the ions that might be formed in the fragmentation pattern for the mass spectrum of phenol and only the formation of singly charged positive are considered for the mass spectrum of phenol.

I've included a stick diagram and table of m/z ions for the mass spectrum of phenol and doing the mass spectrum analysis under standard conditions, databases can be compiled based on complex fingerprint patterns, often involving the relative intensities of many fragment ions, and used to identify compounds including phenol.

In selected cases, where two different fragment ions have the same integer m/z value, I've pointed out that modern mass spectrometers can measure relative ion mass to four decimal places. So, using accurate isotopic masses, I've calculated and compared the accurate ion masses if appropriate for phenol. BUT strictly speaking, 0.0005 should be deducted for singly charged ions to account for the loss of the electron in their formation. I have NOT done this for phenol, but the mass spectrometer software does!

mass spectrum of phenol C6H6O C6H5OH fragmentation pattern of m/z m/e ions for analysis and identification of phenol image diagram doc brown's advanced organic chemistry revision notes

Phenol, C₆H₆O, C₆H₅OH, ,

The molecular structure and naming of aromatic compounds

Interpreting the fragmentation pattern of the mass spectrum of phenol

[M]⁺ is the molecular ion peak with an m/z of 94 corresponding to [C₆H₆O]⁺, the original phenol molecule minus an electron, [C₆H₅OH]⁺.

Unless otherwise stated, C means a ¹²C atom, if not, the isotopic carbon atom ¹³C will be indicated.

The small M+1 peak at m/z 95, corresponds to an ionised phenol molecule with one ¹³C atom in it i.e. an ionised phenol molecule of molecular formula [¹³C¹²C₅H₆O]⁺

Carbon-13 only accounts for ~1% of all carbon atoms (¹²C ~99%), but the more carbon atoms in the molecule, the greater the probability of observing this ¹³C M+1 peak.

Phenol has 6 carbon atoms, so on average, ~1 in 17 molecules will contain a ¹³C atom.

The most abundant ion of the molecule under mass spectrometry investigation (phenol) is usually given an arbitrary abundance value of 100, called the base ion peak, and all other abundances ('intensities') are measured against it (See note above data table of ions).

The base peak ion for the mass spectrum of phenol is m/z 94 ion [C₆H₆O]⁺

Identifying the species giving the most prominent peaks (apart from M) in the fragmentation pattern of phenol.

Unless otherwise indicated, assume the carbon atoms in phenol are the ¹²C isotope.

Some of the possible positive ions, [molecular fragment]⁺, formed in the mass spectrometry of phenol.

The parent molecular ion of phenol: [C₆H₅OH]⁺, m/z 94, which in this case, is also the base peak ion.

m/z value of [fragment]⁺	94	93	66	65	63	55	53
[molecular fragment]⁺	[C₆H₆O]⁺	[C₆H₅O]⁺	[C₅H₆]⁺	[C₅H₅]⁺	[C₅H₃]⁺	[C₃H₃O]⁺	[C₄H₅]⁺

m/z value of [fragment]⁺	51	50 ?	47 ?	40	39	38	37	17
[molecular fragment]⁺	[C₄H₃]⁺	[C₄H₂]⁺	[C₄H₂]⁺	[C₃H₄]⁺	[C₃H₃]⁺	[C₃H₂]⁺	[C₃H]⁺	[OH]⁺

I'm not sure on the identity of some of these ions?

Analysing and explaining the principal ions in the fragmentation pattern of the mass spectrum of phenol

PLEASE NOTE I have found it difficult to find 'authentic' equations to explain mass spectra fragmentation patterns and it is complex chemistry! I've identified the formulae of the ionised fragments on the mass spectrum diagram, but the equations are from the internet or my conjecture as to how the ions might be formed - please take care in using the information, especially for assignments at university or pre-university level.

Atomic masses: H = 1; C = 12 (~1% 13); O = 16

Bond enthalpies = kJ/mol: = 518 (benzene); C-H = 412; C-O = 360; O-H = 463

Possible equations to explain the most abundant ion peaks of phenol (tabulated above)

Formation of m/z 77 ion:

[C₆H₅OH]⁺ ===> [C₆H₅]⁺ + OH

mass change 94 - 17 = 77, the phenyl cation

Scission of the C-O bond.

A characteristic ion in the mass spectra of mono-substituted benzene ring compounds.

Formation of m/z 66 ion:

[C₆H₅OH]⁺ ===> [C₅H₆]⁺ + CO

A complex reaction (from the internet),

mass change 94 - 28 = 66 (M-28)

Loss of CO fragment from parent molecular ion.

The tiny peak at m/z 67 is probably formed in the same way, but containing a ¹³C carbon isotope atom i.e. [¹³C¹²C₅H₅OH]⁺

Formation of m/z 65 ion:

[C₆H₅OH]⁺ ===> [C₅H₅]⁺ + CHO

A complex reaction (from the internet),

mass change 94 - 29 = 65 (M-29 ion peak)

Loss of CHO fragment from parent molecular ion.

The m/z 65 ion is also formed by hydrogen loss from the m/z 66 ion.

The m/z 63 ion may also be formed by H₂ loss from the m/z 65 ion.

Formation of 'some' of the ions with m/z <56 (see diagram and data table)

The m/z 63 to 66 ions (of five carbon atoms) will lose e.g. CH or CO fragments to give the range of m/z 50 to 55 ions of four carbon atoms.

BUT there are ions of similar m/z values containing an oxygen atom e.g. m/z 55 too.

The m/z 47 to 55 ions (of five carbon atoms) will lose e.g. CH or CO fragments to give the range of m/z 37 to 40 ions of four carbon atoms.

Formation of m/z 17 and 77 ions:

[C₆H₅OH]⁺ ===> [OH]⁺ + C₆H₅

Tiny peak, a very low probability of this particular fragmentation reaction.

You can also see a similar sized tiny peak for m/z 77, a common ion in the mass spectra of aromatic benzene ring compounds like phenol, but of very low intensity here in the mass spectrum of phenol.

[C₆H₅OH]⁺ ===> [C₆H₅]⁺ + OH

Key Features of Phenol's Mass Spectrum

Phenol (C₆H₆O, M_r = 94) shows characteristic fragmentation due to the hydroxyl group attached to the aromatic ring:

Molecular ion peak (m/z 94):
Represents intact phenol molecule. Often moderately intense, confirming molecular formula.

Loss of OH radical (m/z 65):
Produces a stable cyclopentadienyl cation (C₅H₅⁺). This is often the base peak (most intense).

Other aromatic fragments:
- m/z 66 (C₆H₆⁺, hydrogenated cyclopentadienyl).
- m/z 39 (C₆H₃⁺, propargyl-type cation).
- Smaller peaks at m/z 77 (phenyl cation, C₆H₆⁺).

Sources: NIST Chemistry WebBook, MassBank Phenol Spectrum, University of Calgary Spectroscopy Notes.

Prominent m/z Ions and Origins

m/z	Ion Formula	Origin / Fragmentation Pathway
94	C₆H₆O⁺	Molecular ion (phenol intact)
77	C₆H₅⁺	Loss of OH radical → phenyl cation
65	C₅H₅⁺	Loss of OH + H → cyclopentadienyl cation
66	C₆H₆⁺	Hydrogenated cyclopentadienyl fragment
39	C₆H₃⁺	Smaller aromatic fragment (propargyl-type cation)

Common Misconceptions

Confusing base peak with molecular ion:
Students often assume the tallest peak is always the molecular ion. For phenol, the base peak is usually m/z 65, not 94.
Overlooking the molecular ion:
Some learners think phenol's molecular ion is absent. It is present and important for deducing formula.
Misidentifying m/z 77:
Students sometimes confuse the phenyl cation (C₆H₅⁺) with benzyl cation (C₆H₅CH₂⁺, m/z 91).

Exam Revision Tips

Always identify the molecular ion (m/z 94): Examiners expect recognition of the parent peak.
Know the base peak (m/z 65): This is diagnostic of phenol and shows aromatic stabilization.
Compare with benzene: Benzene has a molecular ion at m/z 78, so phenol's +16 shift confirms the OH group.
Practice fragment recognition: Be ready to explain why m/z 77 (phenyl cation) and m/z 65 (cyclopentadienyl cation) appear.
Link to structure: Show how loss of OH radical or hydrogen rearrangements stabilize aromatic fragments.
Exam technique: When asked to interpret spectra, always:
1. Identify molecular ion.
2. Spot base peak.
3. Relate fragments to logical structural losses.

QUESTIONS

Advanced A-level chemistry - practise exam questions on the mass spectrum of phenol

A joint AI-doc b re-edit experiment!

Jot down your responses and check out the answers: ANSWERS

If you think there are any errors, please email me asap at chem55555@hotmail.com

I don't mind if students/teachers do a selected printout of these questions and answers.

Q1. What is the molecular ion peak for phenol?

A. m/z = 77 B. m/z = 94 C. m/z = 93 D. m/z = 65

Q2. A strong peak at m/z = 93 is commonly observed. What causes it?

A. Loss of H• from the molecular ion

B. Loss of OH• from the molecular ion

C. Loss of CHO• from the molecular ion

D. Loss of O• from the molecular ion

Q3. A very strong peak at m/z = 77 is characteristic of:

A. CHO⁺ B. C₆H₅⁺ (phenyl cation) C. C₆H₆⁺ D. C₆H₅O⁺

Q4. Which fragment gives rise to the peak at m/z = 65?

A. C₅H₅⁺ B. C₆H₅⁺ C. C₅H₅O⁺ D. C₆H5⁺

Q5. Why is the m/z = 77 peak often more intense than the molecular ion peak?

A. The phenyl cation is very stable

B. The molecular ion does not form

C. The phenyl cation has a higher mass

D. The phenyl cation contains oxygen

Q6. Which peak confirms the presence of oxygen in phenol?

A. m/z = 77 B. m/z = 93 C. m/z = 65 D. m/z = 39

Q7. Which fragment results from loss of OH• from phenol?

A. m/z = 94 B. m/z = 77 C. m/z = 17 D. m/z = 65

Q8. A student sees a peak at m/z = 39. What fragment is this?

A. C₃H₃⁺ B. C₃H₄⁺ C. CHO⁺ D. C₃H₅⁺

Q9. Which combination of peaks confirms phenol rather than benzene?

A. m/z = 77 and 65

B. m/z = 94 and 93

C. m/z = 78 and 77

D. m/z = 65 and 39

Q10. A student claims phenol's mass spectrum should show a peak at m/z = 78 because it contains a benzene ring. Why is this incorrect?

A. Phenol contains no benzene ring

B. The benzene ring fragments before detection

C. Phenol's molecular ion is heavier due to oxygen

D. m/z = 78 is impossible to detect

Jot down your responses and check out the answers: ANSWERS

If you think there are any errors, please email me asap at chem55555@hotmail.com

I don't mind if students/teachers do a selected printout of these questions and answers.

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Links associated with phenol

The H-1 NMR spectrum of phenol

The C-13 NMR spectrum of phenol

The infrared spectrum of phenol

Physical & chemical properties of phenol and some of its derivatives & uses

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ANSWERS
Advanced A-level chemistry - practise exam questions on the mass spectrum of phenol

If you think there are any errors, please email me asap at chem55555@hotmail.com

I don't mind if students/teachers do a selected printout of these questions and answers.

Q1. What is the molecular ion peak for phenol?

A. m/z = 77    B. m/z = 94    C. m/z = 93    D. m/z = 65

Correct answer: B

Explanation: Phenol has molecular formula C₆H₆O, giving a molecular mass of 94, so the molecular ion is C₆H₆O⁺ (m/z = 94).

Common misconception: Students often confuse m/z = 93 (a fragment) with the molecular ion.

Q2. A strong peak at m/z = 93 is commonly observed. What causes it?

A. Loss of H• from the molecular ion

B. Loss of OH• from the molecular ion

C. Loss of CHO• from the molecular ion

D. Loss of O• from the molecular ion

Correct answer: A

Explanation: Phenol often loses a hydrogen radical, forming C₆H₅O⁺ (m/z = 93).

Common misconception: Students sometimes think m/z = 93 is due to losing OH•, but that would give m/z = 17, not 93.

Q3. A very strong peak at m/z = 77 is characteristic of:

A. CHO⁺    B. C₆H₅⁺ (phenyl cation)     C. C₆H₆⁺      D. C₆H₅O⁺

Correct answer: B

Explanation: The phenyl cation (C₆H₅⁺) at m/z = 77 is a classic aromatic fragment.

Common misconception: Students often misassign m/z = 77 to benzene (C₆H₆⁺), but benzene's molecular ion is 78.

Q4. Which fragment gives rise to the peak at m/z = 65?

A. C₅H₅⁺      B. C₆H₅⁺      C. C₅H₅O⁺    D. C₆H5⁺

Correct answer: A

Explanation: The C₅H₅⁺ cation (cyclopentadienyl‑type fragment) appears at m/z = 65, common in aromatic fragmentation.

Common misconception: Students sometimes think m/z = 65 is due to losing OH from phenol, but that would give m/z = 17.

Q5. Why is the m/z = 77 peak often more intense than the molecular ion peak?

A. The phenyl cation is very stable

B. The molecular ion does not form

C. The phenyl cation has a higher mass

D. The phenyl cation contains oxygen

Correct answer: A

Explanation: The phenyl cation is highly stabilised by aromatic resonance, giving a strong peak.

Common misconception: Students sometimes think the molecular ion must be the tallest peak — but the base peak is simply the most stable fragment.

Q6. Which peak confirms the presence of oxygen in phenol?

A. m/z = 77     B. m/z = 93     C. m/z = 65     D. m/z = 39

Correct answer: B

Explanation: The C₆H₅O⁺ (m/z = 93) fragment contains oxygen and is characteristic of phenol.

Common misconception: Students often assume the molecular ion confirms oxygen, but many molecules have mass 94; the fragment pattern is more diagnostic.

Q7. Which fragment results from loss of OH• from phenol?

A. m/z = 94     B. m/z = 77     C. m/z = 17     D. m/z = 65

Correct answer: C

Explanation: Loss of OH• (17) from phenol's molecular ion (94) gives m/z = 77, but the OH• itself is m/z = 17.

Common misconception: Students often think the fragment at 77 is “loss of OH,” but 77 is the remaining aromatic cation, not the OH fragment.

Q8. A student sees a peak at m/z = 39. What fragment is this?

A. C₃H₃⁺     B. C₃H₄⁺     C. CHO⁺     D. C₃H₅⁺

Correct answer: A

Explanation: The C₃H₃⁺ cation (propargyl‑type fragment) appears at m/z = 39, common in aromatic fragmentation.

Common misconception: Students sometimes think m/z = 39 is CHO⁺, but CHO⁺ is m/z = 29.

Q9. Which combination of peaks confirms phenol rather than benzene?

A. m/z = 77 and 65

B. m/z = 94 and 93

C. m/z = 78 and 77

D. m/z = 65 and 39

Correct answer: B

Explanation: Phenol has oxygen‑containing fragments at 94 (M⁺) and 93 (M–1), which benzene lacks.

Common misconception: Students often think benzene and phenol have identical spectra — but benzene's molecular ion is 78, not 94.

Q10. A student claims phenol's mass spectrum should show a peak at m/z = 78 because it contains a benzene ring. Why is this incorrect?

A. Phenol contains no benzene ring

B. The benzene ring fragments before detection

C. Phenol's molecular ion is heavier due to oxygen

D. m/z = 78 is impossible to detect

Correct answer: C

Explanation: Phenol's molecular ion is 94, not 78, because it contains oxygen. The benzene ring contributes to fragmentation patterns (e.g., m/z = 77), but the molecular ion reflects the whole molecule.

Common misconception: Students often confuse benzene's molecular ion (78) with phenol's — but adding oxygen increases the mass.

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