Questions are transcribed from NTA's official question paper. Diagrams are described in words; open the official PDF to see them. Every answer was solved twice independently and is shown only when both agree. Where they did not, we say so.
Match List - I with List - II.
List - I
Mixture of Compounds
A. Diethyl amine + Ethyl amine
B. Acetaldehyde + Acetone
C. Ethanol + Phenol
D. Benzoic acid + Cinnamic acid
List - II
Reagent used to distinguish
I. Bromine water
II. $CHCl_3 + KOH, \Delta$
III. Neutral $FeCl_3$
IV. Ammonical silver nitrate
Choose the correct answer from the options given below:
(1) A-IV, B-II, C-I, D-III
(2) A-IV, B-II, C-III, D-I
(3) A-II, B-IV, C-I, D-III
(4) A-II, B-IV, C-III, D-I
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Answer: Option (4)
Key idea: Distinguishing tests for organic compounds - Carbylamine test, Tollen's test, FeCl₃ test, and Bromine water test
**A. Diethyl amine + Ethyl amine**: Use Carbylamine test ($CHCl_3 + KOH, \Delta$). Primary amine (ethyl amine) gives foul-smelling isocyanide, secondary amine (diethyl amine) doesn't. So A → II
**B. Acetaldehyde + Acetone**: Use Tollen's reagent (Ammonical silver nitrate). Aldehyde gives silver mirror test (positive), ketone doesn't. So B → IV
**C. Ethanol + Phenol**: Use Neutral $FeCl_3$. Phenol gives violet coloration, ethanol doesn't react. So C → III
**D. Benzoic acid + Cinnamic acid**: Use Bromine water. Cinnamic acid has $C=C$ double bond and decolorizes bromine water, benzoic acid (saturated) doesn't. So D → I
Therefore, the correct matching is: A-II, B-IV, C-III, D-I
Consider the three aromatic molecules (P, Q and R) whose structures have been given below:
[Figure: P is N,N-dimethylaniline (benzene ring with N(Me)₂ at para position), Q is para-methyl-N,N-dimethylaniline (benzene ring with N(Me)₂ and Me at para positions), R is 2,6-dimethyl-N,N-dimethylaniline (benzene ring with N(Me)₂ at position 1 and Me groups at positions 2 and 6)]
The correct order regarding the reactivity of these compounds with $Ph-N\equiv N Cl^{(-)}$ under optimum but slightly acidic medium is:
(1) P > Q > R
(2) R > P > Q
(3) R > Q > P
(4)
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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: Azo coupling reaction - electrophilic aromatic substitution with diazonium salts, considering both electronic activation and steric accessibility
Consider the following reactions sequence
[Figure: Benzene ring with NO2 group at top and CH3 group at bottom, reacting with (i) Sn/HCl; OH⁻ (ii) (CH3CO)2O (iii) Br2/AlBr3 (iv) H3O+ to form Major Product (P)]
When the product (P) is subjected to Carius analysis using $\text{AgNO}_3$, 1.0 g of the product (P) will produce _________ g of the precipitate of AgBr. (Nearest Integer)
(Given : molar mass in g mol$^{-1}$ C : 12, H : 1, O : 16, N : 14, Br : 80, Ag : 108)
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Answer: 1
Key idea: Carius analysis for halogen estimation, organic reaction sequence involving reduction, acetylation, electrophilic aromatic substitution, and hydrolysis
Reduce $\text{NO}_2$ to $\text{NH}_2$ using Sn/HCl, then acetylate with $(\text{CH}_3\text{CO})_2\text{O}$ to get p-methylacetanilide
Brominate with $\text{Br}_2/\text{AlBr}_3$ (ortho to acetamido group) then hydrolyze with $\text{H}_3\text{O}^+$ to get Product P: 2-bromo-4-methylaniline
Calculate molar mass of P ($\text{C}_7\text{H}_8\text{BrN}$): $7(12) + 8(1) + 80 + 14 = 186$ g/mol
Match the LIST-I with LIST-II
[Figure: A table with List-I showing Compound column with entries A (cyclohexanol structure with OH group), B (cyclohexylamine structure with NH2 group), C (cyclohexanecarbaldehyde structure with CHO group), D (phenol structure). List-II showing Test column with entries I. Hinsberg's reagent test, II. Phthalein dye test, III. Lucas test, IV. Tollen's test]
Choose the correct answer from the options given below:
(1) A-III, B-I, C-IV, D-II
(2) A-III, B-IV, C-I, D-II
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Answer: Option (1)
Key idea: Identification tests for organic compounds - matching functional groups with their characteristic chemical tests
Identify compound A (cyclohexanol) as a secondary alcohol → matches with Lucas test (III) for distinguishing alcohols
Identify compound B (cyclohexylamine) as a primary amine → matches with Hinsberg's reagent test (I) for distinguishing amines
Identify compound C (cyclohexanecarbaldehyde) as an aldehyde → matches with Tollen's test (IV) which gives silver mirror with aldehydes
Identify compound D (phenol) as an aromatic phenolic compound → matches with Phthalein dye test (II) which forms colored dyes with phenols
Final matching: A-III, B-I, C-IV, D-II which corresponds to option (1)
Identify the incorrect statements.
A. [Figure: benzyl amine structure] is a stronger base than [Figure: aniline structure]
B. [Figure: p-methoxyaniline structure] can be synthesized by Gabriel phthalimide synthesis.
C. [Figure: phenylacetamide structure] $\xrightarrow{\text{Br}_2, \text{NaOH}}$ primary aromatic amine
D. [Figure: p-nitroaniline structure] $\xrightarrow{\text{(i) NaNO}_2, \text{HCl}, 0°\text{C}}$ $\xrightarrow{\text{(ii) } \Delta}$ product will dissolve in NaOH
Choose the correct answer from the options given below:
(1) A and D Only
(2) A and C Only
(3) B and C Only
(4) A and B Only
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Answer: Option (3)
Key idea: Basicity of amines, Gabriel phthalimide synthesis, Hofmann bromamide degradation, and diazotization reactions
**Statement A**: Benzylamine ($C_6H_5CH_2NH_2$) has NH$_2$ on sp$^3$ carbon with localized lone pair, while aniline has NH$_2$ on benzene ring with delocalized lone pair. Benzylamine is more basic. ✓ Correct
**Statement B**: Gabriel phthalimide synthesis works only for primary aliphatic amines using alkyl halides. Aryl halides don't undergo nucleophilic substitution easily, so p-methoxyaniline (aromatic amine) cannot be prepared. ✗ Incorrect
**Statement C**: Hofmann bromamide degradation: $C_6H_5CH_2CONH_2 \xrightarrow{Br_2/NaOH} C_6H_5CH_2NH_2$ (benzylamine). This is a primary aliphatic amine, NOT aromatic amine. ✗ Incorrect
**Statement D**: p-nitroaniline $\xrightarrow{NaNO_2/HCl/0°C}$ diazonium salt $\xrightarrow{\Delta}$ p-nitrophenol. Phenols are acidic and dissolve in NaOH. ✓ Correct
Given below are two statements:
**Statement I:** Heating benzamide with bromine in an ethanolic solution of sodium hydroxide will give benzylamine.
**Statement II:** Nitration of aniline with $HNO_3/H_2SO_4$ at 288 K produces $m$-nitroaniline in higher amount than $o$-nitroaniline (pH adjusted).
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 (4)
Key idea: Hofmann Bromamide Degradation and Electrophilic Aromatic Substitution in Aniline
**Statement I Analysis:** Hofmann bromamide degradation of benzamide: $\text{C}_6\text{H}_5\text{CONH}_2 + \text{Br}_2 + 4\text{NaOH} \rightarrow \text{C}_6\text{H}_5\text{NH}_2$ (aniline)
The product is **aniline** ($\text{C}_6\text{H}_5\text{NH}_2$), NOT benzylamine ($\text{C}_6\text{H}_5\text{CH}_2\text{NH}_2$)
Therefore, **Statement I is FALSE**
**Statement II Analysis:** In acidic medium ($\text{HNO}_3/\text{H}_2\text{SO}_4$), aniline gets protonated to anilinium ion ($\text{C}_6\text{H}_5\text{NH}_3^+$)
The $-\text{NH}_3^+$ group is **meta-directing** (electron-withdrawing), so $m$-nitroaniline is the major product
Therefore, **Statement II is TRUE**
**Answer:** Statement I is false but Statement II is true → Option (4)
Arrange the following compounds according to increasing order of boiling points.
n - $C_4H_9OH$ (A), n - $C_4H_9NH_2$ (B), n - $C_4H_{10}$ (C) and $C_2H_5NHC_2H_5$(D).
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Answer: C < D < B < A
Key idea: Boiling points depend on intermolecular forces: alcohols (O-H) form strongest H-bonds, primary amines (NH₂) form stronger H-bonds than secondary amines (NH), and alkanes have only weak van der Waals forces.
Identify the functional groups: (A) is alcohol, (B) is primary amine, (C) is alkane, (D) is secondary amine
Alkane (C) has only van der Waals forces → lowest boiling point
Secondary amine (D) has one N-H bond → weak hydrogen bonding
Primary amine (B) has two N-H bonds → stronger hydrogen bonding than (D)
Alcohol (A) has O-H bond → strongest hydrogen bonding (O more electronegative than N)
Increasing order of boiling points: $C < D < B < A$
Consider the following reaction.
[Figure: A molecule showing a piperidine ring (6-membered ring with N) with a side chain attached to N. The side chain has a carbonyl group (C=O), then a CH with a methyl branch, then another carbonyl group (C=O) with H attached.]
(i) $\text{NaBH}_4/\text{MeOH}$
(ii) $\text{NaOH(aq)}, \Delta \rightarrow (P)$
(iii) $\text{H}_3\text{O}^+$
The major product (P) formed is:
(1) [Figure: Piperidine ring with N-CH(CH3)-CH2OH side chain] 691121020.
(2) [Figure: Piperidine ring with OH on ring carbon and N-CH(CH3)-CHO side chain] 691121021.
(3) [Figure: Open chain structure HO-C(=O)-(CH2)4-NH-CH(CH3)-CH2OH] 691121022.
(4) [Figure: Open chain structure HO-(CH2)5-NH-CH(CH3)-CH2OH] 691121023.
Show answer and solution
Answer: Option (1)
Key idea: Selective reduction of aldehydes and amides using NaBH4, followed by base-mediated transformations
Identify the starting material: piperidine ring with $N$-CO-CH(CH$_3$)-CHO side chain (amide with aldehyde)
Step (i): NaBH$_4$/MeOH reduces the aldehyde (-CHO) to primary alcohol (-CH$_2$OH), giving $N$-CO-CH(CH$_3$)-CH$_2$OH
Under the combined conditions of NaBH$_4$ and subsequent base treatment, the amide carbonyl can also be reduced
Step (ii): NaOH(aq), $\Delta$ facilitates reduction of amide (N-CO-) to amine (N-CH$_2$-)
Step (iii): H$_3$O$^+$ workup gives final product: piperidine ring with $N$-CH$_2$-CH(CH$_3$)-CH$_2$OH side chain
The major product (P) is piperidine with $N$-CH(CH$_3$)-CH$_2$OH substituent (both carbonyls reduced)
Which statements are True?
A. In Hoffmann bromamide degradation, 4 moles of NaOH and 2 moles of $\text{Br}_2$ are consumed per mole of an amide
B. Hoffmann bromamide reaction is not given by alkyl amides.
C. Primary amines can be synthesized by Hoffmann bromamide degradation.
D. Secondary amide on reaction with $\text{Br}_2$ and NaOH will give secondary amine.
E. The by-products of Hoffmann degradation are $\text{Na}_2\text{CO}_3$, NaBr and $\text{H}_2\text{O}$.
Choose the correct answer from the options given below:
