401. A sample of 2-bromo-2-methylpropane is solvolysed in a mixture containing isotopically labelled water, \(\mathrm{H_2{}^{18}O}\), and ordinary methanol, \(\mathrm{CH_3{}^{16}OH}\).
Assume that the substrate undergoes ionisation to a tert-butyl carbocation, followed by nucleophilic capture by either water or methanol.
The two organic products are isolated before any subsequent oxygen-isotope exchange.
Which observation would be consistent with the stated mechanism?
ⓐ. Both tert-butanol and methyl tert-butyl ether must contain \(\mathrm{^{18}O}\) because water is present in the solvent mixture
ⓑ. Only methyl tert-butyl ether can contain \(\mathrm{^{18}O}\), because the ether contains two carbon–oxygen bonds
ⓒ. Neither product can contain \(\mathrm{^{18}O}\), because oxygen is not present in the original alkyl bromide
ⓓ. Tert-butanol formed by water capture contains \(\mathrm{^{18}O}\), while methyl tert-butyl ether formed by methanol capture retains \(\mathrm{^{16}O}\)
Correct Answer: Tert-butanol formed by water capture contains \(\mathrm{^{18}O}\), while methyl tert-butyl ether formed by methanol capture retains \(\mathrm{^{16}O}\)
Explanation: \(\textbf{Identify the common intermediate:}\)
2-Bromo-2-methylpropane contains bromine bonded to a tertiary carbon.
Under the stipulated ionising conditions, the carbon–bromine bond undergoes heterolytic cleavage:
\[
\mathrm{(CH_3)_3CBr\rightarrow (CH_3)_3C^++Br^-}
\]
The resulting carbocation can be trapped by either oxygen-containing solvent molecule.
\(\textbf{Track oxygen during water capture:}\)
Labelled water supplies the nucleophilic oxygen atom:
\[
\mathrm{(CH_3)_3C^++H_2{}^{18}O
\rightarrow (CH_3)_3C{}^{18}OH_2^+}
\]
Loss of a proton gives labelled tert-butanol:
\[
\mathrm{(CH_3)_3C{}^{18}OH}
\]
The product's oxygen atom therefore originates from the labelled water.
\(\textbf{Track oxygen during methanol capture:}\)
Ordinary methanol also contains an oxygen lone pair capable of attacking the carbocation.
Methanol capture followed by deprotonation produces:
\[
\mathrm{(CH_3)_3C{}^{16}OCH_3}
\]
The ether oxygen originates from methanol, not from the water present elsewhere in the mixture.
\(\textbf{Distinguish proton transfer from oxygen transfer:}\)
The reactions involve proton-transfer steps after formation of carbon–oxygen bonds.
A proton transfer does not automatically exchange the oxygen atom that has become covalently attached to carbon.
The question also explicitly excludes subsequent oxygen-isotope exchange.
\(\textbf{Interpret the two products:}\)
Both products arise from the same tertiary carbocation.
However, they preserve information about which solvent molecule supplied the nucleophilic oxygen.
The isotope label can therefore distinguish the two capture pathways even though they share the same initial ionisation event.
\(\textbf{Integrated conclusion:}\)
Product isotope composition depends on atom origin rather than simply on which labelled substance is present in the reaction vessel.
Labelled water gives labelled tert-butanol, while ordinary methanol gives an ether containing ordinary oxygen.
402. Two \(\alpha,\omega\)-dibromoalkanes are treated separately with sodium in dry ether under highly dilute conditions selected to favour intramolecular Wurtz-type coupling.
P: 1,4-Dibromobutane, \(\mathrm{Br(CH_2)_4Br}\).
Q: 1,5-Dibromopentane, \(\mathrm{Br(CH_2)_5Br}\).
Assume complete intramolecular coupling without competing intermolecular reactions or elimination.
Which statement correctly compares the products and their ring strain?
ⓐ. P gives cyclobutane and Q gives cyclopentane; the four-membered ring has greater angle strain
ⓑ. P gives cyclopentane and Q gives cyclohexane because coupling always adds one carbon atom from sodium
ⓒ. P and Q give but-1-ene and pent-1-ene, respectively, because removal of two bromine atoms necessarily forms a double bond
ⓓ. Both give cyclobutane because the terminal bromine atoms are equivalent in each substrate
Correct Answer: P gives cyclobutane and Q gives cyclopentane; the four-membered ring has greater angle strain
Explanation: \(\textbf{Identify the two reactive carbon termini:}\)
Each starting compound contains two saturated terminal carbons bonded to bromine.
The problem specifies intramolecular formation of a carbon–carbon bond between those terminal carbons.
Sodium participates in the reductive coupling, but it does not become a carbon atom in the organic product.
\(\textbf{Determine the product from P:}\)
P contains four carbon atoms in a single chain.
Joining its two terminal carbons closes a four-membered ring:
\[
\mathrm{Br(CH_2)_4Br+2Na
\xrightarrow{dry\ ether}
C_4H_8+2NaBr}
\]
The cyclic product is cyclobutane.
\(\textbf{Determine the product from Q:}\)
Q contains five carbon atoms.
Joining its terminal carbons produces a five-membered ring:
\[
\mathrm{Br(CH_2)_5Br+2Na
\xrightarrow{dry\ ether}
C_5H_{10}+2NaBr}
\]
The cyclic product is cyclopentane.
\(\textbf{Check carbon and halogen conservation:}\)
The number of carbon atoms remains unchanged during ring closure.
Both bromine atoms leave the organic molecule and are accounted for in sodium bromide.
The product contains two fewer halogen substituents and one additional carbon–carbon bond between previously unconnected chain ends.
\(\textbf{Compare the ring geometries:}\)
Saturated carbon atoms favour approximately tetrahedral bond angles.
A four-membered ring has much greater deviation from the preferred bond-angle geometry than a five-membered ring.
Cyclobutane therefore experiences substantial angle strain.
Cyclopentane can adopt nonplanar conformations that reduce its overall strain.
\(\textbf{Interpret the reaction-condition qualification:}\)
The question explicitly selects highly dilute, intramolecular coupling conditions.
Without that restriction, intermolecular coupling and other reactions could complicate the product mixture.
The ring identities follow from the specified intramolecular connectivity rather than a claim that these reactions must always proceed quantitatively in practice.
\(\textbf{Integrated conclusion:}\)
Terminal dihalides can form cyclic hydrocarbons through intramolecular coupling.
The length of the carbon chain determines the ring size, while the ring geometry influences its strain.
403. But-2-ene undergoes bromine addition to form a vicinal dibromide.
The dibromide is then subjected to controlled removal of one molecule of HBr to isolate 2-bromobut-2-ene as the specified principal intermediate.
This intermediate is subsequently treated with excess sodium amide in liquid ammonia under conditions suitable for removing the remaining HBr.
Assume complete intended conversion without rearrangement.
Which statement correctly identifies the final product and explains why the second elimination requires a stronger base than ordinary dehydrohalogenation of a saturated alkyl bromide?
ⓐ. Butane; the second elimination involves replacement of bromine by hydrogen without changing bond order
ⓑ. But-1-yne; the original internal double bond necessarily migrates to a terminal position
ⓒ. But-2-yne; the second elimination must remove HBr from a vinylic bromide, which resists ordinary elimination conditions
ⓓ. Buta-1,3-diene; the second elimination forms another double bond on a different pair of carbon atoms
Correct Answer: But-2-yne; the second elimination must remove HBr from a vinylic bromide, which resists ordinary elimination conditions
Explanation: \(\textbf{Identify the initial addition reaction:}\)
But-2-ene contains a double bond between carbons 2 and 3:
\[
\mathrm{CH_3CH{=}CHCH_3}
\]
Bromine adds across this double bond, producing 2,3-dibromobutane:
\[
\mathrm{CH_3CH{=}CHCH_3+Br_2
\rightarrow CH_3CHBrCHBrCH_3}
\]
Both bromine atoms are initially bonded to saturated carbon atoms.
\(\textbf{Follow the stipulated first elimination:}\)
Removal of one molecule of HBr generates the specified brominated alkene:
\[
\mathrm{CH_3CHBrCHBrCH_3
\rightarrow CH_3C(Br){=}CHCH_3+HBr}
\]
The intermediate is 2-bromobut-2-ene.
Its remaining bromine atom is now attached directly to a double-bonded carbon.
\(\textbf{Recognise the changed bonding environment:}\)
The original vicinal dibromide has bromine bonded to \(sp^3\)-hybridised carbon.
The intermediate has a vinylic carbon–bromine bond involving an \(sp^2\)-hybridised carbon.
The remaining hydrogen that must be removed is also attached to the neighbouring double-bonded carbon.
\(\textbf{Apply the second elimination:}\)
A sufficiently strong base, such as sodium amide under the specified conditions, removes the remaining HBr:
\[
\mathrm{CH_3C(Br){=}CHCH_3
\xrightarrow{NaNH_2/NH_3(l)}
CH_3C{\equiv}CCH_3}
\]
The new carbon–carbon bond order is three.
The product is but-2-yne.
\(\textbf{Connect the two stages:}\)
The first elimination changes an alkyl carbon–halogen environment into a vinylic carbon–halogen environment.
The second elimination is therefore not merely an identical repetition of the first step.
Its reaction centre and energetic requirements are different.
\(\textbf{Verify carbon connectivity:}\)
The chain retains four carbon atoms throughout the sequence.
The triple bond forms between the same two central carbons that originally constituted the alkene double bond.
It does not migrate to a terminal position.
\(\textbf{Integrated conclusion:}\)
Preparing an alkyne from a vicinal dibromide requires two dehydrohalogenations.
The vinylic halide formed after the first step explains why suitably strong conditions are needed for the second.
404. The allylic bromide \(\mathrm{BrCH_2CH{=}CHCH_3}\) is divided into two portions.
P: Treatment with alcoholic KCN under conditions verified to give direct \(S_\mathrm{N}2\) displacement at the primary bromine-bearing carbon, without substitution at either other carbon.
Q: Solvolysis in water under conditions verified to form a delocalised allylic carbocation that can be captured at either of its two terminal positions.
Assume that the carbon–carbon double bond is not hydrogenated and that elimination is absent.
Which comparison correctly identifies the products?
ⓐ. P gives pent-3-enenitrile; Q gives only but-2-en-1-ol because allylic resonance cannot change the site of nucleophilic capture
ⓑ. P gives but-2-enenitrile; Q gives only but-3-en-2-ol because cyanide does not introduce an additional carbon
ⓒ. P gives pent-3-enenitrile; Q can give both but-2-en-1-ol and but-3-en-2-ol because the two reaction pathways differ
ⓓ. P gives pent-2-enenitrile; Q gives butan-1-ol and butan-2-ol because solvolysis necessarily saturates the double bond
Correct Answer: P gives pent-3-enenitrile; Q can give both but-2-en-1-ol and but-3-en-2-ol because the two reaction pathways differ
Explanation: \(\textbf{Identify the allylic substrate:}\)
Bromine is bonded to a saturated terminal carbon adjacent to a carbon–carbon double bond.
The substrate is therefore an allylic bromide rather than a vinylic bromide.
Its reaction products depend on whether the mechanism is direct substitution or carbocation capture.
\(\textbf{Analyse the direct substitution in P:}\)
The question specifies \(S_\mathrm{N}2\) attack by the carbon end of cyanide at the original bromine-bearing carbon.
The intended conversion is:
\[
\mathrm{BrCH_2CH{=}CHCH_3+KCN
\rightarrow NCCH_2CH{=}CHCH_3+KBr}
\]
The new carbon–carbon bond introduces the nitrile carbon as an additional carbon atom.
Numbering the principal chain from the nitrile carbon gives a five-carbon nitrile with a double bond between carbons 3 and 4.
Thus, P gives pent-3-enenitrile.
No allylic rearrangement has been stipulated for this direct displacement.
\(\textbf{Analyse carbocation formation in Q:}\)
Ionisation of the carbon–bromine bond permits delocalisation of the positive charge:
\[
\mathrm{{}^{+}CH_2CH{=}CHCH_3
\leftrightarrow CH_2{=}CHCH^{+}CH_3}
\]
The two resonance contributors place the electrophilic position at different carbons.
Water can therefore capture the cation at either terminal position of the allylic system.
\(\textbf{Identify the alcohols:}\)
Capture at the carbon that originally bore bromine gives:
\[
\mathrm{HOCH_2CH{=}CHCH_3}
\]
This is but-2-en-1-ol.
Capture at the other allylic terminus gives:
\[
\mathrm{CH_2{=}CHCH(OH)CH_3}
\]
This is but-3-en-2-ol.
The two alcohols retain the original four-carbon skeleton, unlike the cyanide product.
\(\textbf{Integrate mechanism and carbon tracking:}\)
A resonance-stabilised carbocation can allow more than one site of attack, whereas the stipulated direct \(S_\mathrm{N}2\) reaction fixes the bond-forming site.
Only the cyanide pathway adds a carbon atom.
The combined observations identify pent-3-enenitrile for P and two allylic alcohols for Q.
405. Chlorobenzene is subjected to an idealised selective catalytic hydrogenation that completely saturates its aromatic carbon ring while preserving its original carbon–chlorine bond.
The isolated organic product is then compared with the original chlorobenzene for possible reactions with hydroxide.
Assume that the comparison concerns suitable ordinary nucleophilic substitution conditions rather than aromatic substitution under extreme conditions.
Which statement correctly describes the changed carbon–halogen bonding environment?
ⓐ. Hydrogenation gives benzyl chloride with a primary \(sp^3\) carbon–chlorine centre
ⓑ. Hydrogenation gives cyclohexyl chloride with an \(sp^3\) carbon–chlorine centre capable of substitution or elimination
ⓒ. Hydrogenation gives chloromethane with a primary \(sp^3\) carbon–chlorine centre
ⓓ. Hydrogenation gives cyclohexane because its carbon–chlorine bond is cleaved
Correct Answer: Hydrogenation gives cyclohexyl chloride with an \(sp^3\) carbon–chlorine centre capable of substitution or elimination
Explanation: \(\textbf{Identify the original carbon–chlorine bond:}\)
In chlorobenzene, chlorine is bonded directly to an aromatic ring carbon.
That carbon is \(sp^2\)-hybridised.
Lone-pair resonance involving chlorine and the aromatic ring contributes partial double-bond character to the carbon–chlorine bond.
\(\textbf{Apply the specified hydrogenation:}\)
Complete saturation of the ring adds hydrogen across the aromatic pi system.
The problem explicitly requires retention of the carbon–chlorine bond:
\[
\mathrm{C_6H_5Cl+3H_2
\xrightarrow{selective\ catalyst}
C_6H_{11}Cl}
\]
The product is chlorocyclohexane, also called cyclohexyl chloride.
\(\textbf{Identify the new carbon environment:}\)
The chlorine-bearing carbon now belongs to a saturated cyclohexane ring.
It is \(sp^3\)-hybridised and attached to two neighbouring carbon atoms.
The product is therefore a secondary alkyl chloride rather than an aryl chloride.
\(\textbf{Explain the bonding change:}\)
Saturating the benzene ring removes the original aromatic pi-electron system.
The chlorine lone pair no longer interacts with that aromatic system in the same way.
The carbon–chlorine bond consequently has the ordinary saturated-carbon bonding environment instead of the original aryl environment.
\(\textbf{Compare substitution behaviour:}\)
Chlorobenzene resists ordinary saturated-carbon \(S_\mathrm{N}1\) and \(S_\mathrm{N}2\) pathways.
Cyclohexyl chloride, by contrast, can undergo appropriate alkyl-halide reactions under suitable conditions.
However, its secondary reaction centre is sterically less accessible than that of a simple primary chloride.
Elimination may also compete, depending on the reagent and conditions.
\(\textbf{Recognise the experimental restriction:}\)
Catalytic hydrogenation of aromatic halides can involve competing transformations under some practical conditions.
The question avoids this ambiguity by expressly stipulating selective ring hydrogenation without carbon–chlorine bond cleavage.
\(\textbf{Integrated conclusion:}\)
A molecule can retain the same halogen and carbon skeleton while acquiring a different carbon–halogen reaction environment.
Changing the hybridisation and resonance context alters the possible reaction pathways.
406. An unconstrained mixture containing all stereoisomers of 2-bromo-3-methylpentane, \(\mathrm{CH_3CH(Br)CH(CH_3)CH_2CH_3}\), undergoes beta-elimination.
Assume that elimination can occur towards either adjacent beta carbon, that all stereochemically accessible products are represented, and that no rearrangement occurs.
When constitutional and configurational stereoisomers are counted separately, how many distinct alkene structures are possible?
ⓐ. Two, because there are only two different beta carbons
ⓑ. Three, because only the internal alkene can exhibit stereoisomerism
ⓒ. Six, because each constitutional alkene must have three stereoisomers
ⓓ. Four: two configurations of the terminal alkene and E/Z configurations of the internal alkene
Correct Answer: Four: two configurations of the terminal alkene and E/Z configurations of the internal alkene
Explanation: \(\textbf{Locate the elimination centre:}\)
The starting bromide has bromine attached to carbon 2.
Its adjacent beta carbons are carbon 1 and carbon 3.
Both possess hydrogen atoms.
Elimination can therefore form a double bond at either of two positions.
\(\textbf{Identify the terminal alkene:}\)
Removal of a beta-hydrogen from carbon 1 gives:
\[
\mathrm{CH_2{=}CHCH(CH_3)CH_2CH_3}
\]
The constitutional product is 3-methylpent-1-ene.
Its terminal double-bond carbon carries two hydrogen atoms, so it cannot exhibit E/Z isomerism.
\(\textbf{Examine stereogenicity in the terminal alkene:}\)
Carbon 3 remains saturated and is bonded to hydrogen, methyl, ethyl and vinyl groups.
These four substituents are different.
Therefore, 3-methylpent-1-ene contains a stereogenic centre and can exist as two enantiomers.
The starting mixture includes the stereochemical possibilities required to obtain both configurations.
\(\textbf{Identify the internal alkene:}\)
Elimination towards carbon 3 gives:
\[
\mathrm{CH_3CH{=}C(CH_3)CH_2CH_3}
\]
The constitutional product is 3-methylpent-2-ene.
Each double-bonded carbon carries two different substituents.
Consequently, E and Z geometrical isomers are possible.
\(\textbf{Check whether the internal alkene has another stereogenic centre:}\)
Carbon 3 becomes \(sp^2\)-hybridised during this elimination.
The original saturated stereogenic environment at that carbon no longer exists.
Therefore, this alkene contributes two E/Z isomers rather than an additional pair of independently chiral products.
\(\textbf{Count all structures:}\)
The terminal alkene contributes two enantiomers.
The internal alkene contributes two geometrical isomers.
Thus:
\[
N_{\mathrm{total}}=2+2
\]
\[
N_{\mathrm{total}}=4
\]
\(\textbf{Integrated conclusion:}\)
The two elimination directions create different constitutional alkenes.
One retains a stereogenic saturated carbon, while the other introduces a geometrically isomeric double bond.
Both kinds of stereochemistry must be included in the total count.
\(\textbf{Final answer:}\)
\[
N_{\mathrm{distinct\ alkene\ structures}}=4
\]
407. An enantiomerically pure sample of \((R)\)-2-bromobutane undergoes substitution to butan-2-ol through two simultaneous pathways.
The \(S_\mathrm{N}2\) pathway produces only \((S)\)-butan-2-ol.
The \(S_\mathrm{N}1\) pathway produces an exactly racemic alcohol mixture.
The measured specific rotation of the final alcohol mixture is \(+8.0^\circ\), while optically pure \((S)\)-butan-2-ol has a specific rotation of \(+20.0^\circ\) under identical conditions.
Assume quantitative substitution, no elimination and no other optically active products.
What percentages of the alcohol were formed through the \(S_\mathrm{N}2\) and \(S_\mathrm{N}1\) pathways, respectively?
ⓐ. \(40\%\) and \(60\%\)
ⓑ. \(60\%\) and \(40\%\)
ⓒ. \(70\%\) and \(30\%\)
ⓓ. \(20\%\) and \(80\%\)
Correct Answer: \(40\%\) and \(60\%\)
Explanation: \(\textbf{Identify what the rotation measures:}\)
The two enantiomers of butan-2-ol have equal and opposite specific rotations under identical conditions.
An equimolar enantiomeric mixture therefore has zero net rotation.
The observed rotation of a mixture can be used to determine its enantiomeric excess.
\(\textbf{Calculate the observed enantiomeric excess:}\)
The pure \((S)\) alcohol has:
\[
[\alpha]_{\mathrm{pure}\ S}=+20.0^\circ
\]
The mixture has:
\[
[\alpha]_{\mathrm{mixture}}=+8.0^\circ
\]
Using:
\[
ee=\frac{[\alpha]_{\mathrm{mixture}}}{[\alpha]_{\mathrm{pure}\ S}}
\]
\[
ee=\frac{8.0}{20.0}
\]
\[
ee=0.400
\]
The mixture has a \(40.0\%\) enantiomeric excess in favour of \((S)\)-butan-2-ol.
\(\textbf{Connect the excess to the two pathways:}\)
Let \(f\) be the fraction of product formed through \(S_\mathrm{N}2\).
That pathway produces only the \((S)\) enantiomer.
The remaining fraction, \(1-f\), forms through \(S_\mathrm{N}1\) and contributes equal amounts of both enantiomers.
\(\textbf{Express the individual enantiomer fractions:}\)
The total \((S)\) fraction is:
\[
x_S=f+\frac{1-f}{2}
\]
The total \((R)\) fraction is:
\[
x_R=\frac{1-f}{2}
\]
Subtracting gives:
\[
x_S-x_R=f
\]
Therefore, the enantiomeric excess equals the fraction formed through the stereospecific \(S_\mathrm{N}2\) pathway in this model.
\(\textbf{Calculate the mechanism fractions:}\)
\[
f_{S_\mathrm{N}2}=0.400
\]
\[
f_{S_\mathrm{N}1}=1-0.400
\]
\[
f_{S_\mathrm{N}1}=0.600
\]
\(\textbf{Verify the predicted product mixture:}\)
The \(S_\mathrm{N}2\) pathway contributes \(40\%\) of pure \((S)\) alcohol.
The \(60\%\) racemic contribution supplies \(30\%\) of each enantiomer.
The final mixture therefore contains \(70\%\) \((S)\) and \(30\%\) \((R)\) alcohol.
Its enantiomeric excess is:
\[
70\%-30\%=40\%
\]
This agrees with the measured rotation.
\(\textbf{Integrated conclusion:}\)
The measured rotation reveals an excess of one enantiomer, not the percentage of that enantiomer directly.
Under the specified mixture of stereospecific and racemising pathways, the enantiomeric excess provides the \(S_\mathrm{N}2\) contribution.
\(\textbf{Final answer:}\)
\[
\boxed{S_\mathrm{N}2=40\%,\qquad S_\mathrm{N}1=60\%}
\]
408. Two isomeric aromatic fluorides are treated separately with sodium methoxide under conditions intended for activated nucleophilic aromatic substitution:
P: 1-Fluoro-2,4-dinitrobenzene.
Q: 1-Fluoro-3,5-dinitrobenzene.
In both compounds, methoxide is intended to replace fluorine, and both nitro groups are to remain attached to their original aromatic carbons.
Which assessment is most appropriate?
ⓐ. P and Q have identical substitution tendencies because each contains one carbon–fluorine bond and two nitro groups
ⓑ. Q is substantially more activated because two meta nitro groups stabilise the substitution intermediate more effectively than ortho and para nitro groups
ⓒ. P is substantially more activated because its ortho and para nitro groups can stabilise the anionic addition intermediate through resonance, unlike the corresponding meta arrangement in Q
ⓓ. Neither can undergo activated aromatic substitution because a carbon–fluorine bond is always too strong to be displaced
Correct Answer: P is substantially more activated because its ortho and para nitro groups can stabilise the anionic addition intermediate through resonance, unlike the corresponding meta arrangement in Q
Explanation: \(\textbf{Identify the common features:}\)
Both substrates are aromatic fluorides containing two strongly electron-withdrawing nitro groups.
The carbon–fluorine bonds are attached to \(sp^2\)-hybridised aromatic carbons.
The intended reaction is not ordinary saturated-carbon \(S_\mathrm{N}2\) displacement.
\(\textbf{Identify the required aromatic mechanism:}\)
Suitable nucleophilic aromatic substitution can proceed through nucleophilic addition followed by leaving-group elimination.
Methoxide initially attacks the aromatic carbon bearing fluorine.
The resulting anionic addition intermediate requires appropriate stabilisation.
Elimination of fluoride subsequently restores aromaticity.
\(\textbf{Analyse the nitro positions in P:}\)
P contains nitro groups ortho and para to the carbon bearing fluorine.
These groups can participate in stabilising the negatively charged addition intermediate through resonance.
The presence of two appropriately placed withdrawing groups provides particularly favourable electronic activation.
\(\textbf{Analyse the nitro positions in Q:}\)
Q contains nitro groups at the two meta positions relative to fluorine.
The nitro groups still exert electron-withdrawing inductive effects.
However, they do not provide the same direct resonance stabilisation of the characteristic addition intermediate produced by attack at the fluorine-bearing carbon.
\(\textbf{Distinguish number from location:}\)
P and Q contain equal numbers of nitro and fluorine substituents.
They also possess the same molecular formula.
Nevertheless, their different substitution patterns change the electronic stabilisation available to the reaction intermediate.
\(\textbf{Evaluate the carbon–fluorine bond argument:}\)
A strong carbon–fluorine bond does not prevent every aromatic nucleophilic substitution.
In a suitably activated aromatic addition–elimination pathway, fluorine can be displaced.
The reaction conditions and intermediate stabilisation must be considered together.
\(\textbf{Integrated conclusion:}\)
Electron-withdrawing substituents cannot be evaluated merely by counting them.
Their positions relative to the leaving group determine how effectively they stabilise the required aromatic intermediate.
P is consequently the more strongly activated substrate.
409. Two different bromopentanes undergo beta elimination with alcoholic KOH under suitable conditions:
P: 3-Bromopentane, \(\mathrm{CH_3CH_2CH(Br)CH_2CH_3}\)
Q: 2-Bromopentane, \(\mathrm{CH_3CH(Br)CH_2CH_2CH_3}\)
Ignore E/Z stereoisomers and count only distinct constitutional alkene products obtainable by removal of a beta hydrogen. Which comparison is correct?
ⓐ. P forms two constitutional alkenes; Q forms one because Q is less symmetrical
ⓑ. Both P and Q form only pent-1-ene because the terminal alkene is always preferred
ⓒ. P forms one constitutional alkene; Q can form two, with pent-2-ene favoured by the usual Zaitsev tendency
ⓓ. Both P and Q form two constitutional alkenes because each contains two beta carbons
Correct Answer: P forms one constitutional alkene; Q can form two, with pent-2-ene favoured by the usual Zaitsev tendency
Explanation: \(\textbf{Locate beta carbons in P:}\)
3-Bromopentane has bromine on the middle carbon of a symmetric five-carbon chain. Carbon 2 and carbon 4 each provide beta hydrogens. Elimination towards either side forms the same constitutional alkene after correct numbering:
\[
\mathrm{CH_3CH{=}CHCH_2CH_3}
\]
This is pent-2-ene. Its E/Z stereoisomers are excluded by the question, so P gives one constitutional alkene.
\(\textbf{Locate beta carbons in Q:}\)
In 2-bromopentane, the beta carbons are carbon 1 and carbon 3. Removal of a hydrogen from carbon 1 produces pent-1-ene:
\[
\mathrm{CH_2{=}CHCH_2CH_2CH_3}
\]
Removal of a hydrogen from carbon 3 produces pent-2-ene:
\[
\mathrm{CH_3CH{=}CHCH_2CH_3}
\]
These are constitutionally distinct alkenes.
\(\textbf{Use the correct selectivity criterion:}\)
Under the stated usual Zaitsev conditions, pent-2-ene is favoured over pent-1-ene because its double bond is more substituted. That preference affects the product proportions, not the number of constitutionally possible products.
\(\textbf{Integrated conclusion:}\)
The number of beta carbons alone is not the answer. Molecular symmetry makes the two elimination directions equivalent in P, whereas they produce different alkene structures in Q. Thus, P gives one constitutional alkene and Q can give two.
410. Two substituted aromatic chlorides give the following experimental results during hydrolysis:
| Property | P: Chlorobenzene | Q: 4-Nitrochlorobenzene |
|---|
| Measured aromatic C–Cl bond length | \(169\ \mathrm{pm}\) | \(166\ \mathrm{pm}\) |
| Behaviour towards aqueous hydroxide | Requires severe conditions | Undergoes nitro-activated aromatic substitution under less severe conditions |
Treat the bond lengths as experimentally supplied values for these particular samples.
A student concludes that Q cannot react faster because its shorter carbon–chlorine bond must be harder to break.
Which assessment is most defensible?
ⓐ. The conclusion is correct because carbon–halogen bond length alone determines the rate of every substitution reaction
ⓑ. The conclusion is correct because a nitro group at the para position cannot influence the aromatic carbon bearing chlorine
ⓒ. The conclusion is incorrect because Q undergoes ordinary backside \(S_\mathrm{N}2\) displacement at an aromatic \(sp^2\) carbon
ⓓ. The nitro-activated addition intermediate can lower the substitution barrier despite Q's shorter carbon–chlorine bond
Correct Answer: The nitro-activated addition intermediate can lower the substitution barrier despite Q's shorter carbon–chlorine bond
Explanation: \(\textbf{Identify what the structural measurement shows:}\)
The supplied measurements describe carbon–chlorine bond lengths in the original reactant molecules.
Q has the shorter measured bond.
However, a ground-state bond length is not a complete measurement of the activation barrier for a chemical reaction.
\(\textbf{Examine chlorobenzene:}\)
Chlorobenzene contains chlorine bonded directly to an aromatic carbon.
The carbon–chlorine bond has partial double-bond character through resonance.
Ordinary nucleophilic displacement is consequently difficult under mild conditions.
\(\textbf{Identify the structural difference in Q:}\)
Q contains a strongly electron-withdrawing nitro group para to the carbon bearing chlorine.
This group changes the electronic behaviour of the aromatic ring.
Its effect is particularly important when an anionic intermediate can be stabilised through resonance.
\(\textbf{Follow the activated substitution pathway:}\)
Hydroxide can attack the aromatic carbon bearing chlorine under suitable conditions.
The resulting addition intermediate has negative-charge character distributed through the aromatic system.
A para nitro group can stabilise this intermediate.
Elimination of chloride subsequently restores aromaticity.
\(\textbf{Distinguish ground-state stability from transition-state accessibility:}\)
The substitution rate depends on the free-energy barrier between reactants and the relevant transition state.
Stabilisation of intermediates and transition states can make a pathway more favourable even when a bond in the original substrate is relatively short.
Therefore, comparing reactant bond lengths alone is insufficient.
\(\textbf{Reject the ordinary }S_\mathrm{N}2\textbf{ interpretation:}\)
Both substrates contain chlorine bonded to aromatic \(sp^2\)-hybridised carbon atoms.
The nitro group does not convert that carbon into an ordinary primary alkyl-halide centre.
The appropriate explanation involves activated aromatic addition–elimination substitution.
\(\textbf{Integrated conclusion:}\)
Chemical reactivity cannot be predicted exclusively from the strength or length of one bond in the starting molecule.
Electronic stabilisation along a particular reaction pathway can outweigh a simplistic bond-length argument.
Q is more reactive under the specified comparison because its nitro group activates the aromatic substitution mechanism.
411. A chemist wishes to distinguish between two chlorinated compounds:
P: 4-Nitrochlorobenzene, \(\mathrm{4\!-\!O_2NC_6H_4Cl}\).
Q: 4-Nitrobenzyl chloride, \(\mathrm{4\!-\!O_2NC_6H_4CH_2Cl}\).
Each is treated separately with aqueous hydroxide under conditions suitable for its intended nucleophilic substitution, followed by acidification.
Assume complete intended replacement of chlorine and no reduction or displacement of the nitro group.
Which statement correctly identifies the products and the essential mechanistic distinction?
ⓐ. Both give 4-nitrophenol because a para nitro group always forces hydroxyl substitution directly onto the aromatic ring
ⓑ. P gives 4-nitrophenol through activated aromatic substitution, whereas Q gives 4-nitrobenzyl alcohol through substitution at a saturated benzylic carbon
ⓒ. P gives 4-nitrobenzyl alcohol, whereas Q gives 4-nitrophenol because the two compounds contain the same functional groups
ⓓ. Neither undergoes substitution because a nitro group withdraws electrons and therefore deactivates every nucleophilic reaction centre
Correct Answer: P gives 4-nitrophenol through activated aromatic substitution, whereas Q gives 4-nitrobenzyl alcohol through substitution at a saturated benzylic carbon
Explanation: \(\textbf{Identify the chlorine-bearing carbons:}\)
In P, chlorine is attached directly to an aromatic \(sp^2\)-hybridised carbon.
In Q, chlorine is attached to a saturated methylene carbon adjacent to an aromatic ring.
These are chemically different carbon–chlorine bonding environments.
\(\textbf{Examine the electronic activation in P:}\)
The nitro group is para to the aromatic carbon bearing chlorine.
It strongly withdraws electron density and can stabilise the anionic addition intermediate of suitable nucleophilic aromatic substitution.
Under appropriately selected conditions, hydroxide replaces aromatic chlorine through the addition–elimination pathway.
\(\textbf{Identify P's neutral product:}\)
The initial reaction produces the corresponding phenoxide species in basic solution.
Acidification gives:
\[
\mathrm{4\!-\!O_2NC_6H_4OH}
\]
The product is 4-nitrophenol.
The hydroxyl group is bonded directly to the aromatic ring.
\(\textbf{Examine the reaction centre in Q:}\)
The chlorine-bearing carbon is primary and benzylic.
It is accessible to direct nucleophilic attack.
The benzylic environment can also stabilise developing charge in suitable substitution pathways.
Therefore, hydroxide can replace chlorine at the methylene carbon without replacing a ring substituent.
\(\textbf{Identify Q's product:}\)
\[
\mathrm{4\!-\!O_2NC_6H_4CH_2Cl+OH^-
\rightarrow4\!-\!O_2NC_6H_4CH_2OH+Cl^-}
\]
The product is 4-nitrobenzyl alcohol.
The hydroxyl group remains separated from the benzene ring by a methylene carbon.
\(\textbf{Distinguish phenolic and alcoholic hydroxyl:}\)
Although both products contain a hydroxyl group and a para nitro substituent, their hydroxyl-bearing carbons differ.
P gives a phenol.
Q gives a primary benzylic alcohol.
These compounds are not interchangeable in classification or subsequent reaction behaviour.
\(\textbf{Integrate the mechanistic evidence:}\)
P reacts through an electronically activated aromatic substitution mechanism.
Q reacts at an accessible saturated benzylic carbon.
A common nitro group and common chlorine leaving atom do not establish an identical substitution pathway.
\(\textbf{Integrated conclusion:}\)
The precise position of the carbon–chlorine bond determines the relevant reaction mechanism and product functional group.
P gives 4-nitrophenol, whereas Q gives 4-nitrobenzyl alcohol.
412. A secondary alkyl bromide undergoes two parallel nucleophilic substitution pathways in a fixed solvent at constant temperature.
Pathway P is an \(S_\mathrm{N}1\) reaction with rate constant \(k_1\).
Pathway Q is an \(S_\mathrm{N}2\) reaction with rate constant \(k_2\).
The total initial disappearance rate is:
\[
r=[\mathrm{RX}]\left(k_1+k_2[\mathrm{Nu^-}]\right)
\]
The following measurements are obtained:
| Experiment | \([\mathrm{RX}]\) (mol L\(^{-1}\)) | \([\mathrm{Nu^-}]\) (mol L\(^{-1}\)) | Initial rate (mol L\(^{-1}\) min\(^{-1}\)) |
|---|
| I | 0.100 | 0 | 0.00200 |
| II | 0.100 | 0.200 | 0.00800 |
Assume no elimination, rearrangement or other reactions. At \([\mathrm{Nu^-}]=0.300\ \mathrm{mol\,L^{-1}}\) and the same substrate concentration, what fraction of substrate disappearance occurs through the \(S_\mathrm{N}2\) pathway?
ⓐ. \(75.0\%\)
ⓑ. \(66.7\%\)
ⓒ. \(90.0\%\)
ⓓ. \(81.8\%\)
Correct Answer: \(81.8\%\)
Explanation: \(\textbf{Separate the two kinetic contributions:}\)
The overall disappearance rate is the sum of the rates of two independent substitution pathways.
The \(S_\mathrm{N}1\) contribution depends only on substrate concentration.
The \(S_\mathrm{N}2\) contribution depends on both substrate and nucleophile concentrations.
Therefore:
\[
r_{S_\mathrm{N}1}=k_1[\mathrm{RX}]
\]
\[
r_{S_\mathrm{N}2}=k_2[\mathrm{RX}][\mathrm{Nu^-}]
\]
\(\textbf{Determine the first-order rate constant:}\)
In Experiment I, the added nucleophile concentration is zero.
Under the stipulated model, only the \(S_\mathrm{N}1\) contribution remains.
\[
k_1=\frac{r_I}{[\mathrm{RX}]}
\]
\[
k_1=\frac{0.00200}{0.100}
\]
\[
k_1=0.0200\ \mathrm{min^{-1}}
\]
\(\textbf{Isolate the bimolecular contribution:}\)
At the same substrate concentration, Experiment II has the same \(S_\mathrm{N}1\) contribution as Experiment I.
The additional rate is attributed to \(S_\mathrm{N}2\).
\[
r_{S_\mathrm{N}2,II}=0.00800-0.00200
\]
\[
r_{S_\mathrm{N}2,II}=0.00600\ \mathrm{mol\,L^{-1}\,min^{-1}}
\]
Therefore:
\[
k_2=\frac{0.00600}{(0.100)(0.200)}
\]
\[
k_2=0.300\ \mathrm{L\,mol^{-1}\,min^{-1}}
\]
\(\textbf{Apply the new nucleophile concentration:}\)
At \([\mathrm{Nu^-}]=0.300\ \mathrm{mol\,L^{-1}}\):
\[
r_{S_\mathrm{N}1}=(0.0200)(0.100)
\]
\[
r_{S_\mathrm{N}1}=0.00200\ \mathrm{mol\,L^{-1}\,min^{-1}}
\]
\[
r_{S_\mathrm{N}2}=(0.300)(0.100)(0.300)
\]
\[
r_{S_\mathrm{N}2}=0.00900\ \mathrm{mol\,L^{-1}\,min^{-1}}
\]
\(\textbf{Calculate the pathway fraction:}\)
\[
f_{S_\mathrm{N}2}=
\frac{r_{S_\mathrm{N}2}}{r_{S_\mathrm{N}1}+r_{S_\mathrm{N}2}}
\]
\[
f_{S_\mathrm{N}2}=\frac{0.00900}{0.00200+0.00900}
\]
\[
f_{S_\mathrm{N}2}=\frac9{11}=0.81818
\]
\(\textbf{Interpret the mechanistic change:}\)
Increasing nucleophile concentration increases the bimolecular contribution without directly changing the stipulated unimolecular contribution.
Consequently, the fraction undergoing \(S_\mathrm{N}2\) increases even though the substrate, solvent and temperature remain unchanged.
This demonstrates why an overall substitution rate need not follow one elementary rate law when parallel mechanisms operate.
\(\textbf{Final answer:}\)
\[
\boxed{f_{S_\mathrm{N}2}=81.8\%}
\]
413. Two secondary and tertiary alkyl bromides are treated separately with a suitable strong base under conditions favouring ordinary beta elimination:
P: \(\mathrm{CH_3C(Br)(CH_3)CH_2CH_3}\)
Q: \(\mathrm{CH_3CH(Br)C(CH_3)_3}\)
Assume elimination occurs without rearrangement. Which statement correctly compares their accessible major alkene structures?
ⓐ. P preferentially gives 2-methylbut-2-ene; Q can give 3,3-dimethylbut-1-ene because its quaternary beta carbon has no hydrogen
ⓑ. P preferentially gives 2-methylbut-1-ene; Q must form an internal alkene towards its quaternary beta carbon
ⓒ. P cannot eliminate because the bromine-bearing carbon is tertiary; Q cannot eliminate because it is secondary
ⓓ. Both must give only the more substituted internal alkene, regardless of whether the relevant beta carbon bears hydrogen
Correct Answer: P preferentially gives 2-methylbut-2-ene; Q can give 3,3-dimethylbut-1-ene because its quaternary beta carbon has no hydrogen
Explanation: \(\textbf{Identify the elimination requirement:}\)
Ordinary beta elimination requires a hydrogen on a carbon adjacent to the carbon bearing bromine. The base removes such a hydrogen while bromide departs, creating a carbon–carbon double bond. The most substituted potential alkene cannot form if the necessary beta hydrogen is absent.
\(\textbf{Analyse substrate P:}\)
The bromine-bearing carbon is attached to two methyl groups and an ethyl group. A beta hydrogen can be removed from the ethyl group's adjacent \(\mathrm{CH_2}\), producing:
\[
\mathrm{CH_3C(CH_3){=}CHCH_3}
\]
This is 2-methylbut-2-ene, a trisubstituted alkene. Removing a beta hydrogen from one of the methyl groups gives the less substituted 2-methylbut-1-ene. Under the usual Zaitsev preference, 2-methylbut-2-ene predominates.
\(\textbf{Analyse substrate Q:}\)
The carbon immediately next to the bromine-bearing carbon on the branched side is quaternary: \(\mathrm{C(CH_3)_3}\). It has no hydrogen, so an alkene cannot be formed towards that carbon by ordinary beta elimination. The methyl group on the other side does supply beta hydrogens. Elimination there gives:
\[
\mathrm{CH_2{=}CHC(CH_3)_3}
\]
This product is 3,3-dimethylbut-1-ene.
\(\textbf{Integrated conclusion:}\)
P can form a more substituted internal alkene; Q is structurally prevented from eliminating towards its quaternary beta carbon. Beta-hydrogen availability must be checked before applying an alkene-substitution preference.
414. Two primary bromohydrins are separately treated with a suitable base in dry solvent under dilute conditions chosen to favour intramolecular nucleophilic substitution.
P: 4-Bromobutan-1-ol, \(\mathrm{HO(CH_2)_4Br}\).
Q: 5-Bromopentan-1-ol, \(\mathrm{HO(CH_2)_5Br}\).
Assume that the hydroxyl groups are converted into alkoxide nucleophiles and that each alkoxide attacks the carbon bearing bromine within its own molecule.
Which statement correctly identifies the products?
ⓐ. P forms a five-membered cyclic ether, tetrahydrofuran, while Q forms a six-membered cyclic ether, tetrahydropyran
ⓑ. P forms butan-1-ol and Q forms pentan-1-ol because an alkoxide cannot attack a carbon within the same molecule
ⓒ. P forms a four-membered cyclic ether and Q forms a five-membered cyclic ether because the oxygen atom is excluded from ring-size counting
ⓓ. P and Q both produce ethylene oxide because all terminal bromohydrins cyclise to the smallest possible ring
Correct Answer: P forms a five-membered cyclic ether, tetrahydrofuran, while Q forms a six-membered cyclic ether, tetrahydropyran
Explanation: \(\textbf{Identify the two reactive ends:}\)
Both substrates contain a hydroxyl group at one end of a carbon chain and a primary carbon–bromine bond at the other.
Conversion of the hydroxyl group into an alkoxide generates an oxygen-centred nucleophile.
The bromine-bearing carbon remains accessible to backside attack.
\(\textbf{Recognise the intramolecular substitution:}\)
The alkoxide oxygen and electrophilic carbon belong to the same molecule.
The oxygen can therefore form a new carbon–oxygen bond while bromide leaves.
This is an intramolecular Williamson-type ether formation.
\(\textbf{Count the atoms in the ring from P:}\)
P contains four methylene groups between oxygen and the bromine-bearing end.
The new oxygen–carbon bond closes a ring containing those four carbon atoms and one oxygen atom.
Therefore, the ring has five atoms.
\[
\mathrm{HO(CH_2)_4Br
\xrightarrow{base}
(CH_2)_4O}
\]
The product is tetrahydrofuran, also called oxolane.
\(\textbf{Count the atoms in the ring from Q:}\)
Q contains five methylene groups.
Closing the chain produces a ring containing five carbon atoms and one oxygen atom.
Therefore, the ring has six atoms.
\[
\mathrm{HO(CH_2)_5Br
\xrightarrow{base}
(CH_2)_5O}
\]
The product is tetrahydropyran, also called oxane.
\(\textbf{Distinguish ring formation from chain shortening:}\)
Neither reaction removes a carbon atom.
The oxygen atom becomes part of the ring rather than remaining in an external hydroxyl group.
The number of ring atoms therefore exceeds the number of methylene groups by one.
\(\textbf{Interpret the selected reaction conditions:}\)
Dilute conditions can favour appropriate intramolecular reactions relative to collisions between different molecules.
However, ring size and substrate structure also influence the competition between possible pathways.
The question explicitly specifies intramolecular closure as the intended reaction.
\(\textbf{Integrated conclusion:}\)
Correct product prediction requires recognising alkoxide formation, locating the primary carbon–bromine bond and counting every atom in the newly formed ring.
The products are five-membered tetrahydrofuran and six-membered tetrahydropyran.
415. 1,6-Dibromohexane, \(\mathrm{Br(CH_2)_6Br}\), is subjected to two idealised synthetic routes.
Route P: Treatment with sodium in dry ether under conditions giving exclusive intramolecular reductive coupling between the terminal carbons.
Route Q: Complete conversion of both carbon–bromine bonds into carbon–magnesium bonds using excess magnesium in dry ether, followed by treatment with excess dry carbon dioxide and acidic hydrolysis.
Assume that no competing cyclisation occurs during Route Q and no intermolecular coupling occurs during Route P.
Which pair correctly identifies the organic products?
ⓐ. P: Cyclohexane; Q: octanedioic acid
ⓑ. P: Cyclopentane; Q: heptanedioic acid
ⓒ. P: Hex-1-ene; Q: hexanedioic acid
ⓓ. P: Cyclohexane; Q: hexanedioic acid
Correct Answer: P: Cyclohexane; Q: octanedioic acid
Explanation: \(\textbf{Identify the starting carbon skeleton:}\)
1,6-Dibromohexane contains six carbon atoms in an unbranched chain.
Each terminal carbon bears one bromine atom.
The two routes transform these carbon–bromine bonds in different ways.
\(\textbf{Route P — Join the terminal carbons:}\)
The stipulated intramolecular sodium coupling forms a carbon–carbon bond between the two terminal methylene carbons.
\[
\mathrm{Br(CH_2)_6Br+2Na
\rightarrow (CH_2)_6+2NaBr}
\]
Here, \(\mathrm{(CH_2)_6}\) represents a six-membered carbon ring.
The product is cyclohexane.
\(\textbf{Check carbon conservation in P:}\)
The ring contains the original six carbon atoms.
Sodium does not introduce any additional carbon.
The two bromine atoms are removed as bromide-containing inorganic products.
\(\textbf{Route Q — Form the difunctional organomagnesium reagent:}\)
The question specifies complete magnesium insertion at both termini:
\[
\mathrm{Br(CH_2)_6Br+2Mg
\xrightarrow{dry\ ether}
BrMg(CH_2)_6MgBr}
\]
The resulting compound contains two carbon–magnesium bonds.
Each carbon–magnesium centre can react with carbon dioxide.
\(\textbf{Carboxylate both reactive termini:}\)
\[
\mathrm{BrMg(CH_2)_6MgBr+2CO_2
\rightarrow BrMgOOC(CH_2)_6COOMgBr}
\]
Acidic hydrolysis produces:
\[
\mathrm{HOOC(CH_2)_6COOH}
\]
The product is octanedioic acid.
\(\textbf{Count the carbon atoms in Q:}\)
The original chain contains six carbon atoms.
Each carbon dioxide molecule contributes one new carboxyl carbon.
Therefore:
\[
N_C=6+2
\]
\[
N_C=8
\]
The compound is an eight-carbon dicarboxylic acid.
\(\textbf{Distinguish the synthetic strategies:}\)
Route P creates an internal carbon–carbon bond and preserves the original carbon count.
Route Q forms two new carbon–carbon bonds with carbon dioxide and increases the carbon count by two.
Both transformations involve the terminal carbon–bromine positions, but they produce very different carbon frameworks.
\(\textbf{Integrated conclusion:}\)
Tracking the fate of both carbon–bromine bonds establishes the product structures.
Intramolecular coupling gives cyclohexane, whereas double carboxylation gives octanedioic acid.
416. Two substitution series are investigated using the same nucleophile under suitable conditions for the specified mechanisms.
Series I: Hydroxide substitution of fluoromethane and chloromethane through ordinary \(S_\mathrm{N}2\) pathways.
Series II: Methoxide substitution of 1-fluoro-2,4-dinitrobenzene and 1-chloro-2,4-dinitrobenzene through activated aromatic addition–elimination pathways.
Assume comparable conditions within each series.
Which qualitative comparison is most appropriate?
ⓐ. The carbon–fluorine substrate must be faster in both series because fluorine is more electronegative than chlorine
ⓑ. The carbon–chlorine substrate must be faster in both series because chloride is always the better leaving group in every substitution mechanism
ⓒ. Both halogens must show identical reactivity in Series II because the two nitro groups completely determine the rate
ⓓ. Chloromethane is favoured in ordinary \(S_\mathrm{N}2\); activated aromatic fluoride can exceed chloride in addition–elimination
Correct Answer: Chloromethane is favoured in ordinary \(S_\mathrm{N}2\); activated aromatic fluoride can exceed chloride in addition–elimination
Explanation: \(\textbf{Identify the two reaction mechanisms:}\)
Series I involves nucleophilic substitution at saturated methyl carbon.
Series II involves nucleophilic aromatic substitution at an \(sp^2\)-hybridised aromatic carbon.
The corresponding reaction pathways have different electronic requirements.
\(\textbf{Examine the methyl-halide comparison:}\)
Both methyl substrates are sterically accessible.
However, the carbon–fluorine bond is particularly strong.
Fluoride is generally a poor leaving group in ordinary alkyl-halide substitution.
Chloromethane is therefore more suitable than fluoromethane for the stated \(S_\mathrm{N}2\) reaction.
\(\textbf{Identify the electronic activation in Series II:}\)
Both aromatic substrates contain nitro groups ortho and para to the aromatic carbon bearing halogen.
These electron-withdrawing groups stabilise the negatively charged addition intermediate formed during suitable nucleophilic attack.
The reaction proceeds through addition followed by elimination and restoration of aromaticity.
\(\textbf{Explain the distinct role of fluorine:}\)
In activated aromatic addition–elimination reactions, the addition step can strongly influence the overall rate.
The powerful electron-withdrawing inductive effect of fluorine can favour nucleophilic addition to the aromatic carbon.
Consequently, suitably activated aromatic fluorides can be more reactive than analogous aromatic chlorides.
\(\textbf{Distinguish the two leaving-group arguments:}\)
Ordinary methyl-halide \(S_\mathrm{N}2\) substitution depends on the concerted displacement of the leaving group.
Activated aromatic substitution instead involves a distinct addition intermediate.
The rate cannot be inferred from the carbon–halogen homolytic bond enthalpy alone.
\(\textbf{Reject a universal halogen order:}\)
The reactivity order associated with one mechanism cannot be transferred automatically to another.
A stronger carbon–halogen bond does not necessarily imply a slower reaction under every possible substitution pathway.
\(\textbf{Integrated conclusion:}\)
Substrate class, electronic activation and mechanism must be considered together.
The qualitative fluoride–chloride preference can reverse between ordinary alkyl substitution and nitro-activated aromatic substitution.
417. A chemist wishes to prepare 4-bromonitrobenzene as the principal positional isomer from benzene.
Two strategies are proposed.
Route P: Bromination of benzene using \(\mathrm{Br_2/FeBr_3}\), isolation of bromobenzene, followed by controlled mononitration.
Route Q: Mononitration of benzene, isolation of nitrobenzene, followed by bromination using \(\mathrm{Br_2/FeBr_3}\) under suitable conditions.
Assume that the reactions occur and that the usual directing influences govern the substitution patterns.
Which assessment is most appropriate?
ⓐ. Route P favours ortho/para bromonitrobenzene formation and can provide the required para isomer after separation, whereas Route Q favours the meta isomer
ⓑ. Route Q is preferred because a nitro group activates benzene towards para-directed bromination
ⓒ. Both routes favour the same para product because electrophilic aromatic substitution is independent of the order in which substituents are introduced
ⓓ. Neither route can form bromonitrobenzene because bromine and nitro groups cannot coexist on an aromatic ring
Correct Answer: Route P favours ortho/para bromonitrobenzene formation and can provide the required para isomer after separation, whereas Route Q favours the meta isomer
Explanation: \(\textbf{Identify the effect of introducing bromine first:}\)
Bromination of benzene using bromine and a suitable Lewis-acid catalyst produces bromobenzene:
\[
\mathrm{C_6H_6+Br_2\xrightarrow{FeBr_3}C_6H_5Br+HBr}
\]
The bromine atom remains attached directly to an aromatic carbon.
It subsequently influences the orientation of further electrophilic substitution.
\(\textbf{Determine bromine's directing effect:}\)
Bromine withdraws electron density through its inductive effect.
It therefore deactivates the aromatic ring relative to benzene.
However, lone-pair resonance donation gives bromine an ortho/para-directing influence.
\(\textbf{Predict Route-P nitration:}\)
Controlled nitration of bromobenzene gives principally ortho- and para-bromonitrobenzene.
The para isomer is commonly favoured over the ortho isomer by reduced steric crowding.
Nevertheless, the reaction should not be assumed to give a single pure para product without separation.
\(\textbf{Identify the effect of introducing nitro first:}\)
Nitration of benzene produces nitrobenzene:
\[
\mathrm{C_6H_6\xrightarrow{HNO_3/H_2SO_4}C_6H_5NO_2}
\]
The nitro group strongly withdraws electron density by inductive and resonance effects.
It deactivates the ring and is meta directing towards ordinary electrophilic substitution.
\(\textbf{Predict Route-Q bromination:}\)
Subsequent bromination of nitrobenzene therefore favours meta-bromonitrobenzene under conditions allowing the reaction.
This is not the required para isomer.
The directing effect arises from the substituent already present when the second electrophile is introduced.
\(\textbf{Compare the synthetic strategies:}\)
Both routes ultimately introduce one bromine and one nitro group.
However, introducing these substituents in different orders changes the directing group present during the second substitution.
The relative positions in the final products consequently differ.
\(\textbf{Integrated conclusion:}\)
Aromatic synthesis planning requires controlling substitution order as well as reagent identity.
Bromination followed by nitration is the more appropriate route for obtaining the para isomer after suitable separation.
418. Two secondary alkyl chlorides are heated separately with alcoholic KOH under conditions favouring beta-elimination.
P: 3-Chloro-2,2-dimethylpentane, \(\mathrm{(CH_3)_3CCH(Cl)CH_2CH_3}\).
Q: 3-Chloro-2-methylpentane, \(\mathrm{CH_3CH(CH_3)CH(Cl)CH_2CH_3}\).
Assume no rearrangement and that all stereochemically accessible elimination products are considered.
Which statement correctly compares the available elimination directions?
ⓐ. P cannot eliminate because its carbon skeleton contains a quaternary carbon, whereas Q forms only one constitutional alkene
ⓑ. Both form exclusively the same internal alkene because the chlorine-bearing carbon is secondary in each
ⓒ. P eliminates only towards the ethyl side to give 4,4-dimethylpent-2-ene; Q can eliminate in two directions
ⓓ. P and Q each form only a terminal alkene because branching always prevents internal double-bond formation
Correct Answer: P eliminates only towards the ethyl side to give 4,4-dimethylpent-2-ene; Q can eliminate in two directions
Explanation: \(\textbf{Identify the E2 structural requirement:}\)
Beta-elimination involves removal of a hydrogen from a carbon adjacent to the carbon bearing chlorine.
The required beta-hydrogen must also possess suitable stereoelectronic alignment with the leaving group.
The presence or absence of such hydrogens determines the available elimination directions.
\(\textbf{Examine the chlorine-bearing carbon in P:}\)
P has the structure:
\[
\mathrm{(CH_3)_3CCH(Cl)CH_2CH_3}
\]
The chlorine-bearing carbon is attached to a quaternary carbon on one side and a methylene carbon on the other.
Both are formally adjacent beta carbons.
\(\textbf{Evaluate the quaternary side of P:}\)
The quaternary carbon has four carbon–carbon bonds.
It bears no hydrogen.
Therefore, no ordinary beta-elimination can form a double bond between that carbon and the chlorine-bearing carbon.
\(\textbf{Evaluate the other side of P:}\)
The adjacent methylene carbon contains beta-hydrogens.
Elimination can therefore occur towards that side:
\[
\mathrm{(CH_3)_3CCH(Cl)CH_2CH_3
\rightarrow (CH_3)_3CCH{=}CHCH_3+HCl}
\]
The product is 4,4-dimethylpent-2-ene.
Because each double-bond carbon bears two different substituents, E/Z configurations are possible when the required conformations are available.
\(\textbf{Examine the chlorine-bearing carbon in Q:}\)
Q has the structure:
\[
\mathrm{CH_3CH(CH_3)CH(Cl)CH_2CH_3}
\]
The beta carbon on the branched side contains one hydrogen.
The beta carbon on the ethyl side contains two hydrogens.
Both directions are therefore available.
\(\textbf{Identify the constitutional products from Q:}\)
Elimination towards the branched side gives 2-methylpent-2-ene.
Elimination towards the ethyl side gives 4-methylpent-2-ene.
The product ratio depends on alkene stability, steric effects and reaction conditions.
\(\textbf{Integrated conclusion:}\)
Two secondary alkyl chlorides can have different elimination possibilities because their adjacent carbon environments differ.
A quaternary beta carbon blocks one elimination direction, whereas a beta carbon bearing hydrogen permits it.
419. Three hydrocarbons have the same molecular formula, \(\mathrm{C_5H_{10}}\):
P: Cyclopentane.
Q: Methylcyclobutane.
R: Pent-1-ene.
They are subjected separately to two experiments.
Experiment I: Treatment with bromine in a suitable inert solvent in the dark, under conditions that permit ordinary alkene addition but not photochemical alkane substitution.
Experiment II: Controlled free-radical monochlorination with \(\mathrm{Cl_2/h\nu}\), with only constitutionally distinct monochloro products counted.
Which statement correctly identifies the hydrocarbon that does not undergo the ordinary dark bromine-addition reaction and also gives only one constitutional monochloro product?
ⓐ. R, because a terminal carbon–carbon double bond is equivalent to a saturated cycloalkane bond
ⓑ. Q, because its methyl group makes every hydrogen atom equivalent
ⓒ. Both P and Q, because their molecular formulas are identical
ⓓ. P, because it lacks an alkene double bond and all its ring hydrogen positions are constitutionally equivalent
Correct Answer: P, because it lacks an alkene double bond and all its ring hydrogen positions are constitutionally equivalent
Explanation: \(\textbf{Interpret the molecular formula:}\)
The formula \(\mathrm{C_5H_{10}}\) is compatible with both an acyclic alkene and a saturated monocyclic hydrocarbon.
Therefore, the molecular formula alone does not distinguish the three structures.
The reactions must be interpreted using their bonding arrangements.
\(\textbf{Analyse Experiment I for P:}\)
Cyclopentane contains only carbon–carbon single bonds.
It lacks the carbon–carbon double bond required for ordinary bromine addition under the specified conditions.
Therefore, it does not undergo the intended dark alkene-addition reaction.
\(\textbf{Analyse Experiment I for Q:}\)
Methylcyclobutane is also a saturated cycloalkane.
It likewise lacks an alkene double bond.
Under the stated dark conditions, it does not undergo the ordinary alkene-addition reaction.
\(\textbf{Analyse Experiment I for R:}\)
Pent-1-ene contains a carbon–carbon double bond.
Bromine can add across that double bond under suitable conditions:
\[
\mathrm{CH_2{=}CHCH_2CH_2CH_3+Br_2
\rightarrow BrCH_2CHBrCH_2CH_2CH_3}
\]
Thus, R behaves differently from the two saturated cyclic structures.
\(\textbf{Compare monochlorination of P:}\)
Cyclopentane has five equivalent ring carbon positions for constitutional substitution.
Replacing one hydrogen by chlorine at any ring position produces chlorocyclopentane.
Only one constitutional monochloro product is therefore possible.
\(\textbf{Compare monochlorination of Q:}\)
Methylcyclobutane contains several nonequivalent hydrogen environments.
They include the methyl group, the substituted ring carbon and different ring methylene positions.
Consequently, free-radical monochlorination can produce more than one constitutional product.
\(\textbf{Integrate the observations:}\)
Both P and Q satisfy the first observation involving the absence of ordinary dark bromine addition.
Only P satisfies the second observation involving one constitutional monochloro product.
The combination therefore identifies cyclopentane.
\(\textbf{Integrated conclusion:}\)
Reaction-class recognition and molecular symmetry provide independent structural information.
Using both observations distinguishes positional or skeletal isomers that share the same molecular formula.
420. Ethanol is converted into bromoethane using phosphorus tribromide prepared in situ from phosphorus and bromine.
The idealised reactions are:
\[
\mathrm{P_4+6Br_2\rightarrow4PBr_3}
\]
\[
\mathrm{3C_2H_5OH+PBr_3\rightarrow3C_2H_5Br+H_3PO_3}
\]
Initially, the laboratory has \(0.0500\ \mathrm{mol}\) of \(\mathrm{P_4}\), \(0.240\ \mathrm{mol}\) of \(\mathrm{Br_2}\) and \(0.600\ \mathrm{mol}\) of ethanol.
Assume complete intended reactions without side products.
Which pair gives the maximum amount of phosphorus tribromide formed and the corresponding maximum amount of bromoethane?
ⓐ. \(0.200\ \mathrm{mol}\) and \(0.600\ \mathrm{mol}\)
ⓑ. \(0.160\ \mathrm{mol}\) and \(0.480\ \mathrm{mol}\)
ⓒ. \(0.0400\ \mathrm{mol}\) and \(0.120\ \mathrm{mol}\)
ⓓ. \(0.160\ \mathrm{mol}\) and \(0.160\ \mathrm{mol}\)
Correct Answer: \(0.160\ \mathrm{mol}\) and \(0.480\ \mathrm{mol}\)
Explanation: \(\textbf{Identify the two reaction stages:}\)
Phosphorus tribromide is first prepared from elemental phosphorus and bromine.
The resulting reagent then converts ethanol into bromoethane.
The limiting reagent must be determined in the first stage before calculating the second-stage product.
\(\textbf{Determine the phosphorus requirement:}\)
The first balanced equation is:
\[
\mathrm{P_4+6Br_2\rightarrow4PBr_3}
\]
Complete consumption of \(0.0500\ \mathrm{mol}\) \(\mathrm{P_4}\) would require:
\[
n_{\mathrm{required}}(\mathrm{Br_2})=6(0.0500)
\]
\[
n_{\mathrm{required}}(\mathrm{Br_2})=0.300\ \mathrm{mol}
\]
Only \(0.240\ \mathrm{mol}\) bromine is available.
Therefore, bromine is the limiting reagent.
\(\textbf{Calculate the phosphorus tribromide amount:}\)
Six moles of bromine produce four moles of phosphorus tribromide:
\[
n(\mathrm{PBr_3})=\frac46n(\mathrm{Br_2})
\]
\[
n(\mathrm{PBr_3})=\frac46(0.240)
\]
\[
n(\mathrm{PBr_3})=0.160\ \mathrm{mol}
\]
\(\textbf{Determine ethanol consumption in the second stage:}\)
One mole of phosphorus tribromide reacts with three moles of ethanol:
\[
\mathrm{3C_2H_5OH+PBr_3\rightarrow3C_2H_5Br+H_3PO_3}
\]
The ethanol required is:
\[
n_{\mathrm{required}}(\mathrm{C_2H_5OH})=3(0.160)
\]
\[
n_{\mathrm{required}}(\mathrm{C_2H_5OH})=0.480\ \mathrm{mol}
\]
Since \(0.600\ \mathrm{mol}\) ethanol is available, ethanol is in excess.
\(\textbf{Calculate bromoethane formation:}\)
The phosphorus tribromide-to-bromoethane mole ratio is \(1:3\):
\[
n(\mathrm{C_2H_5Br})=3n(\mathrm{PBr_3})
\]
\[
n(\mathrm{C_2H_5Br})=3(0.160)
\]
\[
n(\mathrm{C_2H_5Br})=0.480\ \mathrm{mol}
\]
\(\textbf{Check the sequential limitation:}\)
Although the initially available phosphorus could theoretically produce \(0.200\ \mathrm{mol}\) phosphorus tribromide, bromine limits actual formation to \(0.160\ \mathrm{mol}\).
The bromoethane amount must therefore be calculated from the actual first-stage product.
\(\textbf{Final answer:}\)
\[
\boxed{n(\mathrm{PBr_3})=0.160\ \mathrm{mol},\quad
n(\mathrm{C_2H_5Br})=0.480\ \mathrm{mol}}
\]
