501. Two historically used refrigerants are compared:
P: Chlorodifluoromethane (R-22), \(\mathrm{CHClF_2}\)
Q: Dichlorodifluoromethane (Freon-12), \(\mathrm{CCl_2F_2}\)
Which statement correctly combines their molecular structures with the characteristic ozone-depletion mechanism of chlorine-containing refrigerants?
ⓐ. P and Q contain the same number of chlorine atoms, and fluorine radicals alone explain the classical CFC ozone-depletion cycle
ⓑ. P contains no carbon–hydrogen bond, while Q contains one; both lack carbon–chlorine bonds
ⓒ. Q has one carbon–chlorine bond, whereas P has two; ozone depletion requires the release of molecular chlorine only
ⓓ. P has one C–Cl and one C–H bond; Q has two C–Cl and no C–H bonds; chlorine radicals can deplete ozone
Correct Answer: P has one C–Cl and one C–H bond; Q has two C–Cl and no C–H bonds; chlorine radicals can deplete ozone
Explanation: \(\textbf{Read the structures atom by atom:}\)
The formula \(\mathrm{CHClF_2}\) contains one carbon atom bonded to one hydrogen, one chlorine and two fluorines. Thus, R-22 has one C–H bond and one C–Cl bond. The formula \(\mathrm{CCl_2F_2}\) contains no hydrogen and has two C–Cl bonds, so Freon-12 differs in both hydrogen and chlorine content.
\(\textbf{Connect molecular composition with atmospheric chemistry:}\)
On exposure to sufficiently energetic ultraviolet radiation, suitable chlorine-containing refrigerants can release chlorine radicals. These radicals can take part in catalytic cycles converting ozone into molecular oxygen. Fluorine atoms in the formula do not establish that fluoride or fluorine radicals are the principal species in the familiar chlorine-catalysed ozone-depletion mechanism.
\(\textbf{Avoid overgeneralisation:}\)
Having a C–Cl bond does not imply equal atmospheric persistence or identical ozone-depletion potential for all refrigerants. R-22 is a hydrochlorofluorocarbon with a C–H bond, whereas Freon-12 is a chlorofluorocarbon without one. Their environmental behaviour must not be assumed identical merely because both can contain chlorine.
\(\textbf{Conclusion:}\)
The correct option identifies both molecular structures accurately and links ozone depletion to chlorine-radical chemistry. It does not rely on a repeated limiting-reagent calculation.
502. Two environmental monitoring stations examine atmospheric ozone and ultraviolet radiation.
Station P records a decrease in stratospheric ozone together with an increase in the ultraviolet radiation reaching the ground.
Station Q records a decrease in the local concentration of a chlorofluorocarbon near the ground but continued detection of that compound in the stratosphere.
Consider the following statements.
Statement I: Stratospheric ozone absorbs part of the incoming ultraviolet radiation, so depletion of this layer can increase surface exposure to harmful ultraviolet radiation.
Statement II: A chlorofluorocarbon concentration decrease at one ground-level station proves that every molecule of the compound has undergone chemical decomposition.
Statement III: Relative stability in the lower atmosphere and photochemical decomposition under sufficiently energetic stratospheric ultraviolet radiation are compatible properties of the same chlorofluorocarbon.
Statement IV: Ozone depletion and DDT biomagnification are identical chemical mechanisms because both are associated with persistent organohalogen compounds.
Which combination is correct?
ⓐ. I and II only
ⓑ. II and IV only
ⓒ. I, III and IV only
ⓓ. I and III only
Correct Answer: I and III only
Explanation: \(\textbf{Interpret the ozone measurement:}\)
Ozone in the stratosphere absorbs important portions of incoming ultraviolet radiation.
This absorption contributes to protection of living organisms from excessive ultraviolet exposure.
A decrease in stratospheric ozone can therefore permit more harmful ultraviolet radiation to reach Earth's surface.
Statement I is correct.
\(\textbf{Evaluate the ground-level concentration change:}\)
A local atmospheric concentration depends on emissions, mixing, transport and chemical transformation.
A decrease at one measuring station does not establish the fate of every molecule originally released.
The compound may have been transported to another atmospheric region.
Statement II is incorrect.
\(\textbf{Distinguish atmospheric regions:}\)
Certain chlorofluorocarbons are comparatively unreactive under ordinary conditions in the lower atmosphere.
Their persistence permits transport into the stratosphere.
There, sufficiently energetic ultraviolet radiation can initiate bond cleavage and generate reactive chlorine-containing species.
Statement III is correct.
\(\textbf{Distinguish ozone chemistry from pesticide contamination:}\)
Chlorine radicals can participate in catalytic ozone-destruction cycles.
DDT-related environmental concerns instead involve persistence, fat solubility, biological accumulation and toxicity towards non-target organisms.
The two environmental problems do not involve identical mechanisms.
Statement IV is incorrect.
\(\textbf{Integrate the two station records:}\)
Station P supplies evidence about the protective role of stratospheric ozone.
Station Q illustrates why a local decline in an atmospheric pollutant does not necessarily mean its complete removal from the atmosphere.
Neither observation justifies treating persistence as equivalent to harmlessness.
\(\textbf{Integrated conclusion:}\)
Atmospheric transport, photochemical activation and ultraviolet shielding are connected but distinct processes.
The correct statements are I and III.
503. Pure \((Z)\)-pent-2-ene and pure \((E)\)-pent-2-ene undergo separate additions of bromine in an achiral inert solvent.
Assume that both reactions proceed exclusively by anti addition, and that all stereoisomeric products are isolated without further reaction.
The constitutional product from each alkene is 2,3-dibromopentane.
Which conclusion correctly compares the possible products from the two starting geometrical isomers?
ⓐ. Both starting alkenes yield only the same meso product because each product contains two bromine atoms
ⓑ. The Z alkene gives a racemic pair, whereas the E alkene gives a meso product because anti addition always makes trans alkenes meso
ⓒ. Each starting alkene gives a different enantiomeric pair; no meso form is possible because the two ends of the product's carbon skeleton are different
ⓓ. Both starting alkenes give exactly the same enantiomeric pair because the bromine reagent is achiral
Correct Answer: Each starting alkene gives a different enantiomeric pair; no meso form is possible because the two ends of the product's carbon skeleton are different
Explanation: \(\textbf{Identify the addition mechanism:}\)
Bromination is stipulated to proceed through anti addition across the carbon–carbon double bond.
The two bromine atoms become attached to the two alkene carbons from opposite faces.
Thus, both geometrical isomers form 2,3-dibromopentane without rearrangement.
\(\textbf{Examine the stereogenic centres:}\)
The product contains stereogenic centres at carbons 2 and 3.
Each carbon is attached to bromine, hydrogen and two different carbon-containing groups.
The resulting product configurations must be assessed using the starting alkene geometry as well as the anti-addition requirement.
\(\textbf{Compare the two starting geometries:}\)
The Z and E alkenes have different relative placements of the methyl and ethyl substituents.
Anti addition preserves a specific stereochemical relationship for each starting geometry.
Therefore, the two starting alkenes do not have to furnish the same stereoisomeric products.
\(\textbf{Check for an internal symmetry plane:}\)
In 2,3-dibromobutane, equal methyl groups at the two ends can permit a meso stereoisomer.
In 2,3-dibromopentane, one side of the dibrominated segment terminates in methyl and the other extends to ethyl.
The two ends are constitutionally different, so no internal symmetry operation exchanges the two stereogenic centres to generate a meso form.
\(\textbf{Determine the product sets:}\)
Attack on opposite faces of either alkene occurs equally in an achiral environment.
Each pure geometrical isomer therefore produces an enantiomeric pair as a racemic mixture.
However, the relative stereochemistry of the two centres differs between the Z and E starting alkenes.
The pairs originating from the two geometrical isomers are diastereomerically related to one another.
\(\textbf{Integrated conclusion:}\)
Both reactions form chiral 2,3-dibromopentane products, but the two starting alkene geometries give different racemic pairs.
The absence of a meso product follows from the unsymmetrical carbon skeleton, not merely from the anti-addition rule.
504. Chlorobenzene and nitrobenzene are treated separately with chloromethane and anhydrous aluminium chloride under suitable Friedel–Crafts alkylation conditions.
A student makes the following predictions.
P: Chlorobenzene may undergo ring alkylation to give ortho- and para-chlorotoluene, although its chlorine substituent deactivates the ring relative to benzene.
Q: Nitrobenzene should undergo efficient meta-directed Friedel–Crafts methylation because its nitro group is meta directing.
Which assessment is most appropriate?
ⓐ. Both predictions are correct because every meta-directing substituent activates a benzene ring towards Friedel–Crafts alkylation
ⓑ. P is incorrect because an aromatic chlorine substituent prevents all electrophilic aromatic substitution
ⓒ. P is correct, while Q is incorrect because the strongly deactivated nitrobenzene ring is unsuitable for ordinary Friedel–Crafts alkylation
ⓓ. Both predictions are incorrect because chloromethane cannot provide a methyl group under Friedel–Crafts conditions
Correct Answer: P is correct, while Q is incorrect because the strongly deactivated nitrobenzene ring is unsuitable for ordinary Friedel–Crafts alkylation
Explanation: \(\textbf{Identify the reaction class:}\)
Friedel–Crafts alkylation introduces an alkyl substituent into an aromatic ring through electrophilic aromatic substitution.
A suitable Lewis acid assists generation of the electrophilic species from an alkyl halide.
The reaction requires an aromatic system sufficiently reactive towards the electrophile.
\(\textbf{Examine the electronic effect of chlorine:}\)
Chlorine withdraws electron density through its inductive effect.
It therefore deactivates chlorobenzene relative to benzene.
However, chlorine also donates electron density through lone-pair resonance.
This resonance contribution makes it ortho/para directing in ordinary electrophilic aromatic substitution.
\(\textbf{Predict the stipulated chlorobenzene reaction:}\)
Under suitable conditions permitting Friedel–Crafts alkylation, chlorobenzene can form ortho- and para-chlorotoluene.
The reaction is not guaranteed to produce only one positional isomer.
Steric effects can favour the para product relative to the ortho product.
Therefore, prediction P is chemically appropriate with its stated qualification.
\(\textbf{Examine nitrobenzene:}\)
The nitro group withdraws electron density by both inductive and resonance effects.
It strongly deactivates the aromatic ring towards electrophilic substitution.
Its meta-directing character describes positional preference when an electrophilic substitution is feasible.
\(\textbf{Evaluate the proposed Friedel–Crafts reaction:}\)
Nitrobenzene generally does not undergo ordinary Friedel–Crafts alkylation because of its strong deactivation.
Thus, the fact that nitro is meta directing does not establish efficient formation of meta-nitrotoluene.
Prediction Q is incorrect.
\(\textbf{Distinguish directing effects from reaction feasibility:}\)
A substituent's directing influence and its activating or deactivating effect are separate properties.
Predicting where a substituent would direct an electrophile is not enough if the reaction is strongly disfavoured.
\(\textbf{Integrated conclusion:}\)
Chlorine illustrates deactivation accompanied by ortho/para direction.
Nitro illustrates strong deactivation accompanied by meta direction.
Their different degrees of deactivation determine whether the proposed Friedel–Crafts transformation is appropriate.
505. The two geometrical isomers of 1,2-dichloroethene, \(\mathrm{CHCl{=}CHCl}\), are compared.
In P, both chlorine atoms lie on the same side of the planar carbon–carbon double bond. In Q, the chlorine atoms lie on opposite sides of that double bond. The carbon–chlorine bonds have equal bond-dipole magnitudes in both structures. Each chlorine atom is directly attached to a double-bonded carbon.
Consider the following statements.
Statement I: P can have a non-zero resultant molecular dipole, whereas the corresponding carbon–chlorine dipole contributions in Q cancel by molecular symmetry.
Statement II: P should be classified as an ordinary primary alkyl chloride because each chlorine-bearing carbon has a hydrogen atom.
Statement III: Neither P nor Q possesses a carbon–chlorine bond at an ordinary saturated \(sp^3\)-hybridised alkyl reaction centre.
Statement IV: Ordinary \(S_\mathrm{N}2\) substitution at the chlorine-bearing carbon can be assumed equally feasible for P, Q and chloroethane merely because all contain polar carbon–chlorine bonds.
Which combination is correct?
ⓐ. I, II and III only
ⓑ. I and III only
ⓒ. II, III and IV only
ⓓ. I, III and IV only
Correct Answer: I and III only
Explanation: \(\textbf{Identify the geometrical isomers:}\)
P is cis-1,2-dichloroethene.
Q is trans-1,2-dichloroethene.
Both contain chlorine directly attached to carbons belonging to a carbon–carbon double bond.
\(\textbf{Evaluate molecular polarity:}\)
In the cis isomer, the bond-dipole contributions do not cancel.
The molecule can therefore possess a non-zero permanent dipole moment.
In the symmetrical trans isomer, the corresponding bond-dipole contributions cancel.
Statement I is correct.
\(\textbf{Classify the carbon–chlorine bonds:}\)
A primary alkyl chloride requires chlorine bonded to an appropriate saturated carbon.
The chlorine-bearing carbons in 1,2-dichloroethene are \(sp^2\)-hybridised.
The compounds are vinylic halides, not ordinary primary alkyl chlorides.
Statement II is incorrect.
\(\textbf{Compare the relevant carbon hybridisation:}\)
Both chlorine atoms in P and Q are directly attached to the double-bonded carbon framework.
Their bonding environment differs from that of chlorine in chloroethane, where chlorine is attached to an \(sp^3\)-hybridised carbon.
Statement III is correct.
\(\textbf{Assess the substitution prediction:}\)
An ordinary \(S_\mathrm{N}2\) displacement requires suitable backside approach to a saturated electrophilic carbon.
Vinylic carbon–halogen bonds resist this familiar pathway.
Carbon–chlorine bond polarity alone cannot establish that an ordinary backside displacement is feasible.
Statement IV is incorrect.
\(\textbf{Integrate polarity and reactivity:}\)
The cis and trans isomers have the same atom connectivity and carbon hybridisation but differ in whole-molecule polarity.
Their geometrical difference does not convert their vinylic carbon–chlorine bonds into ordinary alkyl-halide reaction centres.
\(\textbf{Integrated conclusion:}\)
Whole-molecule dipole cancellation and nucleophilic substitution behaviour depend on different structural features.
Statements I and III correctly distinguish the two concepts.
506. Three halogenated methane derivatives are considered:
P: \(\mathrm{CBrClFI}\).
Q: \(\mathrm{CCl_2F_2}\).
R: \(\mathrm{CH_2BrCl}\).
Assume ordinary tetrahedral geometry at carbon, with no isotope-induced distinction between otherwise identical substituents.
Which statement correctly compares chirality and molecular polarity?
ⓐ. P and Q are both chiral because each contains four carbon–halogen bonds, whereas R is achiral and necessarily nonpolar
ⓑ. P is achiral because it has no carbon–hydrogen bond, while Q and R are both chiral
ⓒ. P is chiral; Q and R are achiral but can each possess a nonzero molecular dipole
ⓓ. All three are chiral because every carbon–halogen bond is polar
Correct Answer: P is chiral; Q and R are achiral but can each possess a nonzero molecular dipole
Explanation: \(\textbf{Identify the stereogenicity requirement:}\)
A tetrahedral carbon is a stereogenic centre when its four attached substituents are distinguishable.
The relevant test concerns substituent identity.
It does not require one substituent to be hydrogen or a carbon-containing group.
\(\textbf{Examine P:}\)
The central carbon in P is bonded to bromine, chlorine, fluorine and iodine.
All four attached atoms are different.
Therefore, the tetrahedral carbon is stereogenic.
The molecule can exist as a pair of enantiomers.
\(\textbf{Examine Q:}\)
Q contains two chlorine atoms and two fluorine atoms attached to the same central carbon.
Each pair contains identical substituents.
Therefore, the central carbon does not possess four different attached groups.
Q is achiral.
\(\textbf{Examine R:}\)
R contains two identical hydrogen atoms.
Although bromine and chlorine differ, the two hydrogens prevent its carbon from being a stereogenic centre.
R is also achiral.
\(\textbf{Distinguish chirality from molecular dipole moment:}\)
Molecular polarity depends on the vector sum of bond-dipole contributions.
A molecule may be achiral while still having a non-zero permanent dipole moment.
Consequently, the lack of a stereogenic centre does not imply complete cancellation of bond dipoles.
\(\textbf{Apply the distinction to Q and R:}\)
Neither Q nor R has four identical tetrahedrally arranged bonds.
Their different bond-dipole contributions need not cancel completely.
Thus, each can be achiral and polar.
\(\textbf{Connect the comparison to chemical identity:}\)
Counting halogen atoms is not enough to determine chirality.
The identities and equivalences of the groups attached to the carbon must be established.
\(\textbf{Integrated conclusion:}\)
P illustrates chirality at a carbon bonded entirely to four different halogens.
Q and R illustrate why achirality and absence of a molecular dipole are not equivalent properties.
507. A refrigerator and an aerosol container historically used the same chlorofluorocarbon refrigerant, Freon-12.
A manufacturer presents the following reasoning:
Assertion (A): The chemical stability that made Freon-12 useful in refrigerant and aerosol applications also contributes to its potential for long-range atmospheric transport.
Reason (R): Freon-12 is readily liquefiable under suitable conditions, and this physical property means that its carbon–chlorine bonds must undergo rapid chemical cleavage in the lower atmosphere.
Choose the correct option.
ⓐ. Both A and R are true, and R correctly explains A
ⓑ. Both A and R are true, but R does not explain A
ⓒ. A is false, but R is true
ⓓ. A is true, but R is false
Correct Answer: A is true, but R is false
Explanation: \(\textbf{Evaluate the technological usefulness:}\)
Freon-12 is dichlorodifluoromethane, \(\mathrm{CCl_2F_2}\).
It has historically been used as a refrigerant and aerosol propellant.
Its chemical stability and suitable physical properties contributed to those applications.
\(\textbf{Distinguish liquefaction from reaction:}\)
A gas may be liquefied by suitable changes in temperature and pressure.
Liquefaction changes the physical state without requiring cleavage of carbon–chlorine bonds.
Therefore, ease of liquefaction does not establish rapid chemical decomposition.
\(\textbf{Evaluate atmospheric stability:}\)
Certain chlorofluorocarbons persist in the lower atmosphere because they are comparatively unreactive under ordinary conditions.
That persistence permits atmospheric transport over considerable distances and towards the stratosphere.
Assertion A is correct.
\(\textbf{Identify the stratospheric transformation:}\)
In the stratosphere, sufficiently energetic ultraviolet radiation can initiate photochemical changes.
These can generate chlorine radicals that participate in reactions disturbing the natural ozone balance.
The relevant decomposition conditions differ from ordinary refrigeration or liquefaction.
\(\textbf{Evaluate the reason:}\)
Reason R correctly refers to the physical usefulness of an easily liquefiable substance.
However, it incorrectly infers rapid lower-atmosphere carbon–chlorine bond cleavage from that property.
The reason is therefore false.
\(\textbf{Recognise the property–environment trade-off:}\)
A compound can be technologically attractive because it is chemically stable during ordinary use.
The same stability can become an environmental disadvantage if release permits persistent atmospheric transport.
Useful application properties and environmental safety are not equivalent.
\(\textbf{Integrated conclusion:}\)
The correct inference requires separating physical phase behaviour from chemical stability.
Assertion A is true, whereas Reason R is false.
508. An old industrial cleaning-fluid sample is suspected to be either chloroform, \(\mathrm{CHCl_3}\), or tetrachloromethane, \(\mathrm{CCl_4}\).
An elemental analysis shows that a \(15.4\ \mathrm{g}\) sample contains \(14.2\ \mathrm{g}\) chlorine.
Use atomic masses:
\[
\mathrm{C}=12.0,\quad\mathrm{H}=1.0,\quad\mathrm{Cl}=35.5
\]
Assume the sample is a pure compound and that the analysis is exact to the reported precision.
Which conclusion is most appropriate?
ⓐ. The sample is chloroform because its chlorine mass fraction is about \(92.2\%\); its principal storage concern is conversion into iodoform
ⓑ. The sample is tetrachloromethane because its chlorine mass fraction is about \(92.2\%\); historical cleaning use does not eliminate its toxicity or atmospheric ozone-depletion concerns
ⓒ. The sample is chloroform because its chlorine mass fraction is about \(89.1\%\); its lack of a molecular dipole makes it environmentally persistent
ⓓ. The sample is tetrachloromethane because its chlorine mass fraction is about \(89.1\%\); the absence of hydrogen makes it incapable of atmospheric photochemical transformation
Correct Answer: The sample is tetrachloromethane because its chlorine mass fraction is about \(92.2\%\); historical cleaning use does not eliminate its toxicity or atmospheric ozone-depletion concerns
Explanation: \(\textbf{Identify the analytical quantity:}\)
The mass fraction of chlorine is the chlorine mass divided by the total sample mass.
The supplied measurement can be compared with the theoretical compositions of the two candidate compounds.
\(\textbf{Calculate the sample chlorine fraction:}\)
\[
w_{\mathrm{Cl}}=\frac{m_{\mathrm{Cl}}}{m_{\mathrm{sample}}}
\]
\[
w_{\mathrm{Cl}}=\frac{14.2}{15.4}
\]
\[
w_{\mathrm{Cl}}=0.9221
\]
Thus:
\[
\%\mathrm{Cl}=92.21\%
\]
\(\textbf{Calculate the theoretical composition of chloroform:}\)
\[
M(\mathrm{CHCl_3})=12.0+1.0+3(35.5)
\]
\[
M(\mathrm{CHCl_3})=119.5\ \mathrm{g\,mol^{-1}}
\]
The chlorine contribution is:
\[
m_{\mathrm{Cl,molar}}=3(35.5)=106.5\ \mathrm{g\,mol^{-1}}
\]
Therefore:
\[
\%\mathrm{Cl\ in\ CHCl_3}=\frac{106.5}{119.5}\times100
\]
\[
\%\mathrm{Cl\ in\ CHCl_3}\approx89.12\%
\]
\(\textbf{Calculate the theoretical composition of tetrachloromethane:}\)
\[
M(\mathrm{CCl_4})=12.0+4(35.5)
\]
\[
M(\mathrm{CCl_4})=154.0\ \mathrm{g\,mol^{-1}}
\]
Its chlorine contribution is:
\[
m_{\mathrm{Cl,molar}}=4(35.5)=142.0\ \mathrm{g\,mol^{-1}}
\]
Thus:
\[
\%\mathrm{Cl\ in\ CCl_4}=\frac{142.0}{154.0}\times100
\]
\[
\%\mathrm{Cl\ in\ CCl_4}=92.21\%
\]
\(\textbf{Identify the unknown compound:}\)
The measured mass fraction matches tetrachloromethane.
It does not match chloroform under the supplied pure-compound assumption.
\(\textbf{Connect identity with historical application:}\)
Tetrachloromethane was historically used in cleaning and degreasing applications.
However, it has significant toxicity and contributes to atmospheric ozone-depletion concerns.
Historical usefulness does not establish present-day unrestricted safety.
\(\textbf{Distinguish molecular hydrogen from atmospheric behaviour:}\)
Tetrachloromethane contains no carbon–hydrogen bond.
That structural fact does not prevent every photochemical process or guarantee atmospheric harmlessness.
Its environmental behaviour must be assessed through relevant transport and transformation pathways.
\(\textbf{Final answer:}\)
\[
\boxed{\mathrm{CCl_4},\qquad \%\mathrm{Cl}=92.21\%}
\]
509. Bromoethane is converted into ethylmagnesium bromide using magnesium in dry ether.
The starting bromoethane and the resulting organomagnesium reagent are then compared through two separate reactions:
P: Bromoethane is treated with aqueous hydroxide.
Q: Ethylmagnesium bromide is treated with carbon dioxide, followed by acidic hydrolysis.
Consider the following statements.
Statement I: In bromoethane, carbon bears a partial positive charge relative to bromine and can be attacked by a suitable nucleophile.
Statement II: In ethylmagnesium bromide, the carbon bonded to magnesium has carbanion-like character and can attack an electrophilic carbon.
Statement III: The carbon directly involved in the transformation changes from an electrophilic centre in P to a nucleophilic centre in Q because the bonding partner changes.
Statement IV: Both reactions produce propanoic acid because they begin with the same two-carbon organic fragment.
Which combination is correct?
ⓐ. I, II and IV only
ⓑ. II, III and IV only
ⓒ. I and III only
ⓓ. I, II and III only
Correct Answer: I, II and III only
Explanation: \(\textbf{Examine the original carbon–bromine bond:}\)
Bromine is more electronegative than carbon.
Therefore, the carbon–bromine bond is polarised towards bromine.
The carbon bonded to bromine possesses partial positive character and can behave as an electrophilic centre.
Statement I is correct.
\(\textbf{Identify the reaction in P:}\)
Hydroxide attacks the saturated carbon bearing bromine.
Under suitable substitution conditions:
\[
\mathrm{CH_3CH_2Br+OH^-\rightarrow CH_3CH_2OH+Br^-}
\]
The organic product is ethanol.
The carbon skeleton still contains two carbon atoms.
\(\textbf{Examine magnesium insertion:}\)
Magnesium reacts with bromoethane in dry ether to produce an organomagnesium halide:
\[
\mathrm{CH_3CH_2Br+Mg\xrightarrow{dry\ ether}CH_3CH_2MgBr}
\]
Magnesium is more electropositive than carbon.
The carbon–magnesium bond is polarised towards carbon.
Consequently, the carbon attached to magnesium has nucleophilic, carbanion-like character.
Statement II is correct.
\(\textbf{Follow the reaction in Q:}\)
Carbon dioxide provides an electrophilic carbon atom.
The ethyl group of ethylmagnesium bromide attacks this carbon.
Acidic hydrolysis subsequently gives:
\[
\mathrm{CH_3CH_2MgBr
\xrightarrow[\ H_3O^+\ ]{CO_2}
CH_3CH_2COOH}
\]
The product is propanoic acid.
\(\textbf{Compare the reactive carbon centres:}\)
In bromoethane, the carbon attached to bromine receives an electron pair during nucleophilic substitution.
In ethylmagnesium bromide, the carbon attached to magnesium donates electron density towards an electrophile.
This reversal arises from the different relative electronegativities of bromine, carbon and magnesium.
Statement III is correct.
\(\textbf{Evaluate the product claim:}\)
P gives ethanol, while Q gives propanoic acid.
Carbon dioxide introduces an additional carbon atom only in Q.
Therefore, Statement IV is incorrect.
\(\textbf{Integrated conclusion:}\)
The same organic carbon can participate in opposite polar reaction roles after conversion of a carbon–bromine bond into a carbon–magnesium bond.
The correct statements are I, II and III.
510. 2-Bromo-2,3-dimethylbutane, \(\mathrm{CH_3C(Br)(CH_3)CH(CH_3)CH_3}\), undergoes E2 elimination under two separately controlled conditions.
P: A relatively small, strong base is used under conditions favouring formation of the more substituted alkene.
Q: A sterically bulky, strong base is used under conditions favouring abstraction of the more accessible beta-hydrogen.
Assume that the stipulated regioselectivity is achieved and no rearrangement occurs.
Which statement correctly identifies the principal alkene products and their stereochemical possibilities?
ⓐ. P gives 2,3-dimethylbut-2-ene, while Q gives 2,3-dimethylbut-1-ene; neither principal product exhibits E/Z isomerism
ⓑ. P gives 2,3-dimethylbut-1-ene, while Q gives 2,3-dimethylbut-2-ene; both principal products exhibit E/Z isomerism
ⓒ. Both conditions give exclusively 2,3-dimethylbut-2-ene because a bulky base cannot remove a beta-hydrogen
ⓓ. P gives 2,3-dimethylbut-2-ene, while Q gives 2,3-dimethylbut-1-ene; both principal products exist as E/Z pairs
Correct Answer: P gives 2,3-dimethylbut-2-ene, while Q gives 2,3-dimethylbut-1-ene; neither principal product exhibits E/Z isomerism
Explanation: \(\textbf{Identify the carbon bearing bromine:}\)
The starting compound contains a tertiary carbon–bromine bond.
The carbon bearing bromine is attached to two methyl groups and one neighbouring branched carbon.
An E2 reaction requires removal of a hydrogen from a suitable adjacent beta carbon while bromide departs.
\(\textbf{Locate the two distinct beta-hydrogen environments:}\)
The two methyl groups attached directly to the bromine-bearing carbon are equivalent beta positions.
Each contains accessible beta-hydrogens.
The adjacent methine carbon is another beta position and contains one hydrogen.
Elimination towards these two environments produces different constitutional alkenes.
\(\textbf{Form the more substituted alkene in P:}\)
Removal of the hydrogen from the adjacent methine carbon forms a double bond between the two branched central carbons:
\[
\mathrm{CH_3C(Br)(CH_3)CH(CH_3)CH_3
\xrightarrow[\ \Delta\ ]{small\ base}
(CH_3)_2C{=}C(CH_3)_2}
\]
The product is 2,3-dimethylbut-2-ene.
Its double bond is tetrasubstituted.
\(\textbf{Identify the favoured beta position in Q:}\)
A bulky base has greater difficulty approaching the more crowded methine beta-hydrogen.
Under the stipulated conditions, abstraction from an accessible terminal methyl group becomes favoured.
The resulting double bond is formed between that methyl carbon and the original bromine-bearing carbon.
\(\textbf{Form the less substituted alkene:}\)
\[
\mathrm{CH_3C(Br)(CH_3)CH(CH_3)CH_3
\xrightarrow[\ \Delta\ ]{bulky\ base}
CH_2{=}C(CH_3)CH(CH_3)CH_3}
\]
The product is 2,3-dimethylbut-1-ene.
Its double bond is less substituted than that of the product in P.
\(\textbf{Check E/Z stereoisomerism in P's product:}\)
Each double-bond carbon in 2,3-dimethylbut-2-ene bears two identical methyl groups.
Therefore, neither carbon has the two different substituents required for E/Z isomerism.
\(\textbf{Check E/Z stereoisomerism in Q's product:}\)
The terminal double-bond carbon in 2,3-dimethylbut-1-ene bears two hydrogen atoms.
It therefore also fails the E/Z requirement.
\(\textbf{Integrate regioselectivity and stereochemistry:}\)
The small-base pathway favours the more substituted alkene.
The bulky-base pathway favours the more accessible beta-hydrogen under the stated conditions.
However, neither product forms an E/Z pair, for different structural reasons.
\(\textbf{Integrated conclusion:}\)
Predicting elimination products requires examining both beta-hydrogen accessibility and alkene substitution.
A separate substituent-identity check is then needed before assigning possible geometrical isomerism.
511. 1,2-Dichloroethane undergoes rotation about its central carbon–carbon single bond.
Conformer P is a staggered anti arrangement in which the two carbon–chlorine bonds lie opposite one another when viewed along the carbon–carbon bond, giving a chlorine–carbon–carbon–chlorine dihedral angle of \(180^\circ\). Conformer Q is a staggered gauche arrangement with the corresponding dihedral angle of \(60^\circ\). Treat the two carbon–chlorine bond-dipole vectors as equal in magnitude, and assume the ordinary symmetric molecular geometry of each conformer.
An investigator claims that the molecule must have zero dipole moment at every instant because its molecular formula contains two chlorine atoms.
Which assessment is correct?
ⓐ. Both conformers necessarily have equal non-zero dipoles because rotation around a carbon–carbon single bond cannot alter a vector sum
ⓑ. The anti conformer has cancelling carbon–chlorine dipole contributions, whereas the gauche conformer has a non-zero resultant; the observed molecular polarity can depend on conformational populations
ⓒ. The gauche conformer has zero dipole moment, whereas the anti conformer must be polar because its chlorine atoms are farther apart
ⓓ. The two conformers are constitutional isomers because rotating a carbon–carbon single bond changes which carbon atom is bonded to chlorine
Correct Answer: The anti conformer has cancelling carbon–chlorine dipole contributions, whereas the gauche conformer has a non-zero resultant; the observed molecular polarity can depend on conformational populations
Explanation: \(\textbf{Identify the relevant molecular flexibility:}\)
1,2-Dichloroethane contains a carbon–carbon single bond.
Rotation about this bond changes the relative spatial orientations of the two carbon–chlorine bonds.
It does not break bonds or change the atom connectivity.
Therefore, P and Q are conformers rather than constitutional isomers.
\(\textbf{Examine the bond polarity:}\)
Chlorine is more electronegative than carbon.
Both carbon–chlorine bonds have dipole contributions directed towards chlorine.
The magnitude of an individual bond dipole does not independently determine the resultant molecular dipole.
Vector orientation is also required.
\(\textbf{Analyse the anti arrangement:}\)
The two carbon–chlorine bonds occupy opposite orientations in the symmetric anti conformer.
Their dipole contributions cancel in the stated idealised geometry.
The remaining corresponding bond contributions are likewise related by the molecular symmetry.
Therefore, P has a zero resultant permanent dipole in this model.
\(\textbf{Analyse the gauche arrangement:}\)
Rotation changes the relative directions of the carbon–chlorine bond dipoles.
In the gauche conformer, they are no longer oppositely aligned for cancellation.
Their vector sum is non-zero.
Therefore, Q can possess a permanent molecular dipole.
\(\textbf{Distinguish conformation from configuration:}\)
The anti and gauche arrangements interconvert through rotation around a single bond.
The molecule does not require bond cleavage to pass between them.
This differs from interconversion of E/Z alkene isomers, where free rotation around the double bond is restricted.
\(\textbf{Interpret a bulk polarity measurement:}\)
A sample can contain molecules occupying different conformations.
Their relative populations depend on conformational energies and temperature.
A measured macroscopic property may therefore reflect an ensemble of conformers rather than one permanently fixed geometry.
\(\textbf{Reject the molecular-formula shortcut:}\)
Having two identical chlorine atoms does not guarantee dipole cancellation.
Cancellation depends on molecular geometry at the time of comparison.
\(\textbf{Integrated conclusion:}\)
Bond polarity, conformational geometry and molecular polarity are distinct considerations.
The anti conformer permits cancellation, whereas the gauche conformer can possess a non-zero resultant.
512. An optically inactive primary alkyl bromide, 1-bromo-2-methylpropane, is converted into its Grignard reagent using magnesium in dry ether.
The resulting organomagnesium solution is divided into two equal portions.
P: Quenching with ordinary water, \(\mathrm{H_2O}\).
Q: Quenching with deuterium oxide, \(\mathrm{D_2O}\).
Assume quantitative conversion, no carbon-skeleton rearrangement and complete isotopic incorporation at the carbon originally bonded to magnesium.
Which statement correctly compares the organic products?
ⓐ. Both portions give the same undeuterated alkane because the deuterium of \(\mathrm{D_2O}\) remains permanently bonded to oxygen
ⓑ. P gives 2-methylbutane, while Q gives 2-deuterio-2-methylbutane because quenching always increases the number of carbon atoms
ⓒ. P gives 2-methylpropane, while Q gives 2-methylpropane bearing one deuterium at the terminal carbon that originally carried bromine
ⓓ. P gives 2-methylpropane bearing terminal deuterium, while Q gives undeuterated 2-methylpropane because water and deuterium oxide have opposite isotope-transfer directions
Correct Answer: P gives 2-methylpropane, while Q gives 2-methylpropane bearing one deuterium at the terminal carbon that originally carried bromine
Explanation: \(\textbf{Identify the original carbon–bromine bond:}\)
The starting compound has the structure:
\[
\mathrm{(CH_3)_2CHCH_2Br}
\]
Bromine is attached to a primary saturated carbon.
The substrate contains four carbon atoms in a branched skeleton.
\(\textbf{Form the organomagnesium reagent:}\)
Magnesium insertion in dry ether gives:
\[
\mathrm{(CH_3)_2CHCH_2Br+Mg
\rightarrow (CH_3)_2CHCH_2MgBr}
\]
The carbon originally bearing bromine is now attached to magnesium.
Its carbon–magnesium bond has pronounced carbon-nucleophilic character.
\(\textbf{Identify the role of the quenching reagent:}\)
A Grignard reagent reacts with a protic substance through proton transfer.
Ordinary water supplies hydrogen.
Deuterium oxide can supply deuterium under the stipulated isotope-incorporation conditions.
The reaction replaces the carbon–magnesium bond by a carbon–hydrogen-isotope bond.
\(\textbf{Follow the ordinary-water quench:}\)
\[
\mathrm{(CH_3)_2CHCH_2MgBr+H_2O
\rightarrow (CH_3)_2CHCH_3+MgBrOH}
\]
The organic product is 2-methylpropane.
No carbon atom has been introduced or removed.
\(\textbf{Follow the labelled-water quench:}\)
\[
\mathrm{(CH_3)_2CHCH_2MgBr+D_2O
\rightarrow (CH_3)_2CHCH_2D+MgBrOD}
\]
The product is terminally monodeuterated 2-methylpropane.
Deuterium occupies the position at which the carbon–magnesium bond was broken.
\(\textbf{Distinguish carbon count from isotope incorporation:}\)
The two products have the same carbon connectivity.
The difference is the hydrogen isotope attached to the original reactive carbon.
Deuterium incorporation does not add a carbon atom or change the alkane skeleton.
\(\textbf{Interpret the atom-tracing evidence:}\)
An isotopic quench can reveal the location of a reactive carbon–magnesium bond.
The labelled carbon position is identified by the new carbon–deuterium bond in Q.
\(\textbf{Integrated conclusion:}\)
Conversion of an alkyl bromide into a Grignard reagent creates a carbon centre that can be identified through isotope-specific quenching.
Ordinary water gives the unlabelled alkane, whereas deuterium oxide labels the original carbon–magnesium reaction site.
513. An environmental survey measures DDT in three compartments of a closed experimental ecosystem.
At the beginning, the system contains \(3.00\ \mathrm{mg}\) of DDT in soil and \(1.00\ \mathrm{mg}\) dissolved or suspended in water, with no measurable DDT in the sampled fish.
At a later time, the measured amounts are:
| Compartment | DDT amount at later time |
|---|
| Soil | \(1.20\ \mathrm{mg}\) |
| Water | \(0.300\ \mathrm{mg}\) |
| Fish tissue | \(2.10\ \mathrm{mg}\) |
Assume there are no external inputs or outputs, all DDT-containing compartments are included in the stated measurements, and no unmeasured chemical species retain any of the original DDT.
Which conclusion is supported?
ⓐ. The decline of DDT in soil and water proves that all \(2.50\ \mathrm{mg}\) missing from those compartments was chemically degraded
ⓑ. The measured redistribution into fish tissue accounts for the entire decline in soil and water, so no net degradation is indicated
ⓒ. The system shows a net disappearance of \(0.400\ \mathrm{mg}\) of DDT, while transfer into fish tissue also occurred
ⓓ. The fish tissue must have synthesised \(2.10\ \mathrm{mg}\) of new DDT because no fish contamination was initially measured
Correct Answer: The system shows a net disappearance of \(0.400\ \mathrm{mg}\) of DDT, while transfer into fish tissue also occurred
Explanation: \(\textbf{Identify the required distinction:}\)
A reduction of DDT in one environmental compartment does not necessarily mean that DDT has been chemically degraded.
It may instead have moved into another compartment.
The total measured amount must therefore be calculated before determining net disappearance.
\(\textbf{Calculate the initial system burden:}\)
\[
m_{\mathrm{initial}}=m_{\mathrm{soil}}+m_{\mathrm{water}}+m_{\mathrm{fish}}
\]
\[
m_{\mathrm{initial}}=3.00+1.00+0
\]
\[
m_{\mathrm{initial}}=4.00\ \mathrm{mg}
\]
\(\textbf{Calculate the final burden:}\)
\[
m_{\mathrm{final}}=1.20+0.300+2.10
\]
\[
m_{\mathrm{final}}=3.60\ \mathrm{mg}
\]
\(\textbf{Determine the total disappearance:}\)
\[
m_{\mathrm{disappeared}}=m_{\mathrm{initial}}-m_{\mathrm{final}}
\]
\[
m_{\mathrm{disappeared}}=4.00-3.60
\]
\[
m_{\mathrm{disappeared}}=0.400\ \mathrm{mg}
\]
\(\textbf{Separately evaluate soil and water losses:}\)
\[
\Delta m_{\mathrm{soil}}=3.00-1.20=1.80\ \mathrm{mg}
\]
\[
\Delta m_{\mathrm{water}}=1.00-0.300=0.700\ \mathrm{mg}
\]
Their combined decrease is:
\[
1.80+0.700=2.50\ \mathrm{mg}
\]
\(\textbf{Account for biological transfer:}\)
The fish tissue contains:
\[
m_{\mathrm{fish}}=2.10\ \mathrm{mg}
\]
This amount accounts for part, but not all, of the loss from soil and water.
The remaining difference is:
\[
2.50-2.10=0.400\ \mathrm{mg}
\]
\(\textbf{Interpret the chemical meaning:}\)
Under the stipulated closed-system and complete-measurement assumptions, the \(0.400\ \mathrm{mg}\) deficit is consistent with net transformation or degradation of DDT.
The transfer into fish tissue simultaneously demonstrates why declining soil or water concentrations cannot independently establish degradation.
\(\textbf{Connect the result to persistence:}\)
DDT can remain in environmental and biological compartments because of its persistence and affinity for fatty tissues.
Movement between compartments and chemical degradation are distinct processes.
Both must be considered when interpreting contamination data.
\(\textbf{Final answer:}\)
\[
m_{\mathrm{net\ DDT\ disappearance}}=0.400\ \mathrm{mg}
\]
514. Four identical, freshly prepared chloroform samples are stored for the same period under the conditions shown.
| Sample | Exposure to light | Exposure to atmospheric oxygen | Phosgene detected |
|---|
| P | No | No | No |
| Q | Yes | No | No |
| R | No | Yes | No |
| S | Yes | Yes | Yes |
Assume that all other experimental conditions are identical and that the detection method is sufficiently sensitive to identify phosgene formed during the observation period.
Which inference is justified by the observations and the known oxidation chemistry of chloroform?
ⓐ. Light alone is sufficient for the observed phosgene formation because sample Q contains chloroform
ⓑ. Oxygen alone is sufficient for the observed phosgene formation because sample R is exposed to air
ⓒ. The observations prove that ethanol must be produced whenever chloroform is exposed to both light and oxygen
ⓓ. Light and oxygen together are associated with phosgene formation here; absence of either does not establish chloroform's harmlessness
Correct Answer: Light and oxygen together are associated with phosgene formation here; absence of either does not establish chloroform's harmlessness
Explanation: \(\textbf{Identify the controlled comparison:}\)
The experiment varies two environmental conditions independently.
One variable is exposure to light.
The other is exposure to atmospheric oxygen.
All samples are otherwise treated identically.
\(\textbf{Evaluate the fully protected sample:}\)
P is protected from both conditions.
No phosgene is detected.
This observation is consistent with minimising the storage conditions associated with chloroform oxidation.
\(\textbf{Isolate the effect of light:}\)
Q is exposed to light without atmospheric oxygen.
No phosgene is detected during the observation period.
Therefore, this experiment does not show that light alone is sufficient for the specified oxidation product.
\(\textbf{Isolate the effect of oxygen:}\)
R is exposed to oxygen without light.
No phosgene is detected during the observation period.
Therefore, oxygen alone does not reproduce the observed result under these tested conditions.
\(\textbf{Evaluate the combined exposure:}\)
S is exposed to both light and oxygen.
Phosgene is detected.
The observation is consistent with the familiar oxidation of chloroform:
\[
\mathrm{2CHCl_3+O_2\rightarrow2COCl_2+2HCl}
\]
\(\textbf{Distinguish the data from an unrestricted claim:}\)
The table establishes what happened under the stipulated experimental conditions.
It does not prove that every possible alternative chemical pathway is impossible under all other conditions.
The appropriate conclusion concerns the combined exposure in this particular comparison.
\(\textbf{Evaluate chloroform's intrinsic hazard:}\)
Chloroform has toxic properties independently of the formation of phosgene during storage.
Preventing the detected oxidation reaction does not make inhalation or other exposure to chloroform automatically safe.
\(\textbf{Integrated conclusion:}\)
Controlled comparisons can identify the joint conditions associated with an observed reaction.
The storage result and the toxic properties of the unchanged starting compound must nevertheless be kept separate.
515. A primary alkyl bromide undergoes direct \(S_\mathrm{N}2\) substitution with fluoride or iodide under two separately controlled solvent conditions.
In Experiment P, a strongly hydrogen-bonding polar protic solvent is used.
In Experiment Q, a suitable polar aprotic solvent is used.
Assume that both halide ions are freely dissolved at equal concentrations, the same alkyl bromide is used throughout, precipitation and reverse substitution do not influence the initial-rate measurements, and other reactions are absent.
Which prediction is most consistent with the usual effect of solvent on halide-ion nucleophilicity?
ⓐ. Fluoride must be the faster nucleophile in both solvents because its small ionic radius guarantees unhindered attack
ⓑ. Iodide is generally favoured in the protic solvent; fluoride can become much more nucleophilic in the aprotic solvent
ⓒ. Iodide must be the faster nucleophile in both solvents because it has the highest atomic number and cannot be affected by solvation
ⓓ. Both ions necessarily have identical nucleophilicities in each solvent because their charges are equal
Correct Answer: Iodide is generally favoured in the protic solvent; fluoride can become much more nucleophilic in the aprotic solvent
Explanation: \(\textbf{Identify the reaction requirement:}\)
An \(S_\mathrm{N}2\) reaction requires nucleophilic attack at an accessible saturated carbon while the leaving group departs.
The attacking halide ion supplies an electron pair.
Its effectiveness depends on more than formal ionic charge.
\(\textbf{Examine fluoride in a polar protic solvent:}\)
Fluoride is small and has high charge density.
It can interact strongly with hydrogen-bond-donating solvent molecules.
A strongly solvated fluoride ion must overcome substantial solvent interactions while approaching the electrophilic carbon.
\(\textbf{Examine iodide in the same medium:}\)
Iodide is larger and more polarizable.
Its negative charge is distributed through a larger ionic volume.
It is generally less tightly hydrogen-bond solvated than fluoride in an appropriate polar protic medium.
Consequently, iodide often exhibits greater nucleophilic effectiveness than fluoride under these conditions.
\(\textbf{Change to the polar aprotic solvent:}\)
A suitable polar aprotic solvent can dissolve ionic reagents without strongly hydrogen-bond solvating the anions.
This changes the relative importance of the solvation penalty.
The intrinsic tendency of a less heavily solvated small anion to attack a suitable electrophile can become more apparent.
\(\textbf{Interpret the fluoride–iodide comparison:}\)
Fluoride can become a significantly stronger nucleophile in an appropriate polar aprotic environment.
The familiar polar-protic ordering should therefore not be transferred unchanged to every solvent.
Actual rates still depend on the particular solvent, counterion and substrate.
\(\textbf{Separate nucleophilicity from leaving-group ability:}\)
Fluoride can be a strong nucleophile under suitable conditions while remaining a poor leaving group in ordinary alkyl-halide substitution.
The two descriptions concern different directions and stages of a reaction.
They are not contradictory.
\(\textbf{Use the stated controls:}\)
The question excludes precipitation, different substrate concentrations and reverse reactions as competing explanations.
Therefore, the intended comparison concerns the influence of solvation on the initial nucleophilic attack.
\(\textbf{Integrated conclusion:}\)
Nucleophilic strength is sensitive to solvent environment.
Halide-ion rankings cannot be predicted reliably from atomic number, ionic charge or size alone.
516. An enantiomerically enriched sample of 2-bromo-3-methylbutan-1-ol, \(\mathrm{HOCH_2CH(Br)CH(CH_3)_2}\), is treated with a suitable base under conditions where the hydroxyl group is converted into an alkoxide and intramolecular substitution is the exclusive pathway.
Assume that the alkoxide attacks the adjacent bromine-bearing carbon from the backside, without carbocation formation, elimination or stereochemical scrambling.
Which statement correctly identifies the product and its stereochemical relationship to the original reaction centre?
ⓐ. The product is a three-membered cyclic ether, 2-isopropyloxirane, formed by inversion at the carbon originally bearing bromine
ⓑ. The product is a four-membered cyclic ether, formed with retention because intramolecular substitution cannot proceed by backside attack
ⓒ. The product is 3-methylbut-2-en-1-ol, formed by loss of water rather than displacement of bromide
ⓓ. The product is an open-chain diol, formed because the alkoxide oxygen must attack the methyl groups rather than the bromine-bearing carbon
Correct Answer: The product is a three-membered cyclic ether, 2-isopropyloxirane, formed by inversion at the carbon originally bearing bromine
Explanation: \(\textbf{Identify the two reacting functional groups:}\)
The substrate contains a hydroxyl group at carbon 1.
It also contains bromine attached to the adjacent carbon 2.
Conversion of the hydroxyl group into an alkoxide creates a nucleophile within the same molecule.
\(\textbf{Locate the intramolecular electrophilic carbon:}\)
The carbon bearing bromine is saturated.
It is attached to hydrogen, a hydroxymethyl group and an isopropyl group.
The problem explicitly specifies backside attack at this carbon.
\(\textbf{Determine the atoms in the forming ring:}\)
The alkoxide oxygen is already attached to carbon 1.
Formation of a new bond from that oxygen to carbon 2 closes a ring comprising oxygen, carbon 1 and carbon 2.
Therefore, the ring contains three atoms.
It is an oxirane ring.
\(\textbf{Identify the substituent on the ring:}\)
The bromine-bearing carbon originally carries an isopropyl substituent:
\[
\mathrm{-CH(CH_3)_2}
\]
After ring closure, that substituent remains attached to the same carbon.
The product is 2-isopropyloxirane.
\(\textbf{Determine the stereochemical consequence:}\)
The substitution is stipulated to be an intramolecular \(S_\mathrm{N}2\) event.
The alkoxide attacks opposite the departing bromide.
The tetrahedral arrangement at the reacting carbon therefore inverts.
\(\textbf{Distinguish geometrical inversion from descriptor assignment:}\)
The spatial inversion is a mechanistic result.
Assigning an \(R\) or \(S\) descriptor to the product would additionally require reassessing its CIP substituent priorities.
The question asks for the geometrical relationship rather than assuming the descriptor must simply reverse.
\(\textbf{Evaluate competing structures:}\)
An open-chain diol would require oxygen introduction without closing the ring.
An alkene would require beta-elimination.
Both are excluded by the stipulated exclusive intramolecular substitution pathway.
\(\textbf{Integrated conclusion:}\)
The location of the hydroxyl and bromine groups determines the ring size.
Backside attack creates a three-membered ether and produces inversion at the original bromine-bearing stereogenic centre.
517. Two dibromoalkanes undergo an idealised selective replacement of exactly one bromine atom by iodine.
P: 1,3-Dibromopropane, \(\mathrm{BrCH_2CH_2CH_2Br}\).
Q: 1,2-Dibromopropane, \(\mathrm{BrCH_2CHBrCH_3}\).
Assume that either carbon–bromine bond can be selected for exchange under the appropriate controlled conditions, that only one exchange occurs per molecule, and that no carbon-skeleton rearrangement takes place.
Ignore configurational stereoisomers and count only constitutionally distinct organic products.
Which comparison is correct?
ⓐ. P gives two constitutional products and Q gives one because terminal bromine atoms are never equivalent
ⓑ. Both give one constitutional product because replacing a bromine atom cannot change the carbon skeleton
ⓒ. Both give two constitutional products because each molecule initially contains two bromine atoms
ⓓ. P has two equivalent bromomethyl termini and gives one product; Q has distinct brominated positions and gives two
Correct Answer: P has two equivalent bromomethyl termini and gives one product; Q has distinct brominated positions and gives two
Explanation: \(\textbf{Identify the operation being counted:}\)
Each starting molecule contains two bromine atoms.
Exactly one is replaced by iodine.
However, the number of bromine atoms is not necessarily equal to the number of constitutionally distinct products.
The equivalence or non-equivalence of their positions must be examined.
\(\textbf{Analyse the molecular symmetry of P:}\)
1,3-Dibromopropane has the structure:
\[
\mathrm{BrCH_2CH_2CH_2Br}
\]
Its two terminal bromomethyl groups are constitutionally equivalent.
Exchanging bromine at the left end gives the same structure as exchanging bromine at the right end after reversing the chain numbering.
\(\textbf{Identify P's unique product:}\)
The resulting structure is:
\[
\mathrm{BrCH_2CH_2CH_2I}
\]
It is 1-bromo-3-iodopropane.
Only one constitutional product arises from either equivalent substitution site.
\(\textbf{Analyse the two sites in Q:}\)
1,2-Dibromopropane contains one bromine atom at a primary terminal carbon and another at a secondary carbon.
These carbon–bromine positions are not constitutionally equivalent.
Replacement at each site therefore changes the attachment position of iodine.
\(\textbf{Follow substitution at Q's primary carbon:}\)
\[
\mathrm{BrCH_2CHBrCH_3\rightarrow ICH_2CHBrCH_3}
\]
The product is 2-bromo-1-iodopropane.
\(\textbf{Follow substitution at Q's secondary carbon:}\)
\[
\mathrm{BrCH_2CHBrCH_3\rightarrow BrCH_2CHICH_3}
\]
The product is 1-bromo-2-iodopropane.
These two compounds have the same molecular formula but different halogen attachment positions.
\(\textbf{Distinguish constitutional counting from kinetic preference:}\)
The primary carbon may be more accessible for an ordinary \(S_\mathrm{N}2\) displacement.
However, the question explicitly asks which products are constitutionally possible under separately controlled site-selective conditions.
Relative reaction rates do not erase the constitutional distinction.
\(\textbf{Integrated conclusion:}\)
Substitution-product counting requires identifying symmetry-equivalent reaction sites before considering reaction mechanisms.
P provides one constitutional product, whereas Q can provide two.
518. Two tetrahedral halogen-containing molecules are compared:
P: Tetrachloromethane, \(\mathrm{CCl_4}\).
Q: Tetrafluoromethane, \(\mathrm{CF_4}\).
Both molecules are nonpolar because their four identical carbon–halogen bond dipoles cancel by tetrahedral symmetry.
In a simplified photochemical model, ultraviolet radiation is energetic enough to cleave some carbon–chlorine bonds in P but not energetic enough to cleave carbon–fluorine bonds in Q.
Assume that any released chlorine radicals can enter the specified stratospheric ozone-destruction cycle, and no alternative radical-producing pathway is available for Q.
Which statement best reconciles molecular polarity with environmental reactivity?
ⓐ. Both molecules must exhibit identical ozone-depleting behaviour because they have the same tetrahedral shape and zero dipole moment
ⓑ. Only P can supply chlorine radicals on the stipulated photolysis; zero permanent dipole does not imply identical photoreactivity
ⓒ. Q must release more chlorine radicals because fluorine has a higher electronegativity than chlorine
ⓓ. Neither molecule can undergo bond cleavage because a molecule without a permanent dipole cannot absorb or respond to radiation
Correct Answer: Only P can supply chlorine radicals on the stipulated photolysis; zero permanent dipole does not imply identical photoreactivity
Explanation: \(\textbf{Establish the molecular-polarity similarity:}\)
Both P and Q have four equivalent bonds around a tetrahedral carbon.
The individual carbon–halogen bonds are polar.
However, the four bond-dipole vectors cancel through molecular symmetry.
Therefore, each molecule has zero permanent dipole moment.
\(\textbf{Identify the chemical distinction:}\)
P contains four carbon–chlorine bonds.
Q contains four carbon–fluorine bonds and no chlorine atoms.
The molecules are not chemically interchangeable despite their common tetrahedral symmetry.
\(\textbf{Apply the stipulated photochemical conditions:}\)
The model provides sufficient photon energy for some carbon–chlorine bond cleavage in P.
Such cleavage can release chlorine radicals.
The question expressly excludes the corresponding carbon–fluorine cleavage in Q.
\(\textbf{Connect chlorine radicals to ozone chemistry:}\)
A chlorine radical can react with ozone to form chlorine monoxide and molecular oxygen.
Under suitable subsequent reactions, the chlorine radical is regenerated.
Repeated cycling can contribute to ozone destruction.
\(\textbf{Evaluate the source of chlorine in Q:}\)
Tetrafluoromethane contains no chlorine.
Consequently, it cannot release chlorine radicals directly from its own molecular structure.
The stipulated model also excludes any other radical-producing pathway for Q.
\(\textbf{Distinguish two independent molecular properties:}\)
Permanent dipole moment depends on the vector arrangement of bond polarities.
Photochemical behaviour depends on electronic structure, bond energies, radiation absorption and possible reaction pathways.
Equal permanent dipole moments do not require equal photochemical reactivities.
\(\textbf{Recognise the scope of the conclusion:}\)
The question does not claim that Q lacks every possible environmental effect.
It asks specifically about the stipulated chlorine-radical ozone-depletion pathway.
Only P can supply chlorine radicals through the stated mechanism.
\(\textbf{Integrated conclusion:}\)
Molecular symmetry can make two different substances equally nonpolar without giving them identical chemical or environmental behaviour.
Carbon–halogen identity and the relevant photochemical pathway are decisive.
519. Four hydrocarbons are proposed as products of idealised symmetrical Wurtz coupling using only ONE type of alkyl bromide in each preparation.
Assume that two identical alkyl fragments couple at their original carbon–bromine positions, without rearrangement, elimination or competing reactions.
Which hydrocarbon cannot be obtained through the stipulated symmetrical coupling of a single alkyl-bromide structure?
ⓐ. n-Octane
ⓑ. 2,5-Dimethylhexane
ⓒ. 3,4-Dimethylhexane
ⓓ. 2,3-Dimethylhexane
Correct Answer: 2,3-Dimethylhexane
Explanation: \(\textbf{Identify the structural condition for symmetrical Wurtz coupling:}\)
In a symmetrical Wurtz reaction, two molecules of the same alkyl halide undergo coupling.
The new carbon–carbon bond joins the two carbon atoms that originally carried halogen.
Therefore, the resulting hydrocarbon must be divisible at that newly formed bond into two identical carbon fragments.
\(\textbf{Use the reverse-synthesis test:}\)
Instead of predicting a product from a given bromide, examine each proposed hydrocarbon.
Find a carbon–carbon bond whose cleavage would generate two constitutionally identical alkyl fragments.
If such a bond exists, the corresponding alkyl bromide can provide the proposed structure under the stipulated idealised conditions.
\(\textbf{Examine n-octane:}\)
Its structure is:
\[
\mathrm{CH_3(CH_2)_6CH_3}
\]
Cleaving the central carbon–carbon bond produces two n-butyl fragments.
The corresponding reactant is 1-bromobutane:
\[
\mathrm{2CH_3CH_2CH_2CH_2Br+2Na
\rightarrow CH_3(CH_2)_6CH_3+2NaBr}
\]
Therefore, n-octane is compatible with symmetrical Wurtz coupling.
\(\textbf{Examine 2,5-dimethylhexane:}\)
Its structure is:
\[
\mathrm{CH_3CH(CH_3)CH_2CH_2CH(CH_3)CH_3}
\]
Cleavage of the bond between the central two methylene carbons produces identical 2-methylpropyl fragments.
These correspond to 1-bromo-2-methylpropane.
Therefore, symmetrical coupling can give the proposed hydrocarbon.
\(\textbf{Examine 3,4-dimethylhexane:}\)
Its structure is:
\[
\mathrm{CH_3CH_2CH(CH_3)CH(CH_3)CH_2CH_3}
\]
Cleavage of the central bond joining the two branched carbons produces identical butan-2-yl fragments.
The corresponding reactant is 2-bromobutane.
Although secondary alkyl bromides can undergo competing reactions in practice, the question explicitly assumes exclusive intended coupling.
Therefore, this structure is permitted by the model.
\(\textbf{Examine 2,3-dimethylhexane:}\)
Its structure is:
\[
\mathrm{CH_3CH(CH_3)CH(CH_3)CH_2CH_2CH_3}
\]
The carbon skeleton has an unsymmetrical distribution of branches.
No carbon–carbon bond divides the molecule into two identical four-carbon fragments.
A single symmetrical Wurtz coupling therefore cannot construct this skeleton under the stipulated conditions.
\(\textbf{Distinguish molecular formula from coupling feasibility:}\)
All four proposed hydrocarbons contain eight carbon atoms.
An even carbon count is necessary for symmetrical coupling of identical fragments, but it is not sufficient.
The complete connectivity must also permit decomposition into two identical alkyl fragments.
\(\textbf{Integrated conclusion:}\)
Reverse structural analysis is more informative than carbon counting alone.
Only 2,3-dimethylhexane fails the requirement for symmetrical coupling of one alkyl-bromide structure.
520. A refrigeration system historically operated using Freon-12, \(\mathrm{CCl_2F_2}\).
During maintenance, the recovered refrigerant is cooled and compressed into a sealed container.
Another quantity of the same refrigerant escapes into the atmosphere and eventually reaches the stratosphere, where sufficiently energetic ultraviolet radiation initiates carbon–chlorine bond cleavage.
Consider the following statements.
Statement I: Cooling and compression can change the physical state of the recovered refrigerant without necessarily changing its molecular identity.
Statement II: Sealing recovered refrigerant prevents its immediate release but does not mean that the contained Freon-12 has been chemically destroyed.
Statement III: Photochemical carbon–chlorine bond cleavage changes the molecular identity of the reacting species and can supply chlorine radicals.
Statement IV: Because Freon-12 can be liquefied, its carbon–chlorine bonds must be weaker than the intermolecular attractions responsible for liquefaction.
Which combination is correct?
ⓐ. I, II and III only
ⓑ. I, III and IV only
ⓒ. II and IV only
ⓓ. I, II, III and IV
Correct Answer: I, II and III only
Explanation: \(\textbf{Distinguish physical recovery from chemical transformation:}\)
Freon-12 has historically been used as a refrigerant because of suitable physical properties and chemical stability.
Cooling and compression can cause a gas to become liquid.
A phase change rearranges intermolecular separations without requiring cleavage of the molecule's covalent bonds.
\(\textbf{Evaluate Statement I:}\)
The recovered refrigerant can be liquefied while retaining the formula \(\mathrm{CCl_2F_2}\).
Its carbon–chlorine and carbon–fluorine bonds need not be broken during the phase change.
Statement I is correct.
\(\textbf{Evaluate Statement II:}\)
A sealed recovery container restricts escape of its contents.
However, the refrigerant remains chemically present inside the container.
Preventing release and chemically decomposing a substance are different outcomes.
Statement II is correct.
\(\textbf{Evaluate Statement III:}\)
The refrigerant that reaches the stratosphere can encounter sufficiently energetic ultraviolet radiation.
Photochemical cleavage of a carbon–chlorine bond changes the molecule's bonding arrangement.
It can release a chlorine radical capable of participating in ozone-destruction chemistry.
Statement III is correct.
\(\textbf{Evaluate Statement IV:}\)
Liquefaction concerns overcoming or establishing intermolecular attractions between molecules.
Photochemical bond cleavage concerns covalent bonds within individual molecules.
The ability to liquefy Freon-12 does not establish the claimed ordering of covalent bond energies and intermolecular attractions.
Statement IV is incorrect.
\(\textbf{Interpret the environmental control:}\)
Recovery and containment reduce the opportunity for atmospheric release.
They do not erase the environmental significance of any Freon-12 that escapes and survives long enough for stratospheric transport.
\(\textbf{Integrated conclusion:}\)
Useful phase behaviour, chemical persistence, emission prevention and photochemical decomposition are distinct properties or processes.
Statements I, II and III correctly describe their relationships.
