Haloalkanes and Haloarenes · Lesson 15 of 16
Planning Organic Transformations
“Combine substitution, elimination, addition and coupling into justified routes with complete carbon-skeleton tracking.”
• Plan multistep conversions by identifying which bonds and carbon counts must change. • Use elimination/addition sequences to move halogen or hydroxyl positions. • Use cyanide and suitable carbon nucleophiles to build larger skeletons. • Explain the additional transformation principles needed for the chapter’s conversion problems. • Distinguish a plausible route with separation/work-up from a formal arrow that hides limitations.
Work backward from the target, then check every forward step
A synthetic route is a sequence of chemically supported changes, not a list of reagent names chosen because they appear nearby in a chapter. Compare the starting and target carbon skeletons first. If the carbon count is unchanged, a functional-group swap or position change may suffice. If one carbon must be added, cyanide substitution can provide it. If two fragments must join, identify which is the carbon nucleophile and which is an accessible electrophile. Finally check the fate of every reagent-derived carbon and every functional group.
Several multistep conversion problems use transformations whose full treatment belongs to later functional-group chemistry. The necessary bridge principles are explained here: nitrile hydrolysis retains its carbon as a carboxyl carbon; an acetylide can alkylate a suitable primary halide; a nitrile can react with a Grignard reagent and then hydrolyse to a ketone; and haloform or carbylamine reactions have their own substrate requirements. These are explicit additional steps, not hidden direct conversions.
Move a group through elimination followed by controlled addition
Problem
Change the bromine position without changing three carbons.
- 1.Heat 1-bromopropane with alcoholic base to eliminate HBr and form propene: CH₃CH₂CH₂Br→CH₃CH=CH₂.
- 2.Add HBr without peroxide. The ordinary Markovnikov pathway gives mainly CH₃CHBrCH₃, 2-bromopropane.
- 3.The two steps break and remake different bonds. A simple direct “Br migrates” arrow would conceal the required alkene and addition conditions.
Problem
Convert 2-bromopropane to 1-bromopropane, then to propan-1-ol.
- 1.Eliminate from 2-bromopropane with heated alcoholic base to form propene.
- 2.Use HBr/peroxide so radical anti-Markovnikov addition gives CH₃CH₂CH₂Br.
- 3.Hydrolyse the primary bromide with aqueous hydroxide under substitution-favouring conditions to obtain CH₃CH₂CH₂OH. The same final two steps turn propene into propan-1-ol.
| Position-change target | Route with conditions | Why it works |
|---|---|---|
| But-1-ene→but-2-ene | HBr without peroxide→2-bromobutane; heated alcoholic base→but-2-ene-rich mixture | Addition chooses C2; Zaitsev elimination favours the internal alkene |
| tert-Butyl bromide→isobutyl bromide | Eliminate to 2-methylpropene; add HBr/peroxide | Radical addition places Br at terminal CH₂ |
| 2-Methylprop-1-ene→tert-butyl chloride | HCl under ordinary addition conditions | Tertiary-cation orientation gives (CH₃)₃CCl |
| But-1-ene→1-iodobutane | HBr/peroxide→1-bromobutane; NaI/dry acetone | HBr controls position, then halide exchange changes element |
The butene sequence favours but-2-ene rather than guaranteeing pure material: but-1-ene and E/Z distributions can remain. A practical target may therefore need separation. Likewise a route using HBr/peroxide depends on avoiding conditions that substitute ordinary ionic addition instead. Route validity includes selectivity, not only the existence of a product formula.
One-carbon extension through cyanide
A nitrile R–C≡N contains a carbon supplied by cyanide. On hydrolysis, the nitrile carbon becomes the carboxyl carbon while nitrogen leaves as ammonia or ammonium depending on the medium. In acidic bookkeeping, RCN+2H₂O+H⁺→RCOOH+NH₄⁺. With base the immediate organic product is carboxylate, requiring acidification to isolate the acid. This explains both the extra carbon and the necessary water/heating/work-up conditions.
Problem
Plan C₆H₅CH₂OH→C₆H₅CH₂COOH.
- 1.Replace side-chain OH with Br using suitable HBr or another appropriate alcohol-to-halide reagent, giving C₆H₅CH₂Br.
- 2.Use KCN under suitable substitution conditions to give C₆H₅CH₂C≡N. The new carbon is now bonded to the original benzylic carbon.
- 3.Hydrolyse the nitrile with heated aqueous acid, or base followed by acidification, to obtain 2-phenylethanoic acid. Starting seven carbons become eight; simple oxidation of the original alcohol would not supply that added carbon.
| Starting compound and target | Route | Carbon check |
|---|---|---|
| Ethanol→propanenitrile | Alcohol→bromoethane using suitable reagent; KCN substitution | Two carbons become three |
| Chloroethane→propanoic acid | KCN→propanenitrile; heated hydrolysis and appropriate work-up | The CN carbon becomes the acid carbon |
| Propene→1-nitropropane | HBr/peroxide→1-bromopropane; AgNO₂ substitution | Nitro attachment adds N/O, not a carbon |
| Ethanol→fluoroethane | Suitable conversion to ethyl chloride; Swarts fluoride exchange | Two carbons remain two |
Building an alkyne needs the correct carbon nucleophile
Ethyne has a terminal C–H that can be deprotonated by a sufficiently strong base such as NaNH₂ under suitable dry conditions to give an ethynide nucleophile, HC≡C⁻. A primary alkyl halide can then undergo SN2 carbon–carbon bond formation. Ordinary aqueous hydroxide is not strong enough to prepare the same useful acetylide population, and a tertiary halide would strongly favour elimination rather than clean acetylide substitution.
Problem
Explain where the two additional carbons enter.
- 1.Convert ethanol to bromoethane with a suitable alcohol-halogenation reagent.
- 2.Prepare sodium ethynide from ethyne and NaNH₂ in a suitable dry medium, then react it with the primary ethyl bromide: HC≡CNa+C₂H₅Br→HC≡CCH₂CH₃+NaBr.
- 3.The two-carbon ethynide fragment joins the two-carbon ethyl fragment, producing four-carbon but-1-yne. Ethyne is an explicit external carbon source; the route does not create two carbons from ethanol alone.
Problem
Convert propene to propyne.
- 1.Add Br₂ under ordinary addition conditions to obtain CH₃CHBrCH₂Br, 1,2-dibromopropane.
- 2.Use a sufficiently strong-base double dehydrohalogenation sequence, such as excess NaNH₂ in liquid NH₃ under appropriate anhydrous conditions. Two HBr equivalents are removed to make CH₃C≡CH; excess strong base can also deprotonate that terminal alkyne.
- 3.A final water/proton work-up restores the neutral terminal alkyne. Ordinary alcoholic KOH should not be assumed equally effective at eliminating a second vinylic halide bond.
Ethane to bromoethene can be staged differently: controlled Br₂/light substitution gives bromoethane, alcoholic-base elimination gives ethene, Br₂ addition gives 1,2-dibromoethane, and a controlled single dehydrohalogenation gives CH₂=CHBr. Stopping at a vinyl bromide differs from using excess strong-base conditions to continue toward ethyne. The reagent strength and extent of elimination are therefore part of the target specification.
A nitrile and a Grignard reagent can meet at a ketone
The carbon of C≡N is electrophilic because the bond is polarised toward nitrogen. A carbon nucleophile from a suitable Grignard reagent can add there, giving an imine-derived magnesium species. Acidic hydrolysis replaces the imine nitrogen attachment by C=O, producing a ketone. This is a different role for a nitrile from simply hydrolysing it to an acid. The Grignard stage must precede the aqueous work-up because water would quench the carbon nucleophile.
Problem
Use bromomethane-derived fragments while accounting for all three product carbons.
- 1.Use one portion of CH₃Br with KCN to make CH₃CN. This is ethanenitrile: one carbon comes from the methyl halide and one from cyanide.
- 2.Use another portion with Mg/dry ether to prepare CH₃MgBr. Add it to CH₃CN under appropriate dry conditions, forming an imine-derived magnesium intermediate with two methyl groups on the former nitrile carbon.
- 3.Acidic aqueous work-up hydrolyses that intermediate to (CH₃)₂CO, propanone. The central C=O carbon is the original cyanide carbon; the two methyl groups come from the methyl fragments. Do not add water before the Grignard addition.
Haloform and carbylamine are specific transformations
Isopropyl alcohol contains CH₃CH(OH)– and can first be oxidised to acetone in iodine/alkali chemistry. Repeated iodination at the methyl-ketone position and cleavage yield yellow CHI₃ plus acetate. For the overall alcohol conversion a balanced equation is (CH₃)₂CHOH+4I₂+6NaOH→CHI₃+CH₃COONa+5NaI+5H₂O. The original three-carbon molecule is split into a one-carbon haloform and a two-carbon carboxylate; iodoform is not a simple three-carbon substitution product.
A primary amine such as aniline can undergo the carbylamine transformation with chloroform and alcoholic KOH on heating: C₆H₅NH₂+CHCl₃+3KOH→C₆H₅NC+3KCl+3H₂O. The product is phenyl isocyanide, with N between phenyl and the terminal carbon; it is not benzonitrile C₆H₅CN. A valence-explicit contributor is C₆H₅–N⁺≡C⁻. This reaction class is specific to primary amines in the usual test description, and its identity should be established structurally rather than by recommending an odour observation.
Combine ring direction, side-chain reactivity and coupling
| Transformation | Supported sequence and limitation |
|---|---|
| Toluene→benzyl alcohol | Controlled side-chain halogenation under light→C₆H₅CH₂X; aqueous OH⁻ substitution. Ring-halogenation conditions would give the wrong C–X position. |
| Aniline→chlorobenzene | NaNO₂/HCl at 273–278 K→diazonium; CuCl/HCl replacement. |
| Benzene→4-bromonitrobenzene | Brominate first; nitrate next; isolate the para isomer. |
| Chlorobenzene→p-nitrophenol | Nitrate and isolate para nitro compound; heated-base aryl substitution; acid work-up. |
| Chloroethane→butane; 1-chlorobutane→n-octane | Formal symmetric Wurtz coupling with Na/dry ether using appropriate primary halides. |
| 2-Chlorobutane→3,4-dimethylhexane | Formal sec-butyl Wurtz coupling, qualified by secondary-halide elimination and mixture problems. |
| Benzene→biphenyl | Ring halogenation followed by formal Fittig coupling; preserve two separate rings. |
Problem
A primary C₄H₉Br gives an alkene with alcoholic KOH, then an isomeric bromide with ordinary HBr; its Wurtz product is not n-octane. Identify the structures.
- 1.The two primary C₄H₉Br skeletons are n-butyl and isobutyl bromide. Since coupling does not give n-octane, choose isobutyl bromide, (CH₃)₂CHCH₂Br.
- 2.Elimination gives 2-methylpropene, (CH₃)₂C=CH₂. Ordinary HBr addition gives tert-butyl bromide, (CH₃)₃CBr, a constitutional isomer of the starting halide.
- 3.Coupling the original isobutyl fragments gives 2,5-dimethylhexane, C₈H₁₈. This uses the skeleton information to distinguish a formula shared with n-octane.
Starting from 2-chloropropane rather than 2-bromopropane uses the same position-changing logic: heated alcoholic-base elimination gives propene, HBr/peroxide gives 1-bromopropane, and aqueous hydroxide substitution gives propan-1-ol. The final halogen need not match the initial halogen because the route passes through an alkene and a separately chosen addition reagent.
Quick Quiz
Quiz
Which carbon becomes the carboxyl carbon after nitrile hydrolysis?
Explanation: RCN→RCOOH retains the nitrile carbon and removes nitrogen as ammonia/ammonium.
How can 2-bromopropane be converted toward 1-bromopropane?
Explanation: The alkene creates a new opportunity to control regioselectivity.
Which partner adds two carbons to ethyl bromide to make but-1-yne?
Explanation: SN2 alkylation joins a two-carbon acetylide to the two-carbon ethyl fragment.
Why must aqueous work-up follow the Grignard/nitrile reaction?
Explanation: Keep the carbon-addition stage dry, then hydrolyse the imine-derived intermediate.
What skeleton gives a Wurtz C₈H₁₈ different from n-octane in the stated primary-bromide puzzle?
Explanation: Its coupled product is branched 2,5-dimethylhexane.
Practice Problems
- Plan propene to 1-nitropropane and distinguish the KNO₂ alternative.
- Plan chloroethane to propanoic acid with a balanced hydrolysis representation.
- Give the ethanol-to-but-1-yne sequence and identify the external carbon source.
- Convert tert-butyl bromide to isobutyl bromide without changing the skeleton.
- Explain the carbon origin in bromomethane-to-propanone via nitrile/Grignard chemistry.
- Balance isopropyl alcohol to iodoform under iodine/alkali and explain carbon splitting.
- Give aniline-to-phenylisocyanide chemistry and distinguish it from benzonitrile.
- Plan ethane to bromoethene and state where the sequence must stop.
- Plan benzyl alcohol to 2-phenylethanoic acid and explain why direct oxidation is insufficient.
- Compare formal Wurtz production of butane and 3,4-dimethylhexane.
- Write the final organic product for bromoethane/KCN, sodium phenoxide/ethyl chloride, and 2-methylbut-2-ene/HBr without peroxide.
- Why does a synthesis plan need separations and work-ups explicitly?
HBr/peroxide gives 1-bromopropane. AgNO₂ under the intended substitution conditions gives N-linked CH₃CH₂CH₂NO₂. The KNO₂ comparison instead favours O-linked propyl nitrite, CH₃CH₂CH₂ONO, which is a different connectivity.
Key Takeaways
• Plan a route by tracking changed bonds and carbon counts before choosing reagents. • Elimination followed by regioselective addition can move a substituent’s position. • Cyanide substitution adds a carbon that survives as the nitrile or acid carbon. • Acetylide alkylation needs a strong-base preparation and an accessible electrophile. • Nitrile/Grignard addition must precede aqueous ketone-forming work-up. • Haloform and carbylamine reactions require their specific substrate patterns. • Formal coupling products must be qualified by elimination and mixture limitations. • Product isolation, protonation state and positional separation are part of a complete route.