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Multiple Choice Questions

1. Nitrogen containing compound is [JEE 2006]

(a) CH3-N≡C

(b) CH3-C≡N

(d) CH3-N+≡C-

Answer: (d)

Explanation: Isocyanide test/Carbylamine reaction gives isocyanide $\text{CH}_3\text{N}^+ \equiv \text{C}^-$.

2. In the reaction $\text{CH}_3\text{NH}_2 + \text{CHCl}_3 + \text{KOH} \longrightarrow \text{T} + \text{KCl} + \text{H}_2\text{O}$, the structure of the Product T is [JEE 2010 P-II]

(a) $\text{CH}_3\text{C} \equiv \text{N}$

(b) $\text{CH}_3\text{NH} - \text{CH}_3$

(d) $\text{CH}_3\text{N} \equiv \text{C}$

Answer: (c)

Explanation: $\text{CH}_3\text{NH}_2 + \text{CHCl}_3 + 3\text{KOH} \longrightarrow \text{CH}_3\text{N}^+ \equiv \text{C}^- + 3\text{KCl} + 3\text{H}_2\text{O}$. Product T is methyl isocyanide ($\text{CH}_3\text{N}^+ \equiv \text{C}^-$).

3. A compound with molecular mass 180 is acylated with $\text{CH}_3\text{COCl}$ to get a compound with molecular mass 390. The number of amino groups present per molecule of the former compound is [JEE Main 2013]

(a) 2

(b) 5

(d) 6

Answer: (b)

Explanation: Each acetylation of an $-\text{NH}_2$ group by $-\text{COCH}_3$ replaces $\text{H}$ (mass 1) with $\text{COCH}_3$ (mass 43), increasing the molecular mass by 42 unit.

$$\text{No. of amino groups} = \frac{390 - 180}{42} = \frac{210}{42} = 5$$

4. The gas leaked from a storage tank of the Union Carbide plant in Bhopal gas tragedy was [JEE Main 2013]

(a) Methylisocyanate

(b) Methylamine

(d) Phosgene

Answer: (a)

Explanation: Factual. The gas was methyl isocyanate ($\text{CH}_3\text{NCO}$).

5. In the reaction shown below, the major product(s) formed is/are [JEE Adv. 2014 P-I]

$$\text{Compound} + \text{acetic anhydride} \xrightarrow{\text{CH}_2\text{Cl}_2} \text{Product(s)}$$

(a) Acetylation at primary amine group

(b) Acetylation at acid amide group

(d) No reaction

Answer: (a)

Explanation: Only aliphatic/aromatic amines undergo acetylation and not acid amides, as the lone pair on the amide nitrogen is delocalized into the carbonyl group.

6. For the identification of naphthol using dye test, it is necessary to use [JEE Adv. 2014 P-II]

(a) dichloromethane solution of naphthol

(b) acidic solution of naphthol

(d) alkaline solution of naphthol

Answer: (d)

Explanation: Coupling of diazonium salts with naphthol ($\beta$-naphthol) occurs in alkaline medium to form an insoluble colored azo dye.

7. In the reaction sequence, the product E is [JEE Main 2015]

$$\text{Substituted Aniline} \xrightarrow[0-5^\circ\text{C}]{\text{NaNO}_2 / \text{HCl}} \text{D} \xrightarrow[\Delta]{\text{CuCN} / \text{KCN}} \text{E} + \text{N}_2$$

(a) Cyanide compound with $\text{CH}_3$ retention

(b) Isocyanide compound

(d) Phenol derivative

Answer: (c)

Explanation:

$$\text{Ar-NH}_2 \xrightarrow[0-5^\circ\text{C}]{\text{NaNO}_2 / \text{HCl}} \text{Ar-N}_2^+\text{Cl}^- \text{ (D)} \xrightarrow[\Delta]{\text{CuCN} / \text{KCN}} \text{Ar-CN} \text{ (E)} + \text{N}_2$$

8. The major product of the reaction is [JEE Adv. 2015 P-I]

(a) Monobromo derivative

(b) Dibromo derivative

(d) Tetrabromo derivative

Answer: (c)

Explanation: Bromination of aniline in aqueous solution or with $\text{Br}_2 / \text{H}_2\text{O}$ gives 2,4,6-tribromoaniline precipitate immediately due to the strong activating nature of the $-\text{NH}_2$ group.

9. In the following reactions, the major product W is [JEE Adv. 2015 P-II]

$$\text{Reactant} \xrightarrow{\text{NaOH}} \text{V} \xrightarrow[0^\circ\text{C}]{\text{NaNO}_2, \text{HCl}} \text{W}$$

(a) Azo coupled naphthol derivative

(b) Diazonium salt

(d) Substituted benzene

Answer: (a)

Explanation: Primary aromatic amine forms diazonium salt ($\text{V}$) with $\text{NaNO}_2 / \text{HCl}$ at $0^\circ\text{C}$, which then undergoes azo coupling with $\beta$-naphthol in basic solution ($\text{NaOH}$) to give the orange-red azo dye ($\text{W}$).

10. In the Hofmann bromamide degradation reaction, the number of moles of $\text{NaOH}$ and $\text{Br}_2$ used per mole of amine produced are: [JEE Main 2016]

(a) One mole of $\text{NaOH}$ and one mole of $\text{Br}_2$

(b) Four moles of $\text{NaOH}$ and two moles of $\text{Br}_2$

(d) Four moles of $\text{NaOH}$ and one mole of $\text{Br}_2$

Answer: (d)

Explanation: The balanced chemical equation for Hofmann bromamide degradation reaction is:

$$\text{R-CONH}_2 + \text{Br}_2 + 4\text{NaOH} \longrightarrow \text{R-NH}_2 + \text{Na}_2\text{CO}_3 + 2\text{NaBr} + 2\text{H}_2\text{O}$$

Hence, 1 mole of $\text{Br}_2$ and 4 moles of $\text{NaOH}$ are required per mole of primary amine produced.

11. The product(s) of the following reaction sequence is (are) [JEE Adv. 2016 P-I]

(a) Monobrominated benzene derivative

(b) Tribromobenzene derivative

(d) Phenolic compound

Answer: (b)

Explanation: Acetylation protects $-\text{NH}_2$, followed by bromination, hydrolysis, diazotization with $\text{NaNO}_2/\text{HCl}$, and reduction with $\text{Cu}/\text{HBr}$ or $\text{H}_3\text{PO}_2$ yielding the polybrominated benzene derivative.

12. The nitrogen containing compound produced in the reaction of $\text{HNO}_3$ with $\text{P}_4\text{O}_{10}$ [JEE Adv. 2016 P-II]

(a) can also be prepared by reaction of $\text{P}_4$ and $\text{HNO}_3$

(b) is diamagnetic

(d) reacts with $\text{Na}$ metal producing a brown gas

Answer: (b, d)

Explanation:

$$4\text{HNO}_3 + \text{P}_4\text{O}_{10} \longrightarrow 2\text{N}_2\text{O}_5 + 4\text{HPO}_3$$

The nitrogen-containing compound is $\text{N}_2\text{O}_5$.

(A) $\text{P}_4 + 20\text{HNO}_3 \longrightarrow 4\text{H}_3\text{PO}_4 + 20\text{NO}_2 + 4\text{H}_2\text{O}$. $\text{N}_2\text{O}_5$ cannot be prepared this way. (False)

(B) $\text{N}_2\text{O}_5$ has all paired electrons, so it is diamagnetic. (True)

(C) Structure of $\text{N}_2\text{O}_5$ is $\text{O}_2\text{N-O-NO}_2$ (contains $\text{N-O-N}$ bridge, no $\text{N-N}$ bond). (False)

(D) $\text{N}_2\text{O}_5 + \text{Na} \longrightarrow \text{NaNO}_3 + \text{NO}_2 \uparrow$ (brown gas). (True)

13. Which of the following compounds will form significant amount of meta product during mono-nitration reaction? [JEE Main 2017]

(a) Acetanilide

(b) Chlorobenzene

(d) Aniline

Answer: (d)

Explanation: Nitration is carried out in the presence of conc. $\text{HNO}_3$ + conc. $\text{H}_2\text{SO}_4$. Aniline acts as a base and gets protonated in strongly acidic medium to form the anilinium ion ($\text{C}_6\text{H}_5\text{NH}_3^+$). The anilinium ion is strongly deactivating and meta-directing, leading to ~47% meta-nitroaniline product.

14. The major product of the following reactions is [JEE Adv. 2017 P-II]

$$\text{Aniline derivative} \xrightarrow[0^\circ\text{C}]{\text{NaNO}_2, \text{aq. HCl}} \text{Product}$$

(a) Phenol derivative

(b) Diazonium salt / Coupled product

(d) Nitro compound

Answer: (b)

Explanation: Primary aromatic amine undergoes diazotization at $0^\circ\text{C}$ with $\text{NaNO}_2/\text{HCl}$ to form diazonium ion which couples with phenolic site present in the molecule/solution to yield azo dye product.

15. Which of the following compounds will be suitable for Kjeldahl's method for nitrogen estimation? [JEE Main 2018]

(a) Nitrobenzene

(b) Azo compound

(d) Pyridine

Answer: (c)

Explanation: Kjeldahl's method is not applicable to compounds containing nitrogen in nitro groups, azo groups, or nitrogen present in the ring (e.g., pyridine), as nitrogen in these compounds is not converted to ammonium sulphate under reaction conditions. Hence, aniline is suitable.

16. The predominant form of histamine present in human blood is ($\text{p}K_a\text{, Histidine} = 6.0$) [JEE Main 2018]

(a) Monoprotonated form at aliphatic primary amine

(b) Diprotonated form

(d) Deprotonated anionic form

Answer: (a)

Explanation: At physiological blood pH (~7.4), the aliphatic primary amine group ($\text{p}K_a \approx 9.8$) is completely protonated ($-\text{NH}_3^+$), whereas the imidazole ring nitrogen ($\text{p}K_a \approx 6.0$) remains largely unprotonated.

17. The increasing order of basicity of the following compound [JEE Main 2018]

(a) Amide nitrogen

(b) Aliphatic primary amine

(d) Secondary amine

(a) (d) < (b) < (a) < (c)

(b) (a) < (b) < (c) < (d)

(d) (b) < (a) < (d) < (c)

Answer: (d)

Explanation: Amide nitrogen is least basic due to resonance delocalization with carbonyl group. Guanidine/amidines are most basic as conjugate acid is resonance stabilized. Therefore, basicity order is (b) < (a) < (d) < (c).

18. The compound that does not produce nitrogen gas by the thermal decomposition is [JEE Main 2018]

(a) $(\text{NH}_4)_2\text{SO}_4$

(b) $\text{Ba}(\text{N}_3)_2$

(d) $\text{NH}_4\text{NO}_2$

Answer: (a)

Explanation:

$$(\text{NH}_4)_2\text{Cr}_2\text{O}_7 \xrightarrow{\Delta} \text{N}_2 + 4\text{H}_2\text{O} + \text{Cr}_2\text{O}_3$$

$$\text{NH}_4\text{NO}_2 \xrightarrow{\Delta} \text{N}_2 + 2\text{H}_2\text{O}$$

$$\text{Ba}(\text{N}_3)_2 \xrightarrow{\Delta} \text{Ba} + 3\text{N}_2$$

$$(\text{NH}_4)_2\text{SO}_4 \xrightarrow{\Delta} 2\text{NH}_3 + \text{H}_2\text{SO}_4$$

Only $(\text{NH}_4)_2\text{SO}_4$ evolves $\text{NH}_3$ gas instead of $\text{N}_2$.

19. The compound(s) which generate(s) $\text{N}_2$ gas upon thermal decomposition below $300^\circ\text{C}$ is (are) [JEE Adv. 2018 P-I]

(a) $\text{NH}_4\text{NO}_3$

(b) $(\text{NH}_4)_2\text{Cr}_2\text{O}_7$

(d) $\text{Mg}_3\text{N}_2$

Answer: (b, c)

Explanation:

$\text{NH}_4\text{NO}_3 \xrightarrow{250^\circ\text{C}} \text{N}_2\text{O} + 2\text{H}_2\text{O}$ (gives $\text{N}_2\text{O}$, not $\text{N}_2$).

$(\text{NH}_4)_2\text{Cr}_2\text{O}_7 \xrightarrow{\Delta} \text{N}_2 + \text{Cr}_2\text{O}_3 + 4\text{H}_2\text{O}$ (occurs below $300^\circ\text{C}$).

$\text{Ba}(\text{N}_3)_2 \xrightarrow{180^\circ\text{C}} \text{Ba} + 3\text{N}_2$ (occurs around $180^\circ\text{C}$).

$\text{Mg}_3\text{N}_2 \xrightarrow{>700^\circ\text{C}} 3\text{Mg} + \text{N}_2$.

Thus, below $300^\circ\text{C}$, only $(\text{NH}_4)_2\text{Cr}_2\text{O}_7$ and $\text{Ba}(\text{N}_3)_2$ produce $\text{N}_2$ gas.

20. Aniline reacts with mixed acid (conc. $\text{HNO}_3$ and conc. $\text{H}_2\text{SO}_4$) at 288 K to give P (51 %), Q (47%) and R (2%). The major product(s) of the following reaction sequence is (are) [JEE Adv. 2018 P-II]

$$\text{R} \xrightarrow[(1) \text{Ac}_2\text{O}, \text{pyridine} \quad (2) \text{Br}_2, \text{CH}_3\text{CO}_2\text{H} \quad (3) \text{H}_3\text{O}^+ \quad (4) \text{NaNO}_2, \text{HCl}/273-278\text{K} \quad (5) \text{EtOH}, \Delta]{\text{reaction sequence 1}} \text{S}$$

$$\text{S} \xrightarrow[(1) \text{Sn/HCl} \quad (2) \text{Br}_2/\text{H}_2\text{O (excess)} \quad (3) \text{NaNO}_2, \text{HCl}/273-278\text{K} \quad (4) \text{H}_3\text{PO}_2]{\text{reaction sequence 2}} \text{Major product(s)}$$

(a) 1,2,3-Tribromobenzene

(b) 1,3,5-Tribromobenzene

(d) 1,2,3,4-Tetrabromobenzene

Answer: (d)

Explanation: R is ortho-nitroaniline (2%). Protection of $-\text{NH}_2$ by acetylation, bromination at para-position relative to $-\text{NH}_2$, deprotection, diazotization and deamination with $\text{EtOH}/\Delta$ gives $m$-nitrobromobenzene ($\text{S}$). Reduction of nitro to amine, excess bromination populates open ortho/para positions, followed by diazotization and deamination using $\text{H}_3\text{PO}_2$ yields 1,2,3,4-tetrabromobenzene.

Multiple Choice Questions

21. The major product in the following reaction is : [JEE Main 2019]

$\text{Base} + \text{OH}^- \longrightarrow \text{CH}_3\text{I}$

(a)
Option A

(b)
Option B

(d)
Option D

Answer: (b)

Explanation: Because one double bond is missing in all the given options. So aromaticity is lost in both the rings.

22. The major product obtained in the following reaction is: [JEE Main 2019]

$\text{(i) } \text{CHCl}_3 / \text{KOH} \longrightarrow \text{(ii) } \text{Pd/C/H}_2$

(a) $\text{C}_6\text{H}_5\text{NHCH}_3$

(b) $\text{C}_6\text{H}_5\text{CH}_2\text{NH}_2$

(d) $\text{C}_6\text{H}_5\text{NC}$

Answer: (a)

Explanation: Carbylamine reaction followed by reduction with $\text{Pd/C/H}_2$ converts primary amine to secondary methylamine derivative.

$$\text{R-NH}_2 \xrightarrow{\text{CHCl}_3/\text{KOH}} \text{R-NC} \xrightarrow{\text{Pd/C/H}_2} \text{R-NH-CH}_3$$

23. Major product of the following reaction is: [JEE Main 2019]

$\text{(1) } \text{Et}_3\text{N} \longrightarrow \text{(2) Free radical polymerisation}$

(a)
Option A

(b)
Option B

(d)
Option D

Answer: (d)

Explanation: $\text{NH}_2$ (a) will act as nucleophile as (b) is having delocalised lone pair, leading to the amide monomer followed by free radical polymerisation.

24. The major product of following reaction is: $\text{R}-\text{C}\equiv\text{N} \xrightarrow{(1)\text{ AlH}(i\text{-Bu})_2, (2)\text{ H}_2\text{O}} \text{?}$ [JEE Main 2019]

(a) $\text{RCHO}$

(b) $\text{RCOOH}$

(d) $\text{RCONH}_2$

Answer: (a)

Explanation: DIBAL-H reduces nitriles selectively to imines, which on hydrolysis yield aldehydes:

$$\text{R-C}\equiv\text{N} \xrightarrow{\text{AlH}(i\text{-Bu})_2} \text{R-CH}=\text{NH} \xrightarrow{\text{H}_2\text{O}} \text{RCHO}$$

25. The increasing order of $pK_a$ of the following amino acids in aqueous solution is: Gly, Asp, Lys, Arg [JEE Main 2019]

(a) $\text{Asp} < \text{Gly} < \text{Arg} < \text{Lys}$

(b) $\text{Arg} < \text{Lys} < \text{Gly} < \text{Asp}$

(d) $\text{Asp} < \text{Gly} < \text{Lys} < \text{Arg}$

Answer: (d)

Explanation: Order of acidic strength:

$$\text{Aspartic acid } (\text{HOOC-CH}_2\text{-CH(NH}_2\text{)-COOH}) > \text{Glycine } (\text{NH}_2\text{-CH}_2\text{-COOH}) > \text{Lysine} > \text{Arginine}$$

So $pK_a$ order is: $\text{Asp} < \text{Gly} < \text{Lys} < \text{Arg}$.

26. Arrange the following amines in the decreasing order of basicity: (I) $\text{sp}^3\text{ N}$, (II) $\text{sp}^2\text{ N}$, (III) Delocalised lone pair N [JEE Main 2019]

(a) $\text{I} > \text{II} > \text{III}$

(b) $\text{III} > \text{II} > \text{I}$

(d) $\text{III} > \text{I} > \text{II}$

Answer: (d)

Explanation: Order of basic strength depends on localized vs delocalized lone pairs and hybridization ($\text{sp}^3 > \text{sp}^2$). Thus, $\text{III} > \text{I} > \text{II}$.

27. The correct sequence of amino acids present in the tripeptide given below is : [JEE Main 2019]

(a) Leu - Ser - Thr

(b) Thr - Ser- Leu

(d) Val - Ser - Thr

Answer: (a)

Explanation: Identifying the N-terminal to C-terminal amino acid residues gives Leucine, Serine, and Threonine in order: Leu - Ser - Thr.

28. The major product obtained in the following reaction is: [JEE Main 2019]

$\xrightarrow{(\text{CH}_3\text{CO})_2\text{O / pyridine (1 eqv.)}, \text{room temp.}}$

(a)
Option A

(b)
Option B

(d)
Option D

Answer: (b)

Explanation: Nucleophilic attack by $-\text{NH}_2$ occurs preferentially over $-\text{OH}$ group during acetylation with 1 equivalent of acetic anhydride at room temperature.

29. The major product of the following reaction is: [JEE Main 2019]

$\xrightarrow{(i) \text{Br}_2 / h\nu} \xrightarrow{(ii) \text{KOH(dil)}}$

(a)
Option A

(b)
Option B

(d)
Option D

Answer: (c)

Explanation: Free radical bromination at benzylic position followed by nucleophilic substitution with aqueous $\text{KOH}$ gives the alcohol.

30. The major product formed in the reaction given below will be: [JEE Main 2019]

$\xrightarrow{\text{NaNO}_2, \text{Aq. HCl}, 0-5^\circ\text{C}}$

(a)
Option A

(b)
Option B

(d)
Option D

Answer: (b)

Explanation: Diazotization of primary aliphatic amine leads to carbocation formation, rearrangement, and intramolecular substitution/elimination yielding the secondary amine cyclic product.

31. The correct structure of product 'P' in the following reaction is: [JEE Main 2019]

$\text{Asn} - \text{Ser} + (\text{CH}_3\text{CO})_2\text{O} (\text{excess}) \xrightarrow{\text{NEt}_3} P$

(a)
Option A

(b)
Option B

(d)
Option D

Answer: (a)

Explanation: Asn–Ser is a dipeptide having structure:

$$\text{H}_2\text{N-CH(CONH}_2\text{)-CONH-CH(CH}_2\text{OH)-COOH}$$

Excess acetic anhydride acetylates both free $-\text{NH}_2$ and $-\text{OH}$ groups.

32. The major product of the following reaction is: [JEE Main 2019]

(a)
Option A

(b)
Option B

(d)
Option D

Answer: (b)

Explanation: Reduction with $\text{H}_2/\text{Ni}$ reduces nitrile to primary amine followed by intramolecular condensation with carbonyl group losing $\text{H}_2\text{O}$ to give cyclic amine product.

33. An organic compound is estimated through Dumas method and was found to evolve 6 moles of $\text{CO}_2$, 4 moles of $\text{H}_2\text{O}$ and 1 mole of nitrogen gas. The formula of the compound is: [JEE Main 2019]

(a) $\text{C}_{12}\text{H}_8\text{N}$

(b) $\text{C}_{12}\text{H}_8\text{N}_2$

(d) $\text{C}_6\text{H}_8\text{N}_2$

Answer: (d)

Explanation: Dumas Method:

$$\text{C}_x\text{H}_y\text{N}_z \longrightarrow x\text{CO}_2 + \frac{y}{2}\text{H}_2\text{O} + \frac{z}{2}\text{N}_2$$

$$x = 6, \quad \frac{y}{2} = 4 \implies y = 8, \quad \frac{z}{2} = 1 \implies z = 2$$

Hence, molecular formula is $\text{C}_6\text{H}_8\text{N}_2$.

34. In the following compound, the favourable site/s for protonation is/are :- [JEE Main 2019]

(a) (b), (c) and (d)

(b) (a)

(d) (a) and (d)

Answer: (a)

Explanation: Sites (b), (c) and (d) possess localized lone pair of electrons available easily for protonation.

35. Among the following compounds most basic amino acid is: [JEE Main 2019]

(a) Lysine

(b) Asparagine

(d) Histidine

Answer: (d)

Explanation: Histidine contains an imidazole ring with nitrogen having localized lone pair, making it highly basic among the given amino acids.

36. In the following compounds, the decreasing order of basic strength will be - [JEE Main 2019]

(a) $(\text{C}_2\text{H}_5)_2\text{NH} > \text{C}_2\text{H}_5\text{NH}_2 > \text{NH}_3$

(b) $(\text{C}_2\text{H}_5)_2\text{NH} > \text{NH}_3 > \text{C}_2\text{H}_5\text{NH}_2$

(d) $\text{C}_2\text{H}_5\text{NH}_2 > \text{NH}_3 > (\text{C}_2\text{H}_5)_2\text{NH}$

Answer: (a)

Explanation: Basic strength order in aqueous medium for ethyl substituted amines: $2^\circ > 1^\circ > \text{NH}_3$. Thus, $(\text{C}_2\text{H}_5)_2\text{NH} > \text{C}_2\text{H}_5\text{NH}_2 > \text{NH}_3$.

37. 0.27 g of a long chain fatty acid was dissolved in $100\text{ cm}^3$ of hexane. $10\text{ mL}$ of this solution was added dropwise to the surface of water in a round watch glass. Hexane evaporates and a monolayer is formed. The distance from edge to centre of the watch glass is $10\text{ cm}$. What is the height of the monolayer? [Density of fatty acid $= 0.9\text{ g cm}^{-3}$, $\pi = 3$] [JEE Main 2019]

(a) $10^{-8}\text{ m}$

(b) $10^{-6}\text{ m}$

(d) $10^{-2}\text{ m}$

Answer: (b)

Explanation: Radius of watchglass $r = 10\text{ cm}$

$\Rightarrow \text{Surface area} = \pi r^2 = 3 \times (10\text{ cm})^2 = 300\text{ cm}^2$

Mass of fatty acid in 10 mL solution $= \frac{10}{100} \times 0.27 = 0.027\text{ g}$

$$\text{Volume of fatty acid} = \frac{0.027\text{ g}}{0.9\text{ g/cm}^3} = 0.03\text{ cm}^3$$

$$\text{Height} = \frac{\text{Volume}}{\text{Surface area}} = \frac{0.03\text{ cm}^3}{300\text{ cm}^2} = 0.0001\text{ cm} = 10^{-6}\text{ m}$$

38. The major product obtained in the following reaction: $\xrightarrow{\text{LiAlH}_4 \text{ (excess)}}$ [JEE Main 2019]

(a)
Option A

(b)
Option B

(d)
Option D

Answer: (b)

Explanation: $\text{LiAlH}_4$ reduces amides, esters, and carbonyl groups to corresponding alcohols/amines without reducing isolated $\text{C}=\text{C}$ double bonds.

39. Coupling of benzene diazonium chloride with 1-naphthol in alkaline medium will give [JEE Main 2019]

(a)
Option A

(b)
Option B

(d)
Option D

Answer: (c)

Explanation: Electrophilic aromatic substitution by diazonium cation occurs at position 4 (para) relative to $-\text{OH}$ group in 1-naphthol to give an orange-red azo dye.

Matrix Match/Column Match

40. Match each of the compounds in Column I with its characteristic reaction(s) in Column II. [JEE 2009 P-I]

(a) A$\rightarrow$1, 2, 4, 5; B$\rightarrow$4, 5; C$\rightarrow$1; D$\rightarrow$3

(b) A$\rightarrow$1, 2; B$\rightarrow$2; C$\rightarrow$1, 2, 3; D$\rightarrow$4

(d) A$\rightarrow$1, 2, 3, 4; B$\rightarrow$4; C$\rightarrow$2; D$\rightarrow$1

Answer: (a)

Explanation: Matching characteristic functional group reactions gives A$\rightarrow$1, 2, 4, 5; B$\rightarrow$4, 5; C$\rightarrow$1; D$\rightarrow$3.

41. Match the four starting materials (A, B, C, D) given in Column-I with the corresponding reaction schemes (1, 2, 3, 4) provided in Column-II. [JEE Adv. 2014 P-II]

(a) A $\rightarrow$ 3; B $\rightarrow$ 4; C $\rightarrow$ 2; D $\rightarrow$ 1

(b) A $\rightarrow$ 1; B $\rightarrow$ 2; C $\rightarrow$ 3; D $\rightarrow$ 4

(d) A $\rightarrow$ 3; B $\rightarrow$ 4; C $\rightarrow$ 2; D $\rightarrow$ 1

Answer: (a)

Explanation: Following the step-by-step synthetic conversions for schemes I, II, III, and IV yields: A$\rightarrow$3; B$\rightarrow$4; C$\rightarrow$2; D$\rightarrow$1.

Statement/Assertion and Reason

42. Statement-I: Aniline on reaction with $\text{NaNO}_2/\text{HCl}$ at $0^\circ\text{C}$ followed by coupling with $\beta$-naphthol gives a dark blue coloured precipitate. And
Statement-II: The colour of the compound formed in the reaction of aniline with $\text{NaNO}_2/\text{HCl}$ at $0^\circ\text{C}$ followed by coupling with $\beta$-naphthol is due to the extended conjugation. [JEE 2008 P-II]

(a) Statement-I is True, Statement-II is True; Statement-II is correct explanation for Statement-I

(b) Statement-I is True, Statement-II is True; Statement-II is not a correct explanation for Statement-I

(d) Statement-I is False, Statement-II is True

Answer: (d)

Explanation: Coupling of aniline diazonium salt with $\beta$-naphthol gives a scarlet red dye, not dark blue precipitate. Hence, Statement-I is False and Statement-II is True.

Paragraph

Paragraph for Question Nos. 43 to 45
$\text{RCONH}_2$ is converted into $\text{RNH}_2$ by means of Hofmann bromamide degradation. In this reaction, $\text{RCONHBr}$ is formed from which this reaction has derived its name. Electron donating group at phenyl activates the reaction. Hofmann degradation reaction is an intramolecular reaction [JEE 2006]

43. How can the conversion of (i) to (ii) be brought about?

(a) $\text{KBr}$

(b) $\text{KBr} + \text{CH}_3\text{ONa}$

(d) $\text{Br}_2 + \text{KOH}$

Answer: (d)

Explanation: Conversion of amide to $N$-bromoamide is carried out using $\text{Br}_2 + \text{KOH}$.

44. Which is the rate determining step in Hofmann bromamide degradation?

(a) Formation of (i)

(b) Formation of (ii)

(d) Formation of (iv)

Answer: (d)

Explanation: Migration of the alkyl/aryl group with pair of electrons to electron-deficient nitrogen with simultaneous loss of bromide ion (formation of isocyanate intermediate) is the rate determining step.

45. What are the constituent amines formed when the mixture of (i) and (ii) undergoes Hofmann bromamide degradation?

(a)
Option A

(b)
Option B

(d)
Option D

Answer: (b)

Explanation: Since Hofmann rearrangement is strictly intramolecular, no cross-products are formed.

Paragraph for Question No. 46
There are five students $S_1, S_2, S_3, S_4$ and $S_5$ in a music class and for them there are five seats $R_1, R_2, R_3, R_4$ and $R_5$ arranged in a row, where initially the seat $R_i$ is allotted to the student $S_i, i = 1, 2, 3, 4, 5$. But, on the examination day, the five students are randomly allotted the five seats. [JEE Adv. 2018 P-I]

46. For $i = 1, 2, 3, 4$, let $T_i$ denote the event that the students $S_i$ and $S_{i + 1}$ do not sit adjacent to each other on the day of the examination. Then, the probability of the event $T_1 \cap T_2 \cap T_3 \cap T_4$ is

(a) $\frac{1}{15}$

(b) $\frac{1}{10}$

(d) $\frac{1}{5}$

Answer: (c)

Explanation: The number of total favorable seating arrangements where no $S_i$ and $S_{i+1}$ sit adjacent is 14 out of $5! = 120$.

$$\text{Probability} = \frac{14}{120} = \frac{7}{60}$$

Integer and Subjective

47. $\xrightarrow{(1)\text{ NaOH}, (2)\text{ Q}, \text{R}} T$
Number of atoms of Br in compound 'T' are [JEE Adv. 2019 P-I]

Answer: 4

Explanation: Exhaustive bromination and subsequent synthetic sequence leads to tetrabromo compound 'T'. Thus, number of Br atoms in T is 4.

48. $\text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- \xrightarrow{\text{NaNO}_2 + \text{HCl}} X \longrightarrow Y \text{ (tertiary alcohol + other products)}$.
Find X and Y. Is Y optically active? Write the intermediate steps. [JEE 2005]

Answer: $X = \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^-$, $Y$ is optically active.

Explanation: Rearrangement via carbocation intermediate generates a chiral center yielding an optically active tertiary alcohol product $Y$.