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101. Consider the following compounds:
P: \(\mathrm{CH_3-C(=O)-OCH_3}\)
Q: \(\mathrm{CH_3-C(=O)-NHCH_3}\)
The structural change from P to Q replaces the singly bonded ester oxygen by an \(\mathrm{-NH-}\) unit while retaining the methyl groups and carbonyl framework. Which statement correctly describes the change?
ⓐ. An ether is converted into a primary amine, and the carbonyl pi bond disappears.
ⓑ. An ester is converted into a secondary amide, while the carbonyl pi-bond count remains unchanged.
ⓒ. A carboxylic acid is converted into a secondary amine, while the carbonyl carbon changes from \(sp^2\) to \(sp^3\).
ⓓ. An ester is converted into a primary amide, and the carbonyl carbon changes from \(sp^2\) to \(sp\).
Correct Answer: An ester is converted into a secondary amide, while the carbonyl pi-bond count remains unchanged.
Explanation: \(\textbf{Classify the initial compound:}\)
P contains the arrangement \(\mathrm{-C(=O)-O-C-}\).
The singly bonded oxygen connects the carbonyl carbon to a methyl group.
This identifies an ester functional group.
The compound is methyl ethanoate.
\(\textbf{Examine the modified heteroatom linkage:}\)
In Q, the carbonyl carbon is bonded directly to nitrogen.
The characteristic arrangement is \(\mathrm{-C(=O)-NH-}\).
This identifies an amide functional group.
The nitrogen is attached to the carbonyl carbon, one methyl group and one hydrogen atom.
It is therefore a secondary amide nitrogen.
Q is N-methylethanamide.
\(\textbf{Compare carbonyl bonding:}\)
The C=O double bond is retained in both compounds.
Each contains one sigma bond and one pi bond between the carbonyl carbon and oxygen.
The carbonyl carbon in each structure has three sigma-bonding directions.
It remains \(sp^2\)-hybridised.
\(\textbf{Compare the functional-group consequences:}\)
Replacing the singly bonded ester oxygen with an \(\mathrm{-NH-}\) unit changes the characteristic group from ester to amide.
However, the carbonyl double bond itself is not eliminated.
The presence of nitrogen directly attached to carbonyl carbon is decisive for amide classification.
Q is not an ordinary secondary amine because its nitrogen is part of the amide functional group.
The change also does not create a primary amide because the nitrogen bears one carbon-containing substituent in addition to the acyl group.
The example illustrates that replacing one heteroatom in a characteristic linkage can change functional-group identity without changing the carbonyl carbon's hybridisation or pi-bond count.
102. A neutral, acyclic compound X has molecular formula \(\mathrm{C_4H_8O_2}\). It contains exactly one C=O double bond, no other multiple bond, and an \(\mathrm{-OH}\) group attached to a saturated carbon atom. The carbonyl carbon is directly attached to two carbon atoms. Which functional-group combination is consistent with all these conditions?
ⓐ. A carboxylic acid and an ether
ⓑ. An ester and an alcohol
ⓒ. An aldehyde and an alcohol
ⓓ. A ketone and an alcohol
Correct Answer: A ketone and an alcohol
Explanation: \(\textbf{Interpret the carbonyl attachment condition:}\)
The carbonyl carbon is directly attached to two other carbon atoms.
Therefore, its characteristic arrangement is \(\mathrm{R-C(=O)-R'}\).
This represents a ketone functional group.
A terminal aldehyde would require the carbonyl carbon to be directly attached to hydrogen.
A carboxylic acid or ester would require the carbonyl carbon to be directly attached to another oxygen atom.
Neither condition is compatible with the stated carbonyl connectivity.
\(\textbf{Identify the hydroxyl group:}\)
The molecule also contains an \(\mathrm{-OH}\) group attached to a saturated carbon atom.
This is an alcoholic hydroxyl group.
The oxygen of this hydroxyl group is not directly attached to the carbonyl carbon.
Therefore, the structure contains an alcohol in addition to the ketone.
\(\textbf{Check the molecular formula:}\)
A saturated, acyclic four-carbon compound containing only single bonds would follow the appropriate hydrogen count for its valence-satisfied framework.
The introduction of one carbonyl double bond produces one unit of unsaturation.
For \(\mathrm{C_4H_8O_2}\),
\[
\mathrm{DBE}=\frac{2(4)+2-8}{2}
\]
\[
\mathrm{DBE}=1
\]
The single C=O double bond accounts for this unsaturation.
No ring or additional pi bond is required.
\(\textbf{Verify structural feasibility:}\)
One structure satisfying the conditions is
\[
\mathrm{CH_3-C(=O)-CH(OH)-CH_3}
\]
The carbonyl carbon is bonded to two methyl-containing carbon portions.
The hydroxyl group is attached to an \(sp^3\)-hybridised carbon.
The molecular formula is \(\mathrm{C_4H_8O_2}\).
This structure contains one ketone and one alcohol.
The independent checks involving carbonyl connectivity, hydroxyl attachment and hydrogen deficiency all support the same classification.
Thus, X must contain a ketone and an alcohol under the stated conditions.
103. Three aromatic nitrogen-containing compounds are described:
P: Pyridine, a six-membered aromatic ring containing five carbon atoms and one nitrogen atom.
Q: Pyrrole, a five-membered aromatic ring containing four carbon atoms and one \(\mathrm{-NH-}\) nitrogen atom.
R: Aniline, \(\mathrm{C_6H_5NH_2}\), containing a benzene ring with an external amino group.
Which statement correctly compares their cyclic frameworks and nitrogen environments?
ⓐ. P and Q are aromatic heterocycles; R contains an aromatic carbocyclic ring with an amino substituent.
ⓑ. P is a carbocyclic amine; Q and R are aromatic heterocycles.
ⓒ. P and R are aromatic heterocycles; Q contains a carbocyclic ring with an amino substituent.
ⓓ. P, Q and R are aromatic heterocycles because every molecule contains nitrogen and an aromatic ring.
Correct Answer: P and Q are aromatic heterocycles; R contains an aromatic carbocyclic ring with an amino substituent.
Explanation: \(\textbf{Examine the ring membership in P:}\)
Pyridine contains a six-membered aromatic ring with nitrogen occupying one ring position.
Nitrogen is therefore an integral part of the cyclic framework.
P is consequently an aromatic heterocyclic compound.
Its nitrogen is not an external amino substituent attached to a carbon ring.
\(\textbf{Examine the nitrogen environment in Q:}\)
Pyrrole contains a five-membered ring with four carbon atoms and one nitrogen atom.
The nitrogen forms bonds with two neighbouring ring carbons and is also attached to hydrogen.
Because nitrogen is part of the ring, Q is heterocyclic.
Pyrrole also possesses an aromatic pi-electron system.
One nitrogen lone pair participates in the aromatic delocalisation, contributing to the six-pi-electron arrangement of the ring.
Thus, Q is aromatic and heterocyclic.
\(\textbf{Examine the ring composition in R:}\)
Aniline contains a benzene ring consisting entirely of carbon atoms.
The nitrogen occurs in an external \(\mathrm{-NH_2}\) group directly attached to one aromatic carbon.
Therefore, R possesses an aromatic carbocyclic framework with an amino substituent.
The presence of nitrogen outside the ring does not make the ring heterocyclic.
\(\textbf{Distinguish nitrogen classifications:}\)
Although the nitrogen of pyrrole is bonded to hydrogen, it is not chemically classified as an ordinary primary amine nitrogen.
Its lone pair participates in the aromatic bonding system.
Similarly, the ring nitrogen of pyridine is structurally different from the external amino nitrogen of aniline.
The three molecules therefore have different nitrogen bonding environments even though all contain nitrogen and aromatic frameworks.
The decisive classification criterion is whether nitrogen belongs to the ring or is attached externally to it.
104. A neutral, acyclic compound contains an aldehyde group and a nitrile group in the same molecule. Its structure is \(\mathrm{OHC-CH_2-C{\equiv}N}\). Which statement correctly identifies the numbers of \(sp\)- and \(sp^2\)-hybridised carbon atoms and the total number of pi bonds?
ⓐ. One \(sp\) carbon, one \(sp^2\) carbon and two pi bonds.
ⓑ. Two \(sp\) carbons, one \(sp^2\) carbon and three pi bonds.
ⓒ. One \(sp\) carbon, one \(sp^2\) carbon and three pi bonds.
ⓓ. One \(sp\) carbon, two \(sp^2\) carbons and three pi bonds.
Correct Answer: One \(sp\) carbon, one \(sp^2\) carbon and three pi bonds.
Explanation: \(\textbf{Identify the aldehyde carbon:}\)
The terminal \(\mathrm{-CHO}\) group contains a carbonyl carbon directly attached to hydrogen.
It forms three sigma bonds: one to hydrogen, one to the neighbouring carbon and one as part of the C=O double bond.
The carbonyl carbon is therefore \(sp^2\)-hybridised.
Its remaining unhybridised \(p\) orbital participates in the C=O pi bond.
\(\textbf{Identify the nitrile carbon:}\)
The terminal \(\mathrm{-C{\equiv}N}\) group contains a carbon atom triple-bonded to nitrogen.
The nitrile carbon also forms a single bond to the adjacent methylene carbon.
It therefore has two sigma-bonding directions and is \(sp\)-hybridised.
Two unhybridised \(p\) orbitals remain on this carbon.
These orbitals participate in the two pi bonds of the C≡N triple bond.
\(\textbf{Examine the intervening carbon:}\)
The central \(\mathrm{-CH_2-}\) carbon forms two C–H bonds and two C–C single bonds.
It has four sigma-bonding directions and is \(sp^3\)-hybridised.
Therefore, the molecule contains exactly one \(sp\) carbon and one \(sp^2\) carbon.
\(\textbf{Count the pi bonds:}\)
The aldehyde C=O double bond contributes one pi bond.
The nitrile C≡N triple bond contributes two pi bonds.
Thus,
\[
N_{\pi}=1+2
\]
\[
N_{\pi}=3
\]
\(\textbf{Check functional-group identities:}\)
The molecule contains an aldehyde because the carbonyl carbon bears hydrogen.
It also contains a nitrile because the terminal carbon is triple-bonded to nitrogen.
Neither group requires the intervening methylene carbon to be unsaturated.
The important inference combines functional-group recognition with independent local hybridisation assignments and sigma–pi bond composition.
\(\textbf{Final answer:}\)
\[
\boxed{N_{sp}=1,\quad N_{sp^2}=1,\quad N_{\pi}=3}
\]
105. A benzene ring bears two hydroxyl-containing substituents at para positions. One is an \(\mathrm{-OH}\) group attached directly to the aromatic ring, while the other is a \(\mathrm{-CH_2OH}\) group attached through saturated carbon. Which statement correctly describes the molecule?
ⓐ. It contains two phenolic hydroxyl groups because both are associated with the same aromatic ring.
ⓑ. It contains one phenolic hydroxyl group and one alcoholic hydroxyl group.
ⓒ. It contains two alcoholic hydroxyl groups because both contain O–H bonds.
ⓓ. It contains one phenolic hydroxyl group and one ether linkage.
Correct Answer: It contains one phenolic hydroxyl group and one alcoholic hydroxyl group.
Explanation: The molecule contains two hydroxyl groups, but their immediate carbon–oxygen bonding environments differ.
One hydroxyl oxygen is directly attached to an aromatic-ring carbon.
This arrangement is characteristic of a phenolic hydroxyl group.
The ring carbon carrying this group is \(sp^2\)-hybridised.
The second hydroxyl oxygen is attached to the carbon atom of a \(\mathrm{-CH_2OH}\) substituent.
This carbon is saturated and forms four sigma bonds.
It is therefore \(sp^3\)-hybridised.
The corresponding hydroxyl group is alcoholic rather than phenolic.
Both groups contain O–H bonds, but the presence of an O–H bond does not uniquely determine whether the group is an alcohol or phenol.
The classification depends on the carbon atom directly bonded to oxygen.
The compound therefore contains one phenolic hydroxyl group and one alcoholic hydroxyl group.
No ordinary ether linkage is present because neither oxygen connects two carbon atoms through single bonds.
The two hydroxyl groups are also not structurally equivalent merely because they occur in the same molecule.
A phenolic hydroxyl group can interact electronically with the aromatic ring through resonance.
The alcoholic hydroxyl group on the saturated side-chain carbon is not directly conjugated with the ring in the same manner.
Thus, the two groups may display different chemical behaviour.
The distinction illustrates why multiple identical-looking heteroatom fragments must be classified separately according to their local structural environments.
106. Two compounds contain the same saturated three-carbon framework:
P: \(\mathrm{CH_3CH_2CH_2SH}\)
Q: \(\mathrm{CH_3CH_2CH_2SO_3H}\)
Which statement correctly describes the sulphur-containing groups and the location of their characteristic hydrogen atoms?
ⓐ. P is a sulphonic acid with an O–H bond, while Q is a thiol with an S–H bond.
ⓑ. Both contain S–H bonds because each has one hydrogen associated with sulphur.
ⓒ. P contains a thioether linkage, while Q contains a carboxylic acid group.
ⓓ. P contains a thiol with an S–H bond, while Q contains a sulphonic acid with an O–H bond.
Correct Answer: P contains a thiol with an S–H bond, while Q contains a sulphonic acid with an O–H bond.
Explanation: The two compounds share the same saturated propyl carbon framework.
However, the sulphur-containing substituents have different characteristic bonding arrangements.
In P, sulphur is directly bonded to the terminal propyl carbon and to hydrogen.
The arrangement is \(\mathrm{R-SH}\).
This represents a thiol group.
Its characteristic hydrogen atom is bonded directly to sulphur.
In Q, the terminal substituent is \(\mathrm{-SO_3H}\).
The sulphur atom is attached to the propyl carbon and to three oxygen atoms.
Its conventional structural representation is \(\mathrm{R-S(=O)_2-OH}\).
The acidic hydrogen is therefore bonded to oxygen rather than directly to sulphur.
This identifies a sulphonic acid group.
Q does not contain the characteristic S–H bond of a thiol.
The difference between P and Q is not simply the number of oxygen atoms.
The complete bonding arrangement around sulphur changes from a thiol environment to a sulphonic acid environment.
A thioether would require sulphur to be bonded directly to two carbon-containing groups.
Neither P nor Q possesses that arrangement.
Similarly, a carboxylic acid would require the characteristic \(\mathrm{-C(=O)OH}\) group.
Q instead contains a sulphur-centred acid group.
The two compounds therefore cannot be treated as members of the same thiol homologous series merely because their carbon skeletons are identical.
Functional-group identity must be determined from the complete sulphur bonding environment, including the location of hydrogen.
107. Four neutral compounds are listed:
P: \(\mathrm{CH_3CH_2OH}\)
Q: \(\mathrm{CH_3CH_2SH}\)
R: \(\mathrm{CH_3OCH_3}\)
S: \(\mathrm{CH_3SCH_3}\)
Which statement correctly compares their oxygen- and sulphur-containing functional-group relationships?
ⓐ. P and Q are alcohols, whereas R and S are ethers, because the carbon frameworks are unchanged.
ⓑ. P and R are homologues, whereas Q and S are homologues, because each pair contains the same heteroatom.
ⓒ. P and Q contain analogous heteroatom–hydrogen arrangements, while R and S contain analogous heteroatom linkages between two carbon groups.
ⓓ. P and S are functional isomers, whereas Q and R are functional isomers, because all four contain two carbon atoms.
Correct Answer: P and Q contain analogous heteroatom–hydrogen arrangements, while R and S contain analogous heteroatom linkages between two carbon groups.
Explanation: \(\textbf{Compare the heteroatom–hydrogen arrangements:}\)
P contains the \(\mathrm{C-O-H}\) structural arrangement.
It is an alcohol because oxygen is bonded to saturated carbon and hydrogen.
Q contains the corresponding \(\mathrm{C-S-H}\) arrangement.
It is a thiol because sulphur replaces oxygen in the characteristic heteroatom–hydrogen group.
These structures are analogous in connectivity but belong to different chemical functional-group families.
\(\textbf{Compare the two-carbon heteroatom linkages:}\)
R contains an oxygen atom bonded to two methyl groups.
Its characteristic arrangement is \(\mathrm{C-O-C}\).
It therefore belongs to the ether family.
S contains a sulphur atom bonded to two methyl groups.
Its characteristic arrangement is \(\mathrm{C-S-C}\).
It represents a thioether or sulphide.
\(\textbf{Distinguish functional analogy from homology:}\)
P and Q do not differ by a \(\mathrm{CH_2}\) unit.
They differ by replacement of oxygen with sulphur.
Therefore, they are not consecutive homologues.
Similarly, R and S differ in heteroatom identity rather than by a homologous carbon-chain increment.
P and R share molecular formula \(\mathrm{C_2H_6O}\), making them functional isomers.
Q and S share molecular formula \(\mathrm{C_2H_6S}\), so they are also functional isomers.
However, compounds containing different heteroatoms cannot be structural isomers merely because their carbon counts are equal.
The important distinction is that analogous functional-group arrangements, homologous relationships and structural isomerism represent different classification principles.
The correct comparison follows from identifying both the heteroatom and its immediate carbon or hydrogen attachments.
108. A collection contains four saturated, acyclic monofunctional compounds:
P: \(\mathrm{CH_3CH_2CH_2OH}\)
Q: \(\mathrm{CH_3CH_2CH_2NH_2}\)
R: \(\mathrm{CH_3CH_2CH_2SH}\)
S: \(\mathrm{CH_3CH_2CH_2Cl}\)
The following statements are made:
Statement I: All four compounds have the same three-carbon framework, but their characteristic functional groups differ.
Statement II: P and Q are consecutive homologues because oxygen has been replaced by nitrogen.
Statement III: P contains an alcohol, Q a primary amine, R a thiol and S a chloroalkane.
Statement IV: The molecular formulas of P, Q, R and S are different despite the unchanged carbon framework.
Which combination of statements is correct?
ⓐ. Statements I, III and IV only
ⓑ. Statements I, II and III only
ⓒ. Statements II, III and IV only
ⓓ. Statements I, II, III and IV
Correct Answer: Statements I, III and IV only
Explanation: Statement I is correct.
All four molecules contain the same unbranched three-carbon framework.
The terminal carbon is bonded to a different characteristic heteroatom-containing group in each structure.
Thus, the carbon skeleton is preserved while the functional-group identity changes.
Statement II is incorrect.
A homologous increment requires the addition of one \(\mathrm{CH_2}\) unit within the same characteristic structural family.
P and Q differ by replacement of an oxygen-containing hydroxyl group with a nitrogen-containing amino group.
This is a functional-group change, not a homologous carbon-chain increment.
Their carbon counts are also identical.
Statement III is correct.
P contains an \(\mathrm{-OH}\) group attached to saturated carbon, identifying an alcohol.
Q contains an \(\mathrm{-NH_2}\) group attached to one carbon-containing group, identifying a primary amine.
R contains the \(\mathrm{-SH}\) group characteristic of a thiol.
S contains chlorine bonded to saturated carbon and is a chloroalkane.
Statement IV is also correct.
The molecular formula of P is \(\mathrm{C_3H_8O}\).
Q has molecular formula \(\mathrm{C_3H_9N}\).
R has molecular formula \(\mathrm{C_3H_8S}\).
S has molecular formula \(\mathrm{C_3H_7Cl}\).
The differences in heteroatom identity and associated hydrogen counts produce different complete molecular formulas.
Therefore, the four compounds are not structural isomers of one another.
Structural isomerism requires identical molecular formulas, not merely identical carbon skeletons.
This comparison separates carbon-framework identity, functional-group identity, homologous progression and molecular-formula equality.
Only Statements I, III and IV are correct.
109. A neutral organic compound has the structure \(\mathrm{HOOC-CH_2-CH_2-C{\equiv}N}\). The carboxylic acid group is selected as the principal characteristic group for systematic nomenclature. Which combination correctly describes the compound?
ⓐ. The nitrile is principal, the carboxylic acid is a hydroxy substituent, and the molecule contains two pi bonds.
ⓑ. The carboxylic acid is principal, the nitrile is represented by a cyano prefix, and the molecule contains three pi bonds.
ⓒ. The carboxylic acid is principal, the nitrile is represented by an amino prefix, and the molecule contains two pi bonds.
ⓓ. The nitrile is principal, the carboxylic acid is represented by an oxo prefix, and the molecule contains three pi bonds.
Correct Answer: The carboxylic acid is principal, the nitrile is represented by a cyano prefix, and the molecule contains three pi bonds.
Explanation: \(\textbf{Identify the two characteristic groups:}\)
The left terminal arrangement is \(\mathrm{-C(=O)OH}\), which represents a carboxylic acid.
The right terminal arrangement is \(\mathrm{-C{\equiv}N}\), which represents a nitrile.
The compound therefore contains two distinct characteristic functional groups.
The hydroxyl oxygen of the carboxylic acid is not classified as an independent alcohol group.
Likewise, the nitrile nitrogen is not classified as an amino group.
\(\textbf{Determine the principal group:}\)
Under the applicable functional-group priority rules, the carboxylic acid takes precedence over the nitrile for selection of the principal suffix.
The carboxylic acid group consequently determines the acid ending of the name.
The nitrile group is represented by the prefix cyano when it is not the principal characteristic group.
This does not mean that the nitrile carbon disappears from the molecular structure.
It remains a carbon atom connected to the neighbouring methylene group.
\(\textbf{Examine the multiple bonds:}\)
The carboxylic acid contains one C=O double bond.
That double bond contributes one pi bond.
The nitrile contains one C≡N triple bond.
The triple bond contributes two pi bonds.
Therefore,
\[
N_{\pi}=1+2
\]
\[
N_{\pi}=3
\]
\(\textbf{Check the hybridisation implications:}\)
The carboxyl carbon has three sigma-bonding directions and is \(sp^2\)-hybridised.
The nitrile carbon has two sigma-bonding directions and is \(sp\)-hybridised.
The intervening methylene carbons are \(sp^3\)-hybridised.
The molecule consequently contains different carbon hybridisations even though it has one continuous carbon framework.
Its principal functional-group classification is determined by the carboxylic acid, while its nitrile group remains an additional identifiable characteristic group.
The correct interpretation must satisfy nomenclature priority, functional-group connectivity and multiple-bond accounting simultaneously.
110. Two neutral compounds have the following structures:
P: \(\mathrm{CH_3-CH(NH_2)-COOH}\)
Q: \(\mathrm{H_2N-CH_2-C(=O)-OCH_3}\)
Which statement correctly compares their molecular formulas and functional-group combinations?
ⓐ. They have identical molecular formulas; P contains a primary amine and a carboxylic acid, whereas Q contains a primary amine and an ester.
ⓑ. They have identical molecular formulas; P contains a primary amide and an alcohol, whereas Q contains a primary amide and an ether.
ⓒ. They have different molecular formulas; P contains a primary amine and a carboxylic acid, whereas Q contains a primary amine and an ester.
ⓓ. They have different molecular formulas; P contains a primary amine and an ester, whereas Q contains a primary amine and a carboxylic acid.
Correct Answer: They have identical molecular formulas; P contains a primary amine and a carboxylic acid, whereas Q contains a primary amine and an ester.
Explanation: \(\textbf{Examine the structure of P:}\)
The central carbon of P is attached to a methyl group, an amino group, hydrogen and a carboxylic acid group.
The nitrogen in \(\mathrm{-NH_2}\) is bonded to one saturated carbon and two hydrogen atoms.
It therefore has the bonding environment of a primary amine.
The terminal \(\mathrm{-COOH}\) arrangement is a carboxylic acid.
The complete compound is 2-aminopropanoic acid.
\(\textbf{Examine the structure of Q:}\)
Q also contains an \(\mathrm{-NH_2}\) group attached to saturated carbon.
Its nitrogen therefore has the bonding environment of a primary amine.
However, the carbonyl group appears within the arrangement \(\mathrm{-C(=O)-OCH_3}\).
The singly bonded oxygen is attached to a methyl carbon rather than hydrogen.
This identifies an ester instead of a carboxylic acid.
Q is methyl 2-aminoethanoate.
\(\textbf{Compare their molecular formulas:}\)
For P, the carbon count is three and the hydrogen contributions are three from \(\mathrm{CH_3}\), one from \(\mathrm{CH}\), two from \(\mathrm{NH_2}\) and one from \(\mathrm{COOH}\).
\[
N_H(P)=3+1+2+1=7
\]
For Q, two hydrogens come from \(\mathrm{NH_2}\), two from \(\mathrm{CH_2}\) and three from \(\mathrm{OCH_3}\).
\[
N_H(Q)=2+2+3=7
\]
Both therefore have molecular formula \(\mathrm{C_3H_7NO_2}\).
\(\textbf{Interpret the relationship:}\)
The two structures have identical elemental compositions but different carbonyl–oxygen connectivity.
P contains an acid group with an O–H bond, while Q contains an ester linkage with an O–C bond.
Both retain an amino nitrogen environment.
They are consequently constitutional isomers with different functional-group combinations.
The distinction cannot be established by counting oxygen and nitrogen atoms alone.
The complete local attachments around the carbonyl carbon and singly bonded oxygen must also be examined.
111. Two neutral compounds have molecular formula \(\mathrm{C_4H_7NO}\):
P: A five-membered ring containing four carbon atoms and one nitrogen atom. One ring carbon is a carbonyl carbon directly bonded to the ring nitrogen, and nitrogen bears one hydrogen atom.
Q: A four-membered ring containing four carbon atoms. One ring carbon is a ketone carbonyl carbon, while another ring carbon carries an external \(\mathrm{-NH_2}\) group.
Which statement correctly distinguishes their cyclic and nitrogen-containing functional-group classifications?
ⓐ. P is a carbocyclic ketone containing a primary amine, while Q is a heterocyclic lactam.
ⓑ. Both are heterocyclic amides because each contains nitrogen and a ring carbonyl.
ⓒ. P is a heterocyclic primary amine, while Q is a carbocyclic amide.
ⓓ. P is a heterocyclic cyclic amide, while Q is a carbocyclic compound containing a ketone and a primary amine.
Correct Answer: P is a heterocyclic cyclic amide, while Q is a carbocyclic compound containing a ketone and a primary amine.
Explanation: \(\textbf{Interpret the ring framework of P:}\)
In P, nitrogen occupies one position of the five-membered ring.
The compound is therefore heterocyclic.
The carbonyl carbon is directly bonded to this ring nitrogen.
Consequently, the characteristic arrangement is \(\mathrm{-C(=O)-N-}\), identifying an amide.
Because the amide linkage forms part of a ring, P is a cyclic amide, commonly called a lactam.
The presence of an N–H bond does not turn this amide nitrogen into an independent primary amine group.
\(\textbf{Interpret the ring framework of Q:}\)
In Q, all four ring members are carbon atoms.
The ring is consequently carbocyclic.
The carbonyl carbon is bonded to two neighbouring ring carbon atoms.
It therefore possesses the bonding environment of a ketone.
The \(\mathrm{-NH_2}\) group is attached outside the ring to a saturated carbon.
Its nitrogen is bonded to one carbon-containing group and two hydrogen atoms.
This represents a primary amine group.
\(\textbf{Compare the functional-group combinations:}\)
P contains one amide functional group, while Q contains a ketone and a primary amine.
Both contain one C=O double bond and therefore one carbonyl pi bond.
Both also contain one ring.
Their identical molecular formulas do not establish identical functional-group identities.
The decisive differences concern whether nitrogen belongs to the ring and whether it is directly bonded to the carbonyl carbon.
In P, both conditions are satisfied.
In Q, neither is satisfied.
This comparison demonstrates why ring composition and carbonyl–nitrogen connectivity must be evaluated separately.
112. Consider the molecule \(\mathrm{p-IC_6H_4-CH(Br)-C(=O)F}\). Iodine is bonded directly to a benzene-ring carbon, bromine is bonded to the saturated carbon immediately outside the ring, and fluorine is bonded directly to the carbonyl carbon. Which classification of the three carbon–halogen bonds is correct?
ⓐ. Iodine is aryl, bromine is vinylic, and fluorine belongs to an ordinary fluoroalkane group.
ⓑ. Iodine is benzylic, bromine is aryl, and fluorine is attached to a saturated carbon.
ⓒ. Iodine is aryl, bromine is benzylic, and fluorine belongs to an acyl fluoride group.
ⓓ. Iodine is vinylic, bromine is benzylic, and fluorine belongs to an aryl fluoride group.
Correct Answer: Iodine is aryl, bromine is benzylic, and fluorine belongs to an acyl fluoride group.
Explanation: \(\textbf{Locate the iodine-bearing carbon:}\)
Iodine is bonded directly to a carbon atom belonging to the benzene ring.
That carbon is aromatic and \(sp^2\)-hybridised.
The corresponding carbon–iodine bond is therefore classified as aryl.
The presence of a side chain elsewhere on the ring does not change this attachment.
\(\textbf{Locate the bromine-bearing carbon:}\)
Bromine is attached to the carbon immediately adjacent to the aromatic ring.
This carbon forms single bonds with the ring carbon, the carbonyl carbon, bromine and hydrogen.
It is consequently \(sp^3\)-hybridised.
Because it occupies the saturated carbon position directly next to an aromatic ring, the carbon–bromine bond is benzylic.
It is not vinylic because the bromine-bearing carbon is not part of a C=C double bond.
\(\textbf{Locate the fluorine-bearing carbon:}\)
Fluorine is directly attached to the carbonyl carbon in the \(\mathrm{-C(=O)F}\) arrangement.
This represents an acyl fluoride functional group.
The carbonyl carbon is \(sp^2\)-hybridised, but it does not belong to the aromatic ring.
Therefore, its carbon–fluorine bond is not an aryl carbon–halogen bond.
\(\textbf{Integrate the three classifications:}\)
The molecule contains three different halogens in three different structural environments.
Two halogens are attached to \(sp^2\)-hybridised carbons, but their functional classifications are different.
The third halogen is attached to a saturated benzylic carbon.
The distinction therefore depends on more than halogen identity or carbon hybridisation alone.
The complete connectivity around each halogen-bearing carbon establishes the aryl, benzylic and acyl fluoride classifications.
113. Three neutral compounds have molecular formula \(\mathrm{C_7H_8O}\):
P: Benzyl alcohol, \(\mathrm{C_6H_5CH_2OH}\)
Q: Methoxybenzene, \(\mathrm{C_6H_5OCH_3}\)
R: 4-Methylphenol, \(\mathrm{p-CH_3C_6H_4OH}\)
Which statement correctly compares their oxygen-containing functional groups?
ⓐ. P contains an alcohol, Q contains an aromatic ether, and R contains a phenol.
ⓑ. P contains a phenol, Q contains an aromatic ether, and R contains an alcohol.
ⓒ. P and R contain phenols, while Q contains an aromatic ether.
ⓓ. P and Q contain alcohols, while R contains a phenol.
Correct Answer: P contains an alcohol, Q contains an aromatic ether, and R contains a phenol.
Explanation: \(\textbf{Examine the hydroxyl attachment in P:}\)
In benzyl alcohol, the hydroxyl oxygen is directly bonded to a saturated methylene carbon.
That carbon is attached to the benzene ring but is not itself a ring member.
Therefore, the hydroxyl group is alcoholic rather than phenolic.
The presence of an aromatic ring elsewhere in the molecule does not alter this classification.
\(\textbf{Examine the oxygen linkage in Q:}\)
In methoxybenzene, oxygen is attached directly to an aromatic-ring carbon and to a methyl carbon.
The characteristic arrangement is \(\mathrm{Ar-O-R}\).
This is an aromatic ether linkage.
There is no O–H bond in Q.
Therefore, Q contains neither an alcohol nor a free phenolic hydroxyl group.
\(\textbf{Examine the hydroxyl attachment in R:}\)
In 4-methylphenol, the hydroxyl oxygen is bonded directly to an aromatic-ring carbon.
This is the characteristic environment of a phenolic hydroxyl group.
The methyl substituent elsewhere on the ring does not make the hydroxyl group alcoholic.
\(\textbf{Compare the molecular compositions:}\)
P contains seven carbon atoms, eight hydrogen atoms and one oxygen atom.
Q and R have the same elemental composition.
Thus, all three possess molecular formula \(\mathrm{C_7H_8O}\).
However, their oxygen atoms occupy three distinct functional environments.
P is an alcohol, Q is an ether and R is a phenol.
The three structures therefore demonstrate constitutional differences that cannot be identified from molecular formula alone.
The decisive structural information is whether oxygen is bonded to hydrogen and whether the carbon directly attached to oxygen is aromatic or saturated.
114. Two neutral compounds have molecular formula \(\mathrm{C_2H_5NO_2}\):
P: Nitroethane, \(\mathrm{CH_3CH_2NO_2}\), in which the \(\mathrm{-NO_2}\) group is attached through nitrogen.
Q: Aminoethanoic acid, \(\mathrm{H_2NCH_2COOH}\).
Which statement correctly compares their characteristic functional groups?
ⓐ. P contains a nitrile and an alcohol, while Q contains an amide.
ⓑ. P contains a nitro group, while Q contains a primary amine and a carboxylic acid.
ⓒ. P contains a primary amine and a carboxylic acid, while Q contains a nitro group.
ⓓ. Both contain a nitro group because each has one nitrogen atom and two oxygen atoms.
Correct Answer: P contains a nitro group, while Q contains a primary amine and a carboxylic acid.
Explanation: \(\textbf{Examine the nitrogen environment in P:}\)
Nitroethane contains the characteristic \(\mathrm{-NO_2}\) group.
The nitrogen of this group is directly bonded to the ethyl carbon framework.
Its oxygen-containing bonding arrangement is characteristic of a nitro compound.
It is not an amino group because nitrogen does not possess the \(\mathrm{-NH_2}\) arrangement.
It is also not a nitrile because no C≡N triple bond is present.
\(\textbf{Examine the nitrogen environment in Q:}\)
In aminoethanoic acid, the nitrogen belongs to an \(\mathrm{-NH_2}\) group.
It is attached to a saturated methylene carbon.
The nitrogen therefore has the structural environment of a primary amine.
The second characteristic group is \(\mathrm{-COOH}\).
This is a carboxylic acid group because the carbonyl carbon is directly bonded to a hydroxyl oxygen.
\(\textbf{Compare the molecular formulas:}\)
P contains two carbon atoms, five hydrogen atoms, one nitrogen atom and two oxygen atoms.
Q contains the same elemental composition.
However, the two oxygen atoms in P belong to the nitro group.
In Q, they belong to the carboxylic acid group.
Thus, identical nitrogen and oxygen counts do not establish identical nitrogen-containing functionality.
\(\textbf{Interpret the classification:}\)
P contains one nitro functional group.
Q contains two distinct characteristic groups: an amino group and a carboxylic acid.
They are constitutional isomers rather than homologues.
Their difference arises from complete heteroatom connectivity, not merely from the presence of nitrogen or oxygen.
The comparison demonstrates why a molecular formula must be supplemented by structural information before functional groups can be assigned.
115. Two neutral compounds have molecular formula \(\mathrm{C_3H_7NOS}\):
P: \(\mathrm{CH_3-S-CH_2-C(=O)-NH_2}\)
Q: \(\mathrm{HS-CH_2-C(=O)-NH-CH_3}\)
Which statement correctly distinguishes their sulphur-containing groups and amide nitrogen substitution degrees?
ⓐ. P contains a thiol and a secondary amide, while Q contains a thioether and a primary amide.
ⓑ. P contains a thiol and a primary amide, while Q contains a thioether and a secondary amide.
ⓒ. Both contain thioethers and primary amides because each has one sulphur and one nitrogen atom.
ⓓ. P contains a thioether and a primary amide, while Q contains a thiol and a secondary amide.
Correct Answer: P contains a thioether and a primary amide, while Q contains a thiol and a secondary amide.
Explanation: \(\textbf{Identify the sulphur group in P:}\)
In P, sulphur is bonded directly to two carbon atoms.
One is a methyl carbon and the other is a methylene carbon.
The characteristic arrangement is \(\mathrm{C-S-C}\).
Therefore, P contains a thioether linkage.
No sulphur–hydrogen bond is present in this compound.
\(\textbf{Identify the nitrogen environment in P:}\)
The terminal group is \(\mathrm{-C(=O)-NH_2}\).
Nitrogen is directly attached to the carbonyl carbon and two hydrogen atoms.
This is a primary amide rather than an ordinary primary amine.
\(\textbf{Identify the sulphur group in Q:}\)
Q contains the terminal \(\mathrm{HS-CH_2-}\) arrangement.
Sulphur is directly bonded to hydrogen and carbon.
This represents a thiol group.
\(\textbf{Identify the nitrogen environment in Q:}\)
The amide nitrogen is directly attached to the carbonyl carbon, one methyl carbon and one hydrogen atom.
Thus, it is a secondary amide nitrogen.
The attached methyl group does not make Q an independent secondary amine because nitrogen belongs to the amide linkage.
\(\textbf{Check the molecular relationship:}\)
Both compounds contain three carbon atoms, seven hydrogen atoms, one nitrogen, one oxygen and one sulphur.
They therefore have identical molecular formulas.
However, the positions of the S–H, S–C, N–H and N–C connections differ.
The change affects both the sulphur functional-group classification and the substitution degree of the amide nitrogen.
A complete classification requires examining the atoms directly attached to each heteroatom.
The two molecules consequently provide an example of constitutional isomerism involving simultaneous changes in distinct functional environments.
116. Two neutral compounds have the molecular formula \(\mathrm{C_4H_8O_3}\):
P: \(\mathrm{CH_3-C(=O)-O-CH_2-CH_2OH}\)
Q: \(\mathrm{HOCH_2-CH_2-C(=O)-OCH_3}\)
Which statement correctly compares their functional-group combinations and ester-linkage arrangements?
ⓐ. P contains an ester and an alcohol, whereas Q contains an ether and a carboxylic acid.
ⓑ. Both contain an ester and an alcohol, but the methyl group belongs to the acyl portion in P and the alkoxy portion in Q.
ⓒ. Both contain an ester and an alcohol, but the methyl group belongs to the alkoxy portion in P and the acyl portion in Q.
ⓓ. P contains a carboxylic acid and an ether, whereas Q contains an ester and an alcohol.
Correct Answer: Both contain an ester and an alcohol, but the methyl group belongs to the acyl portion in P and the alkoxy portion in Q.
Explanation: \(\textbf{Identify the complete functional-group arrangements:}\)
P contains the characteristic \(\mathrm{-C(=O)-O-C-}\) arrangement.
Therefore, it possesses an ester functional group.
Its terminal \(\mathrm{-CH_2OH}\) portion contains an alcoholic hydroxyl group.
Consequently, P contains both an ester and an alcohol.
Q also possesses an ester linkage.
Its terminal \(\mathrm{HOCH_2-}\) portion contains an alcoholic hydroxyl group.
Therefore, Q contains the same two characteristic functional-group types.
\(\textbf{Examine the methyl group in P:}\)
In P, the methyl group is directly attached to the carbonyl carbon.
The corresponding acyl portion is \(\mathrm{CH_3-C(=O)-}\).
The singly bonded ester oxygen is attached to a two-carbon chain bearing the hydroxyl group.
\(\textbf{Examine the methyl group in Q:}\)
In Q, the methyl group is directly attached to the singly bonded ester oxygen.
The corresponding alkoxy portion is \(\mathrm{-OCH_3}\).
The hydroxyl-bearing two-carbon chain is attached to the carbonyl carbon.
\(\textbf{Check their molecular compositions:}\)
Each structure contains four carbon atoms, eight hydrogen atoms and three oxygen atoms.
Therefore, they have identical molecular formulas.
Neither possesses a free carboxylic acid \(\mathrm{-COOH}\) group.
The ester oxygen is not separately classified as an ordinary ether functional group.
\(\textbf{Interpret the structural distinction:}\)
The molecules are constitutionally different because the carbon groups attached to the two sides of the ester linkage differ.
However, their principal functional-group family remains the same.
This comparison shows that two structures can possess identical formulas and identical functional-group combinations yet differ in the connectivity of the carbon portions surrounding a characteristic linkage.
Functional-group identification therefore does not completely determine molecular identity.
117. A neutral organic compound has the structure \(\mathrm{HS-CH_2-CH_2-SO_3H}\). Which statement correctly describes its sulphur-containing functional groups and the locations of their characteristic hydrogen atoms?
ⓐ. It contains one thiol with an S–H bond and one sulphonic acid with an O–H bond.
ⓑ. It contains two thiol groups because both sulphur atoms are connected to hydrogen.
ⓒ. It contains one thioether and one carboxylic acid because the molecule contains carbon, oxygen and sulphur.
ⓓ. It contains two sulphonic acid groups because both sulphur atoms belong to one carbon chain.
Correct Answer: It contains one thiol with an S–H bond and one sulphonic acid with an O–H bond.
Explanation: The compound contains two sulphur atoms, but their local bonding environments are different.
The sulphur atom at the left end belongs to the \(\mathrm{HS-CH_2-}\) portion.
It is directly bonded to hydrogen and to a saturated carbon atom.
This arrangement is characteristic of a thiol group.
The corresponding functional-group hydrogen is therefore attached directly to sulphur.
The sulphur atom at the opposite end belongs to the \(\mathrm{-SO_3H}\) portion.
Its conventional bonding arrangement is \(\mathrm{-S(=O)_2-OH}\).
Here, sulphur is attached to the carbon framework and to three oxygen atoms.
The acidic hydrogen of this group is directly attached to oxygen rather than sulphur.
This arrangement is characteristic of a sulphonic acid.
The two sulphur atoms consequently occupy different functional environments even though they belong to the same molecule.
A thioether would require a sulphur atom connecting two carbon-containing groups through single bonds.
The molecule does not contain that characteristic linkage.
A carboxylic acid would require a carbon-centred \(\mathrm{-C(=O)OH}\) group, which is also absent.
The \(\mathrm{-SO_3H}\) group instead contains a sulphur-centred acid arrangement.
The molecule therefore contains one thiol and one sulphonic acid.
The distinction depends on locating the functional-group hydrogen and identifying the atoms bonded directly to sulphur.
This demonstrates why identical element symbols within one structural formula need not represent equivalent functional groups.
118. Three neutral, saturated cyclic compounds are considered:
P: Tetrahydrofuran, a five-membered ring containing four carbon atoms and one ether oxygen atom.
Q: Tetrahydropyran, a six-membered ring containing five carbon atoms and one ether oxygen atom.
R: Cyclopentanol, a five-membered ring containing five carbon atoms and an external alcoholic hydroxyl group.
Which statement correctly compares their molecular formulas and characteristic structural families?
ⓐ. P and R are consecutive cyclic-ether homologues, while Q and R are functional isomers.
ⓑ. P and Q are functional isomers, while Q and R are consecutive cyclic-ether homologues.
ⓒ. P and Q are consecutive cyclic-ether homologues, while Q and R are functional isomers.
ⓓ. All three are consecutive cyclic-ether homologues because each contains one ring and one oxygen atom.
Correct Answer: P and Q are consecutive cyclic-ether homologues, while Q and R are functional isomers.
Explanation: \(\textbf{Determine the molecular formula of P:}\)
Tetrahydrofuran contains four carbon atoms and one oxygen atom within a saturated ring.
Its molecular formula is \(\mathrm{C_4H_8O}\).
The oxygen is bonded to two neighbouring ring carbon atoms.
It therefore forms part of a cyclic ether linkage.
\(\textbf{Determine the molecular formula of Q:}\)
Tetrahydropyran contains five carbon atoms and one ring oxygen.
Its molecular formula is \(\mathrm{C_5H_{10}O}\).
It also contains a cyclic ether linkage.
Compared with P, it contains one additional carbon atom and two additional hydrogen atoms.
Therefore, P and Q differ by one \(\mathrm{CH_2}\) unit.
They retain the same broad saturated cyclic-ether structural family.
They can consequently be treated as consecutive homologues within this cyclic-ether series.
\(\textbf{Determine the molecular formula of R:}\)
Cyclopentanol contains five carbon atoms arranged in a saturated ring.
An external hydroxyl group is attached to one ring carbon.
Its molecular formula is \(\mathrm{C_5H_{10}O}\).
Thus, Q and R have identical molecular formulas.
However, Q contains an ether linkage and a heterocyclic ring.
R contains an alcoholic hydroxyl group and a carbocyclic ring.
They are therefore constitutional isomers belonging to different oxygen-containing functional-group families.
\(\textbf{Check the role of the rings:}\)
All three molecules contain one ring and no carbon–carbon or carbon–oxygen multiple bond.
Their hydrogen deficiencies arise from cyclic connectivity rather than pi bonding.
Nevertheless, identical ring counts and identical oxygen counts do not establish identical functional groups.
The correct classification requires examining whether oxygen is included in the ring or attached externally as \(\mathrm{-OH}\).
This distinction simultaneously explains the homologous relationship between P and Q and the functional isomerism between Q and R.
119. A neutral organic compound has the structure \(\mathrm{H_2N-C(=O)-CH_2-C{\equiv}N}\). The amide is selected as the principal characteristic group for systematic nomenclature. Which statement correctly describes the nitrogen environments and carbon hybridisation?
ⓐ. The molecule contains two primary amines; both nitrogen-bearing carbons are \(sp^2\)-hybridised.
ⓑ. The molecule contains an amide and a nitro group; the carbonyl carbon is \(sp\)-hybridised.
ⓒ. The molecule contains a nitrile as its only characteristic group; both nitrogen atoms belong to the C≡N linkage.
ⓓ. The molecule contains a primary amide and a nitrile; the carbonyl carbon is \(sp^2\)-hybridised and the nitrile carbon is \(sp\)-hybridised.
Correct Answer: The molecule contains a primary amide and a nitrile; the carbonyl carbon is \(sp^2\)-hybridised and the nitrile carbon is \(sp\)-hybridised.
Explanation: \(\textbf{Identify the amide nitrogen:}\)
The left terminal group contains the arrangement \(\mathrm{H_2N-C(=O)-}\).
Nitrogen is bonded directly to the carbonyl carbon and to two hydrogen atoms.
This is the characteristic arrangement of a primary amide.
It is not classified as an ordinary primary amine because the nitrogen is directly attached to a carbonyl carbon.
\(\textbf{Identify the nitrile nitrogen:}\)
At the opposite end of the molecule, nitrogen is triple-bonded to carbon.
The characteristic arrangement is \(\mathrm{-C{\equiv}N}\).
This represents a nitrile functional group.
Thus, the two nitrogen atoms belong to different characteristic groups.
\(\textbf{Assign the carbon hybridisations:}\)
The carbonyl carbon forms three sigma bonds: one to oxygen, one to nitrogen and one to the adjacent methylene carbon.
It is therefore \(sp^2\)-hybridised.
The nitrile carbon forms one sigma bond with the methylene carbon and one sigma bond as part of the C≡N triple bond.
It has two sigma-bonding directions and is therefore \(sp\)-hybridised.
The intervening methylene carbon forms four sigma bonds and is \(sp^3\)-hybridised.
\(\textbf{Examine the multiple bonds:}\)
The amide carbonyl contributes one pi bond.
The nitrile contributes two pi bonds.
The molecule consequently contains three pi bonds in total.
The amide is the specified principal functional group, while the nitrile can be represented using cyano prefix treatment.
The identity of the principal group does not remove the second functional group from the structural classification.
The complete description therefore requires separate recognition of the two nitrogen environments and their associated carbon bonding geometries.
120. A neutral, open-chain compound has the structure \(\mathrm{OHC-CH_2-COOH}\). The carboxylic acid group is selected as the principal characteristic group. Which combination correctly gives the systematic name and number of \(sp^2\)-hybridised carbon atoms?
ⓐ. 3-Hydroxypropanal and one \(sp^2\)-hybridised carbon.
ⓑ. 3-Oxopropanoic acid and two \(sp^2\)-hybridised carbons.
ⓒ. 2-Oxopropanoic acid and two \(sp^2\)-hybridised carbons.
ⓓ. 3-Hydroxypropanoic acid and one \(sp^2\)-hybridised carbon.
Correct Answer: 3-Oxopropanoic acid and two \(sp^2\)-hybridised carbons.
Explanation: \(\textbf{Identify the functional groups:}\)
The left terminal \(\mathrm{-CHO}\) arrangement is an aldehyde.
The right terminal \(\mathrm{-COOH}\) arrangement is a carboxylic acid.
Both groups contain a carbonyl double bond, but their surrounding attachments differ.
The aldehyde carbonyl carbon is directly bonded to hydrogen.
The acid carbonyl carbon is directly bonded to hydroxyl oxygen.
\(\textbf{Select the principal functional group:}\)
The carboxylic acid has higher naming priority than the aldehyde.
Therefore, the carboxylic acid determines the suffix of the name.
The aldehyde carbonyl is represented using the oxo prefix when its carbon is included in the parent chain.
\(\textbf{Number the carbon framework:}\)
The carboxyl carbon receives position 1.
The intervening methylene carbon is position 2.
The terminal aldehyde carbonyl carbon is position 3.
Therefore, the systematic name is 3-oxopropanoic acid.
\(\textbf{Determine carbon hybridisation:}\)
The carboxyl carbon forms three sigma bonds and is \(sp^2\)-hybridised.
The aldehyde carbonyl carbon also forms three sigma bonds and is \(sp^2\)-hybridised.
The intervening methylene carbon has four sigma bonds and is \(sp^3\)-hybridised.
Thus, two carbon atoms are \(sp^2\)-hybridised.
\(\textbf{Check the remaining bonding:}\)
The molecule contains two C=O double bonds.
Each contributes one pi bond.
The carbon framework is open-chain, and there is no C=C double bond.
The presence of an \(\mathrm{-OH}\) portion within \(\mathrm{-COOH}\) does not create an independent alcohol group.
The correct interpretation follows from identifying the two carbonyl environments, applying naming priority and assigning local carbon hybridisation.