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2 April 2026 Shift 1, Q65 · Q65 · MCQ · Organic Chemistry - Some Basic Principles and Techniques · Medium
In IUPAC nomenclature, the correct order of decreasing priority of functional group is:
Check option (4): $-CONH_2, -CHO, -CN, -NH_2, -C\equiv C-$ - Incorrect because nitrile ($-CN$) has higher priority than aldehyde ($-CHO$)
Option (3) is the correct answer
2 April 2026 Shift 1, Q66 · Q66 · MCQ · Organic Chemistry - Some Basic Principles and Techniques · Medium
For the given molecule, "$x$", the preferred site for the attack of the electrophile is: [Figure: A molecule with a carbonyl group (O) connecting two benzene rings through a nitrogen atom (N-H). The left benzene ring has positions labeled 'r' and 'p', and the right benzene ring has positions labeled 'u' and 's']
(1) Predominantly at "r"
(2) "r" and "u"
(3) "p" and "s"
(4) Predominantly at "u"
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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 - Activating and Deactivating Groups in Amides
Given below are two statements:
**Statement I:** In $O_2N$-[benzene ring]-$\stackrel{\oplus}{CH}$-[benzene ring]-$OCH_3$, the carbocation is stabilised by $+R$ effect of $-OCH_3$ group.
**Statement II:** In $O_2N$-[benzene ring]-$\stackrel{\ominus}{CH}$-[benzene ring]-$OCH_3$, the carbanion is stabilised by $-R$ effect of $-NO_2$ group.
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
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Answer: Option (1)
Key idea: Resonance effects on carbocation and carbanion stability; +R and -R effects of substituents
Analyze Statement I: The structure has a carbocation ($CH^+$) between two benzene rings, one with $-NO_2$ and one with $-OCH_3$
For carbocation stability: Need electron donation. The $-OCH_3$ group has $+R$ effect (electron donating through resonance)
The lone pair on oxygen delocalizes through the benzene ring to the carbocation center, stabilizing it. **Statement I is TRUE**
Analyze Statement II: The structure has a carbanion ($CH^-$) between two benzene rings, one with $-NO_2$ and one with $-OCH_3$
For carbanion stability: Need electron withdrawal. The $-NO_2$ group has $-R$ effect (electron withdrawing through resonance)
The negative charge delocalizes into the benzene ring and further into the $-NO_2$ group, stabilizing the carbanion. **Statement II is TRUE**
Both statements correctly identify the stabilizing resonance effects
4 April 2026 Shift 1, Q63 · Q63 · MCQ · Organic Chemistry - Some Basic Principles and Techniques · Medium
Increasing order of electron withdrawing power of following functional groups is:
a. $-CN$
b. $-COOH$
c. $-NO_2$
d. $-I$
(1) $c < b < d < a$
(2) $c < a < b < d$
(3) $d < b < a < c$
(4) $a < b < c < d$
Show answer and solution
Answer: Option (3)
Key idea: Electron Withdrawing Groups - Inductive and Resonance Effects
Identify that electron withdrawing power depends on both inductive (-I) and resonance (-R) effects
Analyze $-COOH$ (b): Moderate EWG - has -I effect but oxygen lone pairs provide +R effect that reduces withdrawing power
Analyze $-CN$ (a): Strong EWG - both strong -I effect (electronegative N) and strong -R effect
Analyze $-NO_2$ (c): Strongest EWG - very strong -I and -R effects due to positive N and two electronegative O atoms
Arrange in increasing order: $d < b < a < c$ or $-I < -COOH < -CN < -NO_2$
4 April 2026 Shift 1, Q65 · Q65 · MCQ · Organic Chemistry - Some Basic Principles and Techniques · Medium
Match the LIST-I with LIST-II
| List-I | List-II |
| Name of reaction | Reagent or catalyst used |
| A. Finkelstein reaction | I. $\text{SbF}_3$ |
| B. Swarts reaction | II. Na, dry ether |
| C. Sandmeyer's reaction | III. NaI |
| D. Fittig reaction | IV. $\text{Cu}_2\text{Cl}_2$ |
Choose the correct answer from the options given below:
(1) A-I, B-IV, C-III, D-II
(2) A-III, B-I, C-IV, D-II
(3) A-IV, B-II, C-I, D-III
(4) A-I, B-III, C-II, D-IV
Show answer and solution
Answer: Option (2)
Key idea: Named Organic Reactions and Their Reagents
Finkelstein reaction: Halogen exchange using $\text{NaI}$ in acetone to convert alkyl chlorides/bromides to iodides. A → III
Swarts reaction: Preparation of alkyl fluorides from alkyl halides using $\text{SbF}_3$ or other metallic fluorides. B → I
Sandmeyer's reaction: Conversion of aryl diazonium salts to aryl halides using copper(I) salts like $\text{Cu}_2\text{Cl}_2$. C → IV
Fittig reaction: Coupling of aryl halides using Na metal in dry ether to form biaryls. D → II
Match: A-III, B-I, C-IV, D-II corresponds to option (2)
Consider the following sequence of reactions to give the major product (X)
[Figure: Benzene ring with three reactions: (i) CH₃Cl / anhydrous AlCl₃, (ii) Cl₂ / FeCl₃, (iii) K₂Cr₂O₇ / H₂SO₄ leading to (X) Major Product]
P g of the major product (X) formed is reacted with NaHCO₃ solution to liberate a gas which occupied 11.2 dm³ at STP.
P = _____ g.
(Given molar mass in g mol⁻¹ H:1, C:12, O:16, Cl:35.5)
Show answer and solution
Answer: 78.25
Key idea: Friedel-Crafts alkylation, electrophilic aromatic substitution, oxidation of alkyl benzenes, and stoichiometry of carboxylic acid reactions
Benzene undergoes Friedel-Crafts alkylation with $\text{CH}_3\text{Cl}/\text{AlCl}_3$ to form toluene
Toluene undergoes electrophilic chlorination with $\text{Cl}_2/\text{FeCl}_3$ to give $p$-chlorotoluene (major product, since $-\text{CH}_3$ is $o/p$ directing)
$p$-Chlorotoluene is oxidized by $\text{K}_2\text{Cr}_2\text{O}_7/\text{H}_2\text{SO}_4$ to give $p$-chlorobenzoic acid (X): $\text{C}_7\text{H}_5\text{ClO}_2$, molar mass $= 156.5$ g/mol
The carboxylic acid reacts with $\text{NaHCO}_3$: $\text{C}_7\text{H}_5\text{ClO}_2 + \text{NaHCO}_3 \rightarrow \text{C}_7\text{H}_4\text{ClO}_2\text{Na} + \text{H}_2\text{O} + \text{CO}_2$
Moles of $\text{CO}_2$ liberated $= \frac{11.2}{22.4} = 0.5$ mol, so moles of X $= 0.5$ mol
2.0 g of a bromo hydrocarbon (X) was subjected to Carius analysis, gave 3.36 g of AgBr. The percentage of carbon in the compound (X) is 26.7%. Total number of carbon atoms in the empirical formula for compound (X) is _____.
(Given molar mass in g mol⁻¹ H:1, C:12, Br : 80, Ag : 108)
Show answer and solution
Answer: 5
Key idea: Empirical Formula Determination using Carius Method
Calculate moles of Br: From 3.36 g AgBr, moles = $\frac{3.36}{188} = 0.01787$ mol, so mass of Br = $0.01787 \times 80 = 1.43$ g
Find % of Br in compound X: $\frac{1.43}{2.0} \times 100 = 71.5\%$
Calculate % of H: Since compound has only C, H, Br: $\% H = 100 - 26.7 - 71.5 = 1.8\%$
Find moles of each element in 2.0 g sample: C = $\frac{0.534}{12} = 0.0445$ mol, H = $\frac{0.036}{1} = 0.036$ mol, Br = $0.01787$ mol
Determine mole ratio by dividing by smallest (0.01787): C:H:Br = 2.5:2:1
Multiply by 2 to get whole numbers: Empirical formula = $C_5H_4Br_2$
4 April 2026 Shift 2, Q61 · Q61 · MCQ · Organic Chemistry - Some Basic Principles and Techniques · Medium
Consider the following molecules/species:
[Figure: Three structures labeled (x), (y), (z). Structure (x) shows a seven-membered ring with one C=O group. Structure (y) shows a molecule with CH₃ groups and C=O. Structure (z) shows CH₃-C(=O)-O⁻]
The correct order of carbon – oxygen double bond length is:
(1) $x > y > z$
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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: C=O bond length depends on resonance and bond order. Resonance delocalizes electrons, reducing double bond character and increasing bond length. Carboxylates have maximum resonance (bond order ~1.5), ketones have pure C=O bonds (bond order ~2).
The following structures are
[Figure: Two stereochemical structures showing a carbon with Br, Me, Cl, and CH3 groups in different spatial arrangements]
(1) enantiomers.
(2) identical molecules.
(3) diastereomers.
(4) meso compounds.
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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: Stereoisomerism - Enantiomers vs Diastereomers
4 April 2026 Shift 2, Q67 · Q67 · MCQ · Organic Chemistry - Some Basic Principles and Techniques · Medium
The strongest conjugate acid will result from:
[Figure: Shows aniline structure with NH₂ group attached to benzene ring]
(1) 6952781510
Show answer and solution
Answer: Option (1)
Key idea: Basicity of amines and conjugate acid strength - The weakest base forms the strongest conjugate acid. Electron-withdrawing groups decrease basicity by reducing electron density on nitrogen.
Recall that conjugate acid strength is inversely related to base strength: weaker base → stronger conjugate acid
Identify that aniline's basicity is affected by substituents on the benzene ring
Electron-withdrawing groups (like $-NO_2$, $-CN$, $-COOH$) decrease electron density on nitrogen through resonance/inductive effects
This reduces the availability of the lone pair on nitrogen, making it a weaker base
The weakest base (aniline with strongest electron-withdrawing group) will form the strongest conjugate acid
Therefore, the compound that is the weakest base gives the strongest conjugate acid
Given below are two statements:
**Statement I:** On the basis of inductive effect, the order of stability of alkyl carbanions is $CH_3^- > CH_3-CH_2^- > (CH_3)_2CH^- > (CH_3)_3C^-$.
**Statement II:** Allyl and benzyl carbanions are more stabilised by inductive effect and not by resonance effect.
In the light of the above statements, choose the *correct* answer from the options given below
(1) Both Statement I and Statement II are correct
(2) Both Statement I and Statement II are incorrect
(3) Statement I is correct but Statement II is incorrect
(4) Statement I is incorrect but Statement II is correct
Show answer and solution
Answer: Option (3)
Key idea: Carbanion stability: inductive vs resonance effects
**Statement I:** Analyze carbanion stability based on inductive effect. Alkyl groups have $+I$ effect (electron-donating).
Since carbanions are negatively charged, electron-donating groups destabilize them by increasing electron density on the negative center.
Order: $CH_3^- > CH_3CH_2^- > (CH_3)_2CH^- > (CH_3)_3C^-$ (fewer alkyl groups = more stable). **Statement I is CORRECT**.
**Statement II:** Allyl carbanion ($CH_2=CH-CH_2^-$) has resonance: $CH_2=CH-CH_2^- \leftrightarrow {}^-CH_2-CH=CH_2$
Benzyl carbanion ($C_6H_5-CH_2^-$) also shows resonance with the benzene ring, delocalizing the negative charge.
Both are stabilized primarily by **resonance**, NOT inductive effect. **Statement II is INCORRECT**.
Answer: Statement I is correct but Statement II is incorrect.
One mole of phenol is treated with dilute HNO$_3$ at 298 K to give a mixture of products. The mixture is separated by steam distillation. The steam volatile compound (X) is separated. The increase in percentage of oxygen in (X) with respect to phenol is _______$\times 10^{-1}$%. (Given molar mass in g mol$^{-1}$ H:1, C:12, N:14, O:16)
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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: Nitration of phenol, steam distillation, and percentage composition calculation
5 April 2026 Shift 2, Q65 · Q65 · MCQ · Organic Chemistry - Some Basic Principles and Techniques · Medium
Complete the following reaction sequence and give the name of major product 'P'.
[Figure: Reaction scheme showing a nitrile group undergoing: (i) OH$^-$/H$_2$O/$\Delta$ (ii) H$_3$O$^+$ (iii) Cl$_2$/Red P (iv) H$_2$O to form P (Major product)]
(1) 2-Chloropropanoic acid
(2)
(3)
(4)
Show answer and solution
Answer: Option (1)
Key idea: Hell-Volhard-Zelinsky (HVZ) reaction for α-halogenation of carboxylic acids and nitrile hydrolysis
Start with propanenitrile ($CH_3CH_2CN$)
Step (i) & (ii): Basic hydrolysis followed by acidification converts nitrile to carboxylic acid: $CH_3CH_2CN \xrightarrow{OH^-/H_2O/\Delta} CH_3CH_2COO^- \xrightarrow{H_3O^+} CH_3CH_2COOH$
Step (iii): Hell-Volhard-Zelinsky reaction - $Cl_2$/Red P causes α-halogenation of carboxylic acid via acyl chloride intermediate
Red P + $Cl_2$ forms $PCl_3$, which converts $COOH$ to $COCl$ and enables enolization for α-chlorination
$R_F$ value for 2-methylpropene in a solvent system (Ethyl acetate + ether) is 0.42. 2-methylpropene is treated with dilute $H_2SO_4$ to give major organic product (X). $R_F$ value for (X) in the same solvent system under identical condition will be:
(1) 0.42
(2) 0.82
(3)
(4)
Show answer and solution
Answer: Option (2)
Key idea: Acid-catalyzed hydration of alkenes and Rf values in TLC based on polarity
Identify the reaction: 2-methylpropene $(CH_3)_2C=CH_2$ reacts with dilute $H_2SO_4$ (acid-catalyzed hydration)
Apply Markovnikov's rule: Water adds to form the more stable tertiary carbocation $(CH_3)_3C^+$
Product X is 2-methylpropan-2-ol (tert-butyl alcohol): $(CH_3)_3COH$
Compare polarity: The alcohol (X) is MORE polar than the alkene due to the $-OH$ group
In the polar solvent system (ethyl acetate + ether) with polar silica gel stationary phase, more polar compounds have better mobile phase solubility
Therefore, the more polar alcohol travels faster and has a HIGHER $R_F$ value: $R_F = 0.82$
Given below are two statements:
**Statement I:** 2,6-diethylcyclohexanone and 6-methyl-2-n-propylcyclohexanone are metamers.
**Statement II:** 2,2,6,6 - tetramethylcyclohexanone exhibits keto-enol tautomerism.
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 (3)
Key idea: Metamerism and Keto-Enol Tautomerism
Check Statement I: Compare molecular formulas of 2,6-diethylcyclohexanone ($C_{10}H_{18}O$) and 6-methyl-2-n-propylcyclohexanone ($C_{10}H_{18}O$)
Identify that both have same formula but different alkyl group distribution: two ethyl groups vs one methyl + one propyl group around the ring
Conclude Statement I is TRUE - these are metamers (differ in alkyl group arrangement with same molecular formula)
Check Statement II: Analyze 2,2,6,6-tetramethylcyclohexanone for $\alpha$-hydrogens needed for keto-enol tautomerism
Note that positions 2 and 6 (the $\alpha$-carbons) each have two methyl groups, leaving NO hydrogens available
Conclude Statement II is FALSE - no $\alpha$-hydrogen means no keto-enol tautomerism possible
Select option (3): Statement I is true but Statement II is false
6 April 2026 Shift 1, Q72 · Q72 · Numerical · Organic Chemistry - Some Basic Principles and Techniques · Medium
Consider the following sequence of reactions. [Figure: Benzene ring - N = N - NH - Benzene ring, followed by (i) HCl (ii) Aniline (in excess) (iii) Warm 40-45°C (iv) Cool, acetic acid, producing Yellow Product (X)] The percentage of nitrogen in the yellow product (X) formed is _______%. (Nearest Integer) (Given Molar mass in g mol$^{-1}$ H:1, C:12, N:14)
Show answer and solution
Answer: 21
Key idea: Rearrangement of diazoaminobenzene to p-aminoazobenzene and percentage composition calculation
Starting compound is diazoaminobenzene: $C_6H_5-N=N-NH-C_6H_5$
Treatment with HCl and excess aniline at 40-45°C causes rearrangement to form p-aminoazobenzene (yellow product X)
Structure of product X: $H_2N-C_6H_4-N=N-C_6H_5$ with molecular formula $C_{12}H_{11}N_3$
Percentage of nitrogen = $\frac{42}{197} \times 100 = 21.32\%$
Nearest integer = 21%
6 April 2026 Shift 2, Q63 · Q63 · Numerical · Organic Chemistry - Some Basic Principles and Techniques · Medium
Shown below is the structure of methyl acetate with three different $\alpha$, $\beta$ and $\gamma$ carbon - oxygen bonds.
[Figure: Structure showing $H_3C-C(=O)-O-CH_3$ with C=O bond labeled as $\alpha$, C-O(carbonyl) bond labeled as $\beta$, and O-CH3 bond labeled as $\gamma$]
The correct order of bond lengths of these bonds is:
Show answer and solution
Answer: α < β < γ
Key idea: Bond lengths in esters: resonance effects and hybridization
Identify the three bonds: $\alpha$ is C=O (carbonyl), $\beta$ is C-O (ester linkage), $\gamma$ is O-CH₃
The $\alpha$ bond is a C=O double bond, which is the shortest (~1.20 Å)
The $\beta$ bond has partial double bond character due to resonance (lone pair on O delocalizes into carbonyl), making it shorter than a normal single bond (~1.35 Å)
The $\gamma$ bond is a regular C-O single bond with no resonance effects, making it the longest (~1.43 Å)
Apply the rule: double bonds < resonance-stabilized single bonds < regular single bonds
Therefore, the correct order is: $\alpha < \beta < \gamma$
6 April 2026 Shift 2, Q72 · Q72 · Numerical · Organic Chemistry - Some Basic Principles and Techniques · Medium
An alkane (Y) requires 8 moles of oxygen for complete combustion and on chlorination with $Cl_2/h\nu$, (Y) gives only one monochlorinated product (Z). The total number of primary carbon atoms in (Y) is _________.
Show answer and solution
Answer: 4
Key idea: Alkane combustion stoichiometry and structural symmetry in organic compounds
Use combustion formula: $C_nH_{2n+2} + \frac{3n+1}{2}O_2 \rightarrow nCO_2 + (n+1)H_2O$. Given 8 moles $O_2$, so $\frac{3n+1}{2} = 8$, giving $n = 5$.
The alkane is $C_5H_{12}$ (pentane isomer).
Since only one monochlorinated product forms, all H atoms must be equivalent. This indicates high symmetry.
Among $C_5H_{12}$ isomers, only neopentane (2,2-dimethylpropane) has this symmetry: $(CH_3)_3C-CH_3$.
Neopentane has 1 central quaternary carbon and 4 primary carbons (the four $-CH_3$ groups).
Total number of primary carbon atoms = 4
8 April 2026 Shift 2, Q63 · Q63 · MCQ · Organic Chemistry - Some Basic Principles and Techniques · Medium
The major product of which of the following reaction is not obtained by rearrangement reaction?
(1) [Figure: Benzene + propyl chloride with Anhyd. AlCl3 arrow pointing to isopropylbenzene]
(2) [Figure: tert-butanol with H3O+ arrow pointing to isobutene]
(3) [Figure: CH3(CH2)4CH3 with Anhyd. AlCl3/HCl arrow pointing to CH3-CH(CH3)(CH2)2CH3 with CH3 branch]
(4) [Figure: neopentyl alcohol with H3O+ arrow pointing to 2-methylbut-2-ene]
Show answer and solution
Answer: Option (2)
Key idea: Carbocation Rearrangements in Organic Reactions
