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Given below are two statements:
Statement (I): Benzyl chloride reacts faster in $S_N1$ mechanism than ethyl chloride.
Statement (II): Ethyl carbocation intermediate is less stabilized by hyperconjugation than benzyl carbocation by resonance.
In the light of the above statements, choose the correct answer from the options given below:
Show answer and solution
Answer: 1
Key idea: SN1 mechanism and carbocation stability - Resonance vs Hyperconjugation
In $S_N1$ mechanism, the rate-determining step is carbocation formation, so reaction rate depends on carbocation stability
Benzyl chloride ($C_6H_5CH_2Cl$) forms benzyl carbocation ($C_6H_5CH_2^+$) which is stabilized by resonance with the benzene ring
Ethyl chloride ($CH_3CH_2Cl$) forms ethyl carbocation ($CH_3CH_2^+$), a primary carbocation stabilized only by weak hyperconjugation
Resonance stabilization is much stronger than hyperconjugation, making benzyl carbocation more stable than ethyl carbocation
Since benzyl carbocation is more stable, benzyl chloride reacts faster in $S_N1$ mechanism - Statement I is TRUE
Statement II correctly explains that ethyl carbocation (hyperconjugation) is less stabilized than benzyl carbocation (resonance) - Statement II is TRUE
Both statements are true and Statement II correctly explains Statement I
Correct statements regarding alkyl halides $(R-X)$ among the following are:
A. Alcohol being less polar solvent as compared to water, alcoholic KOH favours elimination reaction with $R-X$.
B. Order of reactivity towards $S_N1$ mechanism is $C_6H_5-CH_2-Cl > C_6H_5-CHCl-C_6H_5$.
C. Non substituted aryl halides exhibit properties similar to alkyl halides.
D. Vinyl chloride is an example of haloalkene and allyl chloride is an example of haloalkyne.
E. $R-Cl$ can be prepared by reacting $R-OH$ with $SOCl_2$ but $Ar-Cl$ cannot be prepared by reacting $Ar-OH$ with $SOCl_2$.
Choose the correct answer from the options given below:
(1) A, B and C Only
(2) B and D Only
(3) A and E Only
(4) D and E Only
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Answer: Option (3)
Key idea: Reactivity and properties of alkyl halides, aryl halides, and their derivatives; SN1 mechanism; elimination vs substitution reactions
Analyze Statement A: Alcoholic KOH favors elimination because alcohol (less polar) enhances base character of $OH^-$ over nucleophilic character. **Statement A is CORRECT**.
Analyze Statement B: For $S_N1$, carbocation stability matters. $C_6H_5CHClC_6H_5$ forms diphenylmethyl cation (2 phenyl groups) which is MORE stable than benzyl cation from $C_6H_5CH_2Cl$ (1 phenyl group). Given order is wrong. **Statement B is INCORRECT**.
Analyze Statement C: Aryl halides have $C-X$ bond with partial double bond character due to resonance, making them much less reactive than alkyl halides. They don't undergo normal $S_N1$ or $S_N2$ reactions. **Statement C is INCORRECT**.
Analyze Statement D: Vinyl chloride $CH_2=CHCl$ is a haloalkene (correct), but allyl chloride $CH_2=CHCH_2Cl$ is also a haloalkene, NOT a haloalkyne. **Statement D is INCORRECT**.
Analyze Statement E: $R-OH + SOCl_2 \rightarrow R-Cl$ works for alkyl alcohols. However, $Ar-OH + SOCl_2$ does NOT give $Ar-Cl$ because phenolic $C-O$ bond is too strong. **Statement E is CORRECT**.
Only statements A and E are correct, so the answer is option (3).
Identify compounds A and E in the following reaction sequence.
[Figure: Reaction scheme starting with 4-nitro-ethylbenzene, going through steps: (1) Br$_2$/AlBr$_3$ → A, (2) Sn/HCl → B, (3) NaNO$_2$/HCl at 273-278 K → C, (4) C$_2$H$_5$OH → D, (5) (i) KMnO$_4$/KOH then (ii) H$_3$O$^+$ → E]
(1) [Figure: A = 4-bromo-3-nitro-ethylbenzene, E = 3-bromo-benzoic acid]
Show answer and solution
Answer: Option (1)
Key idea: Electrophilic aromatic substitution, nitro group reduction, diazotization, and oxidation of alkyl side chains
Step 1: Bromination with $Br_2/AlBr_3$ on 3-nitro-ethylbenzene gives **A = 4-bromo-3-nitro-ethylbenzene** (Br adds at position para to ethyl and meta to nitro)
Step 2: Reduction with $Sn/HCl$ converts $-NO_2$ to $-NH_2$ giving B = 4-bromo-3-amino-ethylbenzene
Step 3: Diazotization with $NaNO_2/HCl$ at 273-278 K converts $-NH_2$ to $-N_2^+Cl^-$ giving C = diazonium salt
Step 4: Treatment with $C_2H_5OH$ replaces diazonium group with H, giving D = 3-bromo-ethylbenzene
Step 5: Oxidation with $KMnO_4/KOH$ then $H_3O^+$ oxidizes ethyl group to carboxylic acid giving **E = 3-bromo-benzoic acid**
Product C of the following reaction sequence will be
[Figure: Reaction scheme showing aniline ($NH_2$ attached to benzene ring) reacting with $Br_2$/water to give A (Major Product), then A reacting with $NaNO_2$ / HCl at 273-278 K to give B, then B reacting with (i) $HBF_4$ (ii) $NaNO_2$, Cu, $\Delta$ to give C]
(1) 1-Bromo-4-nitrobenzene
(2) 1, 3, 5-Tribromo-2-nitrobenzene
(3) 4-Bromo-1-nitrobenzene
(4) 1, 3, 5-Tribromobenzene
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Answer under review. Our two independent solutions did not agree, so we are not showing an answer yet. Check it against NTA's final answer key.
Key idea: Electrophilic aromatic substitution, diazotization, and deamination reactions of aniline
Match the **LIST-I** with **LIST-II**
| List-I | List-II |
|--------|----------|
| **Reaction** | **Mechanism** |
| A. Williamson Synthesis | I. Electrophilic addition |
| B. Friedel Craft Reaction | II. Free radical substitution |
| C. Bromination of vinyl benzene | III. Nucleophilic substitution |
| D. Chlorination of toluene in light | IV. Electrophilic substitution |
Choose the *correct* answer from the options given below:
Show answer and solution
Answer: A-III, B-IV, C-I, D-II
Key idea: Reaction Mechanisms in Organic Chemistry
**Bromination of vinyl benzene (C):** $Br_2$ adds across $C=C$ double bond in styrene → **Electrophilic addition (I)**
**Chlorination of toluene in light (D):** UV light causes homolytic cleavage of $Cl_2$ → radical chain reaction at methyl group → **Free radical substitution (II)**
[Figure: Benzene ring with vinyl group attached (X)] undergoes reactions: (i) $Br_2 / CHCl_3$ (ii) $NaNH_2$ excess (iii) $CH_3I$ (iv) $H_2, Na/NH_3$ $(l)$ $\longrightarrow$ Major Product (Y)
Compound (X) is subjected to the sequence of reactions as shown above. Molar mass of the major product (Y) formed is __________ $g mol^{-1}$.
(Given molar mass in $g mol^{-1}$ C:12, H: 1, O: 16)
(1) 90
(2) 118
(3) 160
(4) 125
Show answer and solution
Answer: Option (2)
Key idea: Reactions of alkenes and alkynes: bromination, dehydrohalogenation, alkylation of terminal alkynes, and dissolving metal reduction
Starting compound X is styrene: $C_6H_5-CH=CH_2$
Step (i): Bromination with $Br_2/CHCl_3$ gives $C_6H_5-CHBr-CH_2Br$
Step (ii): Excess $NaNH_2$ causes double dehydrohalogenation to form phenylacetylene: $C_6H_5-C≡CH$
Step (iii): $CH_3I$ alkylates the terminal alkyne to give $C_6H_5-C≡C-CH_3$
Step (iv): $Na/NH_3(l)$ reduces the alkyne to trans-alkene: $C_6H_5-CH=CH-CH_3$ (Product Y)
Calculate molar mass of $C_9H_{10}$: $9(12) + 10(1) = 118$ g/mol
An optically active alkyl bromide $C_4H_9Br$, reacts with ethanolic KOH to form major compound [A] which reacts with bromine to give compound [B]. Compound [B] reacts with ethanolic KOH and sodamide to give compound [C]. One molecule of water adds to compound [C] on warming with mercuric sulphate and dilute sulphuric acid at 333 K to form compound [D]. The functional group in compound D will be confirmed by:
(1) Haloform test
(2) Lucas test
(3) Silver mirror test
(4) Benedict test
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Answer: Option (1)
Key idea: Identification of functional groups through chemical tests; Haloform test for methyl ketones
Optically active $C_4H_9Br$ (2-bromobutane) undergoes E2 elimination with ethanolic KOH to give but-2-ene [A]: $CH_3CH=CHCH_3$
[A] reacts with $Br_2$ to form 2,3-dibromobutane [B]: $CH_3CHBr-CHBrCH_3$
[B] undergoes double elimination with ethanolic KOH and sodamide to form but-2-yne [C]: $CH_3C≡CCH_3$
[C] undergoes hydration with $HgSO_4/H_2SO_4$ at 333 K to form butan-2-one [D]: $CH_3COCH_2CH_3$ (a methyl ketone)
Methyl ketones give positive **Haloform test** with $I_2/NaOH$, producing yellow $CHI_3$ precipitate
n-Butane on monochlorination under photochemical condition gives an optically active compound "P". "P" on further chlorination gives dichloro compounds.
The number of dichloro compounds obtained (ignore stereoisomers) is :
(1) 3
(2) 4
(3) 5
(4) 6
Show answer and solution
Answer: Option (2)
Key idea: Free radical halogenation, optical activity, and structural isomerism in haloalkanes
Step 1: Identify that n-butane ($CH_3-CH_2-CH_2-CH_3$) gives an optically active monochlorinated product P, which must have a chiral center.
Step 2: Determine that P is 2-chlorobutane ($CH_3-CHCl-CH_2-CH_3$) since it has a chiral carbon at C-2 with four different groups.
Step 3: Identify all possible positions for the second chlorination: C-1 (terminal $CH_3$), C-2 (already has Cl), C-3 ($CH_2$), and C-4 (terminal $CH_3$).
Step 4: List the dichloro products: 1,2-dichlorobutane ($ClCH_2-CHCl-CH_2-CH_3$), 2,2-dichlorobutane ($CH_3-CCl_2-CH_2-CH_3$), 2,3-dichlorobutane ($CH_3-CHCl-CHCl-CH_3$), and 1,3-dichlorobutane ($ClCH_2-CH_2-CHCl-CH_3$).
Step 5: Count the total number of distinct dichloro compounds (ignoring stereoisomers) = 4.
Given below are two statements :
**Statement I :** Due to increase in van der Waals forces, the order of boiling points is $CH_3CH_2CH_2I > CH_3CH_2I > CH_3I$.
[Figure: Two chemical structures shown. First structure shows para-dichlorobenzene (benzene ring with Cl atoms at positions 1 and 4). Second structure shows ortho-dichlorobenzene (benzene ring with Cl atoms at positions 1 and 2).]
**Statement II :** As [para-dichlorobenzene] is more symmetric, its melting point is higher than [ortho-dichlorobenzene], however its boiling point is lower than [ortho-dichlorobenzene]
In the light of the above statements, choose the correct answer from the options given below :
(1) Both Statement I and Statement II are true
(2) Both Statement I and Statement II are false
(3) Statement I is true but Statement II is false
(4) Statement I is false but Statement II is true
Show answer and solution
Answer: Option (1)
Key idea: Boiling and melting points of haloalkanes and haloarenes; van der Waals forces; molecular symmetry and crystal packing
**Statement I:** As chain length increases in alkyl iodides ($CH_3I$ to $CH_3CH_2CH_2I$), molecular mass and surface area increase, leading to stronger van der Waals forces
The boiling point order $CH_3CH_2CH_2I > CH_3CH_2I > CH_3I$ is correct (102.5°C > 72.3°C > 42.4°C). **Statement I is TRUE**
**Statement II:** Para-dichlorobenzene is highly symmetric, leading to better crystal packing and higher melting point (53.1°C) than ortho isomer (-17.0°C)
Para-dichlorobenzene has zero dipole moment (symmetry), while ortho has dipole moment (~2.5 D)
Ortho-dichlorobenzene has stronger dipole-dipole interactions, giving higher boiling point (180°C) than para (174°C). **Statement II is TRUE**
Both statements are correct, so answer is option (1)
FAQs
Frequently asked questions
How many questions come from Haloalkanes and Haloarenes in JEE Main?
In the 9 shifts of JEE Main 2026 Session 2 we found 9 questions on it, about 1.0 per shift.
Where can I practise Haloalkanes and Haloarenes PYQs for free?
On this page, with solutions. For more, use NTA Abhyas and our free mock tests.
