501. The enolate ion derived from propanone has two useful resonance representations:
P: \(\mathrm{CH_3-C(=O)-CH_2^-}\)
Q: \(\mathrm{CH_3-C(O^-)=CH_2}\)
It undergoes two separately stipulated reactions with bromomethane:
Pathway I: The carbon bearing the negative charge in P attacks methyl carbon.
Pathway II: The oxygen bearing the negative charge in Q attacks methyl carbon.
In both pathways, the C–Br bonding pair moves to bromine, and no subsequent reaction occurs.
Which statement correctly identifies the neutral organic products?
ⓐ. Path I: \(\mathrm{CH_3COCH_2CH_3}\); Path II: \(\mathrm{CH_3C(OCH_3)=CH_2}\); different connectivities.
ⓑ. Both pathways give \(\mathrm{CH_3COCH_2CH_3}\) because resonance contributors cannot undergo attack at different atoms.
ⓒ. Pathway I gives \(\mathrm{CH_3C(OCH_3)=CH_2}\), whereas Pathway II gives \(\mathrm{CH_3COCH_2CH_3}\).
ⓓ. Both pathways give neutral propanone because bromomethane merely transfers a proton to the enolate.
Correct Answer: Path I: \(\mathrm{CH_3COCH_2CH_3}\); Path II: \(\mathrm{CH_3C(OCH_3)=CH_2}\); different connectivities.
Explanation: \(\textbf{Identify the resonance-delocalised nucleophile:}\)
The propanone-derived enolate contains a carbon framework adjacent to a carbonyl group.
Its negative charge can be represented on the alpha carbon or carbonyl oxygen in appropriate resonance contributors.
P and Q therefore describe one delocalised anion rather than two separate constitutional starting substances.
\(\textbf{Identify the electrophile:}\)
Bromomethane has a polar C–Br bond.
The methyl carbon is the specified electrophilic centre.
In either reaction, a nucleophilic electron pair forms a new sigma bond to methyl carbon while bromide departs.
\(\textbf{Follow carbon attack in Pathway I:}\)
The alpha-carbon electron pair forms a new C–C bond.
The carbonyl C=O bond remains in place in the product.
The structure becomes
\[
\mathrm{CH_3-C(=O)-CH_2CH_3}
\]
This compound is butan-2-one.
The organic product is neutral.
\(\textbf{Follow oxygen attack in Pathway II:}\)
The oxygen electron pair forms a new O–C bond with the incoming methyl group.
The C=C bond in the represented oxygen-centred enolate remains present.
The organic product becomes
\[
\mathrm{CH_3-C(OCH_3)=CH_2}
\]
This is an enol ether, 2-methoxyprop-1-ene.
It is also neutral.
\(\textbf{Check composition and charge conservation:}\)
Both reactions incorporate one methyl group from bromomethane.
Both organic products have molecular formula \(\mathrm{C_4H_8O}\).
The initial negative charge is carried by the departing bromide in the final product mixture.
\(\textbf{Distinguish resonance from product formation:}\)
The enolate resonance representations preserve the same starting nuclear framework.
However, making a new bond at carbon rather than oxygen produces different final covalent connectivities.
The two neutral products are consequently constitutional isomers.
No assertion about their relative experimental yields is justified without additional reaction conditions.
The correct answer requires tracking the nucleophilic atom and reconstructing the different carbon–carbon and carbon–oxygen bonds.
502. An organic chloride, \(\mathrm{R-Cl}\), is investigated under two sets of conditions. The following observations are supplied:
| Observation | Experiment P | Experiment Q |
|---|
| Conditions | Irradiation in a relatively nonpolar medium | Dark conditions in an ion-stabilising polar medium |
| Radical-trapping reagent | Suppresses the observed chain reaction; a radical-derived trapped product is detected | No corresponding radical-trapped product detected |
| Charge-separation evidence | No appreciable ionic conductivity attributable to the cleavage products | Ionic products are detected |
Assume that the observed transformations involve cleavage of the C–Cl bond and that the evidence is reliable. Which interpretation is best supported?
ⓐ. P must involve heterolysis because irradiation always produces carbocations; Q must involve homolysis because the reaction occurs in darkness.
ⓑ. Both experiments necessarily involve homolysis because chlorine receives electrons whenever a C–Cl bond breaks.
ⓒ. The radical trap proves that P forms only ionic fragments, while the conductivity in Q proves that only neutral radicals are present.
ⓓ. P supports homolytic radical formation; Q supports heterolytic ion formation, implying different electron distribution.
Correct Answer: P supports homolytic radical formation; Q supports heterolytic ion formation, implying different electron distribution.
Explanation: \(\textbf{Identify the central mechanistic question:}\)
The same C–Cl bond can be considered for different formal electron-distribution pathways.
In homolysis, each separating fragment receives one bonding electron.
In heterolysis, both bonding electrons move to one fragment.
These pathways produce electronically different species.
\(\textbf{Interpret P's radical-trapping result:}\)
A radical-trapping reagent suppresses the observed chain reaction.
A radical-derived trapping product is also detected.
Together, these findings support the involvement of radical intermediates.
For the stipulated C–Cl cleavage, the corresponding homolytic description is
\[
\mathrm{R-Cl\longrightarrow R^{\bullet}+Cl^{\bullet}}
\]
The relatively nonpolar medium and irradiation are consistent with the supplied radical observations.
\(\textbf{Interpret Q's ionic evidence:}\)
The products in Q exhibit evidence of ionic charge separation.
The stipulated electron-pair movement towards chlorine can produce
\[
\mathrm{R-Cl\longrightarrow R^+ + Cl^-}
\]
A suitable polar medium may stabilise such ionic species.
The absence of a detected radical-trapped product is also consistent with an ionic pathway.
\(\textbf{Evaluate the evidence collectively:}\)
Irradiation alone would not prove homolysis under every possible set of conditions.
Likewise, darkness alone would not prove heterolysis.
The radical-trapping and ionic-product observations provide additional mechanistic evidence.
The most defensible interpretation uses all the observations together.
\(\textbf{Distinguish electronic identities:}\)
The radical carbon fragment contains an unpaired electron.
The corresponding carbocation is electron-deficient and positively charged.
These are not interchangeable descriptions merely because the carbon skeleton is the same.
The correct inference is therefore homolytic behaviour in P and heterolytic behaviour in Q, supported by the specified experimental evidence.
503. A secondary carbocation has the structure \(\mathrm{CH_3-CH^+-CH(CH_3)-CH_3}\). Water is available as a nucleophile. Two competing pathways are separately stipulated:
Pathway P: Water immediately attacks the original positively charged carbon, followed by deprotonation.
Pathway Q: A 1,2-hydride shift first transfers hydrogen with its bonding electron pair from the adjacent methine carbon to the positive carbon; water then attacks the rearranged carbocation, followed by deprotonation.
Assume both pathways occur as described, without any additional rearrangement or elimination.
Which statement correctly gives their respective final organic products?
ⓐ. P gives 2-methylbutan-2-ol, while Q gives 3-methylbutan-2-ol; the hydride shift decreases carbocation substitution.
ⓑ. P gives 3-methylbutan-2-ol, while Q gives 2-methylbutan-2-ol; the hydride shift transfers the positive centre to a more substituted carbon before water attacks.
ⓒ. Both pathways give 3-methylbutan-2-ol because migration of hydride changes only the position of a formal charge in a resonance contributor.
ⓓ. P gives 3-methylbutan-1-ol, while Q gives 2-methylbutan-1-ol because water always attacks a terminal carbon.
Correct Answer: P gives 3-methylbutan-2-ol, while Q gives 2-methylbutan-2-ol; the hydride shift transfers the positive centre to a more substituted carbon before water attacks.
Explanation: \(\textbf{Identify the initial carbocation:}\)
The positive charge is located at carbon 2 of the displayed five-carbon framework.
That carbon is directly attached to two other carbon atoms and one hydrogen.
Therefore, the starting ion is a secondary carbocation.
The neighbouring carbon 3 is a methine carbon bearing one hydrogen and two methyl-group connections.
\(\textbf{Follow direct capture in P:}\)
The oxygen lone pair of water forms a sigma bond to the original carbocation carbon.
The immediate oxygen-containing adduct is positively charged because water oxygen now forms three bonds.
Deprotonation restores neutral oxygen.
The resulting alcohol has the structure
\[
\mathrm{CH_3-CH(OH)-CH(CH_3)-CH_3}
\]
The correct systematic name is 3-methylbutan-2-ol.
\(\textbf{Follow the hydride shift in Q:}\)
A hydrogen atom bonded to carbon 3 migrates towards the positive carbon together with its original C–H bonding electron pair.
The original carbocation carbon gains a new C–H bond and becomes neutral.
The carbon that loses hydride becomes positively charged.
The rearranged carbocation is
\[
\mathrm{CH_3-CH_2-C^+(CH_3)-CH_3}
\]
Its positively charged carbon is directly attached to three other carbon atoms.
Therefore, the rearranged ion is tertiary.
\(\textbf{Capture the rearranged ion:}\)
Water attacks the new positive centre.
The resulting oxonium ion then loses a proton.
The neutral product is
\[
\mathrm{CH_3CH_2-C(OH)(CH_3)_2}
\]
Its systematic name is 2-methylbutan-2-ol.
\(\textbf{Check carbon skeleton and charge:}\)
Both products retain five carbon atoms.
The hydride shift rearranges a C–H connection and the location of positive charge, rather than changing the overall carbon count.
Both pathways produce neutral alcohols after deprotonation.
The two products have the same molecular formula but different hydroxyl-bearing carbon environments.
\(\textbf{Interpret the mechanistic distinction:}\)
The hydride shift is a genuine rearrangement because a hydrogen nucleus changes its bonded carbon.
It is not resonance, which would preserve the nuclear sigma framework.
The question specifies the two competing pathways and does not require predicting their experimental product ratio.
The correct answer requires identifying the original cation, reconstructing the rearranged cation and naming the products formed by capture at two different carbon centres.
504. Two organic chlorides are considered:
P: Benzyl chloride, \(\mathrm{C_6H_5CH_2Cl}\).
Q: Chlorobenzene, \(\mathrm{C_6H_5Cl}\).
For each compound, consider two hypothetical cleavage modes in which only its C–Cl bond breaks:
I. Homolysis, with one bonding electron going to each fragment.
II. Heterolysis, with the entire C–Cl bonding electron pair moving to chlorine.
Which statement correctly compares the carbon-containing fragments?
ⓐ. Homolysis gives a benzyl carbocation from P and a phenyl carbocation from Q; heterolysis gives the corresponding radicals.
ⓑ. Both cleavage modes give resonance-stabilised benzyl radicals because both compounds contain a benzene ring.
ⓒ. Homolysis gives benzyl and phenyl carbanions, while heterolysis produces neutral aromatic hydrocarbons.
ⓓ. Homolysis gives benzyl and phenyl radicals; heterolysis gives the respective carbocations, with benzylic resonance only in P.
Correct Answer: Homolysis gives benzyl and phenyl radicals; heterolysis gives the respective carbocations, with benzylic resonance only in P.
Explanation: \(\textbf{Locate the carbon–chlorine bonds:}\)
In P, chlorine is bonded to the methylene carbon directly outside the benzene ring.
In Q, chlorine is bonded directly to an aromatic-ring carbon.
These are different bonding environments, even though both compounds contain a phenyl framework.
\(\textbf{Apply homolysis to P:}\)
Equal division of the C–Cl bonding pair leaves one electron on carbon and one on chlorine.
The carbon-containing fragment is
\[
\mathrm{C_6H_5CH_2^{\bullet}}
\]
This is a benzyl radical.
Its radical-bearing carbon is adjacent to the aromatic pi system and can participate in benzylic resonance.
\(\textbf{Apply homolysis to Q:}\)
Breaking the aromatic C–Cl bond in the same electron-distribution manner formally gives
\[
\mathrm{C_6H_5^{\bullet}}
\]
This is a phenyl radical.
Its unpaired electron is associated with a radical centre on the aromatic-ring carbon.
It must not be classified as a benzyl radical, because no intervening benzylic methylene carbon exists.
\(\textbf{Apply heterolysis to P:}\)
When chlorine receives the complete bonding pair, it becomes \(\mathrm{Cl^-}\).
The carbon fragment is \(\mathrm{C_6H_5CH_2^+}\).
The benzyl carbocation can delocalise electron deficiency through its neighbouring aromatic conjugated system.
\(\textbf{Apply heterolysis to Q:}\)
The corresponding formal carbon fragment is \(\mathrm{C_6H_5^+}\), a phenyl or aryl carbocation.
This does not possess the same ordinary benzylic resonance stabilisation as \(\mathrm{C_6H_5CH_2^+}\).
\(\textbf{Interpret the limitations:}\)
The four fragments are formal consequences of the stipulated electron distributions.
Their formation under real conditions would depend on bond strengths, medium and other energetic factors.
The question does not assert that all four cleavage pathways occur readily.
The essential distinctions are radical versus ionic electron accounting and benzylic versus aryl positioning of the resulting carbon centre.
505. A hypothetical three-carbon cation has the structure \(\mathrm{HO-CH_2-CH_2-CH_2^+}\). The following reaction is stipulated:
Stage I: The lone pair on the hydroxyl oxygen forms an intramolecular bond to the positively charged terminal carbon, giving a cyclic oxonium ion.
Stage II: A base removes the oxygen-bound hydrogen, and the O–H bonding pair remains with oxygen.
No rearrangement, intermolecular attack or elimination occurs.
Which statement correctly identifies the final organic product and explains the charge changes?
ⓐ. Oxirane forms because any intramolecular oxygen attack produces a three-membered ring.
ⓑ. Oxetane forms by four-membered ring closure; oxygen becomes positively charged before deprotonation.
ⓒ. Cyclopropanol forms because oxygen remains outside the newly formed carbon ring.
ⓓ. Butan-1-ol forms because the positive charge contributes an additional carbon atom during ring closure.
Correct Answer: Oxetane forms by four-membered ring closure; oxygen becomes positively charged before deprotonation.
Explanation: \(\textbf{Identify the reacting centres:}\)
The molecule contains a terminal carbocation and an alcoholic oxygen at the opposite end of the three-carbon chain.
The positively charged carbon is electron-deficient.
The neutral hydroxyl oxygen possesses nonbonding electron pairs and can act as a nucleophile.
Both centres belong to the same molecule.
\(\textbf{Follow intramolecular electron donation:}\)
One oxygen lone pair forms a new sigma bond to the positively charged terminal carbon.
The carbocation thereby receives the additional bonding electron pair needed to complete its ordinary tetravalent arrangement.
No carbon atom is added or removed.
\(\textbf{Determine the ring size:}\)
The pre-existing chain contains three connected carbon atoms.
Oxygen is already bonded to the first carbon.
Formation of the new O–C bond to the third carbon closes the sequence
\[
\mathrm{O-C-C-C}
\]
The resulting ring contains four atoms: one oxygen and three carbons.
Therefore, it has the oxetane ring framework.
\(\textbf{Track oxygen's formal charge:}\)
Before attack, the hydroxyl oxygen has two sigma bonds, one to carbon and one to hydrogen.
After forming the new C–O bond, oxygen has three sigma bonds.
It consequently carries positive formal charge in the cyclic oxonium intermediate.
\(\textbf{Apply Stage II:}\)
The base removes the hydrogen attached to oxygen.
The O–H bonding electron pair remains with oxygen.
The ring oxygen now has two C–O sigma bonds and is neutral.
The final organic compound is oxetane.
\(\textbf{Check charge conservation:}\)
The initial organic reactant has charge \(+1\).
Its cyclic oxonium intermediate also has charge \(+1\).
The final ring is neutral, while the proton-accepting base acquires the corresponding proton and positive-charge contribution.
The correct answer requires locating the nucleophile and electrophile, counting ring atoms and following the formal charge through both stages.
506. Two sulphur-containing nucleophiles undergo separately stipulated substitutions with bromoethane:
P: Neutral dimethyl sulphide, \(\mathrm{(CH_3)_2S}\), donates a sulphur lone pair to the bromine-bearing carbon.
Q: Methanethiolate ion, \(\mathrm{CH_3S^-}\), attacks the same carbon through sulphur.
In both reactions, the C–Br electron pair moves completely to bromine and no later reaction occurs.
Which statement correctly compares their organic products?
ⓐ. P produces neutral ethyl methyl sulphide, while Q produces an ethyldimethylsulphonium ion.
ⓑ. Both produce neutral thioethers because sulphur becomes neutral whenever it forms a new carbon–sulphur bond.
ⓒ. P forms an ethyldimethylsulphonium ion; Q forms neutral ethyl methyl sulphide, reflecting sulphur bond counts.
ⓓ. Both produce sulphonium ions because every sulphur atom with two carbon neighbours necessarily carries positive charge.
Correct Answer: P forms an ethyldimethylsulphonium ion; Q forms neutral ethyl methyl sulphide, reflecting sulphur bond counts.
Explanation: \(\textbf{Identify the common electrophile:}\)
Bromoethane contains an electrophilic saturated carbon bonded to bromine.
The stipulated substitution displaces bromide ion.
In each reaction, the incoming electron pair originates from sulphur.
\(\textbf{Examine the initial sulphur in P:}\)
Dimethyl sulphide contains sulphur bonded to two methyl groups.
It is electrically neutral and possesses available nonbonding electron pairs.
A sulphur lone pair can form a third S–C bond with the ethyl group.
\(\textbf{Determine P's product charge:}\)
After attack, sulphur is bonded to two methyl groups and one ethyl group.
In the resulting ordinary sulphonium representation, sulphur carries positive formal charge.
The organic ion is
\[
\mathrm{[(CH_3)_2S(CH_2CH_3)]^+}
\]
Bromide is the counterion.
The original neutral reactants therefore produce a cation–anion pair without changing their total charge.
\(\textbf{Examine the initial sulphur in Q:}\)
Methanethiolate contains sulphur bonded to one methyl group and carries a negative charge.
Its nonbonding electron pair attacks the same electrophilic ethyl carbon.
Formation of the second S–C sigma bond gives
\[
\mathrm{CH_3S-CH_2CH_3}
\]
This is neutral ethyl methyl sulphide, a thioether.
\(\textbf{Check the Q charge balance:}\)
The reactants have combined charge \(-1\).
The organic thioether is neutral.
The departing bromide carries charge \(-1\), preserving the total.
\(\textbf{Integrate the distinction:}\)
Both species behave as nucleophiles because sulphur donates electron density.
However, identical attacking atoms and electrophiles do not guarantee identical product charges.
The initial sulphur valence and formal charge determine whether the newly bonded organic sulphur species is neutral or positively charged.
The correct answer follows from comparing the two starting Lewis structures and tracking their bond counts through substitution.
507. Two acyclic hydrocarbons each contain five carbon atoms:
P: Pent-1-yne, \(\mathrm{HC{\equiv}C-CH_2CH_2CH_3}\).
Q: Pentane, \(\mathrm{CH_3CH_2CH_2CH_2CH_3}\).
A suitable strong base is used under separately specified conditions to remove a proton from either the terminal alkyne carbon of P or a terminal saturated carbon of Q.
Which statement correctly explains the qualitative difference in the carbon-centred conjugate bases, without asserting that both deprotonations occur readily?
ⓐ. The carbanion derived from Q is more stabilised by the local carbon hybridisation because its negatively charged carbon is sp³-hybridised.
ⓑ. Both carbanions are equally stabilised because they contain the same total number of carbon atoms.
ⓒ. The carbanion derived from P must be a carbocation because the terminal triple bond loses one hydrogen nucleus.
ⓓ. P: sp-centred carbanion stabilised by greater s character; Q: sp³-centred without that advantage.
Correct Answer: P: sp-centred carbanion stabilised by greater s character; Q: sp³-centred without that advantage.
Explanation: \(\textbf{Identify the relevant C–H bonds:}\)
P contains a terminal carbon–carbon triple bond with hydrogen attached directly to an sp-hybridised carbon.
Q contains only saturated carbon atoms.
Its terminal carbon–hydrogen bonds occur at sp³-hybridised centres.
These are different local electronic environments.
\(\textbf{Follow formal deprotonation:}\)
A base removes a hydrogen nucleus as \(\mathrm{H^+}\).
The original C–H bonding electron pair remains associated with carbon.
The carbon-containing product consequently bears a negative charge.
This is heterolytic proton removal, not hydrogen-atom abstraction producing a radical.
\(\textbf{Examine the anion from P:}\)
The resulting structure is
\[
\mathrm{^-C{\equiv}C-CH_2CH_2CH_3}
\]
The negatively charged carbon retains the terminal acetylide framework.
Its relevant orbital has substantial s character because the carbon is sp-hybridised.
\(\textbf{Examine the anion from Q:}\)
Deprotonation at a terminal saturated carbon would formally give an ordinary primary alkyl carbanion.
The negatively charged centre is associated with an sp³ carbon environment in the simplified hybridisation model.
Its lone pair does not receive the same high-s-character stabilising influence as the acetylide carbon.
\(\textbf{Connect hybridisation with carbanion stability:}\)
An orbital with greater s character holds its electron density, on average, closer to the nucleus.
This can stabilise negative charge.
Therefore, the sp-centred acetylide anion has an important stabilising advantage over the corresponding ordinary sp³ alkyl carbanion.
\(\textbf{Interpret the reaction condition carefully:}\)
The comparison concerns the electronic structures of the hypothetical conjugate bases.
It does not imply that an arbitrary base will remove a proton from either substrate to an appreciable extent.
Actual deprotonation also depends on acid–base equilibrium and the strength of the chosen base.
The correct conclusion follows from distinguishing proton transfer from homolysis and comparing the local hybridisation of the resulting negative centres.
508. Cyanide ion attacks ethanal under two separately specified conditions:
P: Neutral ethanal, \(\mathrm{CH_3CHO}\), reacts directly with \(\mathrm{CN^-}\).
Q: Ethanal is first protonated at its carbonyl oxygen, giving \(\mathrm{CH_3CH=OH^+}\); cyanide then attacks the carbonyl carbon.
In each addition, cyanide attacks through carbon, the C=O pi-electron pair moves to oxygen, and no subsequent proton transfer occurs.
Which statement correctly compares the immediate organic products?
ⓐ. Both pathways give the same neutral cyanohydrin because carbonyl protonation never affects formal charges.
ⓑ. P gives a neutral cyanohydrin, whereas Q gives an alkoxide ion because the originally protonated oxygen gains another electron pair.
ⓒ. Both pathways give negatively charged alkoxides because cyanide is negatively charged at the beginning of each attack.
ⓓ. P forms a negatively charged cyanohydrin alkoxide; Q forms the neutral cyanohydrin because oxygen was protonated.
Correct Answer: P forms a negatively charged cyanohydrin alkoxide; Q forms the neutral cyanohydrin because oxygen was protonated.
Explanation: \(\textbf{Identify the common bond-forming event:}\)
In both pathways, cyanide carbon attacks the electrophilic carbonyl carbon.
A new C–C sigma bond forms.
The original C=O pi-electron pair moves onto oxygen.
Therefore, the former carbonyl carbon becomes tetrahedral in both immediate products.
\(\textbf{Examine P's initial charge:}\)
Neutral ethanal contains a carbonyl oxygen without an O–H bond.
Cyanide has charge \(-1\).
After attack, the original carbonyl oxygen acquires the displaced pi-electron pair.
It carries negative formal charge.
The immediate product is
\[
\mathrm{CH_3CH(O^-)C{\equiv}N}
\]
Its total charge is \(-1\).
\(\textbf{Examine Q's protonated substrate:}\)
The starting carbonyl oxygen has already formed an O–H bond.
It carries positive formal charge in \(\mathrm{CH_3CH=OH^+}\).
The reacting cyanide has charge \(-1\).
Therefore, the combined initial charge of this second addition is zero.
\(\textbf{Follow Q's electron redistribution:}\)
Cyanide forms the same type of new C–C bond.
The C=O pi-electron pair moves onto the protonated oxygen.
Oxygen then has one C–O sigma bond and one O–H sigma bond.
It becomes neutral.
The immediate product is
\[
\mathrm{CH_3CH(OH)C{\equiv}N}
\]
This is the neutral cyanohydrin.
\(\textbf{Compare the proton positions:}\)
P has no oxygen-bound hydrogen available from the original aldehyde.
Its immediate product must therefore remain an alkoxide until a later protonation occurs.
Q already contains that oxygen-bound hydrogen before cyanide attacks.
No additional proton transfer is needed to obtain its neutral organic product.
\(\textbf{Integrate the distinction:}\)
Protonation before nucleophilic addition changes the formal-charge outcome of the immediate elementary step.
It does not change the carbon skeleton created by cyanide attachment.
The correct conclusion requires following both the nucleophilic bond formation and the earlier oxygen protonation state.
509. A secondary carbocation is represented by \(\mathrm{C_6H_5-CH(CH_3)-CH^+-CH_3}\). Two reaction pathways are considered:
P: Water immediately captures the original carbocation, followed by deprotonation.
Q: A 1,2-hydride shift occurs from the carbon directly bonded to phenyl to the positively charged carbon. Water then captures the rearranged cation, followed by deprotonation.
No further rearrangement or elimination occurs.
Which statement correctly compares the products and the additional stabilisation introduced by Q?
ⓐ. P gives 2-phenylbutan-2-ol and Q gives 3-phenylbutan-2-ol; the hydride shift removes benzylic resonance.
ⓑ. P gives 3-phenylbutan-2-ol, while Q gives 2-phenylbutan-2-ol; Q passes through a tertiary benzylic carbocation.
ⓒ. Both pathways give 3-phenylbutan-2-ol because a hydride shift merely exchanges resonance contributors.
ⓓ. P gives a primary alcohol and Q gives a phenol because hydride migration transfers the phenyl ring to oxygen.
Correct Answer: P gives 3-phenylbutan-2-ol, while Q gives 2-phenylbutan-2-ol; Q passes through a tertiary benzylic carbocation.
Explanation: \(\textbf{Identify the original positive centre:}\)
The starting positive carbon is directly attached to two carbon atoms and one hydrogen.
It is therefore a secondary carbocation.
A saturated carbon lies between this centre and the benzene ring.
The starting positive carbon consequently lacks direct benzylic resonance with the aromatic system.
\(\textbf{Follow direct water capture in P:}\)
Water oxygen donates a lone pair to the original carbocation carbon.
The resulting oxonium intermediate loses a proton.
The neutral product has the structure
\[
\mathrm{C_6H_5-CH(CH_3)-CH(OH)-CH_3}
\]
Numbering the four-carbon chain gives hydroxyl at carbon 2 and phenyl at carbon 3.
Therefore, P produces 3-phenylbutan-2-ol.
\(\textbf{Examine the hydride donor in Q:}\)
The carbon directly bonded to phenyl carries one hydrogen atom.
That hydrogen can migrate with its C–H bonding electron pair to the neighbouring positive centre in the stipulated rearrangement.
The initially positive carbon gains an additional C–H bond.
The carbon losing hydride becomes positively charged.
\(\textbf{Construct the rearranged carbocation:}\)
The rearranged structure is
\[
\mathrm{C_6H_5-C^+(CH_3)-CH_2CH_3}
\]
The positive carbon is now directly attached to phenyl, methyl and ethyl groups.
It has three carbon neighbours and is therefore tertiary.
Its direct attachment to the aromatic ring also provides benzylic resonance-delocalisation opportunities.
\(\textbf{Follow capture of the rearranged centre:}\)
Water attacks the new positive carbon.
Subsequent deprotonation gives
\[
\mathrm{C_6H_5-C(OH)(CH_3)-CH_2CH_3}
\]
This is 2-phenylbutan-2-ol.
\(\textbf{Distinguish the pathways:}\)
P produces an alcohol at the original positively charged carbon.
Q produces an alcohol at the carbon that becomes positively charged after hydride migration.
The rearrangement changes a C–H sigma connection and is not merely resonance.
Both products have the same molecular formula, but their hydroxyl-bearing carbons have different connectivities.
The correct comparison requires identifying the initially non-benzylic cation, reconstructing the tertiary benzylic intermediate and naming the two captured products.
510. A hypothetical C–X bond is examined in the gas phase and in a polar solvent. The following energetic quantities refer to cleavage of the same starting bond, and all solvent-stabilisation values are measured relative to the unchanged solvated reactant.
| Quantity | Value |
|---|
| Gas-phase homolytic cleavage enthalpy | \(+270\ \mathrm{kJ\,mol^{-1}}\) |
| Gas-phase heterolytic cleavage enthalpy, with the electron pair moving to X | \(+880\ \mathrm{kJ\,mol^{-1}}\) |
| Net solvent stabilisation of the radical fragments | \(-40\ \mathrm{kJ\,mol^{-1}}\) |
| Net solvent stabilisation of the ionic fragments | \(-680\ \mathrm{kJ\,mol^{-1}}\) |
Assume these contributions are additive and no other enthalpy terms are relevant. Which statement correctly compares the cleavage enthalpies in the polar solvent?
ⓐ. Homolysis requires \(230\ \mathrm{kJ\,mol^{-1}}\), heterolysis requires \(200\ \mathrm{kJ\,mol^{-1}}\), and heterolysis is thermodynamically less endothermic by \(30\ \mathrm{kJ\,mol^{-1}}\).
ⓑ. Homolysis requires \(310\ \mathrm{kJ\,mol^{-1}}\), heterolysis requires \(1560\ \mathrm{kJ\,mol^{-1}}\), and homolysis remains strongly favoured.
ⓒ. Homolysis requires \(230\ \mathrm{kJ\,mol^{-1}}\), heterolysis requires \(1560\ \mathrm{kJ\,mol^{-1}}\), because solvation stabilisation must be added as a positive cost.
ⓓ. Homolysis requires \(270\ \mathrm{kJ\,mol^{-1}}\), heterolysis requires \(880\ \mathrm{kJ\,mol^{-1}}\), because the solvent affects only activation energy.
Correct Answer: Homolysis requires \(230\ \mathrm{kJ\,mol^{-1}}\), heterolysis requires \(200\ \mathrm{kJ\,mol^{-1}}\), and heterolysis is thermodynamically less endothermic by \(30\ \mathrm{kJ\,mol^{-1}}\).
Explanation: \(\textbf{Identify the gas-phase electronic distinction:}\)
Homolysis divides the C–X bonding electron pair equally between the two fragments.
It formally produces radicals.
Heterolysis transfers the complete bonding pair to X.
It formally produces \(\mathrm{C^+}\)-containing and \(\mathrm{X^-}\) ionic fragments.
\(\textbf{Interpret the solvent terms:}\)
The supplied negative values represent stabilisation of the corresponding product fragments relative to the unchanged solvated reactant.
A negative stabilisation contribution reduces the net cleavage enthalpy.
The values are explicitly specified as additive for this hypothetical comparison.
\(\textbf{Calculate solvated homolysis:}\)
\[
\Delta H_{\mathrm{hom,sol}}=
\Delta H_{\mathrm{hom,gas}}+\Delta H_{\mathrm{solv,radicals}}
\]
\[
\Delta H_{\mathrm{hom,sol}}=270+(-40)
\]
\[
\Delta H_{\mathrm{hom,sol}}=+230\ \mathrm{kJ\,mol^{-1}}
\]
\(\textbf{Calculate solvated heterolysis:}\)
\[
\Delta H_{\mathrm{het,sol}}=
\Delta H_{\mathrm{het,gas}}+\Delta H_{\mathrm{solv,ions}}
\]
\[
\Delta H_{\mathrm{het,sol}}=880+(-680)
\]
\[
\Delta H_{\mathrm{het,sol}}=+200\ \mathrm{kJ\,mol^{-1}}
\]
\(\textbf{Determine the relative enthalpy cost:}\)
\[
\Delta H_{\mathrm{hom,sol}}-\Delta H_{\mathrm{het,sol}}
=230-200
\]
\[
=30\ \mathrm{kJ\,mol^{-1}}
\]
Therefore, heterolytic cleavage has the lower enthalpy requirement in the stipulated polar solvent.
\(\textbf{Interpret the changed-condition result:}\)
In the gas phase, homolysis has a much lower supplied enthalpy requirement.
In the solvent, the larger ionic-fragment stabilisation reverses the relative enthalpy ordering.
This demonstrates how the surrounding medium can affect the thermodynamics of different bond-cleavage modes.
\(\textbf{Check the limitation:}\)
The values are reaction enthalpies, not activation energies.
A lower cleavage enthalpy alone does not prove that the corresponding reaction must proceed faster.
The conclusion is therefore thermodynamic and conditional on the supplied model.
\(\textbf{Final answer:}\)
\[
\boxed{\Delta H_{\mathrm{hom,sol}}=+230,\quad
\Delta H_{\mathrm{het,sol}}=+200\ \mathrm{kJ\,mol^{-1}};\quad
\Delta H_{\mathrm{difference}}=30\ \mathrm{kJ\,mol^{-1}}}
\]
511. Ethoxide ion, \(\mathrm{CH_3CH_2O^-}\), participates in two separately specified reactions:
P: It attacks the carbon of bromomethane, \(\mathrm{CH_3Br}\), while bromide departs.
Q: It removes a hydrogen from carbon 3 of 2-bromobutane while the C–H electron pair forms a C2=C3 double bond and the C2–Br electron pair moves to bromine.
No competing pathway occurs in either case.
Which statement correctly identifies the organic products and explains the distinct role of ethoxide?
ⓐ. P gives methanol, whereas Q gives butan-2-ol; ethoxide acts as a proton donor in both reactions.
ⓑ. P gives ethoxyethane, whereas Q gives 2-methoxybutane; the specified removal of a hydrogen in Q is a substitution at carbon.
ⓒ. P gives ethene, whereas Q gives but-1-ene; both reactions must involve carbocation intermediates.
ⓓ. P: substitution gives methoxyethane (ethoxide as nucleophile); Q: elimination gives but-2-ene (ethoxide as base).
Correct Answer: P: substitution gives methoxyethane (ethoxide as nucleophile); Q: elimination gives but-2-ene (ethoxide as base).
Explanation: \(\textbf{Identify the electron-rich reagent:}\)
Ethoxide contains a negatively charged oxygen with available electron pairs.
It can donate an electron pair directly to an electrophilic carbon.
It can also accept a proton from a suitable molecule.
The specified reacting partner and electron movement determine which role is expressed.
\(\textbf{Follow substitution in P:}\)
Bromomethane contains a polar C–Br bond.
Ethoxide oxygen attacks its methyl carbon.
The carbon–bromine bonding pair moves to bromine.
The new organic product is
\[
\mathrm{CH_3CH_2OCH_3}
\]
This is methoxyethane.
Ethoxide has acted as a nucleophile by forming a new C–O bond.
\(\textbf{Examine the elimination site in Q:}\)
The carbon bearing bromine in 2-bromobutane is carbon 2.
The question specifies removal of hydrogen from carbon 3.
This hydrogen is beta to the carbon bearing bromine.
\(\textbf{Follow the stipulated concerted electron movement:}\)
Ethoxide accepts the beta hydrogen.
The C3–H bonding electron pair forms a pi bond between carbons 2 and 3.
The C2–Br bonding pair moves onto bromine.
The resulting alkene is
\[
\mathrm{CH_3CH=CHCH_3}
\]
This is but-2-ene.
Ethoxide becomes ethanol after accepting the proton.
\(\textbf{Compare the reaction classes:}\)
P replaces bromine with an ethoxy group and is a substitution.
Q removes a beta hydrogen and bromine while forming a C=C bond and is an elimination.
The same anionic reagent therefore acts as a nucleophile or base depending on the specified electron-flow pathway.
\(\textbf{Interpret the limits:}\)
The question does not require predicting the preferred pathway under unspecified experimental conditions.
It defines the elementary transformations explicitly.
Likewise, Q's formula alone does not establish the cis/trans distribution of its alkene product.
The correct answer follows from locating the atom attacked by ethoxide and reconstructing the bonds created and broken.
512. Electrophilic substitution on benzene is represented by the following stipulated sequence:
Stage I: The benzene pi system forms a C–E bond to a monovalent electrophile \(\mathrm{E^+}\), giving a positively charged ring intermediate in which the attacked carbon still carries hydrogen.
Stage II: A monovalent base \(\mathrm{B^-}\) removes that hydrogen as a proton; the C–H bonding pair restores the aromatic pi system.
Assume no rearrangement or alternative reaction occurs.
Consider the following statements.
Statement I: Stage I produces a positively charged sigma complex and temporarily disrupts the original aromatic delocalisation.
Statement II: Stage II restores aromaticity by removal of the hydrogen attached to the electrophile-bearing carbon.
Statement III: The complete reaction is electrophilic addition because the electrophile and hydrogen remain attached to the ring in the final product.
Statement IV: The final aromatic product is \(\mathrm{C_6H_5E}\); the base-derived product is \(\mathrm{BH}\), and the total electrical charge is conserved.
Which combination is correct?
ⓐ. Statements I and III only
ⓑ. Statements II, III and IV only
ⓒ. Statements I, II and IV only
ⓓ. Statements I, II and III only
Correct Answer: Statements I, II and IV only
Explanation: \(\textbf{Identify the first electron source:}\)
Benzene possesses a conjugated aromatic pi-electron system.
The electrophile \(\mathrm{E^+}\) is an electron-pair acceptor.
A ring pi-electron pair can form a new sigma bond to the electrophile in the stipulated first step.
\(\textbf{Examine the Stage I intermediate:}\)
The attacked ring carbon now has a new C–E sigma bond while retaining its original C–H bond.
The ring no longer possesses the original uninterrupted aromatic pi arrangement.
The positive charge is delocalised through suitable resonance contributors of the sigma complex.
Therefore, Stage I produces a positively charged intermediate and temporarily disrupts aromaticity.
Statement I is correct.
\(\textbf{Identify the proton removed in Stage II:}\)
The base removes the hydrogen from the ring carbon that acquired electrophile E.
The electron pair of that C–H bond contributes to restoring the ring pi system.
Aromaticity is consequently regained.
Statement II is correct.
\(\textbf{Determine the final substitution pattern:}\)
The electrophile E remains bonded to the aromatic ring.
The original hydrogen at that position has been removed.
Thus, the net organic transformation replaces a ring hydrogen with E.
The product is \(\mathrm{C_6H_5E}\).
\(\textbf{Evaluate the addition claim:}\)
An addition product would retain the additional attachments without the stated compensating removal of hydrogen.
Here, deprotonation restores the original aromatic ring framework.
The complete reaction is electrophilic aromatic substitution, not net addition.
Statement III is incorrect.
\(\textbf{Check charge conservation:}\)
The starting electrophile has charge \(+1\), and the base has charge \(-1\).
The benzene molecule is neutral.
The final substituted benzene is neutral, and protonation of \(\mathrm{B^-}\) gives neutral \(\mathrm{BH}\).
Therefore, the initial and final total charges are both zero.
Statement IV is correct.
The correct combination follows from distinguishing the temporarily non-aromatic intermediate from the final aromatic substitution product.
513. A carbon-centred radical, \(\mathrm{R^{\bullet}}\), can form the same organic bromide by either of the following separately specified reactions:
P: \(\mathrm{R^{\bullet}+Br_2\longrightarrow RBr+Br^{\bullet}}\)
Q: \(\mathrm{R^{\bullet}+Br^{\bullet}\longrightarrow RBr}\)
Assume both equations represent valid elementary radical reactions under their respective conditions.
Which statement correctly distinguishes their roles in a radical-chain process?
ⓐ. Both are propagation steps because both form the same C–Br sigma bond.
ⓑ. P is a termination step because its organic product is neutral; Q is a propagation step because bromine begins as a radical.
ⓒ. Both are termination steps because the radical originally located on R disappears in both equations.
ⓓ. P: propagation regenerates a bromine radical; Q: termination by radical recombination.
Correct Answer: P: propagation regenerates a bromine radical; Q: termination by radical recombination.
Explanation: \(\textbf{Identify the common organic product:}\)
Both P and Q form a C–Br sigma bond.
The organic product is the neutral bromide \(\mathrm{RBr}\).
However, the other reactants and products differ between the two equations.
The role of a step in a radical chain cannot therefore be decided from the organic product alone.
\(\textbf{Examine P's reactants:}\)
One carbon-centred radical reacts with a neutral bromine molecule.
A bromine atom becomes bonded to carbon.
The remaining bromine atom appears as \(\mathrm{Br^{\bullet}}\).
Thus, the step consumes one radical but also produces another radical.
\(\textbf{Interpret P's chain role:}\)
The bromine radical can participate in a subsequent hydrogen-atom-abstraction or related propagation step.
Therefore, P can help maintain the population of chain-carrying radicals.
It is classified as a propagation step in a suitable radical-chain mechanism.
\(\textbf{Examine Q's reactants:}\)
Q begins with two radicals: \(\mathrm{R^{\bullet}}\) and \(\mathrm{Br^{\bullet}}\).
Their unpaired electrons combine to form the new C–Br sigma bond.
The sole stipulated product is neutral \(\mathrm{RBr}\).
No radical is regenerated.
\(\textbf{Interpret Q's chain role:}\)
Two chain-carrying radicals are removed from the reacting mixture.
Therefore, Q is a radical-combination termination step.
\(\textbf{Compare electron accounting:}\)
In P, homolytic displacement within the bromine reagent supplies a new bromine radical as the carbon radical is consumed.
In Q, the unpaired electrons on two separate radical reactants are paired directly.
Neither reaction requires an ionic carbocation or carbanion intermediate.
\(\textbf{Integrate the mechanistic lesson:}\)
The same stable organic molecule can be formed through different elementary radical pathways.
A propagation step replaces one radical carrier with another.
A combination termination step reduces the radical population.
The correct classification consequently depends on the complete radical balance, not merely on whether a new C–Br bond appears.
514. Three structural isomers of bromopentane are considered:
P: \(\mathrm{BrCH_2CH_2CH_2CH_2CH_3}\)
Q: \(\mathrm{CH_3CH(Br)CH_2CH_2CH_3}\)
R: \(\mathrm{CH_3CH_2CH(Br)CH_2CH_3}\)
In each case, elimination is restricted to removal of bromine and a hydrogen from a carbon directly adjacent to the bromine-bearing carbon. No rearrangement occurs, and geometrical isomers are not counted separately. Which statement correctly identifies the distinct constitutional alkene products possible from each substrate?
ⓐ. P gives pent-1-ene; Q gives pent-1-ene and pent-2-ene; R gives pent-2-ene only.
ⓑ. P gives pent-1-ene and pent-2-ene; Q gives pent-2-ene only; R gives pent-1-ene only.
ⓒ. P gives pent-1-ene only; Q and R both give pent-1-ene and pent-2-ene.
ⓓ. P and R both give pent-1-ene only; Q gives pent-2-ene only.
Correct Answer: P gives pent-1-ene; Q gives pent-1-ene and pent-2-ene; R gives pent-2-ene only.
Explanation: \(\textbf{Identify the elimination requirement:}\)
The bromine-bearing carbon is the alpha carbon for the specified elimination.
A hydrogen must be removed from a directly adjacent beta carbon.
The electrons of the broken beta C–H bond form a pi bond between the alpha and beta carbons while bromide departs.
The location of the available beta carbons therefore determines the possible alkene positions.
\(\textbf{Examine P:}\)
P is 1-bromopentane.
Bromine is attached to carbon 1.
Only carbon 2 is directly adjacent to the bromine-bearing carbon within the carbon skeleton.
Elimination consequently creates a double bond between carbons 1 and 2.
The product is pent-1-ene.
\(\textbf{Examine Q:}\)
Q is 2-bromopentane.
The bromine-bearing carbon has carbon neighbours at positions 1 and 3.
Removal of hydrogen from carbon 1 produces pent-1-ene.
Removal from carbon 3 produces pent-2-ene.
These products differ in double-bond connectivity and are constitutional position isomers.
\(\textbf{Examine R:}\)
R is 3-bromopentane.
Carbons 2 and 4 are the two beta carbons.
However, the unbranched parent skeleton is symmetrical about carbon 3.
Elimination towards carbon 2 and elimination towards carbon 4 therefore describe the same constitutional alkene, pent-2-ene.
\(\textbf{Check the counting convention:}\)
Pent-2-ene can possess geometrical stereoisomers, but the question explicitly excludes counting them separately.
No change in carbon skeleton is permitted.
Thus, P gives one constitutional alkene, Q gives two and R gives one.
The correct answer requires locating the beta carbons and then checking symmetry rather than merely counting the number of available beta-hydrogen atoms.
515. 1-Bromo-1-methylcyclohexane undergoes two separately specified eliminations:
Pathway P: A beta hydrogen is removed from a ring carbon directly adjacent to the bromine-bearing carbon.
Pathway Q: A beta hydrogen is removed from the methyl substituent attached to the bromine-bearing carbon.
In each case, the C–H bonding pair forms a double bond to the carbon that loses bromine. Assume no rearrangement or competing reaction.
Which statement correctly identifies the organic products?
ⓐ. P gives methylenecyclohexane, whereas Q gives 1-methylcyclohexene; both contain an endocyclic double bond.
ⓑ. Both pathways give 1-methylcyclohexene because the methyl substituent is not a ring atom.
ⓒ. P gives 1-methylcyclohexene, whereas Q gives methylenecyclohexane; both retain seven carbon atoms but differ in the position of the double bond.
ⓓ. P gives methylcyclohexane, whereas Q gives cycloheptene because elimination transfers the methyl carbon into the ring.
Correct Answer: P gives 1-methylcyclohexene, whereas Q gives methylenecyclohexane; both retain seven carbon atoms but differ in the position of the double bond.
Explanation: \(\textbf{Identify the alpha carbon:}\)
The bromine-bearing carbon is carbon 1 of the cyclohexane ring.
It is directly attached to ring carbons 2 and 6 and to one methyl carbon.
All three of these neighbouring carbons are potential beta positions under the specified elimination framework.
\(\textbf{Follow ring-directed elimination in P:}\)
A hydrogen is removed from ring carbon 2 or carbon 6.
The corresponding C–H electron pair forms a pi bond to ring carbon 1 as bromide departs.
Because the unsubstituted ring directions are equivalent in the starting molecule, these two alternatives give the same constitutional product.
The new C=C bond lies within the six-membered ring.
The product is 1-methylcyclohexene.
\(\textbf{Follow substituent-directed elimination in Q:}\)
The beta hydrogen is instead removed from the methyl carbon attached to ring carbon 1.
A double bond forms between that methyl carbon and ring carbon 1.
The methyl carbon becomes an exocyclic methylene group, \(\mathrm{=CH_2}\).
The product is methylenecyclohexane.
\(\textbf{Check carbon conservation:}\)
The starting substrate contains six ring carbon atoms and one methyl carbon.
Both eliminations retain all seven carbon atoms.
Neither pathway expands the ring or transfers carbon from one position to another.
\(\textbf{Distinguish the product connectivities:}\)
P contains an endocyclic double bond and a methyl substituent.
Q contains an exocyclic double bond connecting a ring carbon to a methylene carbon.
The products are constitutional isomers, not two resonance representations.
The question specifies both pathways, so no conclusion about their relative amounts is required.
The correct answer follows from identifying the beta hydrogen removed and reconstructing the resulting carbon–carbon multiple bond.
516. An alkyl bromide has molecular formula \(\mathrm{C_5H_{11}Br}\). Under separately controlled elimination conditions, it is observed to give exactly two unbranched constitutional alkenes, pent-1-ene and pent-2-ene, without any carbon-skeleton rearrangement. Geometrical isomers of pent-2-ene are counted as one constitutional product.
Which substrate is consistent with these observations?
ⓐ. 1-Bromopentane
ⓑ. 2-Bromopentane
ⓒ. 3-Bromopentane
ⓓ. 2-Bromo-2-methylbutane
Correct Answer: 2-Bromopentane
Explanation: \(\textbf{Infer the required carbon framework:}\)
Both observed products have an unbranched five-carbon skeleton.
The problem excludes rearrangement.
Therefore, the starting bromide must have a framework capable of retaining that unbranched connectivity through elimination.
This immediately makes a branched starting skeleton unsuitable for both observed products.
\(\textbf{Test 1-bromopentane:}\)
Its bromine-bearing carbon is terminal.
It has only one carbon neighbour from which a beta hydrogen can be removed.
Consequently, elimination gives pent-1-ene.
It cannot generate pent-2-ene through the stipulated ordinary beta elimination without an additional transformation.
\(\textbf{Test 2-bromopentane:}\)
The bromine-bearing carbon is directly bonded to carbons 1 and 3.
Removal of hydrogen from carbon 1 produces a C1=C2 bond.
That product is pent-1-ene.
Removal of hydrogen from carbon 3 produces a C2=C3 bond.
That product is pent-2-ene.
Thus, both required constitutional alkenes are accessible.
\(\textbf{Test 3-bromopentane:}\)
The beta carbons are positions 2 and 4.
They are related by reversal of the unbranched five-carbon chain.
Both possible eliminations consequently give pent-2-ene.
Pent-1-ene is not produced by the stipulated direct elimination.
\(\textbf{Test the branched candidate:}\)
2-Bromo-2-methylbutane has a branched carbon skeleton.
Elimination changes a C–H and C–Br bonding arrangement but does not, by itself, convert its carbon skeleton into an unbranched pentane framework.
Its possible alkenes therefore do not match both observed unbranched products.
\(\textbf{Interpret the reverse inference:}\)
The observed product pair identifies the bromine position more precisely than either product considered alone.
It also demonstrates that counting beta carbons is insufficient without checking whether they lead to distinct constitutional structures.
The only consistent substrate among the options is 2-bromopentane.
517. Cyclopentanol, \(\mathrm{C_5H_{10}O}\), is subjected to a two-stage reaction sequence:
Stage I: Intramolecular elimination of water forms a cyclic alkene without opening the ring.
Stage II: Complete catalytic hydrogenation of the new C=C bond occurs without altering the carbon skeleton.
Which statement correctly gives the Stage I product, the Stage II product and the number of degrees of unsaturation at each stage?
ⓐ. Stage I gives cyclopentane and Stage II gives pentane; the degrees of unsaturation are 1, 1 and 0.
ⓑ. Stage I gives cyclopentene and Stage II gives pentane; the degrees of unsaturation are 1, 2 and 0.
ⓒ. Stage I gives cyclopentene and Stage II gives cyclopentane; the degrees of unsaturation are 1, 1 and 0.
ⓓ. Cyclopentene followed by cyclopentane; respective unsaturation degrees are 1, 2 and 1.
Correct Answer: Cyclopentene followed by cyclopentane; respective unsaturation degrees are 1, 2 and 1.
Explanation: \(\textbf{Identify the original structural restriction:}\)
Cyclopentanol contains a saturated five-membered carbon ring.
The ring itself accounts for one degree of unsaturation.
Its oxygen is present in the alcoholic hydroxyl group rather than in a carbonyl double bond.
\(\textbf{Follow dehydration in Stage I:}\)
The stipulated elimination removes the hydroxyl group and a hydrogen from an adjacent ring carbon as water.
A C=C double bond forms within the ring.
The organic product is cyclopentene, \(\mathrm{C_5H_8}\).
The ring remains intact.
\(\textbf{Determine the intermediate unsaturation:}\)
Cyclopentene contains one ring and one double bond.
Therefore, it has two degrees of unsaturation.
This is also consistent with its molecular formula.
\[
\mathrm{DBE}_{\mathrm{cyclopentene}}
=\frac{2(5)+2-8}{2}
\]
\[
\mathrm{DBE}_{\mathrm{cyclopentene}}=2
\]
\(\textbf{Follow hydrogenation in Stage II:}\)
One molecule of hydrogen adds across the C=C bond.
The double bond becomes a C–C single bond.
The carbon ring is not opened.
The final product is cyclopentane, \(\mathrm{C_5H_{10}}\).
\(\textbf{Determine the final unsaturation:}\)
Cyclopentane remains cyclic but has no carbon–carbon multiple bond.
Its single remaining degree of unsaturation is due entirely to the ring.
\(\textbf{Check atom balance in both stages:}\)
\[
\mathrm{C_5H_{10}O\longrightarrow C_5H_8+H_2O}
\]
\[
\mathrm{C_5H_8+H_2\longrightarrow C_5H_{10}}
\]
The elimination removes water and increases the organic product's degree of unsaturation by one.
The addition of hydrogen subsequently removes that newly formed pi bond but not the ring.
\(\textbf{Final answer:}\)
\[
\mathrm{DBE}_{\mathrm{start}}=1,\quad
\mathrm{DBE}_{\mathrm{Stage\ I}}=2,\quad
\mathrm{DBE}_{\mathrm{Stage\ II}}=1
\]
518. 3-Bromo-3-methylpentane has the structure \(\mathrm{CH_3CH_2-C(Br)(CH_3)-CH_2CH_3}\). Two beta-hydrogen-removal patterns are separately stipulated:
P: Removal of H from either ethyl-group methylene carbon adjacent to the bromine-bearing carbon.
Q: Removal of H from the methyl substituent attached directly to the bromine-bearing carbon.
Which statement correctly identifies the two constitutional alkene products and their eligibility for geometrical isomerism?
ⓐ. P gives 3-methylpent-2-ene, which can exhibit geometrical isomerism; Q gives 2-ethylbut-1-ene, which cannot.
ⓑ. P gives 2-ethylbut-1-ene, which can exhibit geometrical isomerism; Q gives 3-methylpent-2-ene, which cannot.
ⓒ. Both give 3-methylpent-2-ene because elimination cannot occur towards a methyl substituent.
ⓓ. P gives 3-methylpent-1-ene, whereas Q gives 3-methylpent-2-ene; both necessarily exhibit geometrical isomerism.
Correct Answer: P gives 3-methylpent-2-ene, which can exhibit geometrical isomerism; Q gives 2-ethylbut-1-ene, which cannot.
Explanation: \(\textbf{Identify the alpha and beta carbons:}\)
The bromine-bearing carbon is attached to two ethyl groups and one methyl group.
Each directly attached saturated carbon can supply beta hydrogen for the specified elimination.
However, the two ethyl groups are constitutionally equivalent in the starting substrate.
\(\textbf{Follow elimination in P:}\)
Removal of H from the methylene carbon of either ethyl group creates a double bond between that carbon and the original bromine-bearing carbon.
The resulting connectivity is
\[
\mathrm{CH_3CH=C(CH_3)CH_2CH_3}
\]
The systematic name is 3-methylpent-2-ene.
Elimination towards either ethyl group gives the same constitutional alkene.
\(\textbf{Test P for geometrical isomerism:}\)
One doubly bonded carbon bears hydrogen and methyl.
The other bears methyl and ethyl.
Each alkene carbon therefore has two different substituents.
Consequently, P can exhibit two geometrical configurations.
\(\textbf{Follow elimination in Q:}\)
Removal of H from the methyl substituent creates a double bond between that substituent carbon and the original bromine-bearing carbon.
The product is
\[
\mathrm{CH_2=C(CH_2CH_3)_2}
\]
Its longest chain containing the double bond has four carbon atoms.
The parent is but-1-ene, with an ethyl group at carbon 2.
Therefore, the product is 2-ethylbut-1-ene.
\(\textbf{Test Q for geometrical isomerism:}\)
The terminal alkene carbon in Q is \(\mathrm{CH_2}\).
It bears two identical hydrogen substituents.
The other alkene carbon also bears two identical ethyl groups.
Therefore, Q does not satisfy the geometrical-isomerism requirement.
\(\textbf{Interpret the comparison:}\)
Both reactions eliminate the same kinds of atoms and preserve the six-carbon skeleton.
However, the choice of beta carbon changes the resulting multiple-bond position and substituent arrangement.
The two products are constitutional isomers.
The stipulated pathways do not establish which would dominate experimentally.
Their geometrical-isomerism eligibility must be assessed separately from their constitutional identity.
519. A secondary carbocation has the structure \(\mathrm{CH_3-CH^+-CH(CH_3)_2}\). Two different 1,2-shifts are proposed from its neighbouring methine carbon:
Pathway P: Transfer of its attached hydrogen with the C–H bonding electron pair.
Pathway Q: Transfer of one attached methyl group with its C–C bonding electron pair.
In either case, the positively charged carbon receives the migrating group and the carbon losing the group becomes positively charged. Assume no additional rearrangement.
Which statement correctly compares the resulting carbocations?
ⓐ. P gives a secondary carbocation, whereas Q gives a tertiary carbocation; Q is therefore necessarily preferred by substitution degree.
ⓑ. Both pathways give the same tertiary carbocation because hydrogen and methyl migrations are resonance-equivalent.
ⓒ. P produces a tertiary carbocation, whereas Q produces a secondary carbocation; only P provides a clear substitution-degree stabilisation advantage over the starting cation.
ⓓ. P produces a carbanion and Q produces a radical because migration of a sigma-bond electron pair necessarily changes the total ionic charge.
Correct Answer: P produces a tertiary carbocation, whereas Q produces a secondary carbocation; only P provides a clear substitution-degree stabilisation advantage over the starting cation.
Explanation: \(\textbf{Classify the starting ion:}\)
The original positively charged carbon is bonded to a methyl carbon, a neighbouring methine carbon and one hydrogen.
It consequently has two carbon neighbours and is secondary.
The adjacent methine carbon is attached to two methyl groups and the positively charged carbon.
It has one hydrogen and can formally supply either the stipulated hydride or methyl migration.
\(\textbf{Follow hydride migration in P:}\)
Hydrogen moves from the neighbouring methine carbon together with its original bonding electron pair.
The initially positive carbon acquires a new C–H bond and becomes neutral.
The methine carbon that loses hydride becomes positively charged.
The rearranged structure is
\[
\mathrm{CH_3CH_2-C^+(CH_3)_2}
\]
The positive centre is now attached to three carbon groups.
It is therefore tertiary.
\(\textbf{Follow methyl migration in Q:}\)
One methyl group moves from the neighbouring methine carbon to the original positive centre together with its C–C bonding electron pair.
The initially positive carbon becomes neutral.
The carbon losing the methyl group becomes the new positive centre.
A suitable representation is
\[
\mathrm{CH_3-CH(CH_3)-CH^+-CH_3}
\]
The new positively charged carbon is attached to two carbon groups.
It is therefore secondary.
\(\textbf{Compare the electronic outcomes:}\)
P converts a secondary carbocation into a tertiary carbocation.
This gives an ordinary substitution-degree stabilisation advantage through increased alkyl donation and suitable hyperconjugative opportunities.
Q converts the original secondary ion into another secondary ion.
It does not obtain the same clear increase in carbocation substitution degree.
\(\textbf{Distinguish rearrangement from resonance:}\)
Both shifts change which carbon is bonded to a hydrogen or methyl group.
They therefore alter the sigma framework.
Neither is a resonance redistribution with all nuclei fixed.
Both also retain the total positive charge.
\(\textbf{Interpret the mechanistic conclusion:}\)
The stated stability consideration favours P qualitatively.
It does not, by itself, provide a numerical activation barrier or exact product distribution for competing rearrangements.
The correct answer requires reconstructing both migrated structures rather than assuming that every 1,2-shift necessarily generates a more stable carbocation.
520. 3-Bromopropan-1-amine, \(\mathrm{H_2NCH_2CH_2CH_2Br}\), undergoes the following stipulated reaction sequence:
Stage I: A lone pair on its amino nitrogen attacks the bromine-bearing carbon intramolecularly, and the C–Br bonding pair moves to bromine.
Stage II: A base removes one nitrogen-bound hydrogen from the cyclic organic ion, leaving the N–H bonding pair on nitrogen.
No intermolecular reaction, elimination or rearrangement occurs.
Which statement correctly identifies the intermediate and final organic product?
ⓐ. Stage I forms a neutral three-membered ring, and Stage II produces aziridine.
ⓑ. Stage I produces an open-chain secondary amine, and Stage II closes a five-membered ring.
ⓒ. Stage I forms a four-membered cyclic carbanion, and Stage II produces neutral pyrrolidine.
ⓓ. Stage I forms a four-membered cyclic ammonium ion, and Stage II produces neutral azetidine.
Correct Answer: Stage I forms a four-membered cyclic ammonium ion, and Stage II produces neutral azetidine.
Explanation: \(\textbf{Identify the nucleophilic and electrophilic sites:}\)
The molecule contains a neutral amino group with an available nitrogen lone pair.
It also contains a saturated carbon bonded to bromine.
The C–Br bond is polarised, allowing the stipulated nucleophilic substitution at that carbon.
Both sites occur within the same molecule.
\(\textbf{Trace the ring-closing attack:}\)
Nitrogen donates its lone pair to the bromine-bearing terminal carbon.
A new C–N sigma bond forms.
Simultaneously, the C–Br bonding pair moves onto bromine, producing bromide ion.
\(\textbf{Determine the ring size:}\)
The original chain contains three carbon atoms between the original nitrogen attachment and the terminal carbon receiving attack.
Closure therefore creates the sequence
\[
\mathrm{N-C-C-C}
\]
The new ring contains one nitrogen and three carbon atoms.
It is a four-membered nitrogen-containing ring.
\(\textbf{Determine Stage I's nitrogen charge:}\)
The original amino nitrogen has one carbon bond and two N–H bonds.
After forming the new carbon bond, it has four ordinary sigma bonds.
It consequently carries positive formal charge.
Thus, Stage I produces a cyclic ammonium ion with bromide as its counterion.
\(\textbf{Follow deprotonation in Stage II:}\)
The base removes one nitrogen-bound proton.
The original N–H bonding electron pair remains with nitrogen.
Nitrogen then has two C–N bonds, one N–H bond and one lone pair.
It is electrically neutral.
\(\textbf{Identify the final ring:}\)
The neutral four-membered saturated ring containing one nitrogen is azetidine.
The reaction does not produce aziridine, which has a three-membered ring, or pyrrolidine, which has a five-membered ring.
The correct conclusion requires identifying the intramolecular nucleophile, counting ring atoms and tracking nitrogen's formal charge through bond formation and deprotonation.
