3.3 Organic chemistry — coverage pack

Answer

• 3-methylbutan-1-ol

Method: The longest chain containing the OH group has four carbon atoms, so the parent is butanol. Number from the OH end: the OH group is on carbon 1 and the methyl branch is on carbon 3, giving 3-methylbutan-1-ol.

Answer

• CH₃CH(CH₃)CH(CH₂CH₃)CH₂CH₂CH₃

Method: Write a six-carbon parent chain. Attach a methyl group to carbon 2 and an ethyl group to carbon 3, keeping both branch points as CH groups: CH₃CH(CH₃)CH(CH₂CH₃)CH₂CH₂CH₃.

Answer

• 4,4-dimethylpentan-2-ol
• C₇H₁₆O

Method: The longest chain containing the OH group has five carbon atoms. Number from the end nearer the OH group, placing OH on carbon 2 and the two remaining methyl substituents on carbon 4: 4,4-dimethylpentan-2-ol. Counting all atoms gives seven carbons, sixteen hydrogens and one oxygen, so the molecular formula is C₇H₁₆O.

Answer

• It starts at the nitrogen lone pair and ends at the carbon atom where the new C–N bond forms.

Method: A curly arrow follows an electron pair. Ammonia supplies the pair from nitrogen, so the tail is placed on the N lone pair and the arrowhead is placed at the electron-deficient carbon that receives the pair.

Answer

• $\mathrm{CH_3\mathbin{\bullet}+Cl_2\rightarrow CH_3Cl+Cl\mathbin{\bullet}}$; show the radical dots and no electron-pair curly arrows.

Method: The methyl radical removes one chlorine atom from Cl₂, making chloromethane and regenerating a chlorine radical: $\mathrm{CH_3\mathbin{\bullet}+Cl_2\rightarrow CH_3Cl+Cl\mathbin{\bullet}}$. The unpaired electron is represented by a dot on each radical. AQA does not require electron-pair curly arrows for a radical mechanism.

Answer

• Arrow 1 goes from an oxygen lone pair on OH^− to the carbon bonded to iodine; arrow 2 goes from the C–I bond to iodine.

Method: The incoming bond is made using an oxygen lone pair, so the first arrow starts at :OH^− and points to the carbon bearing iodine. The C–I bond then breaks heterolytically, so the second arrow starts in that bond and points to iodine, which takes both bonding electrons and leaves as I^−.

Answer

• position isomerism

Method: Both compounds have the same four-carbon skeleton and the same alcohol functional group, but the OH group occupies a different carbon. They are position isomers.

Answer

• E; the higher-priority group at each alkene carbon is on the opposite side.

Method: On the left carbon, carbon in CH₃ has higher atomic number than hydrogen, so CH₃ has priority. On the right, the first atoms tie as carbon, but CH₂CH₃ next presents C,H,H whereas CH₃ presents H,H,H, so ethyl has priority. Those two higher-priority groups are opposite, hence E.

Answer

• CH₃CH₂CH₂CHO, butanal
• (CH₃)₂CHCHO, 2-methylpropanal
• CH₃COCH₂CH₃, butan-2-one
• The aldehydes are chain isomers; either aldehyde and the ketone are functional-group isomers.

Method: An aldehyde needs a terminal CHO group. A straight four-carbon chain gives butanal, while a branched skeleton gives 2-methylpropanal. A ketone needs an internal C=O group; with four carbons its only position is carbon 2, giving butan-2-one. Changing only the carbon skeleton gives chain isomerism, whereas changing aldehyde to ketone gives functional-group isomerism.

Answer

• different boiling points; fractional distillation

Method: The hydrocarbons vaporise and condense at different temperatures because they have different boiling points. Repeated vaporisation and condensation in a fractionating column is fractional distillation.

Answer

• The C₅ alkane has weaker London forces and a lower boiling point, so it remains gaseous until it reaches the cooler upper region.

Method: The C₁₅ molecule has more electrons and a larger contact area, so its London forces are stronger and more energy is needed to separate its molecules. It therefore has the higher boiling point and condenses in the hotter lower region. The C₅ alkane has weaker London forces, so it rises farther before condensing.

Answer

• Highest to lowest: C₇H₁₆, C₁₂H₂₆, C₁₈H₃₈; separation is a physical change with no covalent bonds broken.

Method: Boiling point increases with chain length because the number of electrons, molecular surface contact and London forces increase. C₁₈H₃₈ therefore condenses first in the hot lower region, C₁₂H₂₆ above it, and C₇H₁₆ highest. Only intermolecular separation and condensation occur, so covalent structures and molecular formulae remain unchanged.

Answer

• $X=\mathrm{C_4H_8}$

Method: Subtract the atoms in C₈H₁₈ from C₁₂H₂₆. The remainder has $12-8=4$ carbon atoms and $26-18=8$ hydrogen atoms, so $X=\mathrm{C_4H_8}$, an alkene.

Answer

• $\mathrm{C_{15}H_{32}\rightarrow C_8H_{18}+C_3H_6+2C_2H_4}$; two ethene molecules; it converts a less-demanded long fraction into demanded fuel and alkene feedstock.

Method: After C₈H₁₈ and C₃H₆, four carbon atoms and eight hydrogen atoms remain, corresponding to two C₂H₄ molecules. The balanced equation is $\mathrm{C_{15}H_{32}\rightarrow C_8H_{18}+C_3H_6+2C_2H_4}$. Cracking adjusts supply to demand by making a shorter motor-fuel alkane and alkenes used as chemical feedstock from a less useful long-chain fraction.

Answer

• Thermal: high temperature and high pressure, with a high percentage of alkenes. Catalytic: high temperature, slight pressure and a zeolite catalyst, mainly motor fuels and aromatic hydrocarbons.

Method: Give both condition sets and link each to its product profile. Thermal cracking is distinguished by high pressure and its alkene-rich output. Catalytic cracking is distinguished by slight pressure plus a zeolite and is directed mainly towards motor-fuel molecules and aromatics.

Answer

• $\mathrm{C_3H_8+5O_2\rightarrow3CO_2+4H_2O}$

Method: Three carbon atoms require 3CO₂, and eight hydrogen atoms require 4H₂O. The products then contain ten oxygen atoms, so five O₂ molecules are needed: $\mathrm{C_3H_8+5O_2\rightarrow3CO_2+4H_2O}$.

Answer

• $\mathrm{2CO+2NO\rightarrow2CO_2+N_2}$; CO is oxidised and NO is reduced.

Method: Pair two CO molecules with two NO molecules to conserve C, N and O: $\mathrm{2CO+2NO\rightarrow2CO_2+N_2}$. Carbon gains oxygen as CO becomes CO₂, so CO is oxidised. NO loses oxygen as its nitrogen forms N₂, so NO is reduced.

Answer

• $2.50\,\mathrm{kg}$ of CaCO₃

Method: Convert 1.60 kg to 1600 g. The amount of SO₂ is $n=1600/64.1=24.96\,\mathrm{mol}$. The equation has a 1:1 ratio, so 24.96 mol of CaCO₃ is required. Its mass is $m=24.96\times100.1=2498\,\mathrm{g}=2.50\,\mathrm{kg}$ to three significant figures.

Answer

• $\mathrm{Cl_2\xrightarrow{UV}2Cl\mathbin{\bullet}}$; homolytic fission

Method: UV radiation breaks the Cl–Cl bond so that one bonding electron goes to each chlorine atom. This homolytic fission gives $\mathrm{Cl_2\xrightarrow{UV}2Cl\mathbin{\bullet}}$.

Answer

• $\mathrm{Cl\mathbin{\bullet}+CH_4\rightarrow HCl+CH_3\mathbin{\bullet}}$
• $\mathrm{CH_3\mathbin{\bullet}+Cl_2\rightarrow CH_3Cl+Cl\mathbin{\bullet}}$

Method: First a chlorine radical abstracts H from methane: $\mathrm{Cl\mathbin{\bullet}+CH_4\rightarrow HCl+CH_3\mathbin{\bullet}}$. The methyl radical then removes Cl from chlorine: $\mathrm{CH_3\mathbin{\bullet}+Cl_2\rightarrow CH_3Cl+Cl\mathbin{\bullet}}$. The regenerated chlorine radical carries the chain forward.

Answer

• $\mathrm{Cl\mathbin{\bullet}+Cl\mathbin{\bullet}\rightarrow Cl_2}$
• $\mathrm{CH_3\mathbin{\bullet}+Cl\mathbin{\bullet}\rightarrow CH_3Cl}$
• $\mathrm{CH_3\mathbin{\bullet}+CH_3\mathbin{\bullet}\rightarrow C_2H_6}$
• Chloromethane can undergo further radical substitution to form more highly chlorinated products.

Method: Termination removes radicals by pairing them: two chlorine radicals form Cl₂, methyl plus chlorine forms CH₃Cl, and two methyl radicals form C₂H₆. Under continued UV exposure, C–H bonds remaining in CH₃Cl can also be substituted, producing CH₂Cl₂, CHCl₃ and CCl₄; the product is therefore a mixture.

Answer

• butanenitrile

Method: CN^− replaces Br and bonds through its carbon atom. That carbon becomes part of the main chain, increasing the carbon count from three to four, so the product is butanenitrile.

Answer

• Arrow from an O lone pair on OH^− to the carbon bonded to Br; arrow from the C–Br bond to Br; products propan-1-ol and Br^−.

Method: Show the C–Br bond polarised toward Br. Draw one curly arrow from a lone pair on :OH^− to the carbon bearing Br, and at the same time a second from the C–Br bond to Br. The new C–O bond gives CH₃CH₂CH₂OH and bromide leaves as Br^−.

Answer

• 1-iodobutane fastest, then 1-bromobutane, then 1-chlorobutane; C–I has the lowest bond enthalpy and C–Cl the highest.

Method: Substitution requires breaking the carbon–halogen bond. Bond enthalpy decreases from C–Cl to C–Br to C–I, so C–I breaks most readily and C–Cl least readily. Although C–Cl is the most polar of the three bonds, the observed order is iodo > bromo > chloro, showing that bond strength is the controlling factor here.

Answer

• propene; OH^− acts as a base

Method: OH^− removes a hydrogen from a carbon next to the C–Br carbon while Br^− leaves. A C=C bond forms, giving propene, and proton acceptance identifies OH^− as a base.

Answer

• Aqueous KOH gives propan-1-ol by nucleophilic substitution; ethanolic KOH gives propene by elimination.
• OH^− donates a lone pair to carbon as a nucleophile in substitution but accepts a proton as a base in elimination.

Method: In aqueous solution, OH^− attacks the carbon bonded to Br and replaces Br^−, so propan-1-ol is favoured; it acts as a nucleophile. In ethanol, OH^− removes a hydrogen from the adjacent carbon while the C–H electrons form C=C and Br^− leaves, so propene is favoured; it acts as a base.

Answer

• but-1-ene and but-2-ene
• O lone pair to a neighbouring H; C–H bond to the C–C bond; C–Br bond to Br.

Method: A hydrogen can be removed from either carbon adjacent to carbon 2. Removal from carbon 1 gives but-1-ene; removal from carbon 3 gives but-2-ene. For either route, draw an arrow from :OH^− to the chosen H, a second from that C–H bond into the adjacent C–C bond to make C=C, and a third from C–Br to Br.

Answer

• It absorbs ultraviolet radiation.

Method: The protective role is absorption of incoming UV radiation, reducing the amount that reaches organisms at Earth's surface.

Answer

• $\mathrm{2O_3\rightarrow3O_2}$; catalyst Cl$\mathbin{\bullet}$; intermediate ClO$\mathbin{\bullet}$.

Method: Add $\mathrm{Cl\mathbin{\bullet}+O_3\rightarrow ClO\mathbin{\bullet}+O_2}$ and $\mathrm{ClO\mathbin{\bullet}+O_3\rightarrow2O_2+Cl\mathbin{\bullet}}$. Cancel Cl$\mathbin{\bullet}$ and ClO$\mathbin{\bullet}$ to leave $\mathrm{2O_3\rightarrow3O_2}$. Chlorine radical is used then regenerated, so it is the catalyst; ClO radical is made then consumed, so it is an intermediate.

Answer

• $\mathrm{CF_2Cl_2\xrightarrow{UV}CF_2Cl\mathbin{\bullet}+Cl\mathbin{\bullet}}$
• The chlorine radical is regenerated in a catalytic cycle and can decompose many ozone molecules.

Method: UV light causes homolytic breaking of one C–Cl bond: $\mathrm{CF_2Cl_2\xrightarrow{UV}CF_2Cl\mathbin{\bullet}+Cl\mathbin{\bullet}}$. The released Cl$\mathbin{\bullet}$ reacts with O₃ to form ClO$\mathbin{\bullet}$, which reacts with another O₃ and reforms Cl$\mathbin{\bullet}$. Because the radical is regenerated rather than used up overall, it repeats the cycle and a small CFC amount can remove much more ozone.

Answer

• A hydrocarbon containing a carbon–carbon multiple bond; ethene contains a C=C bond.

Method: A hydrocarbon contains only carbon and hydrogen. It is unsaturated when it contains a C=C or another carbon–carbon multiple bond; ethene has one C=C double bond.

Answer

• Ethene contains a C=C double covalent bond; it is a centre of high electron density that attracts electrophiles.

Method: Ethane has only a C–C single bond, whereas ethene has a C=C double covalent bond. The double bond is a centre of high electron density, so electron-deficient electrophiles are attracted to it and ethene undergoes addition reactions.

Answer

• The electron-rich C=C attracts an electron-pair acceptor; an electron pair from the double bond forms a new covalent bond, and a second new bond completes addition as C=C becomes C–C.

Method: The C=C bond is a centre of high electron density. An electrophile is attracted to and accepts an electron pair from the double bond, forming a new covalent bond. Subsequent bonding at the other alkene carbon gives two new bonds to the attacking species overall, while the carbon–carbon bond in the saturated product is single.

Answer

• orange to colourless; 1,2-dibromocyclohexane

Method: Br₂ adds across the ring's C=C bond, consuming coloured bromine and forming a dibromo compound. The observation is orange to colourless and the product has Br on the two formerly double-bonded carbons: 1,2-dibromocyclohexane.

Answer

• Arrow from C=C to H in H–Br and from H–Br to Br; secondary CH₃C⁺HCH₃ intermediate; arrow from :Br^− to C⁺; product 2-bromopropane.

Method: Draw a curly arrow from the C=C bond to H and another from the H–Br bond to Br. Attach H to the end carbon so the positive charge is on the middle carbon, giving the more stable secondary carbocation CH₃C⁺HCH₃. Then draw an arrow from a lone pair on Br^− to C⁺, forming 2-bromopropane.

Answer

• Major: 2-bromo-2-methylbutane, formed through a tertiary carbocation.
• Minor: 2-bromo-3-methylbutane, formed through a secondary carbocation.

Method: Proton addition to carbon 3 places the positive charge on carbon 2, which is bonded to three carbon groups and is therefore tertiary. Br^− attack gives 2-bromo-2-methylbutane, the major product. Proton addition to carbon 2 instead places the charge on carbon 3, a secondary carbocation; Br^− attack gives 2-bromo-3-methylbutane, the minor product. The tertiary intermediate is more stable, so its pathway is favoured.

Answer

• [-CH₂-C(CH₃)₂-]_n, with a backbone bond passing through each bracket.

Method: Open the monomer's C=C bond to make the two-carbon C-C backbone. Keep both CH₃ groups attached to the same second carbon, then bracket the unit and continue the backbone bonds through the brackets.

Answer

• Monomer: CH₂=CHCN.
• Poly(prop-2-enenitrile), commonly poly(acrylonitrile).
• Permanent dipole-dipole attractions between polar C≡N groups.

Method: Identify the repeating pair -CH₂-CH(CN)- and restore a C=C bond between those two backbone carbons. The monomer is therefore CH₂=CHCN. Name the polymer by placing the systematic monomer name in poly(...); the polar nitrile bonds give permanent dipole-dipole attraction.

Answer

• C-Cl bonds make neighbouring PVC chains polar, so permanent dipole-dipole attractions act between them.
• In unplasticised PVC these attractions hold chains close together and restrict movement, making the material rigid.
• Plasticiser molecules fit between chains, increase their separation and weaken the total attraction between polymer chains, so the chains slide more easily.
• The polymer backbone contains strong C-C bonds and no hydrolysable functional group, so water does not readily split the chains.

Method: Link the polar C-Cl bonds first to attractions between chains, then link those attractions to restricted chain movement. A plasticiser separates the chains and reduces those effective attractions, increasing flexibility. Finally inspect the backbone: addition polymerisation leaves only robust carbon-carbon links, not ester or amide links that water could hydrolyse.

Answer

• C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂.
• Use yeast at about 30-40 °C and exclude oxygen; an aqueous glucose solution is also required.

Method: Balance six carbon atoms by making two ethanol and two carbon dioxide molecules; this also balances hydrogen and oxygen. A moderate temperature keeps yeast enzymes active without denaturing them, while anaerobic conditions favour fermentation rather than aerobic respiration.

Answer

• The propene pi bond accepts H⁺, forming the more stable secondary carbocation.
• A water lone pair attacks the carbocation to make a protonated alcohol.
• Loss of H⁺ forms propan-2-ol and regenerates the acid catalyst.

Method: Use the electron-rich C=C bond to attack H⁺ in the first step, placing the positive charge on the middle carbon. Water then donates a lone pair to that carbon. Deprotonation gives CH₃CH(OH)CH₃; because the H⁺ consumed first is released last, it is catalytic.

Answer

• $176\,\text{kg}$ of CO₂.
• Growing the glucose crop can remove the same carbon from atmospheric CO₂, so the combustion carbon can form a short carbon cycle.
• The description is not fully justified if cultivation, fertiliser, distillation or transport uses fossil energy; land and food-crop competition also matter.

Method: The amount of ethanol is $92.0/46.0=2.00\,\text{kmol}$. The equation gives twice as much CO₂, so $n(\mathrm{CO_2})=4.00\,\text{kmol}$ and $m=4.00\times44.0=176\,\text{kg}$. Carbon neutrality concerns the whole life cycle, not only this combustion equation.

Answer

• Butanone; the solution changes from orange to green.

Method: Butan-2-ol is secondary because its OH-bearing carbon is bonded to two other carbons. Oxidation therefore forms the ketone butanone, while dichromate(VI) ions are reduced from orange to green chromium(III) ions.

Answer

• Use acidified potassium dichromate(VI), warm gently and distil the aldehyde as it forms.
• (CH₃)₂CHCH₂OH + [O] → (CH₃)₂CHCHO + H₂O.
• On warming, Tollens' reagent gives a silver mirror and Fehling's solution gives a brick-red precipitate.

Method: The starting compound is a primary alcohol, so one oxidation step gives an aldehyde. Gentle heating supplies the reaction rate, and immediate distillation removes the volatile aldehyde from contact with oxidant before it undergoes the second oxidation step. The isolated aldehyde reduces Tollens' silver ions and Fehling's copper(II) ions, giving the stated positive observations.

Answer

• Butan-1-ol gives butanoic acid; butan-2-ol gives butanone; 2-methylpropan-2-ol shows no easy oxidation.
• Add sodium carbonate: butanoic acid effervesces because CO₂ forms, whereas butanone does not react.
• Tollens' reagent would not distinguish these final products because neither gives a positive aldehyde result.

Method: Classify the alcohols by counting carbon groups on the OH-bearing carbon: primary, secondary and tertiary respectively. Reflux with excess oxidant carries the primary alcohol to the acid and the secondary alcohol to the ketone. Then choose a test for the acid functional group, such as carbonate and effervescence, rather than an aldehyde test.

Answer

• Ethene; acid-catalysed elimination (dehydration).
• CH₃CH₂OH → CH₂=CH₂ + H₂O.

Method: Remove OH from one carbon and H from the adjacent carbon, then place a double bond between those two carbons. One ethanol molecule therefore gives ethene and water.

Answer

• 2-Ethylbut-1-ene and 3-methylpent-2-ene.
• A beta hydrogen can be removed from an end ethyl group or from the methyl substituent; the two ethyl groups are equivalent, so removal from either one gives the same 3-methylpent-2-ene structure.

Method: Mark the carbon bearing OH, then inspect every adjacent carbon for a removable H. Forming the double bond towards either equivalent ethyl group gives 3-methylpent-2-ene; forming it towards the methyl group gives CH₂=C(CH₂CH₃)₂, named 2-ethylbut-1-ene from the longest chain containing C=C.

Answer

• An oxygen lone pair attacks H⁺, producing protonated cyclohexanol; the acid O-H bond returns to the acid oxygen if the acid is drawn explicitly.
• The C-O bond pair moves to oxygen so H₂O leaves.
• The adjacent C-H bond pair forms the C=C bond as H⁺ is lost, regenerating the acid catalyst; no base-attack arrow is drawn.
• Cyclohexene or another alkene made by dehydrating a fermentation-derived alcohol can undergo addition polymerisation, replacing a crude-oil-derived monomer route.

Method: First turn poor leaving group OH into neutral water by protonation. Next break C-O heterolytically, keeping that electron pair on oxygen. Finally start the curly arrow at an adjacent C-H bond and end it between the two carbons while H⁺ leaves. Because H⁺ is returned, the sequence is catalytic; coupling alcohol production from biomass with dehydration supplies an alkene feedstock.

Answer

• Shake with bromine water; its orange colour is discharged to colourless.

Method: Use bromine water at room temperature. An alkene adds bromine across C=C, consuming coloured Br₂, so record the specific change orange to colourless.

Answer

• Add sodium carbonate to fresh portions: ethanoic acid effervesces and ethanal does not; the gas is CO₂.
• Warm fresh portions with Tollens' reagent: ethanal forms a silver mirror, while ethanoic acid gives no silver mirror.

Method: Use carbonate to test acidity and Tollens' reagent to test an aldehyde. The two independent positive results cross-check the assignment instead of relying only on the absence of a reaction.

Answer

• Test fresh portions with sodium carbonate: the sample that effervesces CO₂ is ethanoic acid.
• Warm fresh portions of the other samples with Tollens' reagent: the silver-mirror sample is ethanal.
• Shake fresh portions of the remaining two with bromine water: orange to colourless identifies ethene.
• The remaining sample is ethanol; warming it with acidified dichromate confirms this by an orange-to-green change.

Method: Remove the acid first with the selective carbonate test, then remove the aldehyde with Tollens' reagent. Bromine water distinguishes the remaining alkene from alcohol. The final dichromate result is a positive confirmation of ethanol rather than an assignment by elimination alone; fresh portions prevent one reagent changing the next result.

Answer

• $44.02623$

Method: Add the isotope contributions: $2(12.00000)+4(1.00783)+15.99491=44.02623$. Keep the precision supplied because rounding to $44$ would discard the evidence used for formula identification.

Answer

• C₃H₆O₂; its calculated mass is $74.03680$, compared with $74.07321$ for C₄H₁₀O.

Method: For C₄H₁₀O, calculate $4(12.00000)+10(1.00783)+15.99491=74.07321$. For C₃H₆O₂, calculate $3(12.00000)+6(1.00783)+2(15.99491)=74.03680$. The second value differs from the measurement by only $0.00010$.

Answer

• C₄H₈O₂: $88.05246$; C₅H₁₂O: $88.08887$.
• The formula is C₄H₈O₂, with absolute mass error $|88.05260-88.05246|=0.00014$.

Method: Calculate $4(12.00000)+8(1.00783)+2(15.99491)=88.05246$ and $5(12.00000)+12(1.00783)+15.99491=88.08887$. The first candidate is much closer to $88.0526$. Subtract without attaching a sign because the question asks for absolute error: $0.00014$.

Answer

• A C=O bond.

Method: Locate 1718 cm^-1 in the carbonyl range in the Data Booklet. This establishes the presence of C=O but does not by itself distinguish an aldehyde, ketone, acid or ester.

Answer

• The compound is ethyl ethanoate.
• The C=O peak and absence of O-H support an ester, while the matching fingerprint region identifies the particular ester rather than only its functional group.

Method: First use the diagnostic region: C=O is present and an alcohol or acid O-H is absent. Several esters could fit those facts, so compare the complex fingerprint pattern; an identical pattern under the same conditions provides the molecule-specific match.

Answer

• A possible impurity is propan-2-ol or water because the extra broad absorption indicates O-H.
• Bonds in molecules such as CO₂, CH₄ and H₂O absorb outgoing infrared radiation at matching vibrational frequencies.
• Energy is redistributed by re-emission and collisions, so less energy escapes directly to space and the atmosphere warms.

Method: Treat the unexpected O-H band as impurity evidence while the fingerprint still supports propanone as the main component. For the climate explanation, connect the characteristic bond vibrations of greenhouse gases to absorption of terrestrial IR and then to redistribution and retention of energy in the atmosphere.

Answer

• Carbon 2 is chiral; the compound is butan-2-ol.

Method: Carbon 2 is bonded to H, OH, CH₃ and CH₂CH₃, which are four different groups. No other carbon in the structure meets that test.

Answer

• At carbon 2, draw CH₃ and CH₂CH₃ in the plane, with Br on a solid wedge and H on a dashed bond; draw the partner with Br dashed and H wedged.
• They are non-superimposable mirror images and rotate plane-polarised light equally in opposite directions.

Method: Keep the same four groups and reverse the three-dimensional arrangement at the only chiral centre. Swapping the wedge and dash for Br and H produces the mirror configuration; do not change the structural formula or move a group to another carbon.

Answer

• Ethanal forms 2-hydroxypropanenitrile, whose central carbon is attached to H, OH, CN and CH₃, so it is chiral.
• The planar ethanal carbonyl is attacked equally from either face, forming equal amounts of the two enantiomers; their rotations cancel in a racemate.
• Propanone forms 2-hydroxy-2-methylpropanenitrile, whose central carbon has two identical CH₃ groups, so it has no chiral centre and is optically inactive.

Method: Write the four substituents on the former carbonyl carbon for each product. Ethanal gives four different groups and its planar starting group exposes two equally likely faces. Propanone retains two identical methyl groups, so even attack from opposite faces cannot create a pair of enantiomers.

Answer

• Butan-1-ol; CH₃CH₂CH₂CHO + 2[H] → CH₃CH₂CH₂CH₂OH.

Method: Reduction adds two hydrogen equivalents across C=O. Because the starting carbonyl is terminal, the product has a terminal CH₂OH group and is the primary alcohol butan-1-ol.

Answer

• Warm with Tollens' reagent: pentanal gives a silver mirror; pentan-3-one shows no change.
• Warm with Fehling's solution: pentanal gives a brick-red precipitate; pentan-3-one leaves the blue solution unchanged.

Method: Pentanal is an aldehyde and is oxidised by both mild oxidising reagents. Pentan-3-one is a ketone and is not oxidised in either test, so each fresh-portion test assigns the same bottle independently.

Answer

• Nucleophilic addition forms 2-hydroxy-2-methylbutanenitrile.
• A lone pair on the carbon of CN^- attacks the delta-positive carbonyl carbon while the C=O pi pair moves to oxygen; the resulting O^- is then protonated.
• The carbonyl group is planar, so attack occurs equally from either face. The product carbon is bonded to OH, CN, CH₃ and CH₂CH₃, producing a racemic mixture of two enantiomers.
• KCN is highly toxic or poisonous.

Method: Use CN^- as the electron-pair donor and the carbonyl carbon as the electron-pair acceptor. Moving the pi pair to oxygen prevents carbon from exceeding an octet. Protonation gives the hydroxynitrile; checking its four substituents confirms a new chiral centre and equal attack on the two planar faces explains the racemate. Treat cyanide as acutely toxic throughout preparation and disposal.

Answer

• Propyl ethanoate; concentrated sulfuric acid catalyst.
• CH₃COOH + CH₃CH₂CH₂OH ⇌ CH₃COOCH₂CH₂CH₃ + H₂O.

Method: Take propyl from the alcohol and ethanoate from the acid. Join the alcohol oxygen to the acid's acyl carbon, remove water, and show a reversible arrow because esterification establishes an equilibrium.

Answer

• Products: methanol and sodium propanoate.
• CH₃CH₂COOCH₃ + NaOH → CH₃CH₂COONa + CH₃OH.
• Alkaline conditions deprotonate the acid product to its carboxylate salt; acidification is needed to obtain propanoic acid.

Method: Split the ester at the acyl C-O bond. The alkoxy fragment becomes methanol, while the acyl fragment is present as propanoate in excess alkali. Add dilute acid after hydrolysis if the neutral carboxylic acid is required.

Answer

• $2.40\,\text{g}$ of NaOH.
• Glycerol and the sodium salts of long-chain carboxylic acids (soap).

Method: The triglyceride amount is $17.8/890=0.0200\,\text{mol}$. Three ester links require three hydroxide ions, so $n(\mathrm{NaOH})=3(0.0200)=0.0600\,\text{mol}$. Therefore $m=0.0600\times40.0=2.40\,\text{g}$. Breaking all three ester links releases glycerol and three carboxylate salts per triglyceride molecule.

Answer

• Ethyl ethanoate and hydrogen chloride.
• Steamy or misty fumes are observed as HCl meets moist air.

Method: Replace Cl in CH₃COCl by the ethoxy group from ethanol to form CH₃COOCH₂CH₃. The removed proton and chloride form HCl, which produces visible acidic mist in damp air.

Answer

• N-Methylpropanamide and methylammonium chloride.
• The nitrogen lone pair attacks the carbonyl carbon and the C=O pi pair moves to oxygen, giving a tetrahedral intermediate with O^- and N⁺.
• The oxygen pair reforms C=O as Cl^- leaves; a second methylamine molecule removes H⁺, forming CH₃NH₃⁺Cl^-.

Method: Use methylamine as both nucleophile and base. Its first molecule makes the new C-N bond through the charged tetrahedral intermediate: oxygen is negative after receiving the pi pair and nitrogen is positive after donating its lone pair. Collapse expels chloride, then the excess second amine accepts the proton to give the neutral amide and ammonium salt.

Answer

• Ethanoic anhydride route: $180/(180+60.0)\times100=75.0\%$.
• Ethanoyl chloride route: $180/(180+36.5)\times100=83.1\%$.
• Despite its lower atom economy, ethanoic anhydride is safer and easier to handle and produces ethanoic acid rather than corrosive, harmful HCl fumes.

Method: For each stated one-to-one route, divide the desired-product formula mass by the total formula mass of products. This gives $75.0\%$ for the anhydride and $83.1\%$ for the chloride. Atom economy is only one industrial criterion, so compare the hazards and usefulness of the by-products before deciding which reagent is preferable.

Answer

• Benzene is planar and hexagonal; all six C-C bonds have equal length intermediate between a single and double bond; p orbitals overlap to give delocalised pi electrons above and below the ring.

Method: State one structural point at each scale: the whole ring is planar, the six C-C links are equivalent, and the unhybridised p orbitals overlap continuously to delocalise the pi electrons.

Answer

• $156\,\text{kJ mol}^{-1}$

Method: Three isolated double bonds would give $3(-121)=-363\,\text{kJ mol}^{-1}$. Benzene releases only $207\,\text{kJ mol}^{-1}$, so it begins lower in energy by $363-207=156\,\text{kJ mol}^{-1}$. This is the delocalisation stability for the supplied data.

Answer

• The claim is inconsistent with six equal $0.140\,\text{nm}$ bonds; this value lies between the single- and double-bond lengths instead of alternating between them.
• Continuous p-orbital overlap delocalises six pi electrons around the ring and lowers its energy.
• Addition would remove part of this delocalisation, whereas electrophilic substitution temporarily disrupts it but then restores the aromatic ring, so substitution is preferred.

Method: Compare the one measured benzene length with both reference values and note that every ring bond has that same intermediate value. Account for this using delocalisation rather than bond switching. Finally compare the electronic result of the reaction types: addition leaves a less-delocalised product, while substitution replaces H and recovers the stable ring.

Answer

• Concentrated nitric acid and concentrated sulfuric acid; sulfuric acid is the catalyst; the electrophile is NO₂⁺, the nitronium ion.

Method: Use the nitrating mixture of concentrated HNO₃ and H₂SO₄. The stronger sulfuric acid protonates nitric acid, allowing loss of water and formation of NO₂⁺; sulfuric acid is regenerated.

Answer

• 1-Phenylpropan-1-one.
• C₆H₆ + CH₃CH₂COCl → C₆H₅COCH₂CH₃ + HCl.
• The electrophile is CH₃CH₂CO⁺.

Method: AlCl₃ accepts chloride from propanoyl chloride to generate the acylium ion. Replace one benzene H by CH₃CH₂CO-, then combine the removed H with Cl to balance the equation and name the resulting aromatic ketone.

Answer

• CH₃COCl + AlCl₃ → CH₃CO⁺ + AlCl₄^-.
• A curly arrow starts inside the benzene ring and goes to the positive carbon of CH₃CO⁺, forming a C-C bond. In the sigma complex, H and COCH₃ remain bonded to the attacked carbon and the positive charge is delocalised around the other five ring carbons.
• AlCl₄^- removes H⁺; the C-H bond pair moves back into the ring, restoring delocalisation and forming phenylethanone, HCl and AlCl₃.

Method: First use the Lewis acid to remove chloride and create the acylium electrophile. The benzene pi pair then forms the new bond; draw the attacked carbon bonded to both H and the acyl group, with delocalised positive charge on the remaining ring. Finally the C-H electron pair restores the ring as AlCl₄^- supplies chloride to H; this produces HCl and returns AlCl₃, confirming its catalytic role.

Answer

• Excess ammonia in ethanol; heat in a sealed vessel.

Method: Use ethanolic ammonia and heat the sealed reaction mixture. Stating that ammonia is in excess is important because it makes collision with ammonia more likely than collision with the propylamine product, reducing formation of secondary and tertiary amines.

Answer

• Heat bromoethane with potassium cyanide in ethanol under reflux to form propanenitrile; reduce propanenitrile using hydrogen and nickel, or lithium aluminium hydride in dry ether, to form propylamine.

Method: First use ethanolic KCN under reflux. The cyanide carbon joins the two-carbon halogenoalkane, so the intermediate is the three-carbon nitrile propanenitrile. Reduction converts the nitrile group into a primary amine without removing that carbon, giving propylamine. Either H₂/Ni or LiAlH₄ in dry ether is an acceptable reduction route.

Answer

• Heat nitrobenzene under reflux with tin and concentrated hydrochloric acid, then add excess sodium hydroxide to release phenylamine from the phenylammonium salt and separate the organic product.

Method: Sn/HCl reduces the nitro group. In the acidic mixture the amine is protonated, so the immediate product is a phenylammonium salt rather than free phenylamine. Adding excess NaOH removes the proton and liberates phenylamine. The organic product can then be separated from the aqueous mixture and purified, for example by distillation.

Answer

• CH₃CH₂NH₂ + H₂O ⇌ CH₃CH₂NH₃⁺ + OH⁻

Method: The nitrogen lone pair accepts H⁺ from water. Ethylamine therefore becomes ethylammonium, while the water molecule that loses H⁺ becomes OH⁻. Use an equilibrium arrow because ethylamine is a weak base.

Answer

• Ethylamine > ammonia > phenylamine; the ethyl group increases electron density on nitrogen, whereas delocalisation into the benzene ring makes the phenylamine lone pair less available.

Method: Start by comparing lone-pair availability. The electron-releasing ethyl group has a positive inductive effect, so ethylamine accepts H⁺ more readily than ammonia. In phenylamine, the lone pair overlaps with the aromatic π system and is delocalised, so it is less available than the localised lone pair in ammonia.

Answer

• The claim is incorrect. The phenylamine lone pair is delocalised into the ring, whereas the propyl group releases electron density towards nitrogen; propylamine therefore forms the greater hydroxide-ion concentration.

Method: A ring being electron-rich does not by itself make the nitrogen lone pair available. Conjugation lets the phenylamine lone pair spread into the benzene π system, stabilising the unprotonated molecule and reducing its tendency to accept H⁺. The propyl group instead increases electron density at nitrogen through the positive inductive effect. Propylamine is the stronger weak base and shifts its reaction with water further towards alkylammonium and OH⁻ ions.

Answer

• A lone pair of electrons on the nitrogen atom.

Method: A nucleophile donates an electron pair. In an amine, the available electron pair is the lone pair on nitrogen.

Answer

• N-ethylethanamide; nucleophilic addition–elimination; an arrow from the nitrogen lone pair to the carbonyl carbon and an arrow from the C=O π bond to oxygen.

Method: The nitrogen lone pair attacks the δ⁺ carbonyl carbon, so the first curly arrow begins at that lone pair and ends at the carbonyl carbon. To avoid giving carbon five bonds, the C=O π pair moves onto oxygen. Elimination then reforms C=O and removes Cl⁻; deprotonation gives N-ethylethanamide.

Answer

• A tetrabutylammonium salt forms; each amine product retains a nitrogen lone pair until the quaternary ion is reached, and the ion combines a positive hydrophilic head with hydrophobic butyl groups.

Method: Butylamine first attacks another 1-bromobutane molecule. After deprotonation the secondary amine still has a lone pair, so it can attack again; the tertiary amine can also attack, forming the quaternary ion (C₄H₉)₄N⁺ with Br⁻. The permanent positive charge interacts with water, while the hydrocarbon groups interact with grease or non-polar material, giving surfactant behaviour.

Answer

• An amide linkage; water.

Method: A carboxyl group and an amino group condense to form –CONH–. The –OH from the acid and an H from the amine form H₂O.

Answer

• [–NH–(CH₂)₆–NH–CO–(CH₂)₄–CO–]_n; hydrogen bonding.

Method: Remove H from each terminal –NH₂ and OH from each –COOH as the two monomers join repeatedly. This gives the chain fragment –NH–(CH₂)₆–NH–CO–(CH₂)₄–CO– inside repeat brackets. N–H groups donate and carbonyl oxygen atoms accept hydrogen bonds between chains.

Answer

• Ethane-1,2-diol and benzene-1,4-dicarboxylic acid; a polyester; polar C=O and C–O bonds give permanent dipole–dipole attractions, with London forces also present.

Method: Split each –COO– linkage between the acyl carbon and oxygen. Restore H to each chain oxygen to obtain HO–CH₂CH₂–OH, and restore OH to each acyl carbon to obtain HOOC–C₆H₄–COOH in the para arrangement shown. Ester groups identify a polyester. Its polar carbonyl and C–O bonds produce permanent dipoles, so neighbouring chains attract by permanent dipole–dipole forces as well as London forces.

Answer

• A polyalkene has an inert C–C backbone with no hydrolysable polar linkage, whereas a polyester contains ester bonds that can be hydrolysed.

Method: Identify the bond that water, acid or alkali can attack. The polyester has ester links containing an electron-deficient carbonyl carbon; the polyalkene backbone contains only strong C–C bonds and no comparable functional group.

Answer

• The polyamide is more biodegradable because its polar amide links can be hydrolysed, breaking the chain; poly(propene) has a chemically inert carbon–carbon backbone and is non-biodegradable.

Method: Relate biodegradation to bond cleavage. Hydrolysis of –CONH– links divides a polyamide into smaller molecules, so biological processes can eventually break it down. Poly(propene) lacks hydrolysable links: its backbone is made from C–C bonds with C–H and methyl substituents, so it persists much longer.

Answer

• Recycling conserves raw materials but needs clean separated streams; landfill uses land and leaves persistent waste; incineration recovers energy and reduces volume but emits carbon dioxide and may form harmful gases. Sorting makes useful recycling streams possible and separates polymers needing different emission controls.

Method: Landfill requires little processing but occupies land and retains non-biodegradable material. Mechanical recycling reduces demand for new petrochemical feedstock, but mixed or contaminated polymers give poor products and separation costs energy. Incineration handles mixed waste, releases useful energy and greatly reduces volume, but adds CO₂ and can produce acidic or toxic emissions. Sorting raises the quality of recycled material and allows unsuitable fractions to be treated separately, so the justified strategy is usually a combination rather than one universal method.

Answer

• A zwitterion contains both a positive and a negative charge in the same species and has zero overall charge.

Method: For an amino acid, the positive site is normally –NH₃⁺ and the negative site is –COO⁻. The charges cancel, but both must be shown.

Answer

• Acid: CH₃CH(NH₃⁺)COOH; alkali: CH₃CH(NH₂)COO⁻.

Method: In acid, the amino group accepts H⁺ while the carboxyl group remains protonated, giving an overall $+1$ ion. In alkali, OH⁻ removes the carboxyl proton while the amino group is unprotonated, giving an overall $-1$ ion.

Answer

• CH₃CH₂CH(NH₃⁺)COO⁻ + H⁺ → CH₃CH₂CH(NH₃⁺)COOH; it acts as a base.
• CH₃CH₂CH(NH₃⁺)COO⁻ + OH⁻ → CH₃CH₂CH(NH₂)COO⁻ + H₂O; it acts as an acid.

Method: Represent the zwitterion as CH₃CH₂CH(NH₃⁺)COO⁻. HCl protonates –COO⁻ to –COOH, so the zwitterion accepts a proton and behaves as a base. OH⁻ removes a proton from –NH₃⁺, producing –NH₂, H₂O and the carboxylate ion; the zwitterion behaves as an acid.

Answer

• The sequence or order of amino acids in the polypeptide chain.

Method: Primary structure refers only to which amino acids occur and their order along the chain, not to the helix, sheet or overall folding.

Answer

• H₂NCH₂CONHCH(CH₃)COOH; peptide or amide link; water.

Method: Remove OH from glycine's –COOH and H from alanine's –NH₂. Join the glycine carbonyl carbon to the alanine nitrogen, giving –CO–NH– and H₂O. Keep the stated order: glycine is on the N-terminal side and alanine on the C-terminal side.

Answer

• Both peptide links are hydrolysed; hydrogen bonds and sulfur–sulfur bonds can maintain tertiary structure; $R_f=0.59$.

Method: Complete hydrolysis adds the elements of water across each –CONH– link and releases the three amino acids. In proteins, two cysteine residues can form an S–S bond, and polar groups can form hydrogen bonds during folding. For the chromatogram, $R_f=38/64=0.59375$, so $R_f=0.59$ to two decimal places.

Answer

• It increases reaction rate by providing a route with lower activation energy and is regenerated or not consumed overall.

Method: Give both catalytic ideas: the enzyme lowers the activation-energy barrier and is available again after products leave the active site.

Answer

• The active site is three-dimensional and stereospecific; only one enantiomer places its groups in the correct positions to form the required intermolecular interactions at the same time.

Method: Enantiomers have mirror-image arrangements. The active site is itself chiral because it is built from a folded protein. One enantiomer can align complementary charged, polar or non-polar groups with binding groups in the site, whereas the mirror image cannot make the same set of contacts simultaneously.

Answer

• The drug can bind strongly in the active site and block substrate entry; only a suitably oriented stereoisomer makes the required interactions; modelling can screen fit and interactions before laboratory synthesis.

Method: A transition-state-like arrangement may be complementary to the binding groups in the active site, so the drug occupies the site and prevents the normal substrate forming an enzyme–substrate complex. Because the site is stereospecific, the wrong stereoisomer may not align its groups and may bind weakly. A computer model can dock many candidate structures, estimate shape complementarity and interactions, and prioritise the most promising molecules for synthesis and testing.

Answer

• A phosphate group, 2-deoxyribose and one nitrogen-containing base.

Method: List one component from each part of the nucleotide: phosphate, the pentose sugar 2-deoxyribose, and one of A, C, G or T.

Answer

• T–G–C–A–A–T; covalent bonds form the sugar–phosphate backbone, while hydrogen bonds join complementary bases across the strands.

Method: Apply A–T and C–G pairing one position at a time to obtain T–G–C–A–A–T. The continuous backbone within either strand is covalently bonded. The two separate strands associate through hydrogen bonds between paired bases.

Answer

• $49$ hydrogen bonds; C–G pairs form three hydrogen bonds but A–T pairs form two.

Method: The A–T contribution is $8\times2=16$ hydrogen bonds and the C–G contribution is $11\times3=33$. The total is $16+33=49$. Replacing A–T pairs with C–G pairs increases the number of hydrogen bonds between the strands, so more energy is required to overcome the intermolecular attractions.

Answer

• Guanine; a nitrogen atom.

Method: The specification requires the Pt–N interaction to be linked specifically to guanine. The nitrogen donates a lone pair to platinum.

Answer

• Chloride ligands are replaced and Pt forms coordinate bonds to guanine nitrogen atoms; the resulting links distort or hold the DNA so the strands cannot separate or be copied normally.

Method: A guanine nitrogen lone pair replaces a ligand at Pt(II), producing a Pt–N coordinate bond. Formation of more than one such bond links sites in the DNA and changes its shape. The damaged double helix cannot open and act as a normal template, so DNA replication and cell division are inhibited.

Answer

• Cisplatin blocks DNA replication in rapidly dividing tumour cells, but it can also block replication in healthy rapidly dividing cells such as bone-marrow and gut-lining cells; the therapeutic benefit must therefore be weighed against toxicity.

Method: Cancer cells divide rapidly and need repeated DNA replication, so Pt–guanine links can restrict tumour growth. The drug is not perfectly selective: bone-marrow cells and cells replacing the digestive lining also divide frequently and can suffer the same DNA damage. A suitable dose and treatment plan must provide enough anticancer effect while keeping harm to healthy tissue acceptable.

Answer

• It produces less waste and uses a greater fraction of reactant atoms in the desired product; fewer steps can also reduce energy, solvent use or hazardous operations.

Method: Link high atom economy to reduced by-product waste, then link fewer steps to a second environmental or safety benefit such as lower energy demand, less solvent, fewer reagents or fewer purification stages.

Answer

• Steam/H₃PO₄ catalyst gives ethanol; acidified potassium dichromate(VI), warm and distil gives ethanal; acidified potassium dichromate(VI), heat under reflux gives ethanoic acid.

Method: Hydrate ethene with steam using an H₃PO₄ catalyst to make ethanol. Partially oxidise the primary alcohol with acidified K₂Cr₂O₇ and distil the ethanal as it forms. Return ethanal to excess oxidising mixture and heat under reflux to complete oxidation to ethanoic acid.

Answer

• Bromoethane → propanenitrile using ethanolic KCN under reflux → propylamine by reduction with H₂/Ni or LiAlH₄/dry ether → N-propylethanamide using ethanoyl chloride; cyanide contributes its carbon atom.

Method: Use nucleophilic substitution with ethanolic KCN under reflux: CH₃CH₂Br becomes CH₃CH₂CN. The carbon in CN⁻ joins the chain, so the product has three carbons. Reduce propanenitrile with H₂/Ni or LiAlH₄ in dry ether to obtain CH₃CH₂CH₂NH₂. Finally react propylamine with CH₃COCl; nucleophilic addition–elimination gives CH₃CONHCH₂CH₂CH₃.

Answer

• Any two of: it gives one sharp signal, is chemically inert, is volatile, and its signal lies away from most organic signals at $\delta=0$.

Method: All twelve TMS protons are equivalent, so TMS produces a single intense reference peak. It does not normally react with the sample and can be removed easily because it is volatile; its highly shielded signal is assigned $\delta=0$.

Answer

• Ethyl ethanoate, CH₃COOCH₂CH₃; the terminal CH₃ is split by two neighbouring protons into a triplet, CH₂ is split by three into a quartet, and the acyl CH₃ has no adjacent proton and is a singlet.

Method: The 3H triplet and 2H quartet are the paired pattern of an ethyl group: $n=2$ neighbours split CH₃ into three peaks, while $n=3$ neighbours split CH₂ into four. The remaining 3H singlet must have no neighbouring proton. With two oxygen atoms, CH₃COOCH₂CH₃ fits the formula and all three integrations.

Answer

• Ethyl propanoate, CH₃CH₂COOCH₂CH₃.

Method: Two separate 3H-triplet/2H-quartet pairs show two non-equivalent ethyl groups. The formula requires an ester, so placing one ethyl group on the acyl side and one on oxygen gives CH₃CH₂COOCH₂CH₃. Each CH₃ has two adjacent protons and is a triplet; each CH₂ has three adjacent protons and is a quartet. O–CH₂ is further downfield because electronegative oxygen deshields it. The carbonyl carbon and four different alkyl carbons account for five ¹³C signals.

Answer

• $R_f=0.60$

Method: Use $R_f=\text{distance moved by spot}/\text{distance moved by solvent}=4.5/7.5=0.60$. It has no unit because it is a ratio of two distances.

Answer

• Dye P has the larger $R_f$ because it spends a greater proportion of time in the moving solvent and is retained less strongly by the stationary phase.

Method: Movement results from repeated transfer between phases. P is carried further whenever it is in the mobile phase and is not held strongly on the solid, so its average movement is faster. Q spends more time attached to the stationary phase and remains closer to the start line.

Answer

• The first two peaks are consistent with X and Y; the $7.1\,\text{min}$ peak cannot safely be assigned as Z from the near match alone; a mass spectrum supplies fragment and molecular-ion data for comparison with a reference spectrum.

Method: Because conditions are identical, exact retention-time matches make X and Y plausible identities, although another compound could in principle co-elute. The final peak differs from Z's reference time, so it is not justified to label it Z without further evidence. GC first separates the components; the mass spectrometer then records a spectrum for each eluting peak. Matching molecular-ion and fragmentation patterns to standards provides independent structural evidence and makes the assignment stronger.