Haloalkanes have a halogen attached to an sp³ hybridised carbon of an alkyl chain (e.g., CH₃Cl). Haloarenes have a halogen attached directly to an sp² hybridised carbon of an aromatic ring (e.g., chlorobenzene). This distinction in hybridisation is the root cause of major reactivity differences between the two classes.
| Classification Basis | Categories |
|---|---|
| Number of halogens | Mono-, di-, tri-, polyhalogen compounds |
| Hybridisation of C-X carbon | Haloalkanes (sp³), haloarenes (sp²), allylic/vinylic (sp²) |
| Type of carbon bearing X (haloalkanes) | Primary (1°), secondary (2°), tertiary (3°) |
An allylic halide has the halogen on an sp³ carbon directly adjacent to a C=C double bond (e.g., $\text{CH}_2=\text{CH-CH}_2\text{Cl}$). A vinylic halide has the halogen directly attached to a carbon that is itself part of the double bond (e.g., $\text{CH}_2=\text{CHCl}$) — vinylic halides show extremely low reactivity in nucleophilic substitution, similar to haloarenes.
Classify the halogen atom in each of the following as allylic, vinylic, or a simple alkyl halide: (a) CH₂=CH–CH₂–Cl, (b) CH₃–CH=CH–Cl, (c) CH₃–CH₂–Cl.
(a) Allylic — Cl is on an sp³ carbon adjacent to the double bond.
(b) Vinylic — Cl is directly on one of the double-bond carbons.
(c) Simple alkyl halide — no double bond nearby at all.
Halogens are named as substituent prefixes (fluoro, chloro, bromo, iodo) attached to the parent hydrocarbon chain, with the lowest possible locants given to the full set of substituents (including the halogen), named alphabetically alongside any other substituents present.
Give the IUPAC name of $\text{CH}_3-\text{CHBr}-\text{CH}_2-\text{CH}_3$.
The longest chain has 4 carbons (butane). Numbering to give the lowest locant to the bromine substituent (from the end nearer to it), bromine sits at C2.
Name: 2-bromobutane
Since halogens are more electronegative than carbon, the C-X bond is polar, with the carbon carrying a partial positive charge ($\delta^+$) and the halogen a partial negative charge ($\delta^-$). This polarity is exactly why haloalkanes are susceptible to nucleophilic attack at the carbon.
| Property | Trend Down the Halogen Group (F → I) | Reason |
|---|---|---|
| C-X bond length | Increases | Halogen atomic size increases |
| C-X bond enthalpy | Decreases | Longer bond, weaker orbital overlap |
| Dipole moment | Generally decreases slightly | Decreasing electronegativity down the group |
Arrange CH₃F, CH₃Cl, CH₃Br, and CH₃I in order of increasing C-X bond length.
$\text{CH}_3\text{F} < \text{CH}_3\text{Cl} < \text{CH}_3\text{Br} < \text{CH}_3\text{I}$ — bond length increases steadily down the group as the halogen's atomic radius increases.
Alcohols react with various reagents to give alkyl halides: HX (with ZnCl₂ as catalyst, Lucas reagent), PCl₅ or PCl₃, and SOCl₂ (thionyl chloride). Thionyl chloride is generally preferred since the by-products (SO₂ and HCl) are both gases, giving a pure alkyl chloride without further purification needed.
Free radical halogenation of alkanes (using X₂ and UV light/heat) gives a mixture of products for most alkanes, limiting its usefulness for a single pure product except for highly symmetrical starting materials. Electrophilic addition of HX or X₂ to alkenes gives haloalkanes cleanly, following Markovnikov's rule (or anti-Markovnikov with peroxides, for HBr).
Explain why thionyl chloride (SOCl₂) is preferred over PCl₅ for converting an alcohol into an alkyl chloride in the laboratory.
The reaction of an alcohol with SOCl₂ produces the alkyl chloride along with SO₂ and HCl gases as by-products: $\text{R-OH} + \text{SOCl}_2 \rightarrow \text{R-Cl} + \text{SO}_2\uparrow + \text{HCl}\uparrow$. Since both by-products escape as gases, the alkyl chloride product is obtained in a much purer form directly, without needing additional separation steps — unlike with PCl₅, which leaves behind POCl₃ (a liquid) that must be separated from the product.
Alkyl chlorides or bromides react with sodium iodide in dry acetone to give the corresponding alkyl iodide, with NaCl/NaBr precipitating out of the acetone solution (since they're insoluble in it), driving the equilibrium forward: $\text{R-Cl} + \text{NaI} \xrightarrow{\text{dry acetone}} \text{R-I} + \text{NaCl}\downarrow$.
Alkyl chlorides/bromides react with an inorganic fluoride (such as AgF, Hg₂F₂, CoF₂, or SbF₃) to give the corresponding alkyl fluoride — this is the standard method for introducing fluorine, since direct fluorination with F₂ gas is far too vigorous and uncontrollable.
Explain, in terms of solubility, why the Finkelstein reaction proceeds efficiently towards the formation of the alkyl iodide.
Sodium iodide is soluble in dry acetone, but the sodium chloride (or bromide) by-product formed during the reaction is insoluble in acetone and precipitates out of solution as soon as it forms. This continuous removal of a product shifts the reaction's equilibrium (by Le Chatelier's principle) strongly towards further formation of the alkyl iodide, driving the reaction to a high yield.
Arrange 1-chlorobutane, 2-chlorobutane, and 2-chloro-2-methylpropane (all isomers, C₄H₉Cl) in order of increasing boiling point.
2-chloro-2-methylpropane (most branched) $<$ 2-chlorobutane (moderately branched) $<$ 1-chlorobutane (straight chain, least branched) — boiling point decreases as branching increases, since branched molecules have less surface area for van der Waals interactions, so 1-chlorobutane (straight-chain, most surface contact) has the highest boiling point of the three.
SN2 (bimolecular nucleophilic substitution) occurs in a single concerted step — the nucleophile attacks the carbon from the side opposite to the leaving group, while the leaving group departs simultaneously. The rate depends on the concentration of both the substrate and the nucleophile (second order kinetics overall).
Since the nucleophile attacks from the opposite side of the leaving group, the configuration at the carbon is inverted (like an umbrella turning inside out) — this is called Walden inversion. If the substrate is optically active (a single enantiomer), the product will have the inverted configuration.
Explain why methyl halides and primary alkyl halides react readily via the SN2 mechanism, while tertiary alkyl halides essentially do not.
SN2 requires the nucleophile to approach and attack the carbon directly from the side opposite the leaving group. In methyl and primary halides, the carbon is only lightly hindered (few or no bulky alkyl groups nearby), leaving plenty of open space for backside attack. In tertiary halides, three bulky alkyl groups surround the carbon, sterically blocking the approach of an incoming nucleophile from the required backside direction — making SN2 extremely difficult or impossible for tertiary substrates.
SN1 (unimolecular nucleophilic substitution) proceeds in two steps: (1) the C-X bond breaks first (slow, rate-determining step) to form a carbocation intermediate, and (2) the nucleophile then rapidly attacks this carbocation (fast step). The rate depends only on the substrate's concentration, not the nucleophile's (first order kinetics overall).
Since the intermediate carbocation is planar (sp² hybridised), the incoming nucleophile can attack from either face with roughly equal probability, giving a nearly equal mixture of both possible configurations — this results in (partial or complete) racemisation of an optically active substrate.
Explain why tertiary alkyl halides react readily via the SN1 mechanism, while methyl halides essentially do not.
SN1 proceeds through a carbocation intermediate, and its rate depends heavily on how stable that carbocation is. Tertiary alkyl halides form a tertiary carbocation, which is well stabilised by the +I effect and hyperconjugation from three surrounding alkyl groups — making this pathway energetically feasible. A methyl halide would have to form a methyl carbocation, which has no alkyl groups to stabilise it and is extremely unstable — so SN1 essentially never proceeds for methyl (or most primary) halides.
| Factor | Favours SN1 | Favours SN2 |
|---|---|---|
| Substrate type | Tertiary (stable carbocation) | Methyl, primary (less steric hindrance) |
| Nucleophile strength | Weak nucleophile (not rate-determining) | Strong nucleophile (directly involved in rate-determining step) |
| Solvent | Polar protic (stabilises carbocation & leaving group by solvation) | Polar aprotic (doesn't hinder nucleophile's reactivity) |
| Leaving group | Good leaving group helps both mechanisms similarly | Good leaving group helps both mechanisms similarly |
Predict whether the reaction of tert-butyl bromide with water (a weak nucleophile) proceeds mainly via SN1 or SN2, and justify your answer.
SN1. The substrate is tertiary (strongly favouring the formation of a stable tertiary carbocation), and water is a weak nucleophile (unable to effectively drive a concerted SN2 backside attack, especially given the severe steric hindrance a tertiary carbon would present anyway). Both the substrate structure and nucleophile strength strongly favour the SN1 pathway here.
When an alkyl halide is heated with alcoholic KOH, it can undergo β-elimination, removing a halogen from one carbon and a hydrogen from an adjacent (β) carbon to form an alkene, generally following Saytzeff's rule (more substituted alkene favoured).
Alcoholic KOH (a strong base, promotes elimination) and aqueous KOH (a good nucleophile, promotes substitution) give different major products from the same alkyl halide — this is a classic conceptual distinction tested frequently.
Alkyl halides react with magnesium metal in dry ether to form Grignard reagents ($\text{R-MgX}$) — extremely versatile and reactive organometallic compounds, useful in numerous carbon-carbon bond-forming reactions. Grignard reagents react violently with any source of active hydrogen (water, alcohols, amines), so strictly moisture-free (anhydrous) conditions are essential during their preparation and use.
Explain why Grignard reagent preparation must be carried out under strictly anhydrous conditions.
The carbon atom in a Grignard reagent (R-MgX) is highly nucleophilic/basic, and reacts instantly and irreversibly with any trace of water (or any compound with an acidic/active hydrogen), being protonated to simply regenerate the parent alkane (R-H) and destroying the Grignard reagent before it can be used for its intended purpose. Even trace moisture in the glassware or solvent can substantially reduce the yield or completely quench the reagent.
| Compound | Formula | Use / Note |
|---|---|---|
| Chloroform | CHCl₃ | Once used as an anaesthetic; now mainly a solvent and precursor in synthesis (oxidises in air/light to poisonous phosgene, so stored in dark bottles) |
| Iodoform | CHI₃ | Antiseptic; its characteristic odour is used in the iodoform test to detect specific structural features (methyl ketones, ethanol) |
| DDT | Dichlorodiphenyltrichloroethane | Powerful pesticide historically; banned/restricted in many countries due to bioaccumulation and environmental persistence |
| Freons | Chlorofluorocarbons (CFCs) | Once widely used as refrigerants/aerosol propellants; phased out due to their role in stratospheric ozone layer depletion |
CFCs are extremely stable in the lower atmosphere, but on reaching the stratosphere, UV radiation breaks the C-Cl bond to release chlorine free radicals, which catalytically destroy ozone molecules in a chain reaction — a single chlorine radical can destroy thousands of ozone molecules before being deactivated, which is why even small amounts of CFC release cause significant, long-lasting ozone depletion.
Why is chloroform typically stored in dark-coloured, airtight bottles filled up to the brim?
Chloroform slowly oxidises in the presence of both air (oxygen) and light to form phosgene ($\text{COCl}_2$), an extremely toxic gas. Storing it in dark bottles blocks light exposure, and filling to the brim minimises the air space (oxygen) available inside the bottle — both precautions slow down this oxidation and prevent dangerous phosgene accumulation.
In a haloarene (e.g., chlorobenzene), the halogen's lone pair partly delocalises into the aromatic ring through resonance, giving the C-X bond partial double-bond character. This makes the bond: (1) shorter and stronger than a typical C-X single bond, and (2) resistant to nucleophilic attack, since the carbon is sp² hybridised (not easily approached from the back for SN2) and doesn't readily form a stable carbocation by losing the halogen (which would disrupt ring aromaticity, unlike normal SN1 ionisation).
The sp² carbon holds electrons more tightly (higher effective electronegativity due to greater s-character) than an sp³ carbon, strengthening the C-X bond further. Also, the possible carbocation intermediate (if X were to leave, as in SN1) would be a highly unstable aryl cation, since it isn't stabilised by the ring's π system in the way that allylic or benzylic cations are.
Explain, using resonance structures, why the C-Cl bond in chlorobenzene is shorter than the C-Cl bond in an alkyl chloride like ethyl chloride.
In chlorobenzene, one of the chlorine's lone pairs can delocalise into the ring via resonance, creating structures where a C=Cl (partial double bond) character exists at the point of attachment, with a corresponding negative charge shifted onto the ring. This partial double-bond character shortens and strengthens the C-Cl bond compared to a purely single-bonded C-Cl bond in ethyl chloride, where no such resonance delocalisation is possible.
Despite being deactivating (due to the halogen's strong −I effect, which lowers overall electron density in the ring), halogens are ortho, para-directing (due to lone pair donation via resonance, +M effect, specifically stabilising the intermediate carbocation when attack occurs at those positions). So haloarenes react more slowly than benzene itself, but still direct further substitution mainly to ortho/para positions.
While ordinary haloarenes strongly resist nucleophilic substitution (as explained above), the presence of strong electron-withdrawing groups (like $-\text{NO}_2$) at the ortho/para positions relative to the halogen dramatically increases reactivity, since these groups help stabilise the negatively charged intermediate (Meisenheimer complex) formed during nucleophilic attack, making the substitution feasible under milder conditions.
Explain why 2,4-dinitrochlorobenzene reacts readily with aqueous NaOH to give 2,4-dinitrophenol, while chlorobenzene itself does not react under similar mild conditions.
The two strongly electron-withdrawing −NO₂ groups (ortho and para to the chlorine) help stabilise the negatively charged intermediate that forms when hydroxide attacks the ring carbon bearing chlorine — this delocalised, resonance-stabilised intermediate (Meisenheimer complex) makes the reaction pathway energetically much more accessible. In plain chlorobenzene, there are no such electron-withdrawing groups to stabilise this intermediate, so nucleophilic substitution essentially doesn't proceed under similarly mild conditions.