301. Two sulfur-containing nucleophiles react separately with bromomethane under conditions permitting substitution at methyl carbon:
P: Methanethiolate ion, \(\mathrm{CH_3S^-}\).
Q: Dimethyl sulfide, \(\mathrm{CH_3-S-CH_3}\).
In each pathway, a sulfur lone pair forms the new S–C bond and the C–Br bonding pair moves to bromine. No additional proton transfer or bond cleavage occurs. Which statement correctly gives the immediate sulfur-containing products?
ⓐ. P produces neutral dimethyl sulfide, whereas Q produces a trimethylsulfonium cation accompanied by bromide.
ⓑ. P produces a trimethylsulfonium cation, whereas Q produces neutral dimethyl sulfide.
ⓒ. Both produce neutral dimethyl sulfide because sulfur can form only two covalent bonds under all circumstances.
ⓓ. Both produce methanethiolate ion because bromide departure necessarily breaks an existing S–C bond.
Correct Answer: P produces neutral dimethyl sulfide, whereas Q produces a trimethylsulfonium cation accompanied by bromide.
Explanation: \(\textbf{Identify the common electrophile:}\)
Bromomethane contains a polar C–Br bond.
Bromine is more electronegative than carbon.
The methyl carbon consequently has electrophilic character.
In both pathways, the sulfur reagent provides an electron pair that can attack this carbon.
\(\textbf{Follow substitution by methanethiolate in P:}\)
Methanethiolate contains sulfur bonded to one methyl group and carries a negative charge.
A sulfur lone pair forms a new bond to the methyl carbon of bromomethane.
The C–Br bonding pair moves onto bromine.
The resulting sulfur atom is bonded to two methyl groups.
The organic product is
\[
\mathrm{CH_3-S-CH_3}
\]
It is neutral dimethyl sulfide.
Bromide is the separate anionic product.
\(\textbf{Follow substitution by dimethyl sulfide in Q:}\)
Dimethyl sulfide is a neutral molecule with sulfur bonded to two methyl groups.
Its sulfur atom possesses nonbonding electron pairs.
One such pair forms an additional bond to methyl carbon.
Sulfur now has three S–C sigma bonds and carries a positive formal charge.
The organic product is
\[
\mathrm{(CH_3)_3S^+}
\]
It is the trimethylsulfonium cation.
Bromide again forms as the leaving anion.
\(\textbf{Compare the sulfur valence environments:}\)
The starting sulfur in P is anionic and singly substituted.
Formation of another S–C bond gives neutral divalent sulfur in the sulfide product.
The starting sulfur in Q is already neutral and divalent.
Formation of an additional S–C bond gives positively charged trivalent sulfur in the sulfonium product.
\(\textbf{Check charge conservation:}\)
In P, the initial total charge is \(-1\), and the final bromide carries charge \(-1\).
In Q, the initial total charge is zero, while the sulfonium and bromide charges cancel.
Therefore, both substitutions conserve charge.
The difference arises from the initial sulfur electron and bond count, not from any change in the electrophilic methyl carbon.
The example demonstrates how two electron-pair donors can attack the same electrophile yet produce different formal-charge states.
302. A methyl radical, \(\mathrm{CH_3^{\bullet}}\), is involved in two separately described reactions:
P: \(\mathrm{CH_3^{\bullet}+Cl_2\longrightarrow CH_3Cl+Cl^{\bullet}}\)
Q: \(\mathrm{CH_3^{\bullet}+Cl^{\bullet}\longrightarrow CH_3Cl}\)
Which statement correctly compares the radical bookkeeping and implications for a radical-chain process?
ⓐ. P is termination because it consumes methyl radical, whereas Q is propagation because it forms chloromethane.
ⓑ. Both are propagation steps because chloromethane is formed in both.
ⓒ. Both are termination steps because neither contains a radical in the organic product.
ⓓ. P propagates the chain by generating another chlorine radical; Q terminates the chain by radical combination.
Correct Answer: P propagates the chain by generating another chlorine radical; Q terminates the chain by radical combination.
Explanation: \(\textbf{Examine the initial radical population:}\)
In P, the reactants include one methyl radical and one non-radical chlorine molecule.
In Q, the reactants include one methyl radical and one chlorine radical.
Thus, the initial numbers of radical particles differ even though both reactions form chloromethane.
\(\textbf{Analyse electron movement in P:}\)
The methyl radical reacts with the Cl–Cl bond.
A new C–Cl bond forms in chloromethane.
The other chlorine atom emerges as a chlorine radical.
Therefore, the step consumes one radical and generates another.
The chain-carrying radical character remains available for further reaction.
P can consequently function as a propagation step.
\(\textbf{Analyse electron movement in Q:}\)
The methyl radical and chlorine radical each possess an unpaired electron.
Their two unpaired electrons become paired in the newly formed C–Cl sigma bond.
The product is neutral chloromethane.
No new radical is generated by the stated combination.
Therefore, Q consumes two radicals and can function as a termination step.
\(\textbf{Compare the identical organic product:}\)
Both pathways produce the same C–Cl sigma connectivity in chloromethane.
However, their other products differ.
P produces an additional chlorine radical.
Q produces no additional species.
\(\textbf{Interpret the mechanistic consequence:}\)
A product's molecular formula does not uniquely establish whether the elementary step is propagation or termination.
The net number of radicals must also be considered.
In P, the radical population remains unchanged at one radical particle.
In Q, two radical particles are removed.
These different radical balances have different consequences for the continuation of a chain mechanism.
The correct classification therefore depends on following all participating species rather than examining only the organic product.
303. Pyridine and pyrrole are two nitrogen-containing aromatic heterocycles. In pyridine, nitrogen is part of a six-membered aromatic ring; in pyrrole, an N–H unit is part of a five-membered aromatic ring. Which statement correctly compares the involvement of their nitrogen lone pairs in aromaticity and their availability for nitrogen-centred protonation?
ⓐ. The pyridine nitrogen lone pair supplies two electrons to the aromatic sextet, whereas the pyrrole nitrogen lone pair remains entirely outside its aromatic system.
ⓑ. Pyridine's N lone pair lies outside the aromatic sextet and is more available for protonation; pyrrole's lone pair enters its sextet.
ⓒ. Both nitrogen lone pairs lie outside their aromatic systems, so they necessarily have identical protonation behaviour.
ⓓ. The pyrrole nitrogen possesses no lone pair because it forms one N–H bond and two ring bonds.
Correct Answer: Pyridine's N lone pair lies outside the aromatic sextet and is more available for protonation; pyrrole's lone pair enters its sextet.
Explanation: \(\textbf{Examine the nitrogen atom in pyridine:}\)
Pyridine contains a six-membered aromatic ring with one nitrogen atom replacing a ring carbon.
The nitrogen contributes one electron from a p orbital to the aromatic pi system.
Its nonbonding electron pair occupies an orbital lying in the plane of the ring.
That lone pair is not part of the six-electron aromatic pi system.
It therefore remains available for suitable nitrogen-centred protonation.
\(\textbf{Examine the nitrogen atom in pyrrole:}\)
Pyrrole contains a five-membered aromatic ring with one N–H unit.
Its two carbon–carbon double bonds contribute four pi electrons.
The nitrogen lone pair supplies two additional electrons to the conjugated ring system.
The resulting six-electron arrangement satisfies the ordinary aromatic electron-counting condition.
Therefore, the nitrogen lone pair is an important part of pyrrole's aromatic electronic structure.
\(\textbf{Compare the consequences of nitrogen protonation:}\)
In pyridine, protonation can occur through the lone pair that lies outside the aromatic pi system.
The aromatic sextet can remain intact in the resulting pyridinium ion.
In pyrrole, ordinary protonation directly at nitrogen would use the lone pair participating in its aromatic sextet.
Such a change is electronically less favourable in the basic comparison.
\(\textbf{Identify the nucleophilic distinction:}\)
Both nitrogen atoms possess electron pairs in their appropriate neutral Lewis descriptions.
However, those pairs have different roles within their molecular structures.
A lone pair involved in maintaining aromatic conjugation cannot be assumed to be equally available for nitrogen-centred donation as a lone pair outside that system.
This helps explain the substantially greater ordinary basicity of pyridine compared with pyrrole.
The comparison does not imply that pyrrole can never undergo protonation or participate in reactions at other ring positions.
The correct answer follows from identifying which nitrogen electrons contribute to aromaticity and considering what would happen if that pair were used to form a new N–H bond.
304. Isotopically labelled hydroxide, \(\mathrm{^{18}OH^-}\), adds to ethanal, \(\mathrm{CH_3CHO}\), in the following specified two-stage sequence:
Stage I: The labelled hydroxide oxygen forms a bond to the carbonyl carbon, while the original C=O pi electrons move to the original, unlabelled carbonyl oxygen.
Stage II: The negatively charged original carbonyl oxygen receives one proton. No oxygen exchange, dehydration or other reaction occurs.
Which statement correctly identifies the oxygen atoms in the final neutral product?
ⓐ. The labelled oxygen becomes the original carbonyl oxygen, while the unlabelled oxygen leaves as water.
ⓑ. Both oxygen atoms become isotopically labelled because protonation exchanges their nuclear identities.
ⓒ. Labelled oxygen becomes one hydroxyl oxygen; the original carbonyl oxygen becomes the other hydroxyl oxygen.
ⓓ. The labelled oxygen is removed during protonation, leaving only one oxygen atom in the product.
Correct Answer: Labelled oxygen becomes one hydroxyl oxygen; the original carbonyl oxygen becomes the other hydroxyl oxygen.
Explanation: \(\textbf{Identify the two oxygen sources:}\)
Ethanal initially contains one carbonyl oxygen atom of ordinary isotopic composition.
The incoming hydroxide contains a different oxygen atom labelled with oxygen-18.
The isotopic label identifies the nucleus of that particular oxygen atom.
It does not identify an electron pair or a particular formal-charge state.
\(\textbf{Follow the labelled nucleophile in Stage I:}\)
The labelled hydroxide oxygen donates an electron pair to the electrophilic carbonyl carbon.
A new C–O sigma bond forms.
The labelled oxygen remains attached to its original hydrogen.
Therefore, it becomes the oxygen of the newly introduced hydroxyl group.
\(\textbf{Follow the original carbonyl oxygen:}\)
The original C=O pi electrons move towards the original carbonyl oxygen.
That oxygen becomes singly bonded to the former carbonyl carbon.
It acquires a negative formal charge in the immediate addition intermediate.
The oxygen nucleus itself has not moved or been exchanged.
\(\textbf{Apply the protonation in Stage II:}\)
The negatively charged original carbonyl oxygen accepts a proton.
Its charge is neutralised, and it becomes a hydroxyl oxygen.
The labelled oxygen remains part of its previously formed C–O bond.
The final product can be represented by the atom-labelled connectivity
\[
\mathrm{CH_3-CH(^{18}OH)(^{16}OH)}
\]
where \(\mathrm{^{16}O}\) denotes the original unlabelled carbonyl oxygen for illustration.
\(\textbf{Check the product structure:}\)
The former carbonyl carbon is attached to methyl, hydrogen and two hydroxyl groups.
The final molecule is a geminal diol.
The oxygen atoms are attached to the same carbon, but their nuclear isotopic identities remain distinguishable.
The specified protonation changes an O–H bonding state rather than exchanging oxygen atoms.
This demonstrates why curved-arrow electron movement must not be confused with movement of atomic nuclei.
The correct conclusion follows by tracking each oxygen source through the separate bond-formation and proton-transfer stages.
305. A chemist considers two organosulfur reactions at methyl carbon:
P: \(\mathrm{CH_3S^-+CH_3Br\longrightarrow Product\ P+Br^-}\)
Q: \(\mathrm{(CH_3)_2S+CH_3Br\longrightarrow Product\ Q+Br^-}\)
Both are described as electron-pair substitution reactions. Which statement correctly compares the sulphur formal charge and number of S–C bonds in the organic products?
ⓐ. P has neutral sulfur with two S–C bonds, while Q has positively charged sulfur with three S–C bonds.
ⓑ. P has positively charged sulfur with two S–C bonds, while Q has neutral sulfur with three S–C bonds.
ⓒ. Both products have neutral sulfur with three S–C bonds because bromide carries all the reaction charge.
ⓓ. P has sulfur carrying charge \(-1\) with one S–C bond, while Q has sulfur carrying charge \(+1\) with two S–C bonds.
Correct Answer: P has neutral sulfur with two S–C bonds, while Q has positively charged sulfur with three S–C bonds.
Explanation: \(\textbf{Examine the nucleophile in P:}\)
Methanethiolate has one methyl group bonded to sulfur and carries negative charge.
Its sulfur centre contains available nonbonding electron pairs.
It can donate an electron pair towards electrophilic methyl carbon.
\(\textbf{Determine the substitution product in P:}\)
A new S–C bond forms as bromide receives the original C–Br bonding pair.
The resulting organic structure is
\[
\mathrm{CH_3-S-CH_3}
\]
Sulfur now forms two ordinary sigma bonds to carbon.
Its formal charge is zero.
The organic product is dimethyl sulfide.
\(\textbf{Examine the nucleophile in Q:}\)
Dimethyl sulfide is initially neutral.
Its sulfur atom already has two S–C sigma bonds.
It also possesses nonbonding electron pairs.
A sulfur lone pair can form a third S–C bond to the methyl group of bromomethane.
\(\textbf{Determine the substitution product in Q:}\)
The resulting structure is
\[
\mathrm{(CH_3)_3S^+}
\]
The sulfur atom now has three S–C bonds.
Its ordinary formal-charge assignment is \(+1\).
The organic product is a trimethylsulfonium cation.
\(\textbf{Check the reaction charges:}\)
The total charge of the P reactants is \(-1\).
The products are neutral dimethyl sulfide and bromide, giving net charge \(-1\).
The Q reactants have total charge zero.
The products have charges \(+1\) and \(-1\), giving net charge zero.
Therefore, charge conservation is satisfied in both reactions.
The same new bond type can be formed from a neutral or anionic sulfur donor, but the resulting formal-charge state depends on the donor's initial electron and bond count.
This distinction is not determined solely by the fact that bromide leaves.
306. A secondary butyl radical has the structure \(\mathrm{CH_3-CH^{\bullet}-CH_2CH_3}\). Assume that the radical-bearing carbon is effectively planar and that the two faces of its radical orbital are equally accessible in an achiral environment. A bromine radical combines with it to form a C–Br sigma bond, without rearrangement or competing reaction. Which statement correctly describes the stereochemical possibility of the resulting organic product?
ⓐ. The product is a single achiral form of 2-bromobutane because radical combination cannot create a stereogenic carbon.
ⓑ. Only one enantiomer forms because the unpaired electron permanently specifies a tetrahedral configuration before bromine approaches.
ⓒ. Cis- and trans-2-bromobutane form because radical combination creates a restricted carbon–carbon double bond.
ⓓ. Bromine capture can produce the two enantiomers of 2-bromobutane in equal amounts under the stated equal-face conditions.
Correct Answer: Bromine capture can produce the two enantiomers of 2-bromobutane in equal amounts under the stated equal-face conditions.
Explanation: \(\textbf{Examine the starting radical geometry:}\)
The radical-bearing carbon is directly attached to methyl, ethyl and hydrogen groups.
It has three ordinary sigma bonds and one unpaired electron.
The question specifies an effectively planar radical centre.
Therefore, the two faces associated with the radical orbital are available for approach by another radical.
\(\textbf{Identify the incoming species:}\)
The bromine radical contains one unpaired electron.
Radical combination allows the unpaired electrons of the two species to form a new C–Br sigma bond.
The process differs from ordinary nucleophilic substitution because no electron pair is donated from an anionic nucleophile to displace a leaving group.
There is also no C–Br bond being broken in the starting carbon radical.
\(\textbf{Determine the product structure:}\)
Formation of the C–Br bond gives
\[
\mathrm{CH_3-CH(Br)-CH_2CH_3}
\]
The product is 2-bromobutane.
The formerly radical-bearing carbon now possesses four sigma bonds and becomes tetrahedral.
\(\textbf{Evaluate the stereogenic centre:}\)
That carbon is attached to bromine, hydrogen, methyl and ethyl groups.
All four substituents are different.
Therefore, the new tetrahedral carbon is stereogenic.
The product can exist as a pair of non-superimposable mirror-image arrangements.
\(\textbf{Interpret the two approach faces:}\)
Approach of bromine from one face of the planar radical centre gives one tetrahedral arrangement.
Approach from the opposite face gives the mirror-image arrangement.
The question explicitly states that both faces are equally accessible in an achiral environment.
Therefore, neither enantiomer is favoured by the stipulated model.
Equal formation gives a racemic mixture.
\(\textbf{Distinguish radical stereochemistry from alkene geometry:}\)
The reaction creates a new C–Br sigma bond rather than a carbon–carbon double bond.
Therefore, cis–trans alkene isomerism is not involved.
The stereochemical distinction arises from the creation of a tetrahedral carbon with four different substituents.
The result also does not follow from any compulsory inversion rule because no concerted backside displacement at a starting tetrahedral halide has been specified.
The conclusion combines radical electron pairing, change in local carbon geometry and equal-face stereochemical accessibility.
307. Bromoacetone, \(\mathrm{CH_3-C(=O)-CH_2Br}\), contains two potentially electrophilic carbon centres. Methoxide ion, \(\mathrm{CH_3O^-}\), is allowed to react through one of the following separately specified elementary pathways:
Pathway P: Methoxide attacks the carbonyl carbon, and the C=O pi electrons move onto oxygen. No proton transfer occurs.
Pathway Q: Methoxide attacks the bromine-bearing methylene carbon, and the C–Br bonding electron pair moves to bromine.
Which statement correctly identifies the immediate organic products and their charges?
ⓐ. P gives neutral \(\mathrm{CH_3C(OH)(OCH_3)CH_2Br}\); Q gives an anionic ether because methoxide retains its negative charge.
ⓑ. P gives \(\mathrm{CH_3C(O^-)(OCH_3)CH_2Br}\); Q gives neutral \(\mathrm{CH_3COCH_2OCH_3}\), with bromide formed separately.
ⓒ. Both pathways give \(\mathrm{CH_3COCH_2OCH_3}\) because a nucleophile always replaces a halogen when one is present.
ⓓ. P gives \(\mathrm{CH_3C^+(OCH_3)CH_2Br}\); Q gives \(\mathrm{CH_3C(O^-)(OCH_3)CH_2Br}\).
Correct Answer: P gives \(\mathrm{CH_3C(O^-)(OCH_3)CH_2Br}\); Q gives neutral \(\mathrm{CH_3COCH_2OCH_3}\), with bromide formed separately.
Explanation: \(\textbf{Identify the two electrophilic centres:}\)
Bromoacetone contains a carbonyl group and a carbon–bromine bond.
The carbonyl carbon is electrophilic because the C=O bond is polarised towards oxygen.
The bromine-bearing carbon is also potentially electrophilic because bromine withdraws bonding electron density.
These two sites permit different formal nucleophilic reaction pathways.
\(\textbf{Follow methoxide addition in P:}\)
Methoxide possesses a negatively charged oxygen with available nonbonding electron pairs.
An oxygen lone pair forms a new sigma bond to the carbonyl carbon.
Simultaneously, the original C=O pi electrons move onto carbonyl oxygen.
The carbonyl carbon becomes tetrahedral.
The original carbonyl oxygen acquires a negative charge.
The immediate organic intermediate is
\[
\mathrm{CH_3C(O^-)(OCH_3)CH_2Br}
\]
The carbon–bromine bond remains unchanged.
\(\textbf{Follow substitution in Q:}\)
Methoxide attacks the bromine-bearing methylene carbon instead.
Its oxygen forms a new C–O sigma bond.
The original C–Br bonding electron pair moves to bromine.
Bromide ion is consequently displaced.
The organic product is
\[
\mathrm{CH_3COCH_2OCH_3}
\]
This organic product is electrically neutral and retains the original carbonyl double bond.
\(\textbf{Compare the charge distributions:}\)
Both pathways begin with total charge \(-1\).
In P, the negative charge remains on the organic addition intermediate.
In Q, the organic product is neutral and the departing bromide carries the negative charge.
Thus, both pathways conserve electrical charge.
\(\textbf{Interpret the reaction-centre difference:}\)
Carbonyl addition changes C=O bond order without removing bromine.
Substitution changes the C–Br connection without disturbing the carbonyl.
The presence of two electrophilic centres does not mean that every nucleophile must attack the same one.
The question specifies the reacting centre in each case, so no assumption about actual product dominance is required.
The correct answer depends on tracking the electron pair independently through two chemically distinct transformations.
308. A mixture contains \(\mathrm{CH_3^{\bullet}}\), \(\mathrm{C_2H_5^{\bullet}}\) and \(\mathrm{Cl^{\bullet}}\) radicals. Consider the following three possible elementary reactions:
I. \(\mathrm{CH_3^{\bullet}+C_2H_5^{\bullet}\longrightarrow C_3H_8}\)
II. \(\mathrm{C_2H_5^{\bullet}+Cl_2\longrightarrow C_2H_5Cl+Cl^{\bullet}}\)
III. \(\mathrm{Cl^{\bullet}+C_2H_6\longrightarrow HCl+C_2H_5^{\bullet}}\)
Which statement correctly classifies the steps and identifies their implications for the number of radical particles?
ⓐ. I is termination and reduces the number of radicals by two; II and III are propagation steps that each consume one radical and generate another.
ⓑ. I is propagation because propane contains a new C–C bond; II and III are termination because each forms a stable molecule.
ⓒ. All three are propagation because each equation contains an organic compound among its products.
ⓓ. I is initiation because two radicals are present initially; II and III are termination because their reactants contain radicals.
Correct Answer: I is termination and reduces the number of radicals by two; II and III are propagation steps that each consume one radical and generate another.
Explanation: \(\textbf{Examine Step I:}\)
The methyl radical and ethyl radical each contain an unpaired electron.
Their radical-bearing carbon atoms combine to form a new carbon–carbon sigma bond.
The resulting molecule is propane.
Neither product carbon retains an unpaired electron.
Therefore, two radicals are consumed without generation of a new radical.
Step I is a radical-termination process.
\(\textbf{Examine Step II:}\)
An ethyl radical reacts with chlorine.
The C–Cl bond forms while a chlorine radical is generated from the chlorine molecule.
One radical enters the step and one radical emerges.
The chain-carrying radical character is therefore maintained.
Step II is a propagation process.
\(\textbf{Examine Step III:}\)
A chlorine radical abstracts hydrogen from ethane.
An H–Cl bond forms, and an ethyl radical is generated.
Again, one radical is consumed and another is produced.
Therefore, Step III is propagation.
\(\textbf{Compare radical populations:}\)
The net change in radical count for Step I is
\[
\Delta N_{\mathrm{radical}}=0-2=-2
\]
For each of Steps II and III,
\[
\Delta N_{\mathrm{radical}}=1-1=0
\]
\(\textbf{Interpret the mechanism:}\)
Formation of a stable molecule does not automatically make an elementary step a termination step.
Propagation can produce stable molecules while simultaneously regenerating radicals.
Similarly, the presence of radicals among the reactants does not establish initiation.
The decisive feature is whether radicals are created, transferred or eliminated.
The radical population is reduced in Step I but maintained in Steps II and III.
The correct classification therefore requires complete radical bookkeeping rather than inspection of the organic product alone.
309. Ethylbenzene, \(\mathrm{C_6H_5CH_2CH_3}\), contains three different classes of carbon-bound hydrogen positions: aromatic-ring hydrogens, benzylic methylene hydrogens and terminal methyl hydrogens. Exactly one hydrogen atom is removed homolytically, with no subsequent reaction.
Which statement correctly identifies the radical that possesses the characteristic benzylic resonance-delocalisation pathway and the number of hydrogen atoms capable of producing that radical?
ⓐ. Removal of any of the five ring hydrogens produces the same benzylic radical, so five hydrogens qualify.
ⓑ. Removal of any of the three terminal methyl hydrogens produces a benzylic radical, so three hydrogens qualify.
ⓒ. Removal of either methylene hydrogen produces a benzylic radical, so two hydrogens qualify.
ⓓ. Removal of any hydrogen produces a benzylic radical because all carbon atoms belong to one molecule containing benzene.
Correct Answer: Removal of either methylene hydrogen produces a benzylic radical, so two hydrogens qualify.
Explanation: \(\textbf{Identify the benzylic carbon:}\)
The benzylic position is the saturated carbon directly attached to the benzene ring.
In ethylbenzene, this is the methylene carbon of the ethyl substituent.
It lies between the aromatic ring and the terminal methyl group.
\(\textbf{Examine abstraction from the methylene group:}\)
The benzylic methylene carbon contains two hydrogen atoms.
Homolytic removal of either hydrogen produces the carbon-centred radical
\[
\mathrm{C_6H_5CH^{\bullet}CH_3}
\]
The radical-bearing carbon remains directly attached to the aromatic ring.
Its singly occupied orbital can interact with the aromatic pi-electron system.
Therefore, the resulting radical possesses benzylic resonance-delocalisation opportunities.
\(\textbf{Examine abstraction from the terminal methyl group:}\)
Removing a hydrogen from the terminal methyl carbon produces
\[
\mathrm{C_6H_5CH_2CH_2^{\bullet}}
\]
The radical centre is separated from the aromatic ring by a saturated methylene carbon.
It is not directly benzylic.
The same ordinary benzylic resonance pathway is consequently unavailable.
\(\textbf{Examine abstraction from the ring:}\)
Removing an aromatic-ring hydrogen places the radical centre on a carbon belonging to the aromatic ring.
Such a centre is aryl rather than benzylic.
It must not be classified as benzylic merely because the molecule contains a benzene ring.
\(\textbf{Determine the qualifying hydrogen count:}\)
Only the two hydrogens on the benzylic methylene carbon meet the specified requirement.
Thus,
\[
N_{\mathrm{benzylic\ H}}=2
\]
The two hydrogen positions give the same constitutional radical when stereochemical distinctions are ignored.
\(\textbf{Interpret the structural requirement:}\)
The important feature is the direct attachment of the radical-bearing carbon to the aromatic ring.
Neither the total number of molecular hydrogens nor the number of aromatic hydrogens determines that classification.
The correct distinction follows by locating the original C–H bond, reconstructing the resulting radical and testing its orbital connectivity.
\(\textbf{Final answer:}\)
\[
N_{\mathrm{benzylic\ H}}=2
\]
310. A neutral ether, \(\mathrm{CH_3CH_2-O-CH_3}\), is protonated at oxygen to form \(\mathrm{CH_3CH_2-O^+(H)-CH_3}\). The protonated ether is then subjected to a hypothetical heterolytic cleavage in which the O–CH\(_3\) bond breaks and its complete bonding pair remains with oxygen. Which statement correctly gives the immediate fragments and their formal charges?
ⓐ. Neutral methanol and ethyl carbocation are produced because oxygen always retains the ethyl group.
ⓑ. Neutral ethanol and methyl carbocation are produced; the overall positive charge remains on the methyl fragment.
ⓒ. Ethoxide ion and methyl radical are produced because the oxygen initially carries a positive charge.
ⓓ. Methyl anion and protonated ethanol are produced because the C–O bond is polarised towards carbon.
Correct Answer: Neutral ethanol and methyl carbocation are produced; the overall positive charge remains on the methyl fragment.
Explanation: \(\textbf{Identify the protonated ether structure:}\)
The original ether oxygen is attached to an ethyl group and a methyl group.
Protonation adds an O–H bond.
The oxygen now forms three sigma bonds and carries a positive formal charge.
The protonated ether has overall charge \(+1\).
\(\textbf{Determine which bond breaks:}\)
The question specifies cleavage of the O–CH\(_3\) bond.
The bond to the ethyl group remains intact.
Therefore, the resulting oxygen-containing fragment must retain the ethyl group.
\(\textbf{Follow the bonding electron pair:}\)
Both electrons of the O–CH\(_3\) bond move to oxygen.
This is heterolytic cleavage.
The methyl group loses its share of the bonding electron pair.
It consequently becomes a methyl carbocation.
\(\textbf{Examine the oxygen fragment:}\)
Before cleavage, oxygen is bonded to ethyl, methyl and hydrogen.
After losing the methyl connection and receiving the O–C bonding pair, it remains bonded to ethyl and hydrogen.
It has two ordinary sigma bonds and two lone pairs.
Therefore, it is neutral ethanol.
The formal transformation is
\[
\mathrm{CH_3CH_2-O^+(H)-CH_3}
\]
\[
\mathrm{\longrightarrow CH_3CH_2OH+CH_3^+}
\]
\(\textbf{Check charge conservation:}\)
The reactant has charge \(+1\).
Ethanol is neutral, while the methyl carbocation has charge \(+1\).
The total product charge remains \(+1\).
\(\textbf{Distinguish formal cleavage from reaction feasibility:}\)
The question defines a hypothetical cleavage and asks for electron accounting.
It does not assert that formation of a free methyl carbocation from this ether is favourable under ordinary experimental conditions.
The formal fragments follow from the specified bond and electron destination.
A methyl radical would require one-electron division, which is inconsistent with the stated heterolysis.
The correct answer therefore requires tracking the exact bond broken, oxygen's changing formal charge and the conserved overall charge.
311. A neutral compound contains a carbonyl group adjacent to an alkene:
\[
\mathrm{CH_2=CH-C(=O)-CH_3}
\]
A nucleophile, \(\mathrm{Nu^-}\), is allowed to attack under two separately specified pathways:
P: Direct attack at the carbonyl carbon, with the C=O pi electrons moving to oxygen.
Q: Attack at the terminal alkene carbon, with the conjugated pi-electron system shifting so that the original carbonyl oxygen receives the electron pair.
No proton transfer occurs.
Which statement correctly identifies the essential structural difference between the immediate addition intermediates?
ⓐ. P: carbonyl-carbon attack; Q: terminal alkene-carbon attack with enolate delocalisation.
ⓑ. Both pathways must form the new bond at the carbonyl carbon because oxygen is the most electronegative atom.
ⓒ. P gives a saturated carbonyl group without an oxygen charge, whereas Q leaves the original conjugated system completely unchanged.
ⓓ. Q produces an alkyl radical because any addition to a C=C bond requires homolytic cleavage.
Correct Answer: P: carbonyl-carbon attack; Q: terminal alkene-carbon attack with enolate delocalisation.
Explanation: \(\textbf{Identify the conjugated system:}\)
The molecule contains a C=C double bond directly connected to a C=O group.
The relevant sequence is \(\mathrm{C=C-C=O}\).
This alternating multiple-bond arrangement permits conjugative electron movement.
The electrophilic response of the system is therefore not restricted to treating each double bond as completely independent.
\(\textbf{Follow direct carbonyl attack in P:}\)
The nucleophile forms a new sigma bond at the carbonyl carbon.
The C=O pi electrons move onto oxygen.
The carbonyl carbon changes from approximately trigonal-planar to tetrahedral geometry.
The directly formed intermediate contains an alkoxide oxygen.
The original C=C bond remains in place.
\(\textbf{Follow conjugate attack in Q:}\)
The nucleophile forms a new sigma bond at the terminal alkene carbon.
The conjugated pi-electron system redistributes.
The original C=C bond is shifted towards the bond connecting the alkene and carbonyl portions.
The C=O pi electrons move onto oxygen.
The immediate electron distribution can consequently be represented by an oxygen-centred anionic contributor.
\(\textbf{Compare the reaction centres:}\)
P forms the new sigma bond at the carbonyl carbon.
Q forms it at a different carbon belonging to the conjugated system.
The two pathways therefore create different carbon–nucleophile connectivities.
\(\textbf{Check electron accounting:}\)
In both cases, the initial nucleophile has charge \(-1\).
Neither elementary addition step includes proton transfer or departure of a leaving group.
The resulting organic intermediate consequently retains the total negative charge.
Neither pathway requires formation of a free radical.
\(\textbf{Interpret the stated condition:}\)
The question specifies the electron-flow alternatives and does not require predicting which would dominate for an unspecified nucleophile and solvent.
The essential distinction is between direct carbonyl addition and conjugate addition.
Understanding the electron movement requires recognising the conjugated system, identifying the new sigma-bond position and conserving the negative charge.
This illustrates why the location of nucleophilic attack can affect connectivity even when both pathways begin with the same polarised molecule.
312. Two carbon-centred reactive intermediates are formed independently from the same carbon framework:
P: \(\mathrm{CH_3-CH^{\bullet}-CH_2-CH_3}\)
Q: \(\mathrm{CH_3-CH^+-CH_2-CH_3}\)
Each is trapped by a chlorine-containing reagent under separately specified conditions:
P combines with \(\mathrm{Cl^{\bullet}}\).
Q receives an electron pair from \(\mathrm{Cl^-}\).
No rearrangement, elimination or competing reaction occurs.
Which statement is correct?
ⓐ. Both processes form 2-chlorobutane with the same atom-to-atom connectivity, but P involves radical electron pairing whereas Q involves nucleophilic electron-pair donation.
ⓑ. Both processes are homolytic substitutions because chlorine becomes attached to carbon.
ⓒ. P produces a carbocation and Q produces a radical before either can form a C–Cl bond.
ⓓ. P produces 1-chlorobutane, while Q produces 2-chlorobutane, because a radical and carbocation must attach chlorine at different positions.
Correct Answer: Both processes form 2-chlorobutane with the same atom-to-atom connectivity, but P involves radical electron pairing whereas Q involves nucleophilic electron-pair donation.
Explanation: \(\textbf{Compare the initial carbon frameworks:}\)
Both intermediates contain an unbranched four-carbon skeleton.
In both, the reactive carbon occupies position 2.
The difference is electronic rather than constitutional.
P contains an unpaired electron at carbon 2.
Q contains a positive charge and an available orbital at the same carbon.
\(\textbf{Follow trapping of P:}\)
The carbon radical contains one unpaired electron.
The chlorine radical also contains one unpaired electron.
Their two unpaired electrons can pair to form a new C–Cl sigma bond.
The resulting structure is
\[
\mathrm{CH_3-CH(Cl)-CH_2-CH_3}
\]
The product is 2-chlorobutane.
\(\textbf{Follow trapping of Q:}\)
The carbocation is electron-deficient.
Chloride ion possesses available nonbonding electron pairs.
A chloride electron pair forms a new sigma bond to the positively charged carbon.
The resulting structure is again
\[
\mathrm{CH_3-CH(Cl)-CH_2-CH_3}
\]
The organic product is neutral 2-chlorobutane.
\(\textbf{Compare the electron movements:}\)
Radical combination involves pairing of two separately located unpaired electrons.
Carbocation trapping by chloride involves donation of an existing electron pair.
Therefore, the electron-flow descriptions differ even though the final sigma-bond connectivity is identical.
\(\textbf{Check charge conservation:}\)
In P, both reactants are neutral radicals and the product is neutral.
In Q, the carbocation and chloride have charges \(+1\) and \(-1\), giving total charge zero.
The product is also neutral.
\(\textbf{Interpret the mechanistic evidence:}\)
Observing 2-chlorobutane alone does not establish whether the immediate precursor was a radical or a carbocation.
Additional mechanistic evidence would be necessary.
The reaction descriptions explicitly identify the reactant electronic structures, allowing their electron-flow mechanisms to be distinguished.
The correct inference therefore combines reactive-centre location, electron accounting and product connectivity without treating identical products as proof of identical mechanisms.
313. A benzyl radical, \(\mathrm{C_6H_5CH_2^{\bullet}}\), is subjected to two separate formal processes:
P: One electron is removed.
Q: One electron is added.
Assume the benzyl carbon framework remains unchanged and the resulting species retain conjugative interaction with the aromatic ring where electronically permitted.
Which statement correctly identifies the products and their characteristic electron populations?
ⓐ. P forms a benzyl carbanion containing an additional electron pair; Q forms a benzyl carbocation with an empty orbital.
ⓑ. Both processes produce benzyl carbocations because the benzene ring withdraws electrons inductively.
ⓒ. P forms a benzyl carbocation with an electron-deficient benzylic centre; Q forms a benzyl carbanion possessing an additional nonbonding electron pair.
ⓓ. P and Q both produce neutral toluene because electron transfer necessarily adds a hydrogen atom.
Correct Answer: P forms a benzyl carbocation with an electron-deficient benzylic centre; Q forms a benzyl carbanion possessing an additional nonbonding electron pair.
Explanation: \(\textbf{Identify the starting radical:}\)
The benzyl radical has an unpaired electron associated with the carbon directly attached to the benzene ring.
The benzylic carbon forms two C–H bonds and one bond to the aromatic ring.
Its remaining electron is unpaired.
The radical can benefit from interaction with the neighbouring aromatic pi system.
\(\textbf{Follow electron removal in P:}\)
Removing one electron from the neutral radical produces a positively charged species.
The formal transformation is
\[
\mathrm{C_6H_5CH_2^{\bullet}\longrightarrow C_6H_5CH_2^+ + e^-}
\]
The product is a benzyl carbocation.
Its benzylic centre is electron-deficient.
It can interact with the conjugated aromatic framework through an available orbital.
\(\textbf{Follow electron addition in Q:}\)
Adding one electron to the radical creates a negatively charged species.
The formal transformation is
\[
\mathrm{C_6H_5CH_2^{\bullet}+e^-\longrightarrow C_6H_5CH_2^-}
\]
The product is a benzyl carbanion.
The extra electron pairs with the original unpaired electron in the corresponding electronic description.
The negative charge can also possess resonance-delocalisation opportunities through the adjacent aromatic system.
\(\textbf{Compare the resulting intermediates:}\)
The benzyl carbocation has electron-deficient character and may act as an electrophile.
The benzyl carbanion has electron-rich character and may act as a nucleophile.
The radical is electronically distinct from both.
All three species retain the same carbon–hydrogen and carbon–carbon sigma framework.
\(\textbf{Check the transformations:}\)
Electron removal is an oxidation process.
Electron addition is a reduction process.
Neither event requires addition or removal of a hydrogen nucleus.
Thus, neither automatically forms toluene.
The correct answer follows from conserving the nuclear framework while independently tracking electron number, formal charge and available conjugative stabilisation.
314. In a simple carbonyl compound, a nucleophile approaches the electrophilic carbonyl carbon along a direction compatible with formation of a new sigma bond. The carbonyl carbon is initially approximately trigonal planar. Which statement correctly describes the required bond-order change and the immediate geometry when the addition product remains tetrahedral?
ⓐ. The C=O double bond must remain unchanged, and the carbonyl carbon becomes five-coordinate.
ⓑ. The carbonyl pi bond is converted into an additional sigma bond, so the carbonyl carbon remains trigonal planar with four sigma bonds.
ⓒ. The nucleophile donates one electron only, necessarily producing a neutral radical at the carbonyl carbon.
ⓓ. The carbonyl pi electrons shift onto oxygen, the carbon obtains a new sigma bond to the nucleophile and its local geometry becomes approximately tetrahedral.
Correct Answer: The carbonyl pi electrons shift onto oxygen, the carbon obtains a new sigma bond to the nucleophile and its local geometry becomes approximately tetrahedral.
Explanation: \(\textbf{Examine the initial carbonyl bonding:}\)
A carbonyl carbon is connected to oxygen through one sigma bond and one pi bond.
It also forms two additional sigma bonds to its remaining substituents.
Therefore, it has three sigma-bonding directions.
Its local geometry is approximately trigonal planar.
\(\textbf{Identify the incoming electron pair:}\)
An appropriate nucleophile possesses electron density available for donation.
In the stated elementary addition, an electron pair forms a new sigma bond to the carbonyl carbon.
This would increase the carbon's sigma-bond count from three to four.
\(\textbf{Apply the carbon valency requirement:}\)
Carbon cannot ordinarily retain its original C=O double bond while also forming an additional sigma bond without exceeding its normal tetravalency.
Therefore, the carbonyl pi bond must change during the nucleophilic addition.
The C=O pi electrons move towards oxygen.
The original carbon–oxygen bond becomes a single bond in the immediate tetrahedral intermediate.
\(\textbf{Determine the new carbon geometry:}\)
The former carbonyl carbon now forms four sigma bonds.
These bonds are arranged approximately tetrahedrally.
The carbon is consequently described as \(sp^3\)-hybridised in the ordinary introductory treatment of the tetrahedral intermediate.
\(\textbf{Check the electron-flow representation:}\)
The first full-headed arrow begins at the nucleophilic electron source and points towards carbonyl carbon.
The second begins at the C=O pi bond and points towards oxygen.
This arrow arrangement conserves carbon valency and represents electron-pair movement.
It does not require a one-electron radical pathway.
The formal charge on the resulting oxygen depends on the initial charge and protonation state of the reactants.
Therefore, the change from trigonal-planar to tetrahedral geometry is a general structural consequence of the specified addition, whereas the exact charge distribution must be established from the complete reaction.
315. A carbocation \(\mathrm{CH_3-CH^+-CH=CH_2}\) is represented using two resonance contributors:
P: \(\mathrm{CH_3-CH^+-CH=CH_2}\)
Q: \(\mathrm{CH_3-CH=CH-CH_2^+}\)
A nucleophile attacks at carbon 2 in one experiment and at carbon 4 in another, with no rearrangement beyond the resonance-delocalised intermediate and no further reaction. Which statement correctly interprets the resonance contributors and the possible products?
ⓐ. P and Q are separate constitutional carbocation isomers, so they cannot belong to one common intermediate.
ⓑ. P and Q are resonance contributors; attack at distinct positions can yield constitutional isomers.
ⓒ. P and Q are enantiomers, and nucleophilic attack must yield the same molecular connectivity in both experiments.
ⓓ. P and Q represent homolytic cleavage products because the positive charge is shown on different carbons.
Correct Answer: P and Q are resonance contributors; attack at distinct positions can yield constitutional isomers.
Explanation: \(\textbf{Identify the conjugated carbocation:}\)
The positive centre in P is directly adjacent to a C=C bond.
It occupies an allylic position.
The electron-deficient orbital can interact with the neighbouring pi system.
Therefore, the positive charge can be delocalised across the allylic three-carbon portion.
\(\textbf{Interpret contributor P:}\)
P places the positive charge at carbon 2 and the C=C bond between carbons 3 and 4.
The sigma framework remains a connected four-carbon chain.
\(\textbf{Interpret contributor Q:}\)
Q places the C=C bond between carbons 2 and 3 and the positive charge at carbon 4.
No carbon atom has migrated.
The pi electrons and formal positive-charge location are represented differently.
Therefore, P and Q are resonance contributors of the same intermediate.
\(\textbf{Examine capture at carbon 2:}\)
A nucleophile can form a sigma bond at carbon 2 in the formally represented allylic cation.
The resulting structure retains a double bond between carbons 3 and 4.
\(\textbf{Examine capture at carbon 4:}\)
Alternatively, nucleophilic bond formation at carbon 4 corresponds to the resonance description with the positive centre there.
The resulting structure retains a double bond between carbons 2 and 3.
\(\textbf{Compare the organic connectivities:}\)
The two products differ in the position of the nucleophile attachment and the remaining double bond.
They can consequently be constitutional isomers when the nucleophile and remaining structural conditions are appropriate.
This does not require treating P and Q as independently existing carbocation molecules.
One resonance-delocalised intermediate can provide different electrophilic positions.
The correct conclusion distinguishes electronic resonance representations from actual products formed by new sigma-bond attachment at alternative carbons.
316. A hypothetical reaction sequence begins with ethene, \(\mathrm{CH_2=CH_2}\):
Step I: Ethene accepts a proton through donation of its C=C pi-electron pair.
Step II: The resulting ethyl carbocation receives an electron pair from methanol oxygen.
Step III: A base removes one proton from the oxygen of the resulting organic ion.
Which statement correctly identifies the organic species after Step II and the final neutral product after Step III?
ⓐ. Step II: protonated methoxyethane (oxonium ion); Step III: neutral methoxyethane.
ⓑ. After Step II: neutral methoxyethane; after Step III: ethoxide ion.
ⓒ. After Step II: ethene radical cation; after Step III: ethanol.
ⓓ. After Step II: ethyl carbanion; after Step III: dimethyl ether.
Correct Answer: Step II: protonated methoxyethane (oxonium ion); Step III: neutral methoxyethane.
Explanation: \(\textbf{Follow proton addition to ethene:}\)
Ethene contains a C=C pi bond.
The pi electrons can be donated towards the electrophilic proton in the stated formal process.
A new C–H sigma bond forms at one alkene carbon.
The other carbon becomes positively charged.
Because ethene is symmetrical, either initial carbon gives the same ethyl carbocation.
The resulting intermediate is
\[
\mathrm{CH_3CH_2^+}
\]
\(\textbf{Examine methanol as a nucleophile:}\)
Methanol contains an oxygen atom with available nonbonding electron pairs.
One lone pair can form a new bond to the ethyl carbocation carbon.
The positively charged carbon gains a fourth sigma bond and becomes neutral.
\(\textbf{Determine the oxygen bonding state:}\)
Before attack, methanol oxygen is bonded to one methyl group and one hydrogen.
After attack, it is also bonded to the ethyl group.
Oxygen consequently possesses three sigma bonds.
It carries a positive formal charge.
Therefore, the immediate product of Step II is
\[
\mathrm{CH_3CH_2-O^+(H)-CH_3}
\]
This is protonated methoxyethane.
\(\textbf{Follow the final deprotonation:}\)
In Step III, a base removes the proton attached to oxygen.
The original O–H bonding electron pair remains with oxygen.
Oxygen returns to its neutral two-sigma-bond state.
The final organic product is
\[
\mathrm{CH_3CH_2-O-CH_3}
\]
Its systematic name is methoxyethane.
\(\textbf{Check the mechanistic distinctions:}\)
The first step involves donation from a pi bond to an electrophile.
The second involves donation from a neutral oxygen nucleophile to a carbocation.
The third involves proton transfer.
The final ether is neutral only after the oxonium ion is deprotonated.
The sequence does not require formation of a carbanion or a free radical.
The correct answer follows by tracking the location of electron pairs and the changing formal charge on oxygen through consecutive steps.
317. A chiral secondary alcohol, \(\mathrm{CH_3-CH(OH)-CH_2CH_3}\), is subjected to a hypothetical one-electron process that removes an electron from the C–O bonding environment, ultimately producing a carbon-centred radical at carbon 2 without rearranging the carbon skeleton. Assume that the radical centre is effectively planar and can later react with a suitable radical from either face.
Which statement correctly identifies the stereochemical consequence of converting the original tetrahedral centre into the planar carbon radical and subsequently forming a new sigma bond?
ⓐ. The original configuration is necessarily retained because the four-carbon skeleton remains unchanged.
ⓑ. The planar radical lacks fixed handedness; subsequent attack from either face may give different configurations.
ⓒ. The radical becomes a pair of cis–trans isomers because every planar carbon centre contains a double bond.
ⓓ. The carbon radical automatically possesses a complete octet and therefore cannot form another sigma bond.
Correct Answer: The planar radical lacks fixed handedness; subsequent attack from either face may give different configurations.
Explanation: \(\textbf{Identify the starting stereogenic centre:}\)
In butan-2-ol, the hydroxyl-bearing carbon is attached to hydroxyl, hydrogen, methyl and ethyl groups.
These are four different substituents.
The carbon is tetrahedral and can possess two non-superimposable mirror-image configurations.
A pure enantiomer has a defined spatial arrangement at this carbon.
\(\textbf{Interpret the stated radical formation:}\)
The question explicitly specifies conversion of the original stereogenic carbon into an effectively planar carbon-centred radical.
The carbon framework is not rearranged.
However, the local electronic and geometrical environment of carbon 2 has changed.
A planar radical does not possess the same fixed tetrahedral arrangement as the original saturated stereogenic centre.
\(\textbf{Examine the stereochemical information:}\)
The original handedness depends on the tetrahedral disposition of four different substituents.
Once the carbon becomes planar under the stated condition, that original arrangement is no longer maintained as a fixed tetrahedral configuration at the radical centre.
The two faces of the planar centre can be approached separately.
\(\textbf{Follow subsequent sigma-bond formation:}\)
A suitable radical may combine with the carbon-centred radical to produce a new sigma bond.
If the resulting carbon again bears four different groups, it may become stereogenic.
Approach from opposite faces can then produce different spatial configurations.
\(\textbf{Identify the limitation of the inference:}\)
The possibility of forming both stereochemical arrangements does not by itself prove that they will be produced in equal amounts.
Equal proportions require additional conditions, such as equivalent accessibility of both faces in an appropriate environment.
Likewise, a radical need not behave as a freely planar species under every conceivable structural constraint.
Here, effective planarity is explicitly specified.
The correct conclusion is therefore the loss of fixed original tetrahedral handedness at the radical centre and the possibility of different configurations upon subsequent bond formation.
318. In a reaction mixture, a benzyl carbocation, \(\mathrm{C_6H_5CH_2^+}\), is formed in the presence of equal concentrations of two possible electron-pair donors:
P: Water, \(\mathrm{H_2O}\).
Q: Chloride ion, \(\mathrm{Cl^-}\).
Assume that attack by either donor occurs directly at the benzylic carbon without rearrangement. Which statement correctly identifies the immediate organic products and their charge states?
ⓐ. Water gives neutral benzyl alcohol immediately, while chloride gives positively charged benzyl chloride.
ⓑ. Water gives benzyl radical, while chloride gives benzyl carbanion.
ⓒ. Both donors give neutral products immediately because the benzylic carbocation's positive charge disappears upon forming a new sigma bond.
ⓓ. Water gives a positively charged benzyloxonium ion, whereas chloride gives neutral benzyl chloride.
Correct Answer: Water gives a positively charged benzyloxonium ion, whereas chloride gives neutral benzyl chloride.
Explanation: \(\textbf{Identify the electrophilic centre:}\)
The benzyl carbocation contains a positively charged carbon directly attached to a benzene ring.
Its electron-deficient centre can accept a pair of electrons.
The benzylic cation may be resonance-stabilised, but this does not eliminate its capacity for nucleophilic capture.
\(\textbf{Examine attack by water:}\)
Water is neutral and possesses oxygen lone pairs.
One lone pair can form a sigma bond to the benzylic carbon.
The carbon becomes neutral after receiving the electron pair.
However, oxygen now has three sigma bonds: two O–H bonds and one C–O bond.
It therefore carries a positive formal charge.
The immediate organic product is
\[
\mathrm{C_6H_5CH_2OH_2^+}
\]
This is a benzyloxonium ion.
A subsequent deprotonation would be required to give neutral benzyl alcohol.
\(\textbf{Examine attack by chloride:}\)
Chloride contains available nonbonding electron pairs and carries a negative charge.
It can donate an electron pair to the positively charged benzylic carbon.
The resulting C–Cl bond neutralises the original carbocationic centre.
The chlorine atom becomes part of a neutral covalent organic molecule.
The product is
\[
\mathrm{C_6H_5CH_2Cl}
\]
This is benzyl chloride.
\(\textbf{Compare charge accounting:}\)
Water and the carbocation have initial total charge \(+1\).
Their immediate adduct remains positively charged.
Chloride and the carbocation have initial total charge zero.
Their substitution product is neutral.
\(\textbf{Interpret the mechanistic difference:}\)
Both reagents act as electron-pair donors.
Nevertheless, their different initial charges and bonding states lead to different formal-charge outcomes upon bond formation.
Equal reagent concentrations do not establish equal attack rates or product percentages.
The question asks only for the immediate products of the explicitly stated pathways.
The correct conclusion therefore requires following donor electron pairs and recalculating formal charges after bond formation.
319. A neutral organic molecule has two electronically different carbon–chlorine bonds:
P: A carbon–chlorine bond at an ordinary saturated primary alkyl carbon.
Q: A carbon–chlorine bond at an acyl chloride carbonyl carbon, \(\mathrm{-C(=O)Cl}\).
A nucleophile is specified to react through substitution at the carbon bearing chlorine in each case. Which statement correctly compares the necessary local bonding changes?
ⓐ. Both substitutions necessarily proceed through an identical tetrahedral carbocation intermediate.
ⓑ. P permits chloride displacement at saturated carbon; Q permits carbonyl attack, C=O restoration and chloride loss.
ⓒ. Q cannot undergo substitution because the carbonyl carbon already has four valence electrons in its bonds.
ⓓ. P requires breaking a C=O pi bond, while Q cannot change its original carbonyl bond order at any intermediate stage.
Correct Answer: P permits chloride displacement at saturated carbon; Q permits carbonyl attack, C=O restoration and chloride loss.
Explanation: \(\textbf{Compare the electrophilic carbon environments:}\)
In P, chlorine is attached to a saturated primary carbon.
The carbon forms ordinary sigma bonds and is described as \(sp^3\)-hybridised.
In Q, chlorine is attached directly to a carbonyl carbon.
That carbon forms a C=O double bond and has approximately trigonal-planar geometry.
The two C–Cl bonds therefore occur in different structural environments.
\(\textbf{Examine substitution at P:}\)
A suitable nucleophile may attack the electrophilic saturated carbon.
In a direct displacement pathway, a new sigma bond forms as the C–Cl bonding pair moves towards chlorine.
The carbon does not need to become an isolated carbocation.
The elementary substitution replaces the C–Cl sigma connection with a new carbon–nucleophile sigma connection.
\(\textbf{Examine substitution at Q:}\)
The acyl chloride carbonyl carbon is also electrophilic.
A suitable nucleophile may first add to this carbonyl carbon.
During addition, the C=O pi electrons move to oxygen.
A tetrahedral intermediate can result.
Subsequent donation of an oxygen lone pair reforms the carbonyl pi bond.
The C–Cl bond then breaks, releasing chloride.
\(\textbf{Compare local bond-order changes:}\)
The direct saturated-carbon pathway involves replacement of one sigma bond by another.
The carbonyl substitution pathway can involve temporary loss and subsequent restoration of a carbonyl pi bond.
Therefore, the two processes need not share an identical elementary electron-flow pattern.
\(\textbf{Interpret the mechanistic qualification:}\)
The question states that the nucleophile reacts through substitution in both cases.
It does not assert that every saturated alkyl chloride or every acyl chloride reacts by one mechanism under all possible conditions.
The correct structural distinction comes from the local valency and orbital environment of the carbon attached to chlorine.
Identifying only the presence of a C–Cl bond would miss these important differences.
320. An allylic bromide has the structure \(\mathrm{CH_2=CH-CH_2Br}\). Three processes are proposed:
I. Homolytic cleavage of C–Br.
II. Heterolytic cleavage of C–Br with both bonding electrons moving to bromine.
III. Heterolytic cleavage of C–Br with both bonding electrons moving to carbon.
Which statement correctly identifies the carbon-centred products and their available resonance-delocalisation pathways?
ⓐ. I: allylic radical; II: allylic carbocation; III: allylic carbanion; each permits suitably aligned conjugation.
ⓑ. I forms an allylic carbanion, II forms an allylic radical and III forms an allylic carbocation; only II can exhibit resonance.
ⓒ. All three processes give the same allylic radical because the carbon skeleton remains unchanged.
ⓓ. I forms an allylic radical, while II and III produce ordinary saturated ions that cannot interact with the adjacent C=C bond.
Correct Answer: I: allylic radical; II: allylic carbocation; III: allylic carbanion; each permits suitably aligned conjugation.
Explanation: \(\textbf{Locate the reactive carbon:}\)
The carbon bearing bromine is a saturated methylene carbon directly adjacent to the alkene.
It occupies an allylic position.
The neighbouring C=C bond supplies a conjugation-compatible pi system.
The electronic species obtained after cleavage depends on how the bonding electrons are divided.
\(\textbf{Analyse homolysis in I:}\)
One C–Br bonding electron remains with carbon.
The other remains with bromine.
The carbon fragment becomes the allylic radical
\[
\mathrm{CH_2=CH-CH_2^{\bullet}}
\]
Its unpaired electron can be delocalised through the adjacent pi system.
\(\textbf{Analyse heterolysis in II:}\)
Both C–Br bonding electrons move to bromine.
Bromide ion is formed.
The carbon fragment becomes
\[
\mathrm{CH_2=CH-CH_2^+}
\]
This is an allylic carbocation.
The empty orbital at the positively charged carbon can interact with the adjacent pi system.
Therefore, the positive charge can be resonance-delocalised.
\(\textbf{Analyse heterolysis in III:}\)
Both C–Br bonding electrons instead remain with carbon.
The carbon fragment becomes
\[
\mathrm{CH_2=CH-CH_2^-}
\]
This is an allylic carbanion.
Its nonbonding electron pair can participate in conjugation with the neighbouring double bond when orbital alignment permits.
Thus, resonance delocalisation is also possible for this species.
\(\textbf{Check charge conservation:}\)
Homolysis produces two neutral radicals.
The second cleavage produces a carbocation and bromide with cancelling charges.
The third produces a carbanion and a positively charged bromine fragment in the formal electron-accounting description.
Each pathway preserves the original total charge.
\(\textbf{Interpret chemical plausibility:}\)
The electron-pair transfer to carbon in III is a formal alternative and is not asserted to be the preferred real cleavage of an ordinary C–Br bond.
Nevertheless, the resulting carbon-centred species can be identified unambiguously.
The comparison demonstrates that resonance availability and cleavage mode are different questions.
All three products share an allylic sigma framework, but they differ in formal charge, electron population and reactive behaviour.
