601. Four pairs of electronic descriptions are presented:
| Pair | First description | Second description |
|---|
| P | \(\mathrm{CH_2=CH-CH_2^+}\) | \(\mathrm{^+CH_2-CH=CH_2}\) |
| Q | \(\mathrm{CH_3-C(=O)-O^-}\) | \(\mathrm{CH_3-C(-O^-)=O}\), with the formal single and double bonds interchanged between the two original oxygen atoms |
| R | \(\mathrm{CH_3CH_2OH}\) | \(\mathrm{CH_3OCH_3}\) |
| S | \(\mathrm{CH_3CH=CH_2}\) | \(\mathrm{CH_2=CHCH_3}\), obtained by reversing the direction in which the same chain is written |
Assume that P and Q preserve their original nuclear sigma frameworks in the proposed second descriptions. Which statement correctly classifies these pairs?
ⓐ. P and R are resonance pairs; Q and S are constitutional-isomer pairs.
ⓑ. P and Q: resonance pair; R: functional-isomer pair; S: same molecule drawn in reverse.
ⓒ. P and Q are conformational pairs; R is a resonance pair; S is a position-isomer pair.
ⓓ. P and S are constitutional-isomer pairs; Q and R are resonance pairs.
Correct Answer: P and Q: resonance pair; R: functional-isomer pair; S: same molecule drawn in reverse.
Explanation: \(\textbf{Examine P's carbon framework:}\)
Both descriptions contain the same three connected carbon atoms.
Their hydrogen attachments and carbon sigma connections remain unchanged.
Only the C=C pi-electron placement and the location of positive formal charge differ.
Therefore, the drawings are resonance contributors of the allyl cation.
\(\textbf{Examine the carboxylate arrangement in Q:}\)
The carbonyl carbon remains sigma-bonded to the same two oxygen atoms.
Interchanging the formal C=O bond and the negative charge requires redistribution of electron pairs.
No oxygen nucleus changes its attachment to the carbonyl carbon.
Therefore, the two descriptions are carboxylate resonance contributors.
\(\textbf{Examine R:}\)
Ethanol contains the linkage \(\mathrm{C-C-O-H}\).
Dimethyl ether contains the linkage \(\mathrm{C-O-C}\).
Both have molecular formula \(\mathrm{C_2H_6O}\).
However, they belong to different oxygen-containing functional classes.
They are constitutional functional isomers, not resonance contributors.
\(\textbf{Examine S:}\)
Both drawings represent propene.
Writing the same unbranched three-carbon chain from opposite ends does not alter any atom-to-atom connection.
The double bond remains terminal.
Therefore, S does not represent two distinct compounds.
\(\textbf{Integrate the classification criteria:}\)
Resonance requires unchanged nuclear connectivity and redistribution of electrons.
Constitutional isomerism requires identical molecular formulas but different nuclear connectivity.
Reversing the direction of a structural formula can leave the molecule entirely unchanged.
The correct answer must apply these criteria separately instead of assuming that every visually different drawing represents a different isomer.
602. Four species are represented in their ordinary Lewis-electron descriptions:
P: \(\mathrm{CH_3^+}\).
Q: \(\mathrm{CH_3^{\bullet}}\).
R: \(\mathrm{CH_3^-}\).
S: \(\mathrm{CH_4}\).
Consider the following statements.
Statement I: P has three ordinary C–H sigma bonds and an electron-deficient carbon centre.
Statement II: Q contains one unpaired electron and is electrically neutral.
Statement III: R contains a carbon-centred nonbonding electron pair and carries negative formal charge.
Statement IV: S and R possess the same total valence-electron count because both contain four atoms attached to carbon.
Which combination is correct?
ⓐ. Statements I, II and IV only
ⓑ. Statements II, III and IV only
ⓒ. Statements I, III and IV only
ⓓ. Statements I, II and III only
Correct Answer: Statements I, II and III only
Explanation: \(\textbf{Examine the methyl cation P:}\)
The carbon atom is bonded to three hydrogen atoms.
It therefore has three ordinary C–H sigma bonds.
In its conventional electron-deficient representation, it has an available orbital and lacks a complete carbon octet.
The species carries positive charge.
Statement I is correct.
\(\textbf{Examine the methyl radical Q:}\)
The methyl radical also contains three C–H sigma bonds.
However, it possesses one unpaired electron at the carbon centre.
A radical dot represents one electron, not a complete electron pair.
The species is neutral overall.
Statement II is correct.
\(\textbf{Examine the methyl anion R:}\)
The methyl anion contains three C–H sigma bonds.
It also has one nonbonding electron pair associated with carbon.
This electron pair accounts for the ordinary carbanion Lewis description.
The complete species has charge \(-1\).
Statement III is correct.
\(\textbf{Examine methane S:}\)
Methane contains four C–H sigma bonds.
Its carbon has an ordinary complete octet.
It has no corresponding carbon-centred lone pair, unpaired electron or electron-deficient empty orbital.
\(\textbf{Compare the electron counts:}\)
P, Q and R contain three hydrogen atoms, whereas S contains four.
Their total valence-electron populations differ because of their formulas and charges.
Therefore, Statement IV is incorrect.
\(\textbf{Integrate the intermediate identities:}\)
A positively charged carbon, neutral carbon radical and negatively charged carbon can possess the same three C–H sigma connections while differing in nonbonding-electron occupancy.
These differences control their electronic identities and possible chemical roles.
The correct combination is I, II and III.
603. An organic compound contains a carbon–carbon double bond separated from a carbonyl group by one carbon–carbon single bond:
\[
\mathrm{CH_2=CH-C(=O)-CH_3}
\]
Another compound has the structure
\[
\mathrm{CH_2=CH-CH_2-C(=O)-CH_3}
\]
Assume approximately comparable local bond polarities and consider only ordinary direct pi conjugation across the carbon framework.
Which statement correctly compares their electronic structures?
ⓐ. Both contain the same uninterrupted conjugated system because their formulas each contain an alkene and a carbonyl group.
ⓑ. The first compound permits direct alkene–carbonyl conjugation; the saturated methylene carbon in the second interrupts the corresponding ordinary p-orbital pathway.
ⓒ. Neither can exhibit conjugation because the carbonyl oxygen prevents overlap of adjacent carbon p orbitals.
ⓓ. The second compound must have greater direct conjugation because its longer chain provides an additional sigma bond for pi-electron migration.
Correct Answer: The first compound permits direct alkene–carbonyl conjugation; the saturated methylene carbon in the second interrupts the corresponding ordinary p-orbital pathway.
Explanation: \(\textbf{Identify the multiple-bond arrangement in the first compound:}\)
The first compound contains a C=C bond followed by a C–C single bond and a C=O bond.
The relevant carbon atoms possess suitable p orbitals.
Under favourable geometry, these orbitals can form a continuous conjugated system.
The carbonyl oxygen also participates through its p-orbital contribution to the carbonyl pi bond.
\(\textbf{Interpret the conjugative consequence:}\)
The adjacent alkene and carbonyl systems can redistribute pi-electron density through valid resonance contributors.
Electron delocalisation need not remain confined to the individual multiple bonds.
This makes the arrangement an ordinary conjugated unsaturated carbonyl framework.
\(\textbf{Examine the second structure:}\)
In the second compound, a saturated \(\mathrm{-CH_2-}\) group lies between the alkene and carbonyl systems.
This carbon forms four ordinary sigma bonds.
It does not supply the same continuous unhybridised p-orbital path as the directly linked arrangement in the first compound.
\(\textbf{Distinguish transmitted effects:}\)
The second compound may still possess local sigma-bond polarisation and ordinary electronic interactions within each functional region.
However, such effects do not create an uninterrupted direct pi-conjugative pathway between the alkene and carbonyl.
\(\textbf{Evaluate the structural inference:}\)
Simply counting the numbers of pi bonds is insufficient for identifying conjugation.
Their relative connectivity is essential.
An additional saturated carbon can interrupt the specific conjugative interaction even though both compounds retain an alkene and a carbonyl group.
The correct answer therefore follows from reconstructing the continuous sequence of available p orbitals.
604. Consider a compound with the structure \(\mathrm{CH_3-CH(CH_3)-CH_2-C{\equiv}N}\). Its functional classification, systematic name and total number of sp-hybridised atoms are to be determined using the displayed ordinary nitrile connectivity.
Which statement is correct?
ⓐ. It is 3-methylbutanenitrile and contains two sp-hybridised atoms, namely the nitrile carbon and nitrile nitrogen.
ⓑ. It is 2-methylbutanenitrile and contains one sp-hybridised atom because nitrogen does not participate in hybridisation.
ⓒ. It is 3-methylbutanenitrile and contains four sp-hybridised carbon atoms because the triple bond determines the geometry of the entire carbon chain.
ⓓ. It is 2-methylpropanenitrile and contains two sp-hybridised atoms because the nitrile carbon is excluded from parent-chain counting.
Correct Answer: It is 3-methylbutanenitrile and contains two sp-hybridised atoms, namely the nitrile carbon and nitrile nitrogen.
Explanation: \(\textbf{Identify the characteristic group:}\)
The structure contains a terminal \(\mathrm{-C{\equiv}N}\) group.
This is a nitrile.
The nitrile carbon belongs to the carbon framework when naming the ordinary acyclic nitrile.
\(\textbf{Select the principal parent chain:}\)
Starting at the nitrile carbon, the continuous chain passes through the adjacent methylene, the branching methine and one terminal methyl carbon.
Therefore, the principal parent contains four carbon atoms.
The corresponding parent is butanenitrile.
\(\textbf{Apply numbering:}\)
The nitrile carbon receives position 1.
The adjacent methylene is carbon 2.
The branching methine is carbon 3.
The remaining methyl group is consequently a substituent at carbon 3.
The systematic name is 3-methylbutanenitrile.
\(\textbf{Examine the nitrile carbon:}\)
The carbon of the C≡N group has one sigma connection to the adjacent carbon and one sigma connection to nitrogen.
Its two additional C–N bonding interactions are pi bonds.
The two sigma directions are approximately linear.
Therefore, the nitrile carbon is sp-hybridised.
\(\textbf{Examine nitrogen:}\)
The nitrile nitrogen participates in one C–N sigma bond and two C–N pi interactions.
Its lone pair occupies an appropriate hybrid orbital in the ordinary linear description.
It is also sp-hybridised.
\(\textbf{Check the remaining carbon atoms:}\)
The adjacent methylene, branching methine and both methyl carbons possess saturated sigma-bonding environments.
They are sp³-hybridised.
The nitrile triple bond does not impose sp hybridisation on those separate saturated centres.
\(\textbf{Final answer:}\)
\[
N_{\mathrm{sp\ atoms}}=2\quad(\mathrm{nitrile\ C\ and\ N})
\]
605. Two organic compounds contain one nitrogen atom each:
P: \(\mathrm{CH_3CH_2NH_2}\).
Q: \(\mathrm{CH_3C(=O)NH_2}\).
A student argues that both must behave identically as electron-pair donors because each nitrogen has two N–H bonds.
Which statement best evaluates this reasoning?
ⓐ. The argument is correct because the number of N–H bonds uniquely determines nitrogen's electron-donating behaviour.
ⓑ. P: primary amine with available N lone pair; Q: amide with the N lone pair conjugated to C=O.
ⓒ. Q contains a free primary amine and must therefore have stronger ordinary nitrogen lone-pair donation than P.
ⓓ. Neither nitrogen possesses a lone pair because both atoms are bonded to two hydrogen atoms.
Correct Answer: P: primary amine with available N lone pair; Q: amide with the N lone pair conjugated to C=O.
Explanation: \(\textbf{Identify the nitrogen connectivity in P:}\)
P is ethanamine.
Its nitrogen is attached to an ethyl group and two hydrogen atoms.
It possesses one nonbonding electron pair in the ordinary neutral Lewis structure.
Therefore, it is a primary amine.
\(\textbf{Identify the nitrogen connectivity in Q:}\)
Q contains nitrogen directly bonded to a carbonyl carbon.
The characteristic group is \(\mathrm{-C(=O)NH_2}\).
It is therefore an amide rather than an ordinary primary amine.
\(\textbf{Examine the electronic interaction in Q:}\)
The amide nitrogen lone pair can overlap with the neighbouring carbonyl pi system when orbital alignment is favourable.
A resonance contributor can place increased bonding character between carbonyl carbon and nitrogen.
The carbonyl oxygen acquires negative formal charge, while nitrogen acquires positive formal charge in that contributor.
\(\textbf{Compare lone-pair availability:}\)
In P, the nitrogen lone pair is not committed to this particular adjacent carbonyl conjugative interaction.
In Q, the neighbouring carbonyl supplies a resonance-delocalisation pathway for the nitrogen electron pair.
This changes the electronic environment relevant to ordinary lone-pair donation.
\(\textbf{Reject the N–H counting rule:}\)
Both structures contain two nitrogen-bound hydrogen atoms.
However, they differ in the third group attached to nitrogen.
The distinction between an alkyl carbon and a carbonyl carbon is electronically important.
Therefore, identical N–H counts do not establish identical electron-pair behaviour.
\(\textbf{Interpret the conclusion:}\)
Recognising a functional group requires examining the complete local sigma connectivity.
The associated electronic effect must then be considered separately.
The correct answer combines amine-versus-amide classification with resonance availability rather than relying on the number of N–H bonds alone.
606. A hypothetical organic reaction changes the structure \(\mathrm{CH_3-CH(CH_3)-CH_2^+}\) into \(\mathrm{CH_3-C^+(CH_3)-CH_3}\) by transfer of one hydrogen atom together with its C–H bonding electron pair from the central carbon to the terminal positively charged carbon.
Assume that the starting structure and the proposed electron movement are chemically meaningful and that no other reaction occurs. Which statement correctly interprets the transformation?
ⓐ. The positive charge moves by resonance because the hydrogen nuclei remain attached to their original carbon atoms.
ⓑ. The transformation is an addition because it creates one new carbon–hydrogen sigma bond without breaking an existing sigma bond.
ⓒ. The transformation is a 1,2-hydride rearrangement: the original primary carbocation becomes a tertiary carbocation while the positive charge remains conserved.
ⓓ. The transformation is homolysis because the migrating hydrogen moves with one electron rather than an electron pair.
Correct Answer: The transformation is a 1,2-hydride rearrangement: the original primary carbocation becomes a tertiary carbocation while the positive charge remains conserved.
Explanation: \(\textbf{Identify the starting charged carbon:}\)
The original positively charged carbon is terminal.
It is attached to one other carbon atom and two hydrogen atoms.
Therefore, it is a primary carbocation.
\(\textbf{Identify the migrating group:}\)
The neighbouring central carbon contains one hydrogen atom.
The stipulated migration transfers that hydrogen together with its original C–H bonding electron pair.
Such a group is transferred as hydride in formal electron accounting.
\(\textbf{Follow the electron-pair movement:}\)
The electron-deficient terminal carbon receives the migrating hydride.
It forms a new C–H sigma bond and becomes neutral.
The carbon that loses hydride becomes positively charged.
\(\textbf{Classify the rearranged centre:}\)
The resulting positive carbon is bonded directly to three methyl groups.
It is therefore tertiary.
The transformation has converted a primary carbocation into a tertiary carbocation.
\(\textbf{Explain the electronic consequence:}\)
The tertiary centre has greater ordinary alkyl-substitution stabilisation.
Suitable neighbouring C–H sigma bonds can also provide hyperconjugative donation.
These considerations help explain why the specified rearrangement may be favourable.
\(\textbf{Distinguish the reaction class:}\)
A hydrogen nucleus has changed its bonded carbon.
Therefore, the sigma framework has changed.
The process is a rearrangement, not resonance.
It is also not homolysis because the migrating group carries a bonding electron pair, not one unpaired electron.
\(\textbf{Check charge conservation:}\)
The starting species has charge \(+1\).
The rearranged species also has charge \(+1\).
No external atom is added or removed.
The correct description is consequently a 1,2-hydride rearrangement giving a more substituted carbocation.
607. A neutral molecule contains a carbon–carbon double bond and a carbonyl group separated by an sp³-hybridised methylene carbon:
\[
\mathrm{CH_2=CH-CH_2-CHO}
\]
A hypothetical transformation converts it into
\[
\mathrm{CH_3-CH=CH-CHO}
\]
without changing the molecular formula. Which statement correctly explains the relationship between these structures and the resulting change in conjugation?
ⓐ. They are resonance contributors because the molecular formula is unchanged and both contain an alkene and an aldehyde.
ⓑ. Constitutional position isomers; the second allows direct alkene–carbonyl conjugation, the first interrupts it.
ⓒ. They are geometrical stereoisomers because both contain four carbon atoms and the position of their double bond changes.
ⓓ. They are homologues because moving the C=C bond changes the number of methylene units by one.
Correct Answer: Constitutional position isomers; the second allows direct alkene–carbonyl conjugation, the first interrupts it.
Explanation: \(\textbf{Identify the molecular formulas:}\)
Both structures contain four carbon atoms, six hydrogen atoms and one oxygen atom.
Their molecular formula is \(\mathrm{C_4H_6O}\).
They also contain the same aldehyde functional group.
\(\textbf{Examine the first structure:}\)
The first compound contains a terminal C=C bond.
A saturated methylene carbon lies between this alkene and the aldehyde carbonyl carbon.
That saturated carbon interrupts the ordinary continuous p-orbital pathway for direct alkene–carbonyl conjugation.
\(\textbf{Examine the second structure:}\)
The second compound contains the arrangement \(\mathrm{C=C-C=O}\).
Under favourable geometry, the participating atoms possess suitable aligned p orbitals.
Therefore, the alkene and carbonyl groups can interact through a conjugated pi system.
\(\textbf{Determine the structural relationship:}\)
The carbon skeleton remains an unbranched four-carbon chain.
The aldehyde group remains terminal.
However, the carbon–carbon double bond occupies a different position.
Therefore, the compounds are constitutional position isomers.
\(\textbf{Reject the resonance interpretation:}\)
Resonance contributors must preserve the positions of nuclei and their sigma-bond connections.
Changing the location of hydrogen atoms accompanying the alkene shift is not an ordinary resonance transformation.
The two structures are distinct molecular connectivities rather than alternative electron-only drawings of one compound.
\(\textbf{Integrate the electronic distinction:}\)
The change in double-bond position alters which atoms can participate in continuous conjugation.
Thus, position isomerism can change electronic delocalisation opportunities even while the molecular formula and main functional group remain unchanged.
The correct answer requires recognising both the isomeric relationship and the orbital consequence of the changed multiple-bond position.
608. Only neutral, saturated, acyclic compounds containing one sulphur atom and having molecular formula \(\mathrm{C_4H_{10}S}\) are considered. Sulphur may occur either in a thiol group, \(\mathrm{-SH}\), or in a thioether linkage, \(\mathrm{C-S-C}\). Count constitutional isomers separately, but do not count enantiomers as additional constitutional isomers.
Which statement correctly gives the number of thiols, thioethers and constitutional structures containing at least one tetrahedral stereogenic carbon?
ⓐ. Four thiols, three thioethers and one constitutional structure containing a stereogenic carbon.
ⓑ. Three thiols, four thioethers and two constitutional structures containing stereogenic carbons.
ⓒ. Four thiols, two thioethers and two constitutional structures containing stereogenic carbons.
ⓓ. Five thiols, two thioethers and one constitutional structure containing a stereogenic carbon.
Correct Answer: Four thiols, three thioethers and one constitutional structure containing a stereogenic carbon.
Explanation: \(\textbf{Distinguish the sulphur functional groups:}\)
A thiol contains sulphur bonded to hydrogen and carbon.
A thioether contains sulphur bonded directly to two carbon-containing groups.
Both classes may produce saturated acyclic compounds with the same molecular formula.
Therefore, the structures must be enumerated separately by functional class.
\(\textbf{Enumerate the thiols:}\)
The four-carbon skeleton can be unbranched or branched.
The unbranched skeleton gives butane-1-thiol and butane-2-thiol.
The branched skeleton gives 2-methylpropane-1-thiol and 2-methylpropane-2-thiol.
Consequently,
\[
N_{\mathrm{thiols}}=4
\]
\(\textbf{Enumerate the thioethers:}\)
The four carbon atoms must be distributed between two groups attached to sulphur.
The possible carbon distributions are \(1+3\) and \(2+2\).
For \(1+3\), the three-carbon group may be propan-1-yl or propan-2-yl.
This gives two distinct methyl propyl sulphide connectivities.
The \(2+2\) distribution gives diethyl sulphide.
Therefore,
\[
N_{\mathrm{thioethers}}=2+1=3
\]
\(\textbf{Count the complete constitutional set:}\)
\[
N_{\mathrm{constitutional}}=4+3=7
\]
\(\textbf{Examine carbon stereogenicity:}\)
Butane-2-thiol contains an SH-bearing carbon attached to hydrogen, methyl and ethyl groups.
These four substituents are different.
It therefore contains one tetrahedral stereogenic carbon.
The other three thiols do not contain a carbon with four different substituents.
Neither do the three thioethers in this enumeration.
\(\textbf{Interpret the counting convention:}\)
Butane-2-thiol may exist as two enantiomers, but these are not additional constitutional structures.
The correct result follows from combining sulphur functional-group classification, carbon-group distribution and stereogenic-centre analysis.
\(\textbf{Final answer:}\)
\[
\boxed{N_{\mathrm{thiols}}=4,\quad N_{\mathrm{thioethers}}=3,\quad N_{\mathrm{stereogenic\ structures}}=1}
\]
609. Four pairs of organic compounds are listed. Each pair differs in molecular composition by exactly one \(\mathrm{CH_2}\) unit, but the functional-group relationship has not been specified.
| Pair | First compound | Second compound |
|---|
| P | Propan-2-ol | Butan-2-ol |
| Q | Ethanal | Propanone |
| R | Dimethyl ether | Propan-1-ol |
| S | Propene | But-1-ene |
Which pairs represent members of the same respective homologous series?
ⓐ. P, Q and S only
ⓑ. P and S only
ⓒ. Q and R only
ⓓ. P, R and S only
Correct Answer: P and S only
Explanation: \(\textbf{Apply both requirements for homology:}\)
Successive members of a homologous series differ by one \(\mathrm{CH_2}\) unit.
They also possess the same characteristic functional group or corresponding structural family.
The molecular-mass difference alone is not sufficient.
\(\textbf{Evaluate P:}\)
Propan-2-ol has formula \(\mathrm{C_3H_8O}\).
Butan-2-ol has formula \(\mathrm{C_4H_{10}O}\).
Both are saturated acyclic monohydric alcohols.
Their formulas differ by \(\mathrm{CH_2}\).
Therefore, P represents a homologous relationship.
\(\textbf{Evaluate Q:}\)
Ethanal has formula \(\mathrm{C_2H_4O}\).
Propanone has formula \(\mathrm{C_3H_6O}\).
The difference in molecular composition is again \(\mathrm{CH_2}\).
However, ethanal is an aldehyde, whereas propanone is a ketone.
They do not belong to the same aldehyde or ketone homologous series.
Therefore, Q is not an acceptable homologous pair.
\(\textbf{Evaluate R:}\)
Dimethyl ether has formula \(\mathrm{C_2H_6O}\).
Propan-1-ol has formula \(\mathrm{C_3H_8O}\).
Their composition differs by \(\mathrm{CH_2}\), but the first compound is an ether and the second is an alcohol.
Therefore, R is not a homologous pair.
\(\textbf{Evaluate S:}\)
Propene and but-1-ene are both acyclic monoalkenes.
Their formulas are \(\mathrm{C_3H_6}\) and \(\mathrm{C_4H_8}\), respectively.
They differ by one \(\mathrm{CH_2}\) unit.
Therefore, S represents a homologous relationship.
\(\textbf{Integrate the classification:}\)
A mass difference of approximately \(14\ \mathrm{g\,mol^{-1}}\) is necessary for consecutive homologues but does not prove homology by itself.
Functional-group or structural-family continuity must also be established.
Only P and S satisfy both requirements.
610. Four sulphur- or oxygen-containing organic compounds are represented:
P: \(\mathrm{CH_3CH_2SO_3H}\)
Q: \(\mathrm{CH_3CH_2COOH}\)
R: \(\mathrm{CH_3CH_2SH}\)
S: \(\mathrm{CH_3CH_2SCH_3}\)
Consider the following statements.
Statement I: P is an organic sulphonic acid, while Q is a carboxylic acid.
Statement II: R is a thiol and S is a thioether because their sulphur atoms have different direct bonding environments.
Statement III: P and Q contain the same characteristic acid functional group because both structures terminate in an O–H arrangement.
Statement IV: R and S must be classified as different functional classes even though both contain a carbon–sulphur single bond.
Which combination is correct?
ⓐ. Statements I, II and III only
ⓑ. Statements I, III and IV only
ⓒ. Statements II, III and IV only
ⓓ. Statements I, II and IV only
Correct Answer: Statements I, II and IV only
Explanation: \(\textbf{Identify the acid groups:}\)
P contains the characteristic \(\mathrm{-SO_3H}\) group attached to an ethyl carbon framework.
This is the sulphonic acid functional group.
Q contains \(\mathrm{-COOH}\), the carboxylic acid functional group.
These groups differ in their central atoms and local bonding arrangements.
Therefore, Statement I is correct.
\(\textbf{Compare the oxygen-containing endings:}\)
Both acid groups contain an O–H bond.
However, the presence of an O–H bond does not make their complete functional groups identical.
A sulphonic acid contains sulphur as the central atom of its acid group.
A carboxylic acid contains carbon as the central atom of its carboxyl group.
Therefore, Statement III is incorrect.
\(\textbf{Examine R:}\)
The sulphur atom in R is bonded to an ethyl carbon and hydrogen.
Its local functional arrangement is \(\mathrm{C-S-H}\).
This identifies R as a thiol.
\(\textbf{Examine S:}\)
The sulphur atom in S is bonded to an ethyl group and a methyl group.
Its local arrangement is \(\mathrm{C-S-C}\).
This identifies S as a thioether, also called a sulphide.
Statement II is correct.
\(\textbf{Test the shared-bond misconception:}\)
Both R and S contain at least one C–S sigma bond.
However, functional-group classification depends on the complete arrangement around sulphur.
The presence or absence of the S–H bond is decisive here.
Therefore, Statement IV is correct.
\(\textbf{Integrate the structural distinction:}\)
Similar-looking local bonds do not guarantee identical functional classes.
The correct classifications require inspection of the central atom and its directly attached groups.
The correct statements are I, II and IV.
611. A sample contains all configurational stereoisomers of 1,2-dimethylcyclopropane. Both methyl groups are attached to adjacent ring carbons, and ordinary conformational changes are not counted separately.
Which statement correctly describes the number and stereochemical relationships of these compounds?
ⓐ. Two stereoisomers: one cis and one trans, with both individually achiral.
ⓑ. Three stereoisomers: one achiral cis form and a pair of enantiomeric trans forms.
ⓒ. Four stereoisomers: two enantiomeric cis forms and two enantiomeric trans forms.
ⓓ. Three stereoisomers: one achiral trans form and a pair of enantiomeric cis forms.
Correct Answer: Three stereoisomers: one achiral cis form and a pair of enantiomeric trans forms.
Explanation: \(\textbf{Identify the ring restriction:}\)
Cyclopropane contains a three-membered carbon ring.
Two methyl substituents on different ring carbons may lie on the same face or opposite faces of the ring.
These arrangements correspond to cis and trans geometrical relationships.
Ordinary rotation of the molecule in space does not convert one into the other.
\(\textbf{Examine the substituted carbon centres:}\)
Each substituted ring carbon is attached to hydrogen, methyl and two distinguishable directions through the substituted ring.
Each may consequently act as a stereogenic centre.
A preliminary count of two stereogenic centres suggests up to four configurations.
However, internal molecular symmetry must also be considered.
\(\textbf{Analyse the cis arrangement:}\)
In the cis form, both methyl groups occupy the same face of the ring.
The substituents are identical.
The molecular arrangement possesses an internal symmetry relationship that makes its mirror image superimposable.
Therefore, the cis form is achiral.
It corresponds to one meso configurational structure.
\(\textbf{Analyse the trans arrangement:}\)
In the trans form, one methyl group lies on one face and the second lies on the opposite face.
The two possible mirror-related arrangements are not superimposable.
They constitute one enantiomeric pair.
Therefore, the trans relationship contributes two stereoisomers.
\(\textbf{Count the distinct compounds:}\)
\[
N_{\mathrm{cis}}=1
\]
\[
N_{\mathrm{trans}}=2
\]
\[
N_{\mathrm{total}}=1+2=3
\]
\(\textbf{Interpret the symmetry correction:}\)
The expression \(2^n\) provides an upper limit when a molecule contains \(n\) tetrahedral stereogenic centres.
Internal symmetry can reduce the number of distinct stereoisomers.
The correct answer requires combining ring-based geometrical isomerism, mirror-image analysis and the possibility of a meso structure.
\(\textbf{Final answer:}\)
\[
\boxed{N_{\mathrm{total}}=3\quad(1\ \mathrm{meso\ cis}+2\ \mathrm{trans\ enantiomers})}
\]
612. Two nitrogen-containing compounds have molecular formula \(\mathrm{C_2H_3N}\):
P: \(\mathrm{CH_3-C{\equiv}N}\)
Q: \(\mathrm{CH_3-N^+{\equiv}C^-}\)
Assume the displayed charge-separated representation of Q follows ordinary octet and formal-charge conventions.
Which statement correctly identifies their functional classes and relationship?
ⓐ. P is an isocyanide, Q is a nitrile, and they are resonance contributors.
ⓑ. Both are nitriles because each contains a carbon–nitrogen triple bond.
ⓒ. P: nitrile; Q: isocyanide; functional isomers differing in the atom bonded to methyl carbon.
ⓓ. P and Q are homologues because moving nitrogen changes the effective carbon-chain length by one.
Correct Answer: P: nitrile; Q: isocyanide; functional isomers differing in the atom bonded to methyl carbon.
Explanation: \(\textbf{Identify the nitrile connection in P:}\)
P contains the characteristic group \(\mathrm{-C{\equiv}N}\).
The methyl carbon is bonded directly to the carbon of the nitrile group.
Therefore, P is a nitrile.
Its systematic name is ethanenitrile.
\(\textbf{Identify the isocyanide connection in Q:}\)
Q contains the arrangement \(\mathrm{CH_3-N{\equiv}C}\).
The methyl carbon is bonded directly to nitrogen.
This is the characteristic connectivity of an isocyanide, also called an isonitrile.
The conventional charge-separated representation shows positive formal charge on nitrogen and negative formal charge on terminal carbon.
\(\textbf{Check the formal-charge accounting:}\)
Nitrogen in Q is bonded once to methyl and three times to terminal carbon.
Its formal charge is \(+1\).
The terminal carbon has a triple bond to nitrogen and a nonbonding electron pair.
Its formal charge is \(-1\).
The complete molecule is therefore neutral.
\(\textbf{Compare the molecular formulas:}\)
Both compounds contain two carbon atoms, three hydrogen atoms and one nitrogen atom.
Their molecular formulas are identical.
However, the methyl carbon is attached to the nitrile carbon in P and to nitrogen in Q.
Thus, the sigma connectivity differs.
\(\textbf{Distinguish the isomeric relationship:}\)
The compounds belong to different functional classes.
They are therefore constitutional functional isomers.
They cannot be resonance contributors because resonance does not change which nuclei are connected by sigma bonds.
\(\textbf{Interpret the structural test:}\)
The presence of a C≡N-related multiple bond is not sufficient to identify the functional group.
The atom directly attached to the organic carbon framework is decisive.
The correct answer follows from connectivity and charge bookkeeping rather than from molecular formula alone.
613. Three organic chlorides are represented:
P: Vinyl chloride, \(\mathrm{CH_2=CHCl}\).
Q: Allyl chloride, \(\mathrm{CH_2=CHCH_2Cl}\).
R: Benzyl chloride, \(\mathrm{C_6H_5CH_2Cl}\).
In each case, the C–Cl bond undergoes hypothetical homolysis. Consider only the ordinary direct resonance delocalisation available to the resulting carbon-centred radical.
Which statement is correct?
ⓐ. All three form allylic radicals because each starting compound contains a carbon–carbon pi system.
ⓑ. P forms a vinylic radical, while Q and R form radicals capable of ordinary allylic and benzylic resonance, respectively.
ⓒ. P and Q form vinylic radicals, whereas R forms a phenyl radical.
ⓓ. Q and R form carbocations because the chlorine atom receives both C–Cl bonding electrons during homolysis.
Correct Answer: P forms a vinylic radical, while Q and R form radicals capable of ordinary allylic and benzylic resonance, respectively.
Explanation: \(\textbf{Apply the homolytic-cleavage rule:}\)
Homolysis divides the C–Cl bonding electron pair equally.
One electron remains with chlorine, producing a chlorine radical.
The other remains associated with the carbon originally bonded to chlorine.
The carbon-containing fragment therefore becomes a radical rather than a carbocation.
\(\textbf{Examine P:}\)
In vinyl chloride, chlorine is attached directly to an alkene carbon.
Homolysis formally produces \(\mathrm{CH_2=CH^{\bullet}}\).
The radical-bearing carbon belongs to the original C=C framework.
This is a vinylic radical.
It must not be classified as allylic merely because the molecule contains an alkene.
\(\textbf{Examine Q:}\)
In allyl chloride, chlorine is attached to a methylene carbon directly adjacent to the alkene.
Homolysis gives
\[
\mathrm{CH_2=CH-CH_2^{\bullet}}
\]
The radical centre is allylic.
Its unpaired electron can participate in resonance with the neighbouring C=C pi system.
A valid contributor places the unpaired electron at the opposite terminal carbon.
\(\textbf{Examine R:}\)
In benzyl chloride, chlorine is attached to the methylene carbon directly outside the aromatic ring.
Homolysis gives
\[
\mathrm{C_6H_5CH_2^{\bullet}}
\]
This is a benzyl radical.
Its radical-bearing carbon is directly adjacent to the aromatic pi system.
It consequently possesses benzylic resonance-delocalisation opportunities.
\(\textbf{Compare the three descriptors:}\)
Vinyl describes attachment directly to an alkene carbon.
Allyl describes a carbon directly adjacent to an alkene.
Benzyl describes the methylene carbon directly attached to a benzene ring.
Their names encode different connectivities, which determine the corresponding radical electronic environments.
The correct answer requires locating the broken bond before assigning radical type or resonance stabilisation.
614. The following boiling-point data are supplied for four saturated acyclic alcohols measured at the same pressure:
| Compound | Molecular formula | Boiling point |
|---|
| P: Ethanol | \(\mathrm{C_2H_6O}\) | \(78^\circ\mathrm{C}\) |
| Q: Propan-1-ol | \(\mathrm{C_3H_8O}\) | \(97^\circ\mathrm{C}\) |
| R: Butan-1-ol | \(\mathrm{C_4H_{10}O}\) | \(118^\circ\mathrm{C}\) |
| S: 2-Methylpropan-2-ol | \(\mathrm{C_4H_{10}O}\) | \(82^\circ\mathrm{C}\) |
Which interpretation is best supported by the data and the structures?
ⓐ. P, Q, R are consecutive homologues with rising boiling points; R and S are isomers with different boiling points.
ⓑ. P, Q, R and S are four consecutive homologues because every compound is an alcohol.
ⓒ. R and S are homologues because they contain the same functional group and have different boiling points.
ⓓ. The data prove that every branched alcohol must have a lower boiling point than every straight-chain alcohol, regardless of molecular formula.
Correct Answer: P, Q, R are consecutive homologues with rising boiling points; R and S are isomers with different boiling points.
Explanation: \(\textbf{Identify the homologous sequence:}\)
P, Q and R are saturated acyclic monohydric alcohols.
Their molecular formulas differ successively by one \(\mathrm{CH_2}\) unit.
Therefore, they form consecutive members of the same homologous series.
\(\textbf{Examine the measured trend:}\)
The listed boiling points for P, Q and R are \(78^\circ\mathrm{C}\), \(97^\circ\mathrm{C}\) and \(118^\circ\mathrm{C}\).
They increase across this particular sequence.
The growing carbon framework increases the importance of intermolecular dispersion interactions.
All three retain the alcohol functional group capable of intermolecular hydrogen bonding.
These structural changes are consistent with the observed progression.
\(\textbf{Compare R and S:}\)
R and S both have molecular formula \(\mathrm{C_4H_{10}O}\).
They contain the same total number of carbon, hydrogen and oxygen atoms.
However, their carbon connectivity and the substitution environment of the hydroxyl-bearing carbon differ.
They are constitutional isomers rather than homologues.
\(\textbf{Interpret their boiling-point difference:}\)
R boils at \(118^\circ\mathrm{C}\), whereas S boils at \(82^\circ\mathrm{C}\).
The difference demonstrates that identical molecular formulas do not guarantee identical physical properties.
Molecular shape, branching and the resulting intermolecular interactions must also be considered.
\(\textbf{Evaluate the proposed universal trend:}\)
The data establish the relative behaviour of these four particular compounds.
They do not prove a universal boiling-point inequality between every branched and every unbranched alcohol of arbitrary molecular mass.
A compound with a larger carbon framework may have stronger overall intermolecular interactions despite branching.
\(\textbf{Integrate the conclusion:}\)
Homologous relationships depend on both structural family and molecular composition.
Physical properties depend additionally on molecular structure.
The correct answer recognises the supplied homologous progression while avoiding an unsupported generalisation.
615. Hexa-2,4-diene has the structure \(\mathrm{CH_3-CH=CH-CH=CH-CH_3}\). Both internal C=C bonds can independently exhibit E/Z geometrical configurations. Ordinary conformations obtained by rotation about the central single bond are not counted as separate stereoisomers.
How many configurationally distinct geometrical stereoisomers of hexa-2,4-diene exist?
ⓐ. Two
ⓑ. Three
ⓒ. Four
ⓓ. Six
Correct Answer: Three
Explanation: \(\textbf{Identify the two geometrical centres:}\)
Hexa-2,4-diene contains double bonds between carbons 2 and 3 and between carbons 4 and 5.
Each doubly bonded carbon bears two different substituents.
Therefore, both double bonds satisfy the condition for E/Z geometrical isomerism.
\(\textbf{Determine the preliminary count:}\)
Each eligible double bond has two possible configurations.
Treating the two bonds temporarily as distinguishable gives
\[
N_{\mathrm{preliminary}}=2\times2
\]
\[
N_{\mathrm{preliminary}}=4
\]
The nominal combinations are \((E,E)\), \((E,Z)\), \((Z,E)\) and \((Z,Z)\).
\(\textbf{Examine molecular symmetry:}\)
The two ends of hexa-2,4-diene are identical methyl groups.
The carbon chain can be numbered from either end without changing its constitutional identity.
Reversing the chain interchanges the two double-bond positions.
\(\textbf{Identify equivalent configuration labels:}\)
The configuration combination \((E,Z)\) becomes \((Z,E)\) when the same molecular framework is numbered from the opposite end.
These labels consequently describe the same configurational compound.
They are not two independently distinct stereoisomers.
\(\textbf{Count the unique combinations:}\)
The unique configurations are
\[
(E,E),\quad(E,Z),\quad(Z,Z)
\]
Therefore,
\[
N_{\mathrm{unique}}=3
\]
\(\textbf{Distinguish configuration from conformation:}\)
Rotation about the central C–C single bond may produce different spatial conformations.
However, these rotations do not change the E/Z configurations of the two double bonds.
They are excluded from the stated count.
\(\textbf{Interpret the symmetry correction:}\)
Two independently configurable double bonds initially suggest four combinations.
Molecular symmetry makes two of those labels equivalent.
The correct answer therefore requires both E/Z eligibility testing and recognition of equivalent molecular orientations.
\(\textbf{Final answer:}\)
\[
\boxed{N_{\mathrm{geometrical\ stereoisomers}}=3}
\]
616. A simple secondary carbocation has the structure \(\mathrm{CH_3-CH^+-CH_2CH_3}\). Rotation about a sigma bond adjacent to the electron-deficient carbon can alter the alignment of neighbouring C–H bonds relative to the vacant p orbital.
Two statements are proposed:
Statement I: Hyperconjugative stabilisation requires suitable orbital alignment; an adjacent C–H bond is not automatically maximally effective in every molecular orientation.
Statement II: Rotation about an adjacent sigma bond may change the effectiveness of hyperconjugative overlap without changing the carbon skeleton or the carbocation's formal charge.
Which option is correct?
ⓐ. Both statements are correct, and Statement II explains an important geometrical qualification of Statement I.
ⓑ. Both statements are correct, but Statement II contradicts the orbital-overlap requirement in Statement I.
ⓒ. Statement I is correct, but Statement II is incorrect because every sigma-bond rotation requires carbon–carbon bond cleavage.
ⓓ. Statement I is incorrect, but Statement II is correct because hyperconjugation occurs independently of orbital orientation.
Correct Answer: Both statements are correct, and Statement II explains an important geometrical qualification of Statement I.
Explanation: \(\textbf{Identify the electron-accepting orbital:}\)
The positively charged carbon is electron-deficient.
In the ordinary carbocation description, it possesses an available p orbital.
An appropriately oriented sigma bond on a directly neighbouring carbon can donate electron density through hyperconjugation.
\(\textbf{Examine the neighbouring groups:}\)
The positive carbon is attached to methyl and ethyl groups.
Their directly adjacent carbons possess C–H sigma bonds.
These bonds provide potential hyperconjugative donor opportunities.
However, donor eligibility and actual overlap strength are not identical concepts.
\(\textbf{Evaluate Statement I:}\)
An orbital interaction depends on the spatial relationship between participating orbitals.
A sigma bond that has favourable overlap with the vacant p orbital can contribute effectively.
An orientation that reduces that overlap can reduce the corresponding interaction.
Therefore, Statement I is correct.
\(\textbf{Evaluate Statement II:}\)
Rotation around an ordinary adjacent C–C sigma bond changes the orientations of bonds attached to the rotating carbon.
It can consequently alter the alignment of an adjacent C–H bond with the carbocation's vacant p orbital.
Such rotation does not change which carbon atoms are bonded to one another.
It also does not remove the positive charge from the carbocation.
Therefore, Statement II is correct.
\(\textbf{Establish the explanatory relationship:}\)
Statement I specifies the orbital-alignment requirement.
Statement II describes how molecular rotation can change that alignment.
Thus, Statement II provides a geometrical application of the principle in Statement I.
\(\textbf{Interpret the broader limitation:}\)
A count of adjacent C–H bonds is a useful preliminary measure of hyperconjugative opportunities.
It is not an exact measurement of total stabilisation energy in every conformation.
A complete electronic description must consider both connectivity and orbital orientation.
Both statements are correct, and Statement II explains the relevant geometrical qualification.
617. Propadiene, \(\mathrm{CH_2=C=CH_2}\), contains two consecutive carbon–carbon double bonds sharing the central carbon. Consider its ordinary bonding description and assume the central carbon is approximately linear.
Which statement correctly identifies the hybridisation of the central and terminal carbons and the geometrical-isomerism consequence for this unsubstituted molecule?
ⓐ. The central carbon is sp³-hybridised, the terminal carbons are sp²-hybridised, and the molecule must exhibit cis/trans isomerism.
ⓑ. All three carbons are sp²-hybridised because each carbon participates in a C=C bond.
ⓒ. Central C is sp and terminal C atoms are sp²; identical terminal H pairs prevent geometrical isomerism.
ⓓ. The central carbon is sp-hybridised and the terminal carbons are sp³-hybridised; two independent terminal stereogenic centres are present.
Correct Answer: Central C is sp and terminal C atoms are sp²; identical terminal H pairs prevent geometrical isomerism.
Explanation: \(\textbf{Examine the central carbon:}\)
The central carbon forms double bonds to two terminal carbon atoms.
Each double bond contains one sigma bond and one pi bond.
The central carbon consequently has two sigma-bonding directions.
Its ordinary hybridisation is sp.
The sigma-bond arrangement is approximately linear.
\(\textbf{Identify the remaining p orbitals:}\)
An sp-hybridised central carbon retains two unhybridised p orbitals.
These can participate in the two C=C pi bonds.
The two pi-bonding arrangements occupy different perpendicular orbital directions in the conventional cumulative-double-bond description.
\(\textbf{Examine a terminal carbon:}\)
Each terminal \(\mathrm{CH_2}\) carbon forms two C–H sigma bonds and one C–C sigma bond.
It therefore has three sigma-bonding directions.
Its ordinary hybridisation is sp².
The remaining p orbital participates in the corresponding C=C pi bond.
\(\textbf{Test the stereoisomerism condition:}\)
The terminal carbons each carry two identical hydrogen atoms.
They do not have two different substituents at either terminal position.
Therefore, unsubstituted propadiene cannot form a pair of ordinary geometrical stereoisomers differing in those terminal substituent arrangements.
\(\textbf{Integrate the two structural tests:}\)
Hybridisation is determined by the local sigma-bond arrangement and available p orbitals.
Stereoisomerism depends additionally on the identities and spatial relationships of attached substituents.
The presence of two perpendicular pi systems does not automatically create optical or geometrical isomerism when the terminal substituents are identical.
The correct answer follows from evaluating the central carbon, the terminal carbons and the substituent condition separately.
618. Two samples of 2,3-dibromobutane are prepared:
P: An equimolar mixture of the \((2R,3R)\) and \((2S,3S)\) configurations.
Q: A pure sample of the meso \((2R,3S)\) configuration.
Assume ideal measurements and no chemical interconversion of configurations. Which statement correctly compares their optical activity and stereochemical composition?
ⓐ. P and Q are both optically inactive; P is a racemic mixture whose enantiomeric rotations cancel, whereas Q is an achiral meso compound possessing internal symmetry.
ⓑ. P is optically active because each molecule is individually chiral, whereas Q is inactive because it contains no stereogenic carbons.
ⓒ. Both P and Q are racemic mixtures because both show zero optical rotation.
ⓓ. P and Q contain the same molecular species because all three configurations have the same molecular formula.
Correct Answer: P and Q are both optically inactive; P is a racemic mixture whose enantiomeric rotations cancel, whereas Q is an achiral meso compound possessing internal symmetry.
Explanation: \(\textbf{Identify the stereogenic framework:}\)
2,3-Dibromobutane contains two carbon atoms bearing bromine, hydrogen and distinct carbon-chain directions.
These carbons can act as stereogenic centres.
The molecule consequently permits more than one configurational arrangement.
\(\textbf{Examine sample P:}\)
The \((2R,3R)\) and \((2S,3S)\) forms are non-superimposable mirror images.
They constitute an enantiomeric pair.
Each pure enantiomer can rotate plane-polarised light.
However, the two enantiomers have equal and opposite specific rotations under identical conditions.
\(\textbf{Apply P's composition:}\)
P contains the two enantiomers in equal molar proportions.
Their optical rotations cancel in the ideal mixture.
Therefore, P is optically inactive as a bulk sample.
This is racemic compensation.
\(\textbf{Examine sample Q:}\)
The meso configuration contains stereogenic carbons.
However, the complete molecule possesses internal symmetry.
Its mirror image is superimposable on the original molecular arrangement.
Therefore, Q is achiral despite containing stereogenic centres.
\(\textbf{Compare the reasons for zero rotation:}\)
P is an equimolar mixture of two individually chiral molecular species.
Q consists of one achiral meso configurational species.
Both can have zero observed optical rotation, but for different stereochemical reasons.
\(\textbf{Reject the formula-based inference:}\)
All configurations possess identical molecular formulas and sigma connectivity.
That does not make their three-dimensional configurations identical.
Likewise, zero optical rotation does not by itself establish that a sample is racemic.
The correct conclusion requires distinguishing compensation between enantiomers from molecular achirality caused by internal symmetry.
619. Two compounds have molecular formula \(\mathrm{C_3H_6BrCl}\):
P: \(\mathrm{BrCH_2-CH(Cl)-CH_3}\).
Q: \(\mathrm{ClCH_2-CH(Br)-CH_3}\).
Assume that all tetrahedral stereogenic carbon configurations are included, but ordinary conformations are not counted separately.
Which statement correctly describes their constitutional relationship and the total number of configurational compounds represented by P and Q together?
ⓐ. P and Q are enantiomers of one another, and together represent two compounds.
ⓑ. P and Q are position isomers, each with one stereocentre and two enantiomers: four forms altogether.
ⓒ. P and Q are identical compounds written in opposite directions, and together represent two configurations.
ⓓ. P and Q are homologues because bromine and chlorine occupy different positions, giving four constitutional structures.
Correct Answer: P and Q are position isomers, each with one stereocentre and two enantiomers: four forms altogether.
Explanation: \(\textbf{Compare the molecular formulas:}\)
Both structures contain three carbon atoms, six hydrogen atoms, one bromine atom and one chlorine atom.
Therefore, they satisfy the molecular-formula condition for constitutional isomerism.
\(\textbf{Examine P's connectivity:}\)
In P, bromine is attached to the terminal methylene carbon.
Chlorine is attached to the neighbouring secondary carbon.
The secondary carbon is bonded to H, Cl, methyl and bromomethyl groups.
These four substituents are different.
Therefore, P contains one tetrahedral stereogenic carbon.
\(\textbf{Examine Q's connectivity:}\)
In Q, chlorine is attached to the terminal methylene carbon.
Bromine is attached to the secondary carbon.
The secondary carbon is bonded to H, Br, methyl and chloromethyl groups.
Again, all four substituents are different.
Therefore, Q also contains one stereogenic carbon.
\(\textbf{Determine the constitutional relationship:}\)
P and Q differ in which halogen is attached to each of the two substituted carbon positions.
This changes the atom-to-atom connectivity.
They are consequently constitutional position isomers rather than enantiomers.
\(\textbf{Count configurations within each structure:}\)
Each compound has one stereogenic carbon.
In the absence of other stereogenic elements or symmetry restrictions, each has two enantiomeric configurations.
\[
N_P=2
\]
\[
N_Q=2
\]
\(\textbf{Combine the configuration counts:}\)
Since the two constitutional structures are different, none of P's configurations is identical to one of Q's configurations.
\[
N_{\mathrm{total}}=2+2
\]
\[
N_{\mathrm{total}}=4
\]
\(\textbf{Interpret the classification:}\)
Two molecules cannot be enantiomers if their underlying sigma connectivities differ.
Enantiomerism must be evaluated within a fixed constitutional framework.
The correct answer requires checking both the halogen attachments and the stereogenic centres before combining the counts.
\(\textbf{Final answer:}\)
\[
\boxed{N_{\mathrm{configurational\ compounds}}=4}
\]
620. Four neutral compounds have molecular formula \(\mathrm{C_4H_8O}\): P is butanal, \(\mathrm{CH_3CH_2CH_2CHO}\); Q is butan-2-one, \(\mathrm{CH_3COCH_2CH_3}\); R is but-3-en-1-ol, \(\mathrm{HOCH_2CH_2CH=CH_2}\); and S is 3-methoxyprop-1-ene, \(\mathrm{CH_2=CHCH_2OCH_3}\). Consider the following statements.
Statement I: The molecular formula corresponds to one degree of unsaturation.
Statement II: P and Q differ in the connectivity of their carbonyl functional groups.
Statement III: R and S each contain a C=C bond but belong to different oxygen-containing functional classes.
Statement IV: The common formula requires every compound to contain a C=O bond.
Which combination is correct?
ⓐ. Statements I, II and III only
ⓑ. Statements I, III and IV only
ⓒ. Statements II and IV only
ⓓ. Statements I, II, III and IV
Correct Answer: Statements I, II and III only
Explanation: Equal molecular formulas do not require equal connectivities or equal functional-group types.
\(\textbf{Determine the unsaturation count:}\)
For a neutral molecule containing carbon, hydrogen and oxygen, the degree-of-unsaturation relation is
\[
\mathrm{DBE}=\frac{2C+2-H}{2}
\]
Substitution for \(\mathrm{C_4H_8O}\) gives
\[
\mathrm{DBE}=\frac{2(4)+2-8}{2}=1
\]
Statement I is correct because all four structures have one double bond and no ring.
P contains an aldehyde carbonyl at the terminal carbon, whereas Q contains a ketone carbonyl between two carbon groups.
They have the same formula but different carbonyl connectivities, making Statement II correct.
R has a terminal C=C bond together with an alcoholic \(\mathrm{-OH}\) group attached to a saturated carbon.
S also has a terminal C=C bond, but its oxygen atom is bonded to two carbon groups in an ether linkage.
Thus Statement III correctly distinguishes the oxygen functional classes while recognising the shared alkene bond.
Neither R nor S contains a C=O bond.
Statement IV is consequently incorrect.
The example shows why a molecular formula and unsaturation count must be combined with connectivity to identify a functional group.
