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301. Two separate preparations of chloroethane use an excess of ethanol.
Experiment P begins with \(0.100\ \mathrm{mol}\) of phosphorus trichloride.
Experiment Q begins with \(0.100\ \mathrm{mol}\) of thionyl chloride.
Assume complete intended conversion of each chlorinating reagent and no side reactions.
The relevant equations are:
\[
\mathrm{3C_2H_5OH+PCl_3\rightarrow3C_2H_5Cl+H_3PO_3}
\]
\[
\mathrm{C_2H_5OH+SOCl_2\rightarrow C_2H_5Cl+SO_2+HCl}
\]
Which statement correctly compares the theoretical chloroethane amounts and the by-product-related purification advantage?
ⓐ. Both produce \(0.100\ \mathrm{mol}\) chloroethane; P gives only gaseous by-products
ⓑ. P gives \(0.300\ \mathrm{mol}\) chloroethane; Q gives \(0.100\ \mathrm{mol}\) and gaseous by-products
ⓒ. P produces \(0.100\ \mathrm{mol}\) and Q produces \(0.300\ \mathrm{mol}\) chloroethane; Q gives phosphorous acid
ⓓ. Both produce \(0.300\ \mathrm{mol}\) chloroethane because each reagent contains chlorine
Correct Answer: P gives \(0.300\ \mathrm{mol}\) chloroethane; Q gives \(0.100\ \mathrm{mol}\) and gaseous by-products
Explanation: \(\textbf{Identify the shared organic transformation:}\)
Both reactions replace the hydroxyl group of ethanol by chlorine.
The intended organic product is chloroethane.
However, the number of alcohol molecules converted per molecule of chlorinating reagent differs.
\(\textbf{Evaluate the phosphorus trichloride stoichiometry:}\)
The balanced equation is:
\[
\mathrm{3C_2H_5OH+PCl_3\rightarrow3C_2H_5Cl+H_3PO_3}
\]
One mole of phosphorus trichloride can produce three moles of chloroethane under the stated idealisation.
Therefore:
\[
n_P(\mathrm{C_2H_5Cl})
=3n(\mathrm{PCl_3})
\]
\[
n_P(\mathrm{C_2H_5Cl})=3(0.100)
\]
\[
n_P(\mathrm{C_2H_5Cl})=0.300\ \mathrm{mol}
\]
\(\textbf{Evaluate the thionyl chloride stoichiometry:}\)
The balanced equation is:
\[
\mathrm{C_2H_5OH+SOCl_2\rightarrow C_2H_5Cl+SO_2+HCl}
\]
One mole of thionyl chloride produces one mole of chloroethane:
\[
n_Q(\mathrm{C_2H_5Cl})
=n(\mathrm{SOCl_2})
\]
\[
n_Q(\mathrm{C_2H_5Cl})=0.100\ \mathrm{mol}
\]
\(\textbf{Compare the inorganic by-products:}\)
The phosphorus trichloride reaction produces phosphorous acid, \(\mathrm{H_3PO_3}\).
The thionyl chloride reaction produces sulphur dioxide and hydrogen chloride.
Both are gases under ordinary product-isolation conditions and can escape from the reaction mixture.
This helps explain why thionyl chloride is often preferred for preparing relatively pure alkyl chlorides.
\(\textbf{Distinguish yield capacity from purification:}\)
P gives more moles of chloroethane per mole of the specified chlorinating reagent.
Q offers a more convenient gaseous by-product profile.
These are different criteria for choosing a preparation method.
\(\textbf{Final answer:}\)
\[
n_P(\mathrm{C_2H_5Cl})=0.300\ \mathrm{mol},\quad
n_Q(\mathrm{C_2H_5Cl})=0.100\ \mathrm{mol}
\]
302. Two reactions begin with bromoethane:
Route P: Treatment with sodium nitrite under conditions favouring oxygen-linked nitrite formation, followed by hydrolysis of the isolated product.
Route Q: Treatment with silver nitrite under conditions favouring nitrogen-linked nitro-compound formation, followed by complete reduction of the isolated nitro compound.
Assume selective reactions without competing transformations.
Which pair correctly identifies the organic products obtained after the second stage of each route?
ⓐ. P gives ethanamine; Q gives ethanol
ⓑ. P gives nitroethane; Q gives ethyl nitrite
ⓒ. P gives ethanoic acid; Q gives ethanol
ⓓ. P gives ethanol; Q gives ethanamine
Correct Answer: P gives ethanol; Q gives ethanamine
Explanation: \(\textbf{Identify the ambident nucleophile:}\)
The nitrite ion contains both nitrogen and oxygen atoms capable of serving as the atom through which linkage to an organic group occurs.
Consequently, different reagents and conditions can produce compounds with different connectivity.
The products need not have different molecular formulas to be chemically distinct.
\(\textbf{Route P — Oxygen-linked substitution:}\)
Under the stipulated conditions, sodium nitrite gives the oxygen-linked alkyl nitrite:
\[
\mathrm{C_2H_5Br+NaNO_2
\rightarrow C_2H_5ONO+NaBr}
\]
The product is ethyl nitrite.
Its ethyl group is attached to oxygen rather than directly to nitrogen.
\(\textbf{Hydrolyse the alkyl nitrite:}\)
Under suitable hydrolysis conditions, the alkyl nitrite can regenerate the corresponding alcohol:
\[
\mathrm{C_2H_5ONO+H_2O
\rightleftharpoons C_2H_5OH+HNO_2}
\]
The intended organic product is ethanol.
\(\textbf{Route Q — Nitrogen-linked substitution:}\)
Silver nitrite favours formation of a nitroalkane:
\[
\mathrm{C_2H_5Br+AgNO_2
\rightarrow C_2H_5NO_2+AgBr}
\]
In nitroethane, the ethyl carbon is bonded directly to nitrogen.
This linkage differs from that in ethyl nitrite.
\(\textbf{Reduce the nitro compound:}\)
Suitable reduction converts the nitro group into an amino group:
\[
\mathrm{C_2H_5NO_2
\xrightarrow{reduction}
C_2H_5NH_2}
\]
The product is ethanamine.
\(\textbf{Compare the chemical consequences:}\)
Ethyl nitrite and nitroethane have the same elemental composition but different atom connectivity.
The oxygen-linked compound can undergo the specified hydrolysis to an alcohol.
The nitrogen-linked compound can be reduced to a primary amine.
\(\textbf{Integrated conclusion:}\)
Ambident nucleophile selectivity can determine the outcome of a later functional-group transformation.
The two routes give ethanol and ethanamine, respectively, despite beginning with the same bromoalkane.
303. Two aryl chlorides are compared:
P: Chlorobenzene.
Q: 4-Nitrochlorobenzene.
Each is treated with aqueous sodium hydroxide under appropriately selected conditions, and the resulting alkaline reaction mixture is subsequently acidified.
For P, the conditions are \(623\ \mathrm{K}\) and approximately \(300\ \mathrm{atm}\). For Q, conditions suitable for nitro-activated nucleophilic aromatic substitution are used.
Which statement correctly compares their reactivity and the effect of the final acidification?
ⓐ. P cannot undergo substitution under any conditions, whereas Q gives chlorobenzene after acidification
ⓑ. Both form phenol because the nitro group is necessarily removed during hydroxide substitution
ⓒ. P requires severe conditions to form phenoxide, while Q benefits from para-nitro activation; acidification gives phenol from P and 4-nitrophenol from Q
ⓓ. Q reacts only through an ordinary \(S_\mathrm{N}2\) mechanism at the aromatic carbon because the nitro group converts it into an \(sp^3\)-hybridised carbon
Correct Answer: P requires severe conditions to form phenoxide, while Q benefits from para-nitro activation; acidification gives phenol from P and 4-nitrophenol from Q
Explanation: \(\textbf{Identify the common bonding environment:}\)
Both substrates contain chlorine attached directly to an aromatic carbon.
The carbon–chlorine bond possesses characteristics different from those of an ordinary primary alkyl chloride.
Consequently, ordinary mild hydroxide substitution is generally unfavourable for unactivated chlorobenzene.
\(\textbf{Analyse chlorobenzene under severe conditions:}\)
Chlorobenzene can be converted into sodium phenoxide by heating with aqueous sodium hydroxide at approximately \(623\ \mathrm{K}\) and \(300\ \mathrm{atm}\):
\[
\mathrm{C_6H_5Cl+2NaOH
\xrightarrow[300\,atm]{623\,K}
C_6H_5ONa+NaCl+H_2O}
\]
These severe conditions illustrate the resistance of the ordinary aryl carbon–chlorine bond to nucleophilic substitution.
\(\textbf{Identify the role of acidification for P:}\)
Sodium phenoxide is converted into phenol by treatment with acid:
\[
\mathrm{C_6H_5ONa+HCl\rightarrow C_6H_5OH+NaCl}
\]
Thus, the isolated neutral organic product is phenol.
\(\textbf{Analyse the nitro-substituted substrate:}\)
Q contains a nitro group para to chlorine.
The nitro group strongly withdraws electron density and stabilises the negatively charged intermediate formed during suitable aromatic addition–elimination substitution.
This activation permits hydroxide replacement under conditions less severe than those required for unactivated chlorobenzene.
\(\textbf{Track the nitro group:}\)
The intended nucleophilic substitution replaces chlorine, not the nitro group.
The initial product in alkaline solution is the corresponding nitrophenoxide species.
Acidification gives 4-nitrophenol.
\(\textbf{Distinguish the two reaction explanations:}\)
The difference is not caused by converting Q into an ordinary saturated alkyl halide.
Both starting carbon–chlorine bonds remain attached to aromatic carbons.
The nitro group instead changes the electronic stabilisation available to the aromatic substitution intermediate.
\(\textbf{Integrated conclusion:}\)
P demonstrates the forcing conditions needed for an unactivated haloarene.
Q demonstrates the enhanced reactivity associated with a suitably positioned electron-withdrawing group.
The respective neutral products after acidification are phenol and 4-nitrophenol.
304. Ethylbenzene is subjected to a three-stage synthesis:
Step I: Controlled benzylic monochlorination using \(\mathrm{Cl_2/h\nu}\).
Step II: Hydrolysis of the isolated benzylic chloride under conditions producing the corresponding alcohol without rearrangement.
Step III: Oxidation of the isolated alcohol under conditions suitable for conversion of a secondary alcohol into a ketone.
Which final product is expected?
ⓐ. Benzoic acid, because every benzylic oxidation removes the entire ethyl group
ⓑ. Acetophenone, because benzylic chlorination followed by substitution gives a secondary alcohol that can be oxidised to a ketone
ⓒ. Benzaldehyde, because the two-carbon side chain necessarily loses one carbon during hydrolysis
ⓓ. 4-Chloroacetophenone, because photochemical chlorination initially introduces chlorine para to the ethyl group
Correct Answer: Acetophenone, because benzylic chlorination followed by substitution gives a secondary alcohol that can be oxidised to a ketone
Explanation: \(\textbf{Identify the original side chain:}\)
Ethylbenzene contains the group \(\mathrm{-CH_2CH_3}\) attached to a benzene ring.
The carbon directly adjacent to the aromatic ring is the benzylic carbon.
It possesses hydrogen atoms susceptible to appropriate radical substitution.
\(\textbf{Step I — Benzylic chlorination:}\)
Under controlled photochemical conditions, chlorine can replace one benzylic hydrogen:
\[
\mathrm{C_6H_5CH_2CH_3+Cl_2
\xrightarrow{h\nu}
C_6H_5CH(Cl)CH_3+HCl}
\]
The product is 1-chloroethylbenzene.
Chlorine is attached to a saturated benzylic carbon rather than directly to the aromatic ring.
\(\textbf{Step II — Hydrolysis of the benzylic chloride:}\)
Under the stipulated conditions, hydroxide replaces chlorine:
\[
\mathrm{C_6H_5CH(Cl)CH_3+OH^-
\rightarrow C_6H_5CH(OH)CH_3+Cl^-}
\]
The product is 1-phenylethanol.
Its hydroxyl-bearing carbon is attached to a phenyl group and a methyl group.
It is therefore a secondary alcohol.
\(\textbf{Step III — Oxidation of the secondary alcohol:}\)
A secondary alcohol can be oxidised to a ketone under suitable conditions:
\[
\mathrm{C_6H_5CH(OH)CH_3
\xrightarrow{[O]}
C_6H_5COCH_3}
\]
The resulting compound is acetophenone, also called 1-phenylethanone.
\(\textbf{Check carbon conservation:}\)
The initial molecule contains the benzene ring and a two-carbon side chain.
The stipulated substitution, hydrolysis and secondary-alcohol oxidation steps do not remove either carbon of that side chain.
The product therefore remains an eight-carbon organic compound.
\(\textbf{Distinguish oxidation conditions:}\)
Strong oxidation of an alkylbenzene under different conditions can produce benzoic acid.
However, that is not the transformation specified in Step III.
The question explicitly oxidises the isolated secondary alcohol under ketone-forming conditions.
\(\textbf{Integrated conclusion:}\)
Following the identity and classification of each intermediate is essential.
The sequence proceeds through a benzylic chloride and a secondary benzylic alcohol before giving acetophenone.
305. Bromobenzene undergoes the following two experiments:
P: Treatment with heavy water, \(\mathrm{D_2O}\), under mild conditions without magnesium.
Q: Treatment with magnesium in dry ether, followed by complete quenching of the resulting organomagnesium reagent with excess \(\mathrm{D_2O}\).
Here D denotes deuterium.
Assume that magnesium insertion and subsequent isotope transfer proceed as intended in Q.
Which statement correctly predicts the principal aromatic products and the carbon at which deuterium becomes attached?
ⓐ. P gives deuterobenzene, whereas Q gives benzene because magnesium removes the isotope label
ⓑ. Both experiments give deuterobenzene because water readily replaces aromatic bromine
ⓒ. P gives phenol, whereas Q gives bromobenzene with deuterium attached to an unchanged bromine-bearing carbon
ⓓ. P largely retains bromobenzene, whereas Q gives deuterobenzene with D attached to the aromatic carbon originally bearing bromine
Correct Answer: P largely retains bromobenzene, whereas Q gives deuterobenzene with D attached to the aromatic carbon originally bearing bromine
Explanation: \(\textbf{Identify the original carbon–halogen environment:}\)
In bromobenzene, bromine is attached directly to an aromatic \(sp^2\)-hybridised carbon.
Ordinary mild nucleophilic displacement at that carbon is unfavourable.
Water does not become an efficient aromatic bromine-replacement reagent merely because it contains the heavier hydrogen isotope.
\(\textbf{Analyse Experiment P:}\)
The reaction mixture contains bromobenzene and heavy water but no reagent that converts the aryl carbon–bromine bond into an organometallic bond.
Under the specified mild conditions, the aromatic bromine remains attached.
The principal organic compound is therefore unchanged bromobenzene.
\(\textbf{Analyse the first stage of Q:}\)
Magnesium inserts into the aromatic carbon–bromine bond under suitable dry-ether conditions:
\[
\mathrm{C_6H_5Br+Mg
\xrightarrow{dry\ ether}
C_6H_5MgBr}
\]
The resulting compound is phenylmagnesium bromide.
Its carbon–magnesium bond is strongly polarised towards carbon.
The phenyl carbon therefore possesses substantial nucleophilic and basic character.
\(\textbf{Follow the deuterium-transfer reaction:}\)
Heavy water contains O–D bonds that can supply deuterium during proton-transfer-type quenching.
The Grignard reagent reacts with \(\mathrm{D_2O}\):
\[
\mathrm{C_6H_5MgBr+D_2O
\rightarrow C_6H_5D+MgBrOD}
\]
The organic product is deuterobenzene.
\(\textbf{Identify the labelled position:}\)
The carbon–magnesium bond occupied the same aromatic position as the original carbon–bromine bond.
Deuterium replaces the magnesium-containing group at that carbon.
Therefore, the label appears at the aromatic carbon that originally bore bromine.
\(\textbf{Distinguish the two mechanisms:}\)
Experiment P attempts no effective ordinary substitution at an unactivated aryl carbon.
Experiment Q first changes the bond polarity and reactivity through organomagnesium formation, then introduces deuterium by quenching.
The formation of deuterobenzene is not evidence that ordinary heavy water directly displaces aromatic bromide.
\(\textbf{Integrated conclusion:}\)
The sequence illustrates a change from an electrophilic carbon–halogen environment to a nucleophilic carbon–magnesium environment.
The isotope label reveals the carbon at which that organometallic transformation occurred.
306. 5-Bromopent-1-ene, \(\mathrm{BrCH_2CH_2CH_2CH{=}CH_2}\), undergoes the following sequence:
Step I: Treatment with alcoholic KCN under conditions giving exclusive substitution at the terminal carbon bearing bromine.
The isolated nitrile is then divided into two portions.
Portion P: Complete reduction of the nitrile group using \(\mathrm{LiAlH_4}\) under conditions preserving the carbon–carbon double bond.
Portion Q: Selective catalytic hydrogenation of the carbon–carbon double bond under conditions preserving the nitrile functional group.
Which pair correctly identifies the final organic products?
ⓐ. P: Hex-5-en-1-amine; Q: hexanenitrile
ⓑ. P: Pent-4-en-1-amine; Q: pentanenitrile
ⓒ. P: Hexanenitrile; Q: hex-5-en-1-amine
ⓓ. P: Hexan-1-amine; Q: hex-5-en-1-ol
Correct Answer: P: Hex-5-en-1-amine; Q: hexanenitrile
Explanation: \(\textbf{Identify the starting functional groups:}\)
5-Bromopent-1-ene contains five carbon atoms.
It has a primary carbon–bromine bond at one end and a carbon–carbon double bond at the other.
The specified first reaction replaces bromide without consuming the alkene.
\(\textbf{Step I — Extend the carbon skeleton:}\)
Cyanide attacks through its carbon atom:
\[
\mathrm{BrCH_2CH_2CH_2CH{=}CH_2+KCN
\rightarrow NCCH_2CH_2CH_2CH{=}CH_2+KBr}
\]
The nitrile carbon becomes part of the principal carbon chain.
The product contains six carbon atoms and is hex-5-enenitrile.
\(\textbf{Portion P — Selective nitrile reduction:}\)
Lithium aluminium hydride can reduce an appropriate nitrile group to a primary amine.
The nitrile carbon becomes a methylene carbon attached to nitrogen:
\[
\mathrm{NCCH_2CH_2CH_2CH{=}CH_2
\xrightarrow{LiAlH_4,\ workup}
H_2NCH_2CH_2CH_2CH_2CH{=}CH_2}
\]
The carbon–carbon double bond is retained under the stipulated conditions.
The product is hex-5-en-1-amine.
\(\textbf{Portion Q — Selective alkene hydrogenation:}\)
In Q, the reaction conditions are explicitly selected to hydrogenate the carbon–carbon double bond while retaining the nitrile group.
Addition of hydrogen converts the terminal alkene into a saturated carbon chain:
\[
\mathrm{NCCH_2CH_2CH_2CH{=}CH_2+H_2
\xrightarrow{selective\ catalyst}
NCCH_2CH_2CH_2CH_2CH_3}
\]
The product is hexanenitrile.
\(\textbf{Compare the functional-group outcomes:}\)
P converts the nitrile into an amine while preserving the alkene.
Q converts the alkene into an alkane while preserving the nitrile.
Neither reaction changes the six-carbon skeleton established during cyanide substitution.
\(\textbf{Check the carbon-number inference:}\)
The starting bromoalkene contains five carbon atoms.
Substitution by carbon-bound cyanide increases this count to six.
All later products therefore retain six carbon atoms.
\(\textbf{Integrated conclusion:}\)
The first stage illustrates carbon-chain extension through nucleophilic substitution.
The second stage illustrates how different reducing conditions can target different functional groups.
The correct products are hex-5-en-1-amine from P and hexanenitrile from Q.
307. Three chlorinated methane derivatives are compared using the following physical data:
Each compound is considered separately in contact with water of density \(1.00\ \mathrm{g\,mL^{-1}}\). Assume that the organic liquids remain as separate phases.
Which conclusion correctly relates molecular symmetry, liquid density and boiling point?
| Compound | Density at room temperature (g mL\(^{-1}\)) | Normal boiling point (°C) |
|---|---|---|
| P: \(\mathrm{CH_2Cl_2}\) | 1.336 | 39.6 |
| Q: \(\mathrm{CHCl_3}\) | 1.489 | 61.2 |
| R: \(\mathrm{CCl_4}\) | 1.595 | 76.7 |
ⓐ. R is nonpolar by tetrahedral symmetry, yet sinks below water and has the highest boiling point of the three
ⓑ. R has the greatest molecular dipole moment because it contains the greatest number of polar carbon–chlorine bonds
ⓒ. P forms the upper liquid layer because it contains fewer chlorine atoms than Q and R
ⓓ. Q has the highest boiling point because its molecular dipole moment must be greater than that of the symmetrical R molecule
Correct Answer: R is nonpolar by tetrahedral symmetry, yet sinks below water and has the highest boiling point of the three
Explanation: \(\textbf{Distinguish bond polarity from molecular polarity:}\)
All three compounds contain polar covalent carbon–chlorine bonds.
However, a molecular dipole moment is the vector sum of the individual bond-dipole contributions.
The number of polar bonds alone does not determine the resultant molecular dipole.
\(\textbf{Examine tetrachloromethane:}\)
In \(\mathrm{CCl_4}\), four equivalent carbon–chlorine bonds are arranged tetrahedrally.
Their dipole contributions cancel because of the molecular symmetry.
Consequently, tetrachloromethane has zero permanent molecular dipole moment despite possessing four polar bonds.
\(\textbf{Determine the liquid-layer positions:}\)
A liquid denser than water forms the lower layer when the liquids are sufficiently immiscible.
The supplied densities satisfy:
\[
1.336\gt 1.00
\]
\[
1.489\gt 1.00
\]
\[
1.595\gt 1.00
\]
Thus, each chlorinated liquid forms the lower layer in its respective water-contact experiment.
\(\textbf{Interpret the boiling-point data:}\)
The boiling points increase from dichloromethane to chloroform to tetrachloromethane.
The larger molecules contain more electrons and are generally more polarizable.
Their London dispersion interactions become stronger.
These interactions can produce a higher boiling point even when the molecule has no permanent dipole moment.
\(\textbf{Separate three physical properties:}\)
Density depends on mass per unit volume.
Boiling point depends on the energetic requirements for separating molecules into the vapour phase.
Molecular dipole moment depends on the distribution and geometry of electrical charge.
These quantities cannot be ranked simply by counting chlorine atoms or polar bonds.
\(\textbf{Integrated conclusion:}\)
R is a nonpolar molecule with polar bonds.
It is denser than water and has the highest boiling point in the given comparison.
This illustrates why molecular geometry and intermolecular forces must be considered independently.
308. Two haloalkanes have the following densities at \(25^\circ\mathrm{C}\):
\[
\rho(\mathrm{1\!-\!chloropropane})=0.890\ \mathrm{g\,mL^{-1}}
\]
\[
\rho(\mathrm{1\!-\!bromopropane})=1.335\ \mathrm{g\,mL^{-1}}
\]
A mixture is prepared from \(30.0\ \mathrm{mL}\) of 1-chloropropane and \(20.0\ \mathrm{mL}\) of 1-bromopropane.
Additional 1-chloropropane is then introduced.
Assume that the two haloalkanes are completely miscible with each other, that their volumes are additive, and that the resulting organic liquid is immiscible with water of density \(1.000\ \mathrm{g\,mL^{-1}}\).
What is the minimum additional volume of 1-chloropropane required to bring the organic mixture to the density of water?
ⓐ. \(20.0\ \mathrm{mL}\)
ⓑ. \(25.0\ \mathrm{mL}\)
ⓒ. \(30.9\ \mathrm{mL}\)
ⓓ. \(40.0\ \mathrm{mL}\)
Correct Answer: \(30.9\ \mathrm{mL}\)
Explanation: \(\textbf{Identify the physical condition:}\)
A liquid mixture has the same density as water when its total mass divided by its total volume equals \(1.000\ \mathrm{g\,mL^{-1}}\).
The problem specifies volume additivity, so the mixture's total volume is the sum of the component volumes.
\(\textbf{Calculate the initial chloropropane mass:}\)
Using:
\[
m=\rho V
\]
\[
m_{\mathrm{chloride}}=(0.890)(30.0)
\]
\[
m_{\mathrm{chloride}}=26.70\ \mathrm{g}
\]
\(\textbf{Calculate the initial bromopropane mass:}\)
\[
m_{\mathrm{bromide}}=(1.335)(20.0)
\]
\[
m_{\mathrm{bromide}}=26.70\ \mathrm{g}
\]
\(\textbf{Determine the initial mixture density:}\)
\[
m_0=26.70+26.70
\]
\[
m_0=53.40\ \mathrm{g}
\]
\[
V_0=30.0+20.0=50.0\ \mathrm{mL}
\]
\[
\rho_0=\frac{53.40}{50.0}
\]
\[
\rho_0=1.068\ \mathrm{g\,mL^{-1}}
\]
The initial organic mixture is therefore denser than water.
\(\textbf{Introduce the additional volume:}\)
Let \(x\) be the additional volume of 1-chloropropane, measured in millilitres.
The added mass is:
\[
m_{\mathrm{added}}=0.890x
\]
The final volume is:
\[
V_{\mathrm{final}}=50.0+x
\]
The final mass is:
\[
m_{\mathrm{final}}=53.40+0.890x
\]
\(\textbf{Apply the equal-density condition:}\)
\[
\rho_{\mathrm{final}}=\frac{53.40+0.890x}{50.0+x}
\]
Set:
\[
1.000=\frac{53.40+0.890x}{50.0+x}
\]
\[
50.0+x=53.40+0.890x
\]
\[
x-0.890x=3.40
\]
\[
0.110x=3.40
\]
\[
x=\frac{3.40}{0.110}
\]
\[
x=30.91\ \mathrm{mL}
\]
\(\textbf{Check the direction of the change:}\)
Additional 1-chloropropane lowers the mixture density because its density is less than that of water.
At the calculated volume, the mixture has the same density as water under the stated idealisation.
A slightly larger addition would make the organic phase less dense than water.
\(\textbf{Final answer:}\)
\[
\boxed{V_{\mathrm{additional}}=30.9\ \mathrm{mL}}
\]
309. The following experimental data are supplied for three methyl halides:
Which statement best reconciles all three trends?
| Compound | Dipole moment (D) | C–X bond enthalpy (kJ mol\(^{-1}\)) | Normal boiling point (°C) |
|---|---|---|---|
| \(\mathrm{CH_3Cl}\) | 1.860 | 351 | −24.2 |
| \(\mathrm{CH_3Br}\) | 1.830 | 293 | 3.6 |
| \(\mathrm{CH_3I}\) | 1.636 | 234 | 42.5 |
ⓐ. The boiling-point order must follow the dipole-moment order because dipole–dipole attractions are the only intermolecular forces present
ⓑ. Iodomethane has the strongest carbon–halogen bond because it has the highest boiling point
ⓒ. Chloromethane should have the highest boiling point because its carbon–halogen bond is strongest
ⓓ. Boiling points rise with increasing polarizability and dispersion forces, despite decreasing C–X bond enthalpy and dipole moment
Correct Answer: Boiling points rise with increasing polarizability and dispersion forces, despite decreasing C–X bond enthalpy and dipole moment
Explanation: \(\textbf{Read the bond-enthalpy trend:}\)
The supplied carbon–halogen bond enthalpies decrease from chloromethane to bromomethane to iodomethane.
This trend is consistent with increasing halogen size and increasing bond length.
The carbon–iodine bond is therefore the weakest of the three in the stated comparison.
\(\textbf{Read the molecular dipole-moment trend:}\)
The dipole moments also decrease:
\[
1.860\gt 1.830\gt 1.636\ \mathrm{D}
\]
Chloromethane possesses the largest molecular dipole moment among the three listed compounds.
\(\textbf{Read the boiling-point trend:}\)
The boiling points increase:
\[
-24.2\lt 3.6\lt 42.5^\circ\mathrm{C}
\]
Iodomethane consequently has the highest normal boiling point despite possessing the smallest dipole moment and weakest carbon–halogen bond in the table.
\(\textbf{Identify the controlling intermolecular effect:}\)
As the halogen changes from chlorine to bromine to iodine, the molecular mass and electronic polarizability increase.
The larger, more polarizable molecules experience stronger London dispersion attractions.
These attractions contribute substantially to the energy required for vaporisation.
\(\textbf{Distinguish intramolecular and intermolecular energy:}\)
Carbon–halogen bond enthalpy describes the energy associated with breaking a bond within an isolated gaseous molecule.
Ordinary boiling separates molecules from one another without cleaving their carbon–halogen covalent bonds.
Therefore, a weaker carbon–iodine bond does not imply that iodomethane must boil at a lower temperature.
\(\textbf{Integrated conclusion:}\)
Dipole moments and bond strengths are insufficient to predict the boiling-point order independently.
The increased molecular mass and polarizability of the heavier methyl halides account for the observed rise in boiling point.
This is why iodomethane is the highest-boiling member of the supplied series.
310. Fluoromethane and chloromethane are compared under similar conditions favouring nucleophilic substitution by the same nucleophile. Which statement correctly relates carbon–halogen bond polarity, bond strength and substitution behaviour?
ⓐ. The C–F bond is less polar than C–Cl, and fluoromethane usually substitutes faster
ⓑ. The C–F bond is stronger than C–Cl, but fluoromethane reacts faster because its greater bond polarity is sufficient to overcome the poorer fluoride leaving group
ⓒ. The C–Cl bond is more polar than C–F, so chloromethane cannot undergo nucleophilic substitution
ⓓ. The C–F bond is more polar and stronger than C–Cl, while chloromethane generally undergoes ordinary nucleophilic substitution more readily
Correct Answer: The C–F bond is more polar and stronger than C–Cl, while chloromethane generally undergoes ordinary nucleophilic substitution more readily
Explanation: \(\textbf{Separate bond polarity from bond cleavage:}\)
Both methyl halides have a polar carbon–halogen bond because the halogen is more electronegative than carbon. Fluorine is more electronegative than chlorine, so the C–F bond normally has greater bond polarity.
\(\textbf{Compare bond strengths:}\)
The C–F bond is relatively short and strong, whereas the C–Cl bond is longer and easier to break under comparable ordinary substitution conditions. The higher polarity of the C–F bond therefore does not mean that fluorine leaves more readily as a halide ion.
\(\textbf{Evaluate the substitution reaction:}\)
Methyl halides can undergo an \(S_\mathrm{N}2\) pathway with a suitable nucleophile. Both substrates are unhindered, so the important distinction here is leaving-group ability and carbon–halogen bond strength. Chloride is normally a better leaving group than fluoride, and chloromethane is generally more reactive in ordinary nucleophilic substitution.
\(\textbf{Integrated conclusion:}\)
Electronegativity and bond polarity describe electron distribution, whereas substitution rate also depends on bond cleavage, leaving-group ability and conditions. Therefore, stronger C–F polarity is compatible with lower ordinary substitution reactivity.
311. Chlorobenzene, \(\mathrm{C_6H_5Cl}\), and (trifluoromethyl)benzene, \(\mathrm{C_6H_5CF_3}\), undergo separately controlled electrophilic mononitration.
Assume substitution takes place only on the aromatic ring under conditions suitable for the respective substrates.
Which pair correctly predicts their characteristic orienting effects and the underlying bonding distinction?
ⓐ. Chlorobenzene is meta directing because chlorine withdraws inductively; (trifluoromethyl)benzene is ortho/para directing because fluorine has lone pairs
ⓑ. Both are ortho/para directing because both molecules contain carbon–halogen bonds
ⓒ. Chlorobenzene directs ortho/para by chlorine lone-pair resonance; the ring-bound \(\mathrm{-CF_3}\) group is meta directing
ⓓ. Both are meta directing because any substituent containing halogen atoms necessarily withdraws electron density without resonance donation
Correct Answer: Chlorobenzene directs ortho/para by chlorine lone-pair resonance; the ring-bound \(\mathrm{-CF_3}\) group is meta directing
Explanation: \(\textbf{Identify the directly attached substituents:}\)
Chlorobenzene contains chlorine bonded directly to an aromatic carbon.
In (trifluoromethyl)benzene, the aromatic carbon is bonded to the carbon atom of a \(\mathrm{-CF_3}\) group.
The fluorine atoms are attached to that substituent carbon rather than directly to the benzene ring.
\(\textbf{Analyse chlorobenzene:}\)
Chlorine withdraws electron density through its inductive effect.
This makes chlorobenzene less reactive than benzene towards ordinary electrophilic substitution.
However, chlorine also possesses lone pairs that can interact directly with the aromatic \(\pi\)-electron system.
Resonance donation stabilises the appropriate intermediates associated with ortho and para attack.
Therefore, chlorine is deactivating but ortho/para directing.
\(\textbf{Analyse the trifluoromethyl group:}\)
The \(\mathrm{-CF_3}\) group strongly withdraws electron density through its inductive effect.
Its fluorine lone pairs cannot participate in the same direct halogen-to-ring resonance interaction as chlorine in chlorobenzene.
The group therefore behaves as a strongly deactivating, meta-directing substituent in ordinary electrophilic aromatic substitution.
\(\textbf{Predict the nitration patterns:}\)
Mononitration of chlorobenzene principally gives ortho- and para-nitrochlorobenzene.
Mononitration of (trifluoromethyl)benzene favours the meta-nitro product under suitable conditions.
\(\textbf{Distinguish substituent composition from attachment:}\)
It would be incorrect to classify every compound containing fluorine as an aryl fluoride.
The fluorine atoms of \(\mathrm{-CF_3}\) are bonded to an aliphatic substituent carbon.
The atom directly attached to the ring determines which resonance interactions are possible.
\(\textbf{Integrated conclusion:}\)
Inductive withdrawal occurs in both compounds, but direct chlorine-to-ring resonance distinguishes chlorobenzene.
The bonding arrangement explains why the two halogen-containing substituents produce different nitration orientations.
312. A mixture contains approximately equal amounts of the ortho-, meta- and para-isomers of dichlorobenzene.
The following physical data are supplied:
The chemist cools the mixture to \(20^\circ\mathrm{C}\) and considers separating the components by formation of a crystalline solid, followed by filtration.
Which assessment is most appropriate?
| Isomer | Melting point (°C) | Boiling point (°C) |
|---|---|---|
| 1,2-Dichlorobenzene | −17.5 | 180.5 |
| 1,3-Dichlorobenzene | −24.8 | 173.0 |
| 1,4-Dichlorobenzene | 53.5 | 174.1 |
ⓐ. 1,2-Dichlorobenzene is the expected crystalline component because its boiling point is highest
ⓑ. All three isomers should crystallise equally because their molecular formulas and numbers of chlorine atoms are identical
ⓒ. 1,3-Dichlorobenzene is expected to crystallise because the meta substitution pattern necessarily gives the strongest crystal lattice
ⓓ. 1,4-Dichlorobenzene is the expected crystalline component; its comparatively high melting point is associated with its symmetrical packing in the solid state
Correct Answer: 1,4-Dichlorobenzene is the expected crystalline component; its comparatively high melting point is associated with its symmetrical packing in the solid state
Explanation: \(\textbf{Identify the structural relationship:}\)
The three compounds are positional isomers with the same molecular formula.
Each contains two chlorine atoms attached to an aromatic ring.
Their differences arise from the relative positions of those chlorine atoms.
\(\textbf{Compare their melting points:}\)
The ortho and meta isomers have melting points below \(0^\circ\mathrm{C}\).
At \(20^\circ\mathrm{C}\), both are above their individual melting points.
The para isomer melts at approximately \(53.5^\circ\mathrm{C}\).
At \(20^\circ\mathrm{C}\), it is below its melting point and can exist as a crystalline solid.
\(\textbf{Identify the separation opportunity:}\)
Cooling a suitable liquid mixture can cause a component with a sufficiently high melting point to crystallise preferentially.
The solid can then be separated from the remaining liquid by filtration under appropriate conditions.
The given temperatures therefore support preferential crystallisation of 1,4-dichlorobenzene.
\(\textbf{Explain the para isomer's melting behaviour:}\)
The para isomer has a comparatively symmetrical molecular structure.
This symmetry permits more efficient packing in the crystal lattice.
Favourable crystal packing contributes to its unusually high melting point relative to the other two positional isomers.
\(\textbf{Distinguish melting from boiling:}\)
The boiling points of the three isomers are relatively close.
The ortho isomer has the highest listed boiling point but the para isomer has the highest melting point.
Therefore, a boiling-point ranking cannot be used directly to predict which component crystallises at a particular temperature.
\(\textbf{Recognise the experimental qualification:}\)
The conclusion concerns the preferred component for crystallisation under the stated idealised comparison.
Actual separation efficiency also depends on the composition and solid–liquid equilibrium of the mixture.
The individual melting-point data do not alone establish quantitative recovery.
\(\textbf{Integrated conclusion:}\)
The substantial difference in melting points provides a stronger basis for selecting the para isomer by cooling than the relatively small differences in boiling points.
The expected crystalline component is 1,4-dichlorobenzene.
313. Ethanol and bromoethane are compared at approximately atmospheric pressure.
Which explanation best accounts for the contrast despite the greater molar mass of bromoethane?
| Compound | Molar mass (g mol\(^{-1}\)) | Normal boiling point (°C) | Behaviour in water |
|---|---|---|---|
| Ethanol | 46.1 | 78.4 | Miscible |
| Bromoethane | 108.97 | 38.4 | Only slightly soluble |
ⓐ. The carbon–bromine bond is nonpolar, so bromoethane possesses no intermolecular attractions
ⓑ. Ethanol's O–H group forms strong hydrogen bonds, whereas bromoethane's polar C–Br bond cannot provide equivalent interactions
ⓒ. Ethanol has the higher boiling point because its carbon–oxygen bond is always stronger than every carbon–bromine bond
ⓓ. Bromoethane is less soluble only because its molecular formula contains fewer hydrogen atoms than ethanol
Correct Answer: Ethanol's O–H group forms strong hydrogen bonds, whereas bromoethane's polar C–Br bond cannot provide equivalent interactions
Explanation: \(\textbf{Identify the functional groups:}\)
Ethanol contains a hydroxyl group attached to a saturated carbon.
Bromoethane contains bromine attached to a saturated carbon.
Both molecules possess polar covalent bonds, but the intermolecular interactions available to them differ.
\(\textbf{Examine ethanol's intermolecular forces:}\)
The O–H group allows ethanol molecules to form hydrogen bonds with one another.
The oxygen atom can also participate in hydrogen bonding with water molecules.
These interactions are important in ethanol's boiling behaviour and its miscibility with water.
\(\textbf{Examine bromoethane's intermolecular forces:}\)
The carbon–bromine bond is polar.
Bromoethane therefore exhibits intermolecular attractions, including dipole-related and London dispersion forces.
However, it lacks an O–H or N–H group capable of providing hydrogen bonding comparable to that of ethanol.
Its polar carbon–bromine bond does not make it a hydrogen-bonding analogue of ethanol.
\(\textbf{Explain the boiling-point contrast:}\)
Bromoethane has a much greater molar mass than ethanol.
Nevertheless, the strong hydrogen-bonding network among ethanol molecules contributes to its higher boiling point.
The observed data show that molecular mass is not the only factor governing boiling temperature.
\(\textbf{Explain the water-solubility contrast:}\)
Dissolving an organic compound in water requires disruption and replacement of intermolecular attractions.
Ethanol forms favourable hydrogen bonds with water, making mixing energetically favourable.
Bromoethane cannot compensate for disruption of water's hydrogen-bond network to the same extent.
Consequently, its water solubility is much lower.
\(\textbf{Distinguish bond enthalpy from intermolecular forces:}\)
The strength of a particular covalent bond within a molecule is not the same as the strength of intermolecular attractions between different molecules.
Ordinary boiling does not require breaking carbon–oxygen or carbon–bromine covalent bonds.
\(\textbf{Integrated conclusion:}\)
Hydrogen bonding explains both ethanol's relatively high boiling point and its extensive miscibility with water.
The comparison demonstrates why a heavier haloalkane can boil below a lighter alcohol.
314. Two samples contain stereoisomeric 2-bromobutane.
Sample P contains optically pure \((R)\)-2-bromobutane.
Sample Q contains an equimolar mixture of \((R)\)- and \((S)\)-2-bromobutane.
A chemist proposes ordinary fractional distillation using an achiral apparatus to separate Q into its two pure enantiomers.
Which statement correctly evaluates the proposal?
ⓐ. Fractional distillation necessarily separates the enantiomers because their optical rotations have opposite signs
ⓑ. Sample Q must have a higher boiling point than either pure enantiomer because its overall optical rotation is zero
ⓒ. The enantiomers have identical boiling points in an achiral environment; an enantioselective method is needed for resolution
ⓓ. Sample P is achiral because an optically pure substance contains only one configuration
Correct Answer: The enantiomers have identical boiling points in an achiral environment; an enantioselective method is needed for resolution
Explanation: \(\textbf{Identify the stereoisomeric relationship:}\)
\((R)\)-2-Bromobutane and \((S)\)-2-bromobutane are non-superimposable mirror images.
They possess the same molecular formula, atom connectivity and corresponding bond lengths.
Their ordinary physical properties are identical in an achiral environment, apart from properties sensitive to handedness.
\(\textbf{Examine optical activity:}\)
Each pure enantiomer rotates plane-polarised light.
Under identical measurement conditions, their rotations have equal magnitudes and opposite signs.
An equimolar mixture has zero net optical rotation because the individual contributions cancel.
\(\textbf{Interpret boiling behaviour:}\)
Enantiomers have identical ordinary boiling points in an achiral environment.
Their corresponding vaporisation properties are also identical under those conditions.
Therefore, ordinary fractional distillation does not provide the differential volatility needed to separate them.
\(\textbf{Evaluate the proposed separation:}\)
Fractional distillation is effective when components differ sufficiently in volatility.
Opposite signs of optical rotation do not establish different boiling points.
Consequently, an achiral distillation apparatus cannot resolve the two enantiomers simply by collecting different boiling fractions.
\(\textbf{Identify a suitable resolution principle:}\)
A chiral environment can distinguish enantiomers through diastereomeric interactions.
Suitable methods may involve a chiral stationary phase or conversion into separable diastereomeric derivatives.
The important feature is the introduction of a chiral discrimination mechanism.
\(\textbf{Distinguish chirality from optical cancellation:}\)
Sample Q is optically inactive because the enantiomeric rotations cancel.
This does not make the individual molecules achiral.
Sample P contains chiral molecules of one configuration and can exhibit optical activity.
\(\textbf{Integrated conclusion:}\)
The problem requires distinguishing molecular chirality, bulk optical rotation and ordinary phase-transition properties.
An equimolar enantiomeric mixture cannot be resolved by conventional fractional distillation alone.
315. Three isomeric alkyl bromides have the following approximate normal boiling points:
An equimolar mixture is considered for fractional distillation. The isolated fractions would subsequently be compared under conditions favouring ordinary \(S_\mathrm{N}1\) substitution.
Which statement correctly combines the expected distillation sequence and substitution preference?
| Compound | Structure | Boiling point (°C) |
|---|---|---|
| P | 1-Bromobutane | 101.6 |
| Q | 2-Bromobutane | 91.2 |
| R | 2-Bromo-2-methylpropane | 73.3 |
ⓐ. R distils first, Q second and P last; R is also the most favourable for ordinary \(S_\mathrm{N}1\) substitution because it can form a tertiary carbocation
ⓑ. P distils first because it is primary; P is also fastest in ordinary \(S_\mathrm{N}1\) substitution
ⓒ. R distils last because tertiary alkyl bromides have the strongest London dispersion forces; P is the most favourable for \(S_\mathrm{N}1\)
ⓓ. Q distils first because it is secondary and forms a racemic sample; all three have identical \(S_\mathrm{N}1\) tendencies
Correct Answer: R distils first, Q second and P last; R is also the most favourable for ordinary \(S_\mathrm{N}1\) substitution because it can form a tertiary carbocation
Explanation: \(\textbf{Identify the shared molecular composition:}\)
P, Q and R are constitutional isomers with the formula \(\mathrm{C_4H_9Br}\).
They therefore have the same molecular mass.
Their different boiling points arise from differences in molecular shape and associated intermolecular interactions.
\(\textbf{Interpret the distillation sequence:}\)
At a given pressure, the component with the lowest boiling point is the most volatile.
The supplied boiling points satisfy:
\[
73.3\lt 91.2\lt 101.6^\circ\mathrm{C}
\]
Therefore, the expected order of increasing boiling point is:
\[
\mathrm{R\lt Q\lt P}
\]
R is expected in the earliest boiling fraction, followed by Q and then P.
\(\textbf{Explain the structural trend:}\)
Greater branching generally reduces the effective contact surface between haloalkane molecules.
This reduces the strength of their London dispersion attractions compared with a less branched isomer of the same molecular formula.
R is the most highly branched compound and has the lowest boiling point.
\(\textbf{Identify the substitution classes:}\)
P is a primary alkyl bromide.
Q is secondary.
R is tertiary.
The degree of substitution at the bromine-bearing carbon is therefore different in each compound.
\(\textbf{Predict the ordinary }S_\mathrm{N}1\textbf{ preference:}\)
An ordinary \(S_\mathrm{N}1\) mechanism involves ionisation to form a carbocation.
R can form a comparatively stable tertiary carbocation.
Q can form a secondary carbocation.
P would ordinarily form an unstable primary carbocation.
Thus, R is structurally most favourable for the stated ionisation-controlled pathway.
\(\textbf{Avoid conflating the physical and kinetic rankings:}\)
The boiling-point order follows the supplied physical data and is associated with intermolecular attractions.
The \(S_\mathrm{N}1\) preference follows carbocation stability.
The two rankings happen to identify R as important in this comparison, but for different chemical reasons.
\(\textbf{Integrated conclusion:}\)
Molecular branching influences both physical properties and reaction-centre structure.
However, boiling point and substitution rate must be reasoned out independently.
R distils first and is also the most favourable ordinary \(S_\mathrm{N}1\) substrate.
316. A primary alkyl bromide is hydrolysed using aqueous hydroxide under two experimental arrangements.
Experiment P: The haloalkane and water form two liquid phases and are left almost unstirred.
Experiment Q: The same amounts are mixed vigorously, increasing contact between the phases.
The intrinsic elementary substitution pathway in the reacting region is assumed to be \(S_\mathrm{N}2\), and temperature remains unchanged.
The observed conversion is faster in Q.
Which interpretation is most defensible?
ⓐ. Interfacial mixing increases the intrinsic \(S_\mathrm{N}2\) rate constant by creating a more substituted carbon centre
ⓑ. Interfacial mixing increases the elementary rate constant in direct proportion to stirring speed
ⓒ. Interfacial mixing converts the reaction from \(S_\mathrm{N}2\) to \(S_\mathrm{N}1\) without any change of solvent composition
ⓓ. Interfacial mixing can accelerate observed conversion by improving reactant contact without changing the elementary \(S_\mathrm{N}2\) mechanism
Correct Answer: Interfacial mixing can accelerate observed conversion by improving reactant contact without changing the elementary \(S_\mathrm{N}2\) mechanism
Explanation: \(\textbf{Identify the physical difficulty:}\)
Many haloalkanes are only slightly soluble in water.
Consequently, an aqueous reaction mixture can contain a separate organic liquid phase.
The concentrations available for reaction at a particular location may differ from the total analytical concentrations initially placed in the vessel.
\(\textbf{Examine Experiment P:}\)
The two liquids are left almost unstirred.
The contact area between them may be relatively small.
Transfer of haloalkane into the aqueous region, or transport of reagents to the interface, can limit the observed overall conversion.
\(\textbf{Examine Experiment Q:}\)
Vigorous stirring disperses one liquid into the other and can increase interfacial contact.
It may reduce mass-transfer limitations and improve the availability of reactants to the regions in which substitution occurs.
The measured conversion can therefore increase even if the elementary chemical reaction remains unchanged.
\(\textbf{Distinguish transport from reaction kinetics:}\)
The intrinsic elementary \(S_\mathrm{N}2\) rate law involves the substrate and nucleophile concentrations in the actual reacting environment.
An observed bulk conversion rate may also depend on phase contact and material transport.
The intrinsic rate constant and the observed transport-influenced rate are not automatically the same quantity.
\(\textbf{Reject a compulsory mechanism change:}\)
Stirring does not change the substitution level of the carbon bearing bromine.
It also does not require formation of a carbocation.
Therefore, increased stirring cannot by itself be used to establish a change from \(S_\mathrm{N}2\) to \(S_\mathrm{N}1\).
\(\textbf{Integrated conclusion:}\)
Physical solubility and phase behaviour can influence how efficiently a nucleophilic substitution proceeds in a real reaction vessel.
Improved mass transfer provides a chemically reasonable explanation for the observed faster conversion without changing the stipulated elementary pathway.
317. Two haloalkanes listed in the NCERT density table are considered:
P: 1-Bromopropane, density \(1.335\ \mathrm{g\,mL^{-1}}\).
Q: Dichloromethane, density \(1.336\ \mathrm{g\,mL^{-1}}\).
At the specified temperature, exactly \(30.0\ \mathrm{mL}\) of each pure liquid is measured separately.
Use molar masses:
\[
M(P)=123.0\ \mathrm{g\,mol^{-1}}
\]
\[
M(Q)=84.9\ \mathrm{g\,mol^{-1}}
\]
Which statement correctly compares the masses, numbers of molecules and expected layer positions when each liquid is separately contacted with water of density \(1.00\ \mathrm{g\,mL^{-1}}\)?
ⓐ. P has the greater mass and contains more molecules; only P forms the lower layer
ⓑ. Q has the greater mass but contains fewer molecules; only Q forms the lower layer
ⓒ. Q has a slightly greater mass and contains substantially more molecules; both liquids form lower layers
ⓓ. The equal volumes contain equal numbers of molecules because their densities are nearly identical
Correct Answer: Q has a slightly greater mass and contains substantially more molecules; both liquids form lower layers
Explanation: \(\textbf{Distinguish equal volume from equal amount:}\)
Equal liquid volumes do not generally contain equal numbers of molecules.
The number of moles depends on the mass and molar mass of each substance.
The density is needed to convert the measured volume into mass.
\(\textbf{Calculate the mass of P:}\)
\[
m_P=\rho_PV_P
\]
\[
m_P=(1.335)(30.0)
\]
\[
m_P=40.05\ \mathrm{g}
\]
\(\textbf{Calculate the mass of Q:}\)
\[
m_Q=\rho_QV_Q
\]
\[
m_Q=(1.336)(30.0)
\]
\[
m_Q=40.08\ \mathrm{g}
\]
Thus, Q has a slightly greater mass.
\(\textbf{Determine the amount of P:}\)
\[
n_P=\frac{m_P}{M_P}
\]
\[
n_P=\frac{40.05}{123.0}
\]
\[
n_P=0.326\ \mathrm{mol}
\]
\(\textbf{Determine the amount of Q:}\)
\[
n_Q=\frac{m_Q}{M_Q}
\]
\[
n_Q=\frac{40.08}{84.9}
\]
\[
n_Q=0.472\ \mathrm{mol}
\]
\(\textbf{Compare the number of molecules:}\)
The number of molecules is proportional to the amount in moles:
\[
N=nN_A
\]
Since:
\[
0.472\gt 0.326
\]
Q contains substantially more molecules than P in the equal-volume samples.
\(\textbf{Determine the layer positions:}\)
Both densities exceed the density of water:
\[
1.335\gt 1.00
\]
\[
1.336\gt 1.00
\]
Assuming separate organic phases form as specified, both liquids are denser than water and occupy the lower layer.
\(\textbf{Integrated conclusion:}\)
The two samples have nearly equal masses because their densities are almost identical.
However, their molar masses differ substantially, so their molecular populations differ.
Q has slightly greater mass, more molecules and the same lower-layer position as P.
\(\textbf{Final answer:}\)
\[
m_Q=40.08\ \mathrm{g}\gt m_P=40.05\ \mathrm{g},
\quad n_Q=0.472\ \mathrm{mol}\gt n_P=0.326\ \mathrm{mol}
\]
318. Two primary alkyl bromides are compared under conditions suitable for ordinary \(S_\mathrm{N}2\) substitution and, separately, under strongly ionising conditions:
P: 1-Bromopropane, \(\mathrm{CH_3CH_2CH_2Br}\).
Q: Allyl bromide, \(\mathrm{CH_2{=}CHCH_2Br}\).
Which interpretation best accounts for the special reactivity of Q?
ⓐ. Q is vinylic because bromine is present in a molecule containing a double bond, so ordinary backside attack is prohibited
ⓑ. P forms a resonance-stabilised carbocation, whereas Q must form a primary carbocation without resonance
ⓒ. Q is more reactive only because it contains fewer hydrogen atoms; the adjacent double bond does not affect any substitution pathway
ⓓ. Q has an accessible primary reaction centre and can benefit from allylic resonance stabilisation, whereas P lacks the corresponding adjacent \(\pi\)-system
Correct Answer: Q has an accessible primary reaction centre and can benefit from allylic resonance stabilisation, whereas P lacks the corresponding adjacent \(\pi\)-system
Explanation: \(\textbf{Identify the substitution classes:}\)
Both substrates contain bromine attached to a primary saturated carbon.
The carbon bearing bromine is \(sp^3\)-hybridised in each compound.
Therefore, neither substrate is a vinylic bromide.
\(\textbf{Identify the additional structural feature in Q:}\)
In allyl bromide, the bromine-bearing carbon is adjacent to a carbon–carbon double bond.
This carbon is allylic.
The neighbouring \(\pi\)-electron system can interact with developing charge in suitable reaction intermediates or transition states.
\(\textbf{Examine ordinary }S_\mathrm{N}2\textbf{ substitution:}\)
Both substrates provide relatively accessible primary reaction centres.
Backside nucleophilic attack is therefore geometrically possible in each.
In Q, the adjacent \(\pi\)-system can provide additional stabilisation to the developing electronic distribution in the substitution transition state.
This contributes to the enhanced reactivity commonly associated with allylic halides.
\(\textbf{Examine carbocation-forming conditions:}\)
Ionisation of P would ordinarily give an unstabilised primary propyl carbocation.
This is highly unfavourable.
Ionisation of Q would instead give an allylic carbocation:
\[
\mathrm{CH_2{=}CHCH_2^+
\leftrightarrow {}^+CH_2CH{=}CH_2}
\]
The positive charge is delocalised over the allylic system.
This resonance stabilisation makes Q more suitable for a carbocation-containing pathway than an ordinary primary bromide.
\(\textbf{Distinguish allylic and vinylic structures:}\)
A vinylic halide contains the halogen attached directly to a double-bonded carbon.
In Q, bromine is attached to the neighbouring saturated carbon.
The difference is decisive for ordinary substitution reactivity.
\(\textbf{Integrated conclusion:}\)
Formal primary classification is not the entire explanation of substrate reactivity.
The electronic environment adjacent to the carbon–halogen bond also matters.
Allyl bromide combines an accessible reaction centre with resonance-related stabilisation not available in 1-bromopropane.
319. The following two compounds have different molecular formulas and molar masses:
P: Bromomethane, \(\mathrm{CH_3Br}\), molar mass \(94.9\ \mathrm{g\,mol^{-1}}\).
Q: Chloroethane, \(\mathrm{C_2H_5Cl}\), molar mass \(64.5\ \mathrm{g\,mol^{-1}}\).
Their normal boiling points are \(3.6^\circ\mathrm{C}\) and \(12.3^\circ\mathrm{C}\), respectively.
A student claims that P must have the higher boiling point because its molecule is heavier and bromine is more polarizable than chlorine.
Which evaluation is most appropriate?
ⓐ. The claim is correct because boiling point depends exclusively on molar mass
ⓑ. The claim is correct because the energy required to break a carbon–bromine bond determines the heat of vaporisation
ⓒ. The claim conflicts with the supplied boiling points: Q boils higher because overall intermolecular interactions and molecular shape matter
ⓓ. The claim is incorrect because chloroethane forms ordinary intermolecular O–H hydrogen bonds
Correct Answer: The claim conflicts with the supplied boiling points: Q boils higher because overall intermolecular interactions and molecular shape matter
Explanation: \(\textbf{Compare the molecular masses:}\)
Bromomethane has the greater molar mass.
The heavier bromine atom also contributes considerable electronic polarizability.
These characteristics can favour stronger London dispersion interactions when other structural factors are comparable.
\(\textbf{Compare the actual boiling-point data:}\)
The measured boiling points are:
\[
T_b(P)=3.6^\circ\mathrm{C}
\]
\[
T_b(Q)=12.3^\circ\mathrm{C}
\]
Therefore:
\[
T_b(Q)\gt T_b(P)
\]
The observed order directly contradicts the student's universal mass-based prediction.
\(\textbf{Identify the missing structural consideration:}\)
Chloroethane contains a two-carbon chain, whereas bromomethane contains only one carbon atom.
Molecular size, shape and contact geometry influence dispersion interactions.
The complete electronic distribution also affects polar intermolecular forces.
Consequently, molecular mass by itself does not determine the boiling point across compounds with different alkyl groups and different halogens.
\(\textbf{Distinguish the useful within-series trend:}\)
For a fixed alkyl group, the usual boiling-point order is:
\[
\mathrm{RI\gt RBr\gt RCl\gt RF}
\]
This trend concerns comparable compounds with the same alkyl group.
It should not be applied as an exceptionless rule when both the alkyl skeleton and halogen identity change simultaneously.
\(\textbf{Reject the covalent-bond explanation:}\)
Boiling separates molecules rather than breaking their internal carbon–halogen bonds.
A carbon–bromine bond's dissociation enthalpy is therefore not the direct energy requirement for vaporisation.
\(\textbf{Reject the hydrogen-bond explanation:}\)
Chloroethane contains no oxygen atom and no O–H bond.
Ordinary O–H hydrogen bonding cannot explain its boiling point.
\(\textbf{Integrated conclusion:}\)
The appropriate physical prediction must consider the full molecular structure and the relevant intermolecular interactions.
The supplied experiment shows that the lighter chloroethane has the higher boiling point.
320. Benzene and chlorobenzene are compared at \(25^\circ\mathrm{C}\).
The following approximate physical data are supplied:
Assume that water has density \(1.00\ \mathrm{g\,mL^{-1}}\) and that each organic compound is contacted with water separately.
Which statement best reconciles the observed physical behaviour with the presence of a polar carbon–chlorine bond in chlorobenzene?
| Property | Benzene | Chlorobenzene |
|---|---|---|
| Density (g mL\(^{-1}\)) | 0.876 | 1.106 |
| Boiling point (°C) | 80.1 | 131.7 |
| Solubility in water (g L\(^{-1}\), approximately) | 1.8 | 0.5 |
ⓐ. Chlorobenzene has the higher boiling point and forms the lower organic layer, but its carbon–chlorine bond polarity does not guarantee greater solubility in water
ⓑ. Chlorobenzene must be more water-soluble than benzene because every molecule containing a polar bond forms stronger hydrogen bonds with water
ⓒ. Benzene forms the lower organic layer because its molecular formula contains no chlorine
ⓓ. Benzene has the higher boiling point because it contains no strongly polar carbon–halogen bond
Correct Answer: Chlorobenzene has the higher boiling point and forms the lower organic layer, but its carbon–chlorine bond polarity does not guarantee greater solubility in water
Explanation: \(\textbf{Identify the bonding distinction:}\)
Benzene contains carbon and hydrogen but no carbon–halogen bond.
Chlorobenzene contains chlorine directly attached to an aromatic carbon.
Its carbon–chlorine bond is polar covalent.
However, local bond polarity does not determine every physical property of the complete molecule.
\(\textbf{Compare the boiling points:}\)
Chlorobenzene has the higher boiling point:
\[
131.7\gt 80.1^\circ\mathrm{C}
\]
It has greater molar mass and stronger overall intermolecular attractions under the supplied comparison.
Its increased polarizability and dipole-related interactions contribute to its higher boiling point.
\(\textbf{Compare the densities:}\)
The density of benzene is:
\[
0.876\ \mathrm{g\,mL^{-1}}\lt 1.00\ \mathrm{g\,mL^{-1}}
\]
Therefore, benzene forms the upper organic layer when contacted with water.
The density of chlorobenzene is:
\[
1.106\ \mathrm{g\,mL^{-1}}\gt 1.00\ \mathrm{g\,mL^{-1}}
\]
Therefore, chlorobenzene forms the lower organic layer.
\(\textbf{Compare aqueous solubilities:}\)
The supplied data show:
\[
S_{\mathrm{benzene}}\approx1.8\ \mathrm{g\,L^{-1}}
\]
\[
S_{\mathrm{chlorobenzene}}\approx0.5\ \mathrm{g\,L^{-1}}
\]
Chlorobenzene is therefore less soluble in water under these conditions despite possessing a polar carbon–chlorine bond.
\(\textbf{Explain why polarity alone is insufficient:}\)
Dissolution requires consideration of interactions broken within the pure substances and interactions formed in the mixture.
Water possesses a strong hydrogen-bonding network.
Chlorobenzene does not provide hydrogen-bonding interactions sufficient to make it extensively miscible with water.
Its relatively large aromatic, hydrophobic portion remains important.
\(\textbf{Integrated conclusion:}\)
A polar carbon–halogen bond can influence molecular properties without producing universal increases in water solubility.
The boiling point, density and solubility comparisons must be evaluated independently.
Chlorobenzene boils higher, forms the lower layer and is less water-soluble than benzene under the supplied conditions.