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Haloalkanes and Haloarenes Set-3 MCQ Online Practice Test | Class 12 | Quiz Tweek

Haloalkanes and Haloarenes Set-3

Practice Important MCQs for Haloalkanes and Haloarenes — Chemistry, Class 12.

The chapter “Haloalkanes and Haloarenes” is crucial because it builds the core understanding of reaction mechanisms (SN1/SN2/SNAr/E1/E2/benzyne), stereochemical outcomes, and how structure and substituents control reactivity and product formation. These concepts are frequently asked in CBSE boards and are also central to competitive exams, where reasoning-based questions on carbocation/transition-state stability and steric + electronic effects are common.

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Q1. Arrange the following haloalkanes in order of increasing rate of bimolecular nucleophilic substitution (SN2) with iodide ion ( $I^-$ ) in acetone at 25°C: $\mathrm{CH_3Cl},\ \mathrm{CH_3CH_2Cl},\ \mathrm{(CH_3)_2CHCl},\ \mathrm{(CH_3)_3CCl}$ .

(A)

$\mathrm{(CH_3)_3CCl < (CH_3)_2CHCl < CH_3Cl < CH_3CH_2Cl}$

(B)

$\mathrm{(CH_3)_3CCl < CH_3CH_2Cl < (CH_3)_2CHCl < CH_3Cl}$

(C)

$\mathrm{(CH_3)_3CCl < (CH_3)_2CHCl < CH_3CH_2Cl < CH_3Cl}$

(D)

$\mathrm{(CH_3)_3CCl < CH_3CH_2Cl < CH_3Cl < (CH_3)_2CHCl}$

Q2. Which of the following aryl chlorides will undergo nucleophilic aromatic substitution by methoxide ( $\mathrm{CH_3O^-}$ in methanol) most rapidly via the addition–elimination (Meisenheimer) pathway?

(A)

$\mathrm{p\text{-}ClC_6H_4NO_2}$ (para‑chloronitrobenzene)

(B)

$\mathrm{C_6H_5Cl}$ (chlorobenzene)

(C)

$\text{1-chloro-2,4-dimethylbenzene}$

(D)

$\text{1-chloro-2,4-dimethoxybenzene}$

Q3. Predict the correct order of rates for solvolysis (hydrolysis in aqueous ethanol) at 25°C (fastest → slowest) of the following bromides: $\mathrm{C_6H_5CH_2Br}$ (benzyl bromide), $\mathrm{(CH_3)_3CBr}$ (tert‑butyl bromide), $\mathrm{CH_3CH_2CH_2Br}$ (n‑propyl bromide), $\mathrm{C_6H_5Br}$ (bromobenzene). Choose the correct sequence.

(A)

$\mathrm{(CH_3)_3CBr > CH_3CH_2CH_2Br > C_6H_5CH_2Br > C_6H_5Br}$

(B)

$\mathrm{(CH_3)_3CBr > C_6H_5CH_2Br > CH_3CH_2CH_2Br > C_6H_5Br}$

(C)

$\mathrm{(CH_3)_3CBr > C_6H_5CH_2Br > C_6H_5Br > CH_3CH_2CH_2Br}$

(D)

$\mathrm{C_6H_5CH_2Br > (CH_3)_3CBr > CH_3CH_2CH_2Br > C_6H_5Br}$

Q4. Solvolysis (ethanol) of (R)-2-bromo-1-phenylpropane gives substitution product with predominant retention of configuration at the stereocentre, whereas solvolysis of (R)-2-bromobutane gives a racemic mixture. Which statement best explains this observation?

(A)

The benzylic substrate forms a fully planar classical carbocation stabilized by resonance, and planar carbocations always give retention on solvolysis.

(B)

The adjacent phenyl ring participates (anchimeric assistance) to form a bridged phenonium-type intermediate; nucleophilic attack on this non-classical intermediate leads predominantly to retention of configuration.

(C)

The benzylic substrate reacts exclusively by an SN2 pathway (backside attack) leading to inversion that appears as retention due to a subsequent rapid second inversion.

(D)

The phenyl group converts the pathway to an E1 elimination followed by electrophilic addition of solvent across the double bond, which enforces retention of stereochemistry.

Q5. When $\mathrm{1\text{-}chloro\text{-}2,4\text{-}dinitrobenzene}$ and $\mathrm{chlorobenzene}$ are each treated separately with excess $\mathrm{NaNH_2}$ in liquid ammonia at high temperature, which of the following correctly describes their typical courses of reaction?

(A)

$\mathrm{1\text{-}chloro\text{-}2,4\text{-}dinitrobenzene}$ undergoes nucleophilic aromatic substitution by addition–elimination (single clean substitution at Cl); $\mathrm{chlorobenzene}$ undergoes reaction via a benzyne intermediate giving a mixture of ortho/para substituted products.

(B)

$\mathrm{chlorobenzene}$ undergoes addition–elimination giving a single substitution product, while $\mathrm{1\text{-}chloro\text{-}2,4\text{-}dinitrobenzene}$ forms benzyne and gives isomeric mixtures.

(C)

Both substrates react by the benzyne mechanism under these conditions and give similar isomeric mixtures.

(D)

Both substrates undergo addition–elimination (Meisenheimer) smoothly to give single, identical positional products.

Q6. Among the following alkyl halides, which will undergo the fastest bimolecular nucleophilic substitution ( $S_N2$ ) with $\,\mathrm{CN^-}$ in dry dimethyl sulfoxide ( $\mathrm{DMSO}$ )?

(A)

$\mathrm{CH_3Cl}$

(B)

$\mathrm{CH_3Br}$

(C)

$\mathrm{CH_3I}$

(D)

$\mathrm{CH_3F}$

Q7. In aqueous ethanol (a polar protic medium), which of the following haloalkanes will hydrolyse to the corresponding alcohol most rapidly via an $S_N1$ pathway?

(A)

$\mathrm{C_6H_5CH_2Cl}$ (benzyl chloride)

(B)

$\mathrm{(CH_3)_3CCl}$ (tert‑butyl chloride)

(C)

$\mathrm{CH_3CH_2CH_2CH_2Cl}$ (1‑chlorobutane)

(D)

$\mathrm{C_6H_5Cl}$ (chlorobenzene)

Q8. Optically active $(R)$ ‑2‑bromobutane is treated with excess sodium ethoxide ( $\mathrm{NaOEt}$ ) in ethanol ( $\mathrm{EtOH}$ ) at 25°C. The isolated substitution product 2‑ethoxybutane is obtained as 75% (S) and 15% (R); elimination (2‑butene) is 10%. Which mechanistic explanation best accounts for this stereochemical distribution?

(A)

Pure $S_N2$ substitution (back‑side attack) with competing $E2$ elimination only

(B)

Pure $S_N1$ (carbocation) mechanism giving racemate plus elimination

(C)

Predominant $E2$ followed by nucleophilic addition to the alkene (anti‑Markovnikov)

(D)

Predominant $S_N2$ (inversion) with a minor $S_N1$ pathway causing partial racemisation and a small $E2$ contribution

Q9. Assertion (A): $p$ ‑Nitrochlorobenzene undergoes nucleophilic aromatic substitution with methoxide ( $\mathrm{MeO^-}$ ) at room temperature to give $p$ ‑nitroanisole via an addition–elimination (Meisenheimer) mechanism.
Reason (R): The strongly electron‑withdrawing nitro group at the para position stabilises the anionic Meisenheimer intermediate by delocalisation of negative charge, lowering the activation energy for the nucleophilic addition step.

(A)

A is true, R is false.

(B)

Both A and R are true and R correctly explains A.

(C)

Both A and R are true but R does not explain A.

(D)

A is false, R is true.

Q10. Chlorobenzene does not undergo nucleophilic substitution with hydroxide ion under normal conditions, yet treatment with $\mathrm{NaNH_2}$ at high temperature followed by acid workup can afford aniline. Which option best explains this difference in reactivity?

(A)

Under strongly basic, high‑temperature conditions $\mathrm{NaNH_2}$ induces deprotonation at an ortho/adjacent carbon and elimination of $\mathrm{Cl^-}$ to generate a benzyne intermediate; nucleophilic addition of $\mathrm{NH_2^-}$ to benzyne followed by protonation yields aniline.

(B)

$\mathrm{NaNH_2}$ coordinates to chlorine making $\mathrm{Cl}$ a much better leaving group so direct $S_N2$ on the aryl carbon occurs.

(C)

High temperature converts the aromatic ring into a non‑aromatic carbocation (aryl cation), which is then trapped by $\mathrm{NH_3}$ to give aniline.

(D)

Chlorobenzene reacts slowly with $\mathrm{OH^-}$ at room temperature; $\mathrm{NaNH_2}$ simply speeds up the same addition–elimination pathway that $\mathrm{OH^-}$ would follow.

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