NeetCrackers Chemistry Practice
Practice Set

Organic Compounds Containing Halogens — Practice Questions with Answers

30 fully solved questions covering nomenclature, preparation, SN1/SN2 mechanisms, and reactions of haloalkanes and haloarenes.

R
Questions written and solutions verified by Roshan, science educator with 5 years of experience teaching NEET and JEE aspirants. Checked against NCERT and previous years' papers. Last reviewed September 2026.
10MCQs · 1 mark
10Short Answer · 3 marks
10Long Answer · 5 marks
90Total Marks

📑 On This Page

  1. Section A — 10 MCQs (1 mark each)
  2. Section B — 10 Short Answer Questions (3 marks each)
  3. Section C — 10 Long Answer Questions (5 marks each)
Section A

🎯 Multiple Choice Questions

10 questions · 1 mark each · Correct answer highlighted with explanation

Q1 An SN1 reaction mechanism proceeds via the formation of a(n):
Correct Answer: (a). SN1 reactions proceed through a two-step mechanism, with a carbocation intermediate formed in the rate-determining first step.
Q2 An SN2 reaction mechanism is characterised by:
Correct Answer: (a). SN2 proceeds in one concerted step, with the nucleophile attacking from the side opposite the leaving group, resulting in inversion of configuration (Walden inversion).
Q3 Tertiary alkyl halides generally favour the:
Correct Answer: (a). Tertiary alkyl halides favour SN1, since they readily form a stable tertiary carbocation, while their steric bulk hinders the backside attack required for SN2.
Q4 Haloarenes (aryl halides) are generally much less reactive towards nucleophilic substitution compared to haloalkanes, primarily because of:
Correct Answer: (a). Resonance donation of the halogen lone pair into the ring gives the C-X bond partial double-bond character, making it stronger and shorter, and hence more resistant to nucleophilic substitution.
Q5 Grignard reagents are prepared by reacting an alkyl halide with:
Correct Answer: (a). Grignard reagents ($RMgX$) are prepared by reacting an alkyl (or aryl) halide with magnesium turnings in anhydrous (dry) ether.
Q6 Racemisation is characteristically associated with the:
Correct Answer: (a). SN1 proceeds via a planar carbocation intermediate, which can be attacked by the nucleophile from either face, typically resulting in a racemic (or partially racemised) mixture of products.
Q7 DDT (dichlorodiphenyltrichloroethane) was historically used primarily as a(n):
Correct Answer: (a). DDT was widely used historically as a potent insecticide, though its use is now restricted/banned in many countries due to environmental persistence and bioaccumulation concerns.
Q8 The rate of an SN2 reaction depends on the concentration of:
Correct Answer: (a). SN2 is a second-order (bimolecular) reaction overall, with rate depending on the concentration of both the substrate and the nucleophile.
Q9 Chloroform ($CHCl_3$) is stored in dark-coloured bottles, completely filled, primarily to prevent:
Correct Answer: (a). Chloroform slowly undergoes air oxidation (especially in the presence of light) to form the highly toxic gas phosgene — storage in dark, completely filled bottles minimises air and light exposure, reducing this decomposition.
Q10 Freons (chlorofluorocarbons, CFCs) have been implicated in causing:
Correct Answer: (a). CFCs released into the atmosphere break down under UV radiation, releasing chlorine free radicals that catalytically destroy stratospheric ozone molecules.
Section B

✍️ Short Answer Questions

10 questions · 3 marks each · Full model answers included

Q1. Distinguish between the SN1 and SN2 mechanisms in terms of number of steps, rate law, and stereochemical outcome.
3 Marks
Answer

SN1: Two-step mechanism (carbocation intermediate formed first, then attacked by nucleophile); rate law depends only on substrate concentration (first-order overall); typically results in racemisation (loss of stereochemical purity).

SN2: Single-step, concerted mechanism (bond breaking and forming simultaneously); rate law depends on both substrate and nucleophile concentration (second-order overall); results in inversion of configuration (Walden inversion) at the reacting carbon.

Q2. Explain why haloarenes (like chlorobenzene) show much lower reactivity towards nucleophilic substitution compared to haloalkanes.
3 Marks
Answer

In haloarenes, one of the halogen's lone pairs can be delocalised (via resonance) into the adjacent aromatic ring's pi system — this donation gives the carbon-halogen bond significant partial double-bond character, making it considerably shorter and stronger than a typical single C-X bond in a haloalkane.

Additionally, the aromatic ring's electron-rich character somewhat repels an incoming nucleophile, and any potential nucleophilic attack directly on the ring carbon would require temporarily disrupting the ring's valuable aromatic stability. This combination of a stronger C-X bond and electronic/steric hindrance from the aromatic system is precisely why haloarenes are considerably less reactive towards nucleophilic substitution than analogous haloalkanes.

Q3. Explain why tertiary alkyl halides react predominantly via SN1, while primary alkyl halides react predominantly via SN2.
3 Marks
Answer

Tertiary alkyl halides readily form a comparatively stable tertiary carbocation (due to substantial hyperconjugative/inductive stabilisation from three adjacent alkyl groups), favouring the two-step SN1 pathway; additionally, their considerable steric bulk around the reactive carbon hinders the direct backside nucleophilic attack required for SN2.

Primary alkyl halides would form a highly unstable primary carbocation if SN1 were attempted (with minimal stabilisation available), strongly disfavouring this pathway; their comparatively unhindered (sterically open) reactive carbon readily permits the backside nucleophilic attack required for the single-step SN2 mechanism, which is strongly favoured instead.

Q4. Describe the preparation and one synthetic use of a Grignard reagent.
3 Marks
Answer

Preparation: An alkyl (or aryl) halide is reacted with magnesium metal turnings in dry (anhydrous) ether: $R-X + Mg \xrightarrow{\text{dry ether}} R-MgX$ (a Grignard reagent).

Synthetic use: Grignard reagents are highly versatile carbon nucleophiles, widely used to form new carbon-carbon bonds — for example, reaction with formaldehyde ($HCHO$) followed by aqueous acidic workup produces a primary alcohol with one additional carbon atom compared to the original Grignard reagent's alkyl group, a valuable method for carbon chain extension in organic synthesis.

Q5. Why must Grignard reagents be prepared and used under strictly anhydrous (moisture-free) conditions?
3 Marks
Answer

Grignard reagents are extremely strong bases and highly reactive carbon nucleophiles — the carbon atom bonded to magnesium carries significant negative (carbanion-like) character.

Even trace amounts of water present would immediately react with the Grignard reagent, rapidly protonating (quenching) it to form the corresponding simple alkane and destroying the valuable reagent before it can be used for its intended synthetic purpose: $RMgX + H_2O \rightarrow RH + Mg(OH)X$ — this extreme moisture sensitivity is precisely why strictly anhydrous conditions (dry glassware, dry solvents, inert atmosphere) are essential throughout both preparation and use of Grignard reagents.

Q6. Explain why alkyl halides, despite the C-X bond having significant polarity, do not typically dissolve in (or react readily with) water.
3 Marks
Answer

Although the C-X bond itself is polar, alkyl halides are generally poorly soluble in water because they cannot form hydrogen bonds with water molecules (lacking an appropriately positioned, sufficiently acidic hydrogen atom to donate, or a sufficiently basic site relative to water's own hydrogen-bonding network) — the energy required to break existing water-water hydrogen bonds to accommodate the alkyl halide is not adequately compensated for by new interactions with the non-hydrogen-bonding alkyl halide molecule.

Regarding reactivity, alkyl halides do not typically react appreciably with pure, neutral water at ordinary conditions because water is a comparatively weak nucleophile, and simple hydrolysis of most common alkyl halides proceeds only very slowly under such mild conditions without a more forcing catalyst, stronger nucleophile, or more vigorous reaction conditions.

Q7. Explain the term "Finkelstein reaction," and state its practical significance in the preparation of alkyl iodides.
3 Marks
Answer

The Finkelstein reaction involves treating an alkyl chloride or bromide with sodium iodide dissolved in dry acetone: $R-Cl(\text{or }Br) + NaI \xrightarrow{\text{dry acetone}} R-I + NaCl(\text{or }NaBr)$.

Significance: This reaction is driven forward specifically because sodium chloride and sodium bromide (unlike sodium iodide) are insoluble in dry acetone — as they precipitate out of the reaction mixture, this continuously shifts the equilibrium towards the products (per Le Chatelier's principle), providing a practical, efficient method for converting more readily available alkyl chlorides/bromides into the often more synthetically useful, more reactive alkyl iodides.

Q8. Explain why (R)-2-bromobutane undergoes predominantly (though not exclusively) inversion of configuration when reacting via SN2 with hydroxide ion, while extensive racemisation would be expected via SN1.
3 Marks
Answer

In the SN2 mechanism, the hydroxide nucleophile directly attacks the carbon from the side exactly opposite the leaving bromide group (backside attack) in a single, concerted step — this specific spatial requirement means the incoming group always ends up on the opposite side from where the leaving group departed, resulting in a complete, predictable inversion of the stereocentre's configuration.

In the SN1 mechanism, by contrast, the leaving group departs first, forming a flat (planar), $sp^2$-hybridised carbocation intermediate — this planar intermediate can be attacked by the incoming nucleophile with roughly equal probability from either face, typically resulting in a racemic (or significantly racemised) mixture of both possible stereochemical outcomes, rather than the clean, predictable inversion characteristic of SN2.

Q9. Explain the environmental concern associated with chlorofluorocarbons (CFCs, or "Freons"), including the basic mechanism of ozone depletion.
3 Marks
Answer

CFCs are highly stable, unreactive compounds under normal atmospheric conditions, allowing them to persist long enough to diffuse upward into the stratosphere, where they are exposed to intense UV radiation.

This UV radiation causes CFC molecules to undergo homolytic C-Cl bond cleavage, releasing highly reactive chlorine free radicals ($Cl^\bullet$) — these radicals catalytically destroy ozone molecules ($Cl^\bullet + O_3 \rightarrow ClO^\bullet + O_2$, followed by further steps regenerating $Cl^\bullet$), meaning a single chlorine radical can destroy many thousands of ozone molecules before being eventually removed from the stratosphere — this catalytic destruction cycle is the fundamental mechanism behind CFC-induced ozone layer depletion, a major environmental concern due to ozone's crucial role in absorbing harmful UV radiation.

Q10. Explain why vinyl halides (halogen directly attached to a double-bonded carbon) are generally unreactive towards typical nucleophilic substitution reactions, similar to haloarenes.
3 Marks
Answer

In a vinyl halide (e.g., $CH_2=CH-Cl$), the halogen's lone pair can be delocalised via resonance into the adjacent pi bond system, similar to the situation in haloarenes — this delocalisation gives the C-X bond partial double-bond character, shortening and strengthening it compared to a typical single C-X bond found in saturated alkyl halides.

Additionally, the carbon bearing the halogen is $sp^2$ hybridised in a vinyl halide, making it more electronegative than a typical $sp^3$ carbon — this increased electronegativity further strengthens the C-X bond and reduces the carbon's susceptibility to nucleophilic attack. This combination of resonance stabilisation and $sp^2$ hybridisation effects is precisely why vinyl halides, like haloarenes, are generally quite unreactive towards conventional nucleophilic substitution reactions.

Section C

📘 Long Answer / Numerical Questions

10 questions · 5 marks each · Detailed step-by-step solutions

Q1. Write the complete mechanism (with intermediate structures) for the SN1 hydrolysis of tert-butyl bromide with water, and explain the rate-determining step.
5 Marks
Answer

Step 1 (rate-determining, slow): tert-Butyl bromide ionises, with the C-Br bond breaking heterolytically to form a stable, planar tertiary carbocation and a bromide ion: $(CH_3)_3C-Br \rightarrow (CH_3)_3C^+ + Br^-$

Step 2 (fast): Water (a weak nucleophile, but present in excess as solvent) attacks the planar carbocation from either face, forming a protonated (oxonium) alcohol intermediate: $(CH_3)_3C^+ + H_2O \rightarrow (CH_3)_3C-OH_2^+$

Step 3 (fast, deprotonation): A second water molecule removes a proton from the oxonium intermediate, giving the final product, tert-butyl alcohol: $(CH_3)_3C-OH_2^+ + H_2O \rightarrow (CH_3)_3C-OH + H_3O^+$

The rate-determining step is Step 1 (the slow, unimolecular ionisation) — this is why the overall reaction rate depends only on the concentration of the alkyl halide substrate, not on the nucleophile (water) concentration, consistent with first-order SN1 kinetics.

Q2. Explain, with the mechanism, how a Grignard reagent reacts with carbon dioxide to eventually form a carboxylic acid, and identify the product formed from $CH_3MgBr$.
5 Marks
Answer

Step 1: The carbon atom of the Grignard reagent (bearing significant carbanion-like character) attacks the electrophilic carbon of carbon dioxide, forming a magnesium carboxylate salt intermediate: $CH_3-MgBr + CO_2 \rightarrow CH_3-COO-MgBr$

Step 2 (acidic workup): Treatment with dilute aqueous acid (e.g., dilute $HCl$ or $H_2SO_4$) protonates the carboxylate, releasing the free carboxylic acid: $CH_3-COO-MgBr + H^+ \rightarrow CH_3-COOH + MgBrX$

Product from $CH_3MgBr$: Acetic acid ($CH_3COOH$) — this reaction is a valuable synthetic method for extending a carbon chain by exactly one carbon atom, converting a Grignard reagent directly into a carboxylic acid.

Q3. Compare the relative rates of SN2 hydrolysis for methyl bromide, ethyl bromide, isopropyl bromide, and tert-butyl bromide, explaining the trend using steric hindrance.
5 Marks
Answer

SN2 reactions require the incoming nucleophile to approach and attack the carbon directly from the backside, opposite the leaving group — this approach is significantly hindered by bulky substituent groups attached to (or near) the reacting carbon.

Relative rate order (SN2): Methyl > Ethyl > Isopropyl > tert-Butyl (i.e., methyl bromide reacts fastest via SN2, tert-butyl bromide reacts by far the slowest, essentially not proceeding via SN2 at all in practice).

This trend directly reflects increasing steric hindrance around the reactive carbon, moving from a completely unhindered methyl group, through increasingly substituted (primary, secondary, tertiary) carbons — the progressively greater steric bulk increasingly blocks the necessary backside nucleophilic approach required specifically for the SN2 mechanism.

Q4. Describe, with a balanced equation, the preparation of chloroform (trichloromethane) from acetone (or ethanol) using bleaching powder, and name this classic reaction.
5 Marks
Answer

This is known as the haloform reaction. Using ethanol as the starting material with bleaching powder (calcium oxychloride, a source of $Cl_2$/hypochlorite):

Step 1: Ethanol is first oxidised to acetaldehyde, then further chlorinated at the alpha carbon (three times) to give trichloroacetaldehyde.

Step 2: The trichloroacetaldehyde intermediate is then cleaved by hydroxide ion (from the bleaching powder mixture), releasing chloroform and a formate salt.

Overall equation: $CH_3CH_2OH + 4Cl_2 \rightarrow CCl_3CHO + 5HCl$ (formation of the trichloro intermediate), followed by cleavage: $CCl_3CHO + Ca(OH)_2 \rightarrow CHCl_3 + (HCOO)_2Ca$ (simplified representation) — the overall haloform reaction converts a methyl ketone or specific alcohol (with an adjacent $CH_3-CH(OH)-$ or $CH_3-CO-$ group) into a haloform (like chloroform) plus a carboxylic acid salt.

Q5. Explain, with reasoning, why (R)-2-butanol reacting with $SOCl_2$ (via the $S_N i$ mechanism) gives predominantly retention of configuration, unlike typical SN1 or SN2 reactions.
5 Marks
Answer

The reaction of an alcohol with thionyl chloride ($SOCl_2$) proceeds via a special mechanism called $S_Ni$ (substitution nucleophilic internal), which is mechanistically distinct from both standard SN1 and SN2.

In this mechanism, the oxygen of the alcohol first attacks sulfur, forming a chlorosulfite ester intermediate, releasing HCl. This intermediate then decomposes intramolecularly — the departing $-OSOCl$ group and the incoming chloride (released internally, from the same molecule, without fully diffusing away) remain positioned on the same face of the molecule during this internal ion-pair collapse, rather than the chloride approaching from the opposite face as would happen in standard SN2.

Since the chloride effectively delivers itself from the same side that the original leaving group departed from (rather than from the opposite side, as in true SN2), this specific $S_Ni$ pathway results in overall retention of configuration at the stereocentre, distinctly different from the inversion expected in typical SN2 reactions.

Q6. Explain, with the mechanism, why treatment of an alkyl halide with alcoholic KOH tends to favour elimination (forming an alkene), while aqueous KOH favours substitution.
5 Marks
Answer

In an alcoholic solvent (like ethanol), hydroxide ion is comparatively poorly solvated and behaves as a strong, relatively "naked" base — under these conditions, it preferentially removes a proton from a carbon adjacent (beta) to the leaving group, initiating an elimination (E1 or E2) reaction pathway that forms an alkene, rather than attacking the carbon bearing the leaving group directly.

In an aqueous solvent (water), hydroxide ion is much more heavily and effectively solvated (surrounded by hydrogen-bonded water molecules), significantly reducing its basicity/reactivity as a proton-abstracting base, while its nucleophilicity (ability to attack an electrophilic carbon directly) is comparatively less affected — under these aqueous conditions, hydroxide preferentially acts as a nucleophile, favouring the substitution pathway (SN1 or SN2) over elimination.

This differing balance between hydroxide's basic (elimination-promoting) versus nucleophilic (substitution-promoting) character, depending specifically on the solvent used, is precisely why alcoholic KOH favours elimination while aqueous KOH favours substitution for the same alkyl halide substrate.

Q7. Explain, using resonance structures, why the C-Cl bond length in chlorobenzene (approximately 169 pm) is shorter than the C-Cl bond length in a typical alkyl chloride like methyl chloride (approximately 178 pm).
5 Marks
Answer

In chlorobenzene, one of chlorine's lone pairs of electrons can be delocalised into the adjacent aromatic ring's pi system through resonance — this can be represented by additional resonance structures where the chlorine forms a formal double bond to the ring carbon, with the corresponding negative charge shifted onto specific ring carbon positions (ortho and para).

This resonance delocalisation gives the C-Cl bond significant partial double-bond character, in addition to its underlying sigma bond — since double bonds are inherently shorter than single bonds, this partial double-bond character shortens the overall C-Cl bond in chlorobenzene compared to the purely single-bonded C-Cl bond present in methyl chloride, where no such resonance delocalisation with an adjacent pi system is possible.

Q8. Predict and explain the major product formed when 2-bromobutane reacts with sodium ethoxide (a strong, moderately bulky base) via the E2 mechanism, applying Zaitsev's rule.
5 Marks
Answer

2-Bromobutane ($CH_3-CHBr-CH_2-CH_3$) has two different sets of beta-hydrogens available for E2 elimination — those on C1 (giving but-1-ene upon elimination) and those on C3 (giving but-2-ene upon elimination).

According to Zaitsev's rule, elimination reactions preferentially form the more highly substituted (generally more stable) alkene as the major product, when a genuine choice between regiochemical outcomes exists.

Major product: But-2-ene ($CH_3-CH=CH-CH_3$), since this is a disubstituted alkene (more stable, more substituted double bond), compared to the monosubstituted but-1-ene ($CH_2=CH-CH_2-CH_3$) that would form from elimination towards the terminal carbon — but-2-ene forms as the predominant product under standard E2 conditions with a moderately sized base like ethoxide.

Q9. Explain, with a mechanistic outline, the Sandmeyer reaction for converting an aromatic diazonium salt into an aryl halide.
5 Marks
Answer

The Sandmeyer reaction converts an aryl diazonium salt (formed from an aromatic primary amine via diazotisation) into an aryl halide (specifically chloride or bromide), using the corresponding cuprous halide as a catalyst.

Reaction: $ArN_2^+ + CuCl \rightarrow ArCl + N_2 + Cu^+$ (regenerated, continuing the catalytic cycle)

The reaction proceeds via a radical-type mechanism, where the copper(I) catalyst facilitates the loss of nitrogen gas ($N_2$) from the diazonium salt, generating an aryl radical intermediate, which then combines with the halide from the copper(I) halide catalyst to form the final aryl halide product — this reaction is particularly valuable synthetically, since direct electrophilic halogenation of certain aromatic rings can be difficult to control regioselectively, while the Sandmeyer reaction proceeds specifically at the exact position where the original amino/diazonium group was located.

Q10. Explain why methyl chloride, despite being a primary halide with essentially no steric hindrance, cannot react via the SN1 mechanism to any significant extent.
5 Marks
Answer

The SN1 mechanism critically depends on the formation of a reasonably stable carbocation intermediate in its rate-determining first step.

If methyl chloride were to attempt ionisation via SN1, it would need to form a methyl carbocation ($CH_3^+$) — this represents the least stable possible type of carbocation, since it has no adjacent alkyl groups whatsoever to provide any hyperconjugative or inductive stabilisation of the resulting positive charge.

Since this methyl carbocation intermediate would be prohibitively unstable and high in energy, the SN1 pathway for methyl chloride has an extremely high activation energy barrier, making it essentially inaccessible under normal reaction conditions — methyl chloride instead reacts exclusively (and very readily, given its minimal steric hindrance) via the SN2 mechanism, precisely because the SN1 pathway is simply not energetically viable for this substrate.

📚 Continue Studying