Answer

(b) The structure of given amines are as follows

Hence, triethylamine is tertiary amine. The correct choice is (b).

• (a) 1-methylcyclohexylamine: This is a primary amine because the nitrogen atom is bonded to only one alkyl group (the 1-methylcyclohexyl group).

• (c) tert-butylamine: This is a primary amine because the nitrogen atom is bonded to only one alkyl group (the tert-butyl group).

• (d) N-methylaniline: This is a secondary amine because the nitrogen atom is bonded to two groups: one methyl group and one phenyl group (from aniline).

Answer

(d) IUPAC name of $\mathrm{CH} _{2}=\mathrm{CHCH} _{2}^{2} \mathrm{NHCH} _{3}$ is $\mathrm{N}$-methylprop-2-en-1-amine Hence, option (d) is correct.

• Option (a) allyl methylamine: This name does not follow the IUPAC nomenclature rules. “Allyl” is a common name and not a systematic IUPAC name. Additionally, it does not specify the position of the methylamine group on the carbon chain.

• Option (b) 2-amino-4-pentene: This name incorrectly identifies the structure. The compound does not have an amino group at the 2-position of a pentene chain. The correct structure has an amine group attached to the first carbon of a propene chain with a methyl group attached to the nitrogen.

• Option (c) 4-aminopent-1ene: This name suggests that the amine group is on the fourth carbon of a pentene chain, which is incorrect. The correct structure has the amine group attached to the first carbon of a propene chain with a methyl group attached to the nitrogen.

Thinking Process

This problem is based on concept of basic strength of various types of amine depending upon inductive effect, resonance and solvation.

Answer

(c)

Since, $+I$ effect and solvation increases basic character while $-I$ effect and resonance decreases basic character. Hence, correct choice is (c).

• (a) $\mathrm{CH}_3\mathrm{NH}_2$: The basicity of methylamine ($\mathrm{CH}_3\mathrm{NH}_2$) is lower than that of dimethylamine ($\left(\mathrm{CH}_3\right)_2\mathrm{NH}$) because dimethylamine has two electron-donating methyl groups, which increase the electron density on the nitrogen atom more effectively than a single methyl group.

• (b) $\mathrm{NCCH}_2\mathrm{NH}_2$: The presence of the electron-withdrawing cyano group ($\mathrm{NC}$) decreases the electron density on the nitrogen atom, making it less basic compared to dimethylamine.

• (d) $\mathrm{C}_6\mathrm{H}_5\mathrm{NHCH}_3$: The phenyl group ($\mathrm{C}_6\mathrm{H}_5}$) is electron-withdrawing through resonance and inductive effects, which decreases the electron density on the nitrogen atom, reducing its basicity compared to dimethylamine.

Answer

(a) Aniline is weakest Bronsted base among the given four compounds due to resonance present in case of aniline.

Resonating structure of aniline

Hence, lone pair of nitrogen are less available for donation to the acid.

• (b) $\mathrm{CH} _{3} \mathrm{CH} _{2} \mathrm{NH} _{2}$: Ethylamine is a stronger Bronsted base compared to aniline because it lacks resonance stabilization. The lone pair on the nitrogen is more available for donation to an acid.

• (c) $\mathrm{(CH} _{3} \mathrm{) _{2} NH}$: Dimethylamine is a stronger Bronsted base compared to aniline because it also lacks resonance stabilization. The lone pair on the nitrogen is more available for donation to an acid.

• (d) $\mathrm{CH} _{3} \mathrm{NH} _{2}$: Methylamine is a stronger Bronsted base compared to aniline because it lacks resonance stabilization. The lone pair on the nitrogen is more available for donation to an acid.

Answer

(c) $S_{N} 1$ reaction proceeds through formation of carbocation. Hence, more stable be the carbocation more reactivity towards $\mathrm{S} _{\mathrm{N}} 1$ mechanism.

Hence, the reaction will proceed through $\mathrm{S} _{\mathrm{N}} 1$ mechanism when, $\mathrm{C} _{6} \mathrm{H} _{5} \mathrm{CH} _{2} \mathrm{Br}$ is the substrate. because on ionisation it gives a resonance stabilised carbocation $\left(\mathrm{C} _{6} \mathrm{H} _{5} \mathrm{CH} _{2}\right)$.

• (a) $\mathrm{CH} _{3} \mathrm{Br}$: The carbocation formed, $\mathrm{CH} _{3}^{\oplus}$, is a primary carbocation, which is highly unstable and does not favor the $\mathrm{S} _{\mathrm{N}} 1$ mechanism.

• (b) $\mathrm{C} _{6} \mathrm{H} _{5} \mathrm{Br}$: The carbocation formed, $\mathrm{C} _{6} \mathrm{H} _{5}^{\oplus}$, is not stable because the phenyl cation is highly unstable due to the lack of resonance stabilization.

• (d) $\mathrm{C} _{2} \mathrm{H} _{5} \mathrm{Br}$: The carbocation formed, $\mathrm{C} _{2} \mathrm{H} _{5}^{\oplus}$, is a primary carbocation, which is not stable and does not favor the $\mathrm{S} _{\mathrm{N}} 1$ mechanism.

Answer

(b) Aryl nitro compound can’t be converted into amine using $\mathrm{LiAlH} _{4}$ in ether.

Hence, option (b) is the correct choice.

• (a) $\mathrm{H}_2$ (excess) / Pt: This reagent is effective for reducing aryl nitro compounds to amines. The hydrogenation process using $\mathrm{H}_2$ in the presence of a platinum catalyst is a well-known method for this transformation.

• (c) $\mathrm{Fe}$ and $\mathrm{HCl}$: This combination is also effective for reducing aryl nitro compounds to amines. The iron acts as a reducing agent in the acidic medium provided by hydrochloric acid, facilitating the reduction process.

• (d) $\mathrm{Sn}$ and $\mathrm{HCl}$: Similar to the iron and hydrochloric acid combination, tin and hydrochloric acid are effective for reducing aryl nitro compounds to amines. Tin serves as the reducing agent in the acidic environment, enabling the reduction.

Answer

(c) In order to prepare $1^{\circ}$ amine from an alkyl halide with simultaneous addition of one $\mathrm{CH} _{2}$ group in the carbon chain. The reagent used as a source of nitrogen is $\mathrm{KCN}$. Chemical transformation can be shown as

• (a) Sodium amide, $\mathrm{NaNH}_2$: Sodium amide is typically used for deprotonation reactions and the formation of alkynes from vicinal dihalides. It does not add a $\mathrm{CH}_2$ group to the carbon chain and is not suitable for the preparation of primary amines with an additional carbon atom.

• (b) Sodium azide, $\mathrm{NaN}_3$: Sodium azide is used to convert alkyl halides to azides, which can then be reduced to primary amines. However, this process does not involve the addition of a $\mathrm{CH}_2$ group to the carbon chain.

• (d) Potassium phthalimide, $\mathrm{C}_6\mathrm{H}_4(\mathrm{CO})_2\mathrm{N}^{-}\mathrm{K}^{+}$: Potassium phthalimide is used in the Gabriel synthesis to prepare primary amines from alkyl halides. This method does not add a $\mathrm{CH}_2$ group to the carbon chain.

Answer

(d) Source of nitrogen in Gabriel phthalimide synthesis is potassium phthalimide.

• (a) Sodium azide, $\mathrm{NaN}_3$: Sodium azide is not used in the Gabriel synthesis of amines. It is typically used in the synthesis of azides and in the preparation of other nitrogen-containing compounds, but not in the Gabriel synthesis.

• (b) Sodium nitrite, $\mathrm{NaNO}_2$: Sodium nitrite is not involved in the Gabriel synthesis of amines. It is commonly used in diazotization reactions and in the preparation of diazonium salts, but it does not serve as a source of nitrogen in the Gabriel synthesis.

• (c) Potassium cyanide, $\mathrm{KCN}$: Potassium cyanide is not used in the Gabriel synthesis of amines. It is often used in nucleophilic substitution reactions and in the synthesis of nitriles, but it does not provide the nitrogen source in the Gabriel synthesis.

Answer

(c)

Chemical transformation can be shown as

While other given set of reactants give primary amine only.

• (a) $2^{\circ} \mathrm{R}-\mathrm{Br}+\mathrm{NH} _{3}$: This reaction typically leads to the formation of a primary amine ($1^{\circ}$ amine) because the ammonia ($\mathrm{NH}_3$) will replace the bromine atom, resulting in a primary amine.

• (b) $2{ }^{\circ} \mathrm{R}-\mathrm{Br}+\mathrm{NaCN}$ followed by $\mathrm{H} _{2} / \mathrm{Pt}$: This sequence of reactions will first form a nitrile ($\mathrm{R-CN}$) and then reduce it to a primary amine ($1^{\circ}$ amine) upon hydrogenation.

• (d) $1^{\circ} \mathrm{R}-\mathrm{Br}(2 \mathrm{~mol})+$ potassium phthalimide followed by $\mathrm{H} _{3} \mathrm{O}^{+}$/ heat: This is the Gabriel synthesis, which is specifically designed to produce primary amines ($1^{\circ}$ amines) from alkyl halides.

Answer

(d) The best reagent tor converting 2-phenylpropanamide into 2- phenylpropanamine is $\mathrm{LiAlH} _{4}$ in ether. Reaction is as given below

• (a) excess $\mathrm{H}_2$: Excess hydrogen ($\mathrm{H}_2$) typically requires a catalyst such as palladium, platinum, or nickel to reduce compounds. However, it is not effective for reducing amides to amines. This method is more suitable for reducing alkenes, alkynes, and nitro groups.

• (b) $\mathrm{Br}_2$ in aqueous $\mathrm{NaOH}$: Bromine in aqueous sodium hydroxide is used in the Hofmann rearrangement, which converts primary amides to primary amines with one fewer carbon atom. This reaction would not be suitable for converting 2-phenylpropanamide to 2-phenylpropanamine as it would result in a different product.

• (c) iodine in the presence of red phosphorus: Iodine in the presence of red phosphorus is typically used for the reduction of nitro compounds to amines. It is not effective for reducing amides to amines. This reagent is not suitable for the conversion of 2-phenylpropanamide to 2-phenylpropanamine.

Answer

(b) The best reagent for converting 2-phenylpropanamide into 1-phenylethanamine is by $\mathrm{NaOH} / \mathrm{Br} _{2}$ and chemical transformation can be done as

This occurs due to intramolecular migration of alkyl group. It is an example of Hofmann bromamide reaction.

• (a) excess $\mathrm{H}_2 / \mathrm{Pt}$: This reagent is typically used for hydrogenation reactions, which reduce double or triple bonds in alkenes or alkynes, or for reducing nitro groups to amines. It is not suitable for converting amides to amines.

• (c) $\mathrm{NaBH}_4 /$ methanol: Sodium borohydride is a mild reducing agent that is commonly used to reduce aldehydes and ketones to alcohols. It is not effective for reducing amides to amines.

• (d) $\mathrm{LiAlH}_4 / ether: Lithium aluminum hydride is a strong reducing agent that can reduce amides to amines. However, it is not the best reagent for this specific transformation because it requires more stringent conditions and can lead to over-reduction or other side reactions. The Hofmann bromamide reaction using $\mathrm{NaOH} / \mathrm{Br}_2$ is more specific and efficient for this conversion.

Answer

(b) Hofmann bromamide degradation is shown by $\mathrm{Ar}-\mathrm{C}-\mathrm{NH} _{2}$ by which amide is converted into amine via undergoing intramolecular migration of phenyl group.

• (a) $\mathrm{ArNH} _{2}$: This option is incorrect because Hofmann bromamide degradation specifically involves amides, not amines. $\mathrm{ArNH} _{2}$ is an amine, not an amide.

• (c) $\mathrm{ArNO} _{2}$: This option is incorrect because nitro compounds ($\mathrm{ArNO} _{2}$) do not participate in Hofmann bromamide degradation. The reaction requires an amide group.

• (d) $\mathrm{ArCH} _{2} \mathrm{NH} _{2}$: This option is incorrect because Hofmann bromamide degradation involves the conversion of an amide to an amine. $\mathrm{ArCH} _{2} \mathrm{NH} _{2}$ is an amine with an alkyl group, not an amide.

Answer

(d) The correct increasing order of basic strength is as follows

Greater the electron density towards ring, greater will be its basic strength.

Electron withdrawing group decreases basic strength while electron donating group increases basic strength.

• Option (a) II < III < I: This option is incorrect because it suggests that compound II is less basic than compound III. However, compound II has an electron-donating group which increases its basic strength compared to compound III, which has an electron-withdrawing group.

• Option (b) III < I < II: This option is incorrect because it suggests that compound I is more basic than compound II. Compound II has a stronger electron-donating group compared to compound I, making it more basic than compound I.

• Option (c) III < II < I: This option is incorrect because it suggests that compound I is the most basic. However, compound II has a stronger electron-donating group compared to compound I, making compound II more basic than compound I.

Answer

(c) Methylamine reacts with $\mathrm{HNO} _{2}$ (nitrous acid) to form methanol.

• Option (a) $\mathrm{CH} _{3}-\mathrm{O}-\mathrm{N}=\mathrm{O}$: This compound is methyl nitrite, which is not formed in the reaction between methylamine and nitrous acid. The reaction typically leads to the formation of methanol and nitrogen gas, not methyl nitrite.

• Option (b) $\mathrm{CH} _{3}-\mathrm{O}-\mathrm{CH} _{3}$: This compound is dimethyl ether, which is not a product of the reaction between methylamine and nitrous acid. The reaction does not involve the formation of an ether linkage.

• Option (d) $\mathrm{CH} _{3} \mathrm{CHO}$: This compound is acetaldehyde, which is not formed in the reaction between methylamine and nitrous acid. The reaction leads to the formation of methanol, not an aldehyde.

Answer

(b) Chemical reaction takes place during reaction of methylamine with nitrous acid is as follows

• Option (a) $\mathrm{NH}_3$: Ammonia ($\mathrm{NH}_3$) is not evolved in this reaction. When methylamine reacts with nitrous acid, the primary amine group is converted into a diazonium salt, which then decomposes to release nitrogen gas ($\mathrm{N}_2$) and not ammonia.

• Option (c) $\mathrm{H}_2$: Hydrogen gas ($\mathrm{H}_2$) is not produced in this reaction. The reaction between methylamine and nitrous acid involves the formation and decomposition of a diazonium intermediate, leading to the release of nitrogen gas ($\mathrm{N}_2$), not hydrogen.

• Option (d) $\mathrm{C}_2\mathrm{H}_6$: Ethane ($\mathrm{C}_2\mathrm{H}_6$) is not a product of this reaction. The reaction mechanism does not involve any carbon-carbon bond formation that would result in the production of ethane. The primary product of the decomposition of the diazonium salt is nitrogen gas ($\mathrm{N}_2$).

Answer

(c) Nitration of benzene using a mixture of conc. $\mathrm{H} _{2} \mathrm{SO} _{4}$ and conc. $\mathrm{HNO} _{3}$ proceeds as

This reaction is known as electrophilic substitution reaction.

• (a) $\mathrm{NO}_2$: This species is not the electrophile in the nitration reaction. The actual electrophile that initiates the reaction is $\mathrm{NO}_2^+$, not $\mathrm{NO}_2$.

• (b) $\mathrm{NO}^+$: This species is not involved in the nitration of benzene. The correct electrophile is $\mathrm{NO}_2^+$, which is generated from the reaction between conc. $\mathrm{H}_2\mathrm{SO}_4$ and conc. $\mathrm{HNO}_3$.

• (d) $\mathrm{NO}_2^-$: This species is an anion and not an electrophile. The nitration reaction requires an electrophile, which is $\mathrm{NO}_2^+$, not an anion like $\mathrm{NO}_2^-$.

Answer

(c) Aromatic nitro compound on reaction with $\mathrm{Fe}$ and $\mathrm{HCl}$ gives aromatic primary amine as shown below

• (a) Aromatic oxime: Reduction of aromatic nitro compounds using Fe and HCl does not produce aromatic oximes. Oximes are typically formed by the reaction of hydroxylamine with aldehydes or ketones, not by the reduction of nitro compounds.

• (b) Aromatic hydrocarbon: Reduction of aromatic nitro compounds using Fe and HCl does not lead to the formation of aromatic hydrocarbons. Aromatic hydrocarbons are typically formed by the removal of functional groups or by dehydrogenation processes, not by the reduction of nitro groups.

• (d) Aromatic amide: Reduction of aromatic nitro compounds using Fe and HCl does not yield aromatic amides. Aromatic amides are usually formed by the reaction of aromatic amines with carboxylic acids or their derivatives, not by the reduction of nitro compounds.

Answer

(b) Greater will be the strength of base, greater will be its reactivity towards dilute $\mathrm{HCl}$. Hence, $\left(\mathrm{CH} _{3}\right) _{2} \mathrm{NH}$ has highest basic strength as it has highest reactivity.

• Option (a): $\mathrm{CH}_3\mathrm{NH}_2$ has a lower basic strength compared to $\left(\mathrm{CH}_3\right)_2\mathrm{NH}$ because it has only one alkyl group, which provides less electron-donating effect to the nitrogen atom, making it less reactive towards dilute HCl.

• Option (c): $\left(\mathrm{CH}_3\right)_3\mathrm{N}$ has a lower basic strength compared to $\left(\mathrm{CH}_3\right)_2\mathrm{NH}$ because the steric hindrance caused by the three methyl groups around the nitrogen atom reduces its ability to donate electrons, making it less reactive towards dilute HCl.

• Option (d): $\mathrm{NH}_3$ has the lowest basic strength among the given options because it has no alkyl groups to donate electron density to the nitrogen atom, making it the least reactive towards dilute HCl.

Answer

(a) Acid anhydride on reaction with primary amine produces amide as

• (b) Imide: Imides are typically formed from the reaction of an anhydride with a secondary amine, not a primary amine. The reaction between an acid anhydride and a primary amine does not provide the necessary conditions for imide formation.

• (c) Secondary amine: Secondary amines are not formed in this reaction. Instead, the primary amine reacts with the acid anhydride to form an amide and a carboxylic acid, not another amine.

• (d) Imine: Imines are formed from the reaction of primary amines with aldehydes or ketones, not with acid anhydrides. The reaction between an acid anhydride and a primary amine leads to the formation of an amide, not an imine.

Answer

(b)

This reaction is called Gattermann reaction. In this reaction, $\mathrm{Cl}, \mathrm{Br}$ and $\mathrm{CN}$ can be introduced into the benzene ring by simply treating diazonium salts with $\mathrm{HCl}, \mathrm{HBr}, \mathrm{KCN}$, respectively in presence of copper powder instead of using Cu (I) salts.

• Sandmeyer reaction: This reaction involves the substitution of an aryl diazonium salt with a halide or a cyanide ion using copper(I) salts (CuCl, CuBr, CuCN). The given reaction uses copper powder and HCl, which is characteristic of the Gattermann reaction, not the Sandmeyer reaction.

• Claisen reaction: This reaction typically refers to the Claisen condensation, which is a carbon-carbon bond-forming reaction between esters or between an ester and a carbonyl compound under basic conditions. It does not involve diazonium salts or the formation of aryl halides.

• Carbylamine reaction: This reaction involves the formation of isocyanides (carbylamines) by the reaction of primary amines with chloroform and a base. It does not involve diazonium salts or the formation of aryl halides.

Answer

(b) Best method of preparing primary amines from alkyl halides without changing the number of carbon atoms in the chain is Gabriel phthalimide synthesis

• (a) Hofmann bromamide reaction: This method is used to convert amides to amines with one less carbon atom. Therefore, it changes the number of carbon atoms in the chain, which does not meet the requirement of the question.

• (c) Sandmeyer reaction: This reaction is typically used to convert aryl diazonium salts to aryl halides, cyanides, or other substituted aromatic compounds. It is not suitable for preparing primary amines from alkyl halides.

• (d) Reaction with $\mathrm{NH}_3$: Direct reaction of alkyl halides with ammonia can lead to a mixture of primary, secondary, and tertiary amines, as well as quaternary ammonium salts. This lack of selectivity makes it an inefficient method for preparing primary amines specifically.

Answer

(d) Nitrobenzene will not undergo azo coupling reaction with benzene diazonium chloride while other three undergo diazo coupling reaction very easily. Diazonium cation is a weak $E^{+}$and hence reacts with electron rich compounds cotaining electron donating group i.e., $-\mathrm{OH} _{1}-\mathrm{NH} _{2}$ and $-\mathrm{OCH} _{3}$ groups and not with compounds containing electron withdrawing group, i.e., $\mathrm{NO} _{2}$ etc.

• Aniline: Aniline contains an electron-donating amino group (-NH₂), which makes the benzene ring electron-rich and facilitates the azo coupling reaction with benzene diazonium chloride.

• Phenol: Phenol contains an electron-donating hydroxyl group (-OH), which increases the electron density on the benzene ring and promotes the azo coupling reaction with benzene diazonium chloride.

• Anisole: Anisole contains an electron-donating methoxy group (-OCH₃), which enhances the electron density on the benzene ring and supports the azo coupling reaction with benzene diazonium chloride.

Answer

(c) Phenol is weakest Bronsted base as phenol after loosing $\mathrm{H}^{+}$produces least stable conjugate acid among the compounds.

Oxygen has more electronegative than $\mathrm{N}$. So, $\mathrm{O}-\mathrm{H}$ bond is more polar and it has highest value of acidic character. Since, phenol is more acidic that alcohol, therefore, phenol has the least tendency to accept a proton and hence it is weak Bronsted base. Hence, phenol is least basic among given four choices.

• Option (a) Aniline: Aniline is a stronger Bronsted base compared to phenol because the nitrogen atom in aniline has a lone pair of electrons that can readily accept a proton. Nitrogen is less electronegative than oxygen, making the lone pair on nitrogen more available for protonation.

• Option (b) Ethanol: Ethanol is a stronger Bronsted base compared to phenol because the oxygen atom in ethanol has a lone pair of electrons that can accept a proton. However, ethanol is less acidic than phenol, making it more likely to act as a base.

• Option (d) Ammonia: Ammonia is a stronger Bronsted base compared to phenol because the nitrogen atom in ammonia has a lone pair of electrons that can readily accept a proton. Nitrogen is less electronegative than oxygen, making the lone pair on nitrogen more available for protonation.

Answer

(d) Aniline is a weaker base than $\mathrm{NH} _{3}$ due to delocalization of lone pair of electrons of the $\mathrm{N}$-atom over the benzene ring. pyrrole is not more basic because the lone pair of electrons on the $\mathrm{N}$-atom is donated towards aromatic sextet formation.

Therefore, pyrrolidine is strongest base as lone pair of nitrogen does not involved in resonance and also due to presence of two alkyl ring residue, basic strength becomes high among given four compounds.

• Aniline is a weaker base than $\mathrm{NH}_3$ due to delocalization of the lone pair of electrons of the $\mathrm{N}$-atom over the benzene ring.
• Pyrrole is not more basic because the lone pair of electrons on the $\mathrm{N}$-atom is donated towards aromatic sextet formation.
• Ammonia ($\mathrm{NH}_3$) is less basic than pyrrolidine because it lacks the additional alkyl groups that increase electron density on the nitrogen, making pyrrolidine a stronger base.

Answer

(a)

Basic strength of the above species can be explained on the basis of electronegativity of central atom and its proton accepting tendency. Here, amide ion is most basic among given compounds due to presence of negative charge and two pair of electrons on nitrogen atom.

• Option (b): $\mathrm{OH}^{-}>\mathrm{NH} _{2}^{-}>\mathrm{H} _{2} \mathrm{O}>\mathrm{NH} _{3}$

  This option is incorrect because $\mathrm{NH} _{2}^{-}$ is more basic than $\mathrm{OH}^{-}$ due to the lower electronegativity of nitrogen compared to oxygen, which makes $\mathrm{NH} _{2}^{-}$ a stronger base.

• Option (c): $\mathrm{NH} _{3}>\mathrm{H} _{2} \mathrm{O}>\mathrm{NH} _{2}^{-}>\mathrm{OH}^{-}$

  This option is incorrect because neutral molecules like $\mathrm{NH} _{3}$ and $\mathrm{H} _{2} \mathrm{O}$ are less basic than their corresponding anions ($\mathrm{NH} _{2}^{-}$ and $\mathrm{OH}^{-}$). Therefore, $\mathrm{NH} _{3}$ and $\mathrm{H} _{2} \mathrm{O}$ cannot be more basic than $\mathrm{NH} _{2}^{-}$ and $\mathrm{OH}^{-}$.

• Option (d): $\mathrm{H} _{2} \mathrm{O}>\mathrm{NH} _{3}>\mathrm{OH}^{-}>\mathrm{NH} _{2}^{-}$

  This option is incorrect because $\mathrm{H} _{2} \mathrm{O}$ and $\mathrm{NH} _{3}$ are neutral molecules and are less basic than their corresponding anions ($\mathrm{OH}^{-}$ and $\mathrm{NH} _{2}^{-}$). Therefore, $\mathrm{H} _{2} \mathrm{O}$ and $\mathrm{NH} _{3}$ cannot be more basic than $\mathrm{OH}^{-}$ and $\mathrm{NH} _{2}^{-}$.

Answer

(b) $1^{\circ}$ and $2^{\circ}$ amines have higher boiling points due to intermolecular $\mathrm{H}$-bonding but less votatile than $3^{\circ}$ amines and hydrocarbons of comparable molecular mass. Further, because of polar $\mathrm{C}-\mathrm{N}$ bonds, $3^{\circ}$ amines are more polar than hydrocorbons which are almost non-polar. Hence, due to weak dipole-dipole interactions, $3^{\circ}$ amines have higher boiling point (i.e., less volatile) than hydrocarbons.

In other words, hydrocarbons are more volatile among given compounds as amine are less volatile than hydrocarbon.

• (a) II: This option is incorrect because $1^{\circ}$ amines have higher boiling points due to intermolecular hydrogen bonding, making them less volatile compared to hydrocarbons.

• (c) I: This option is incorrect because $2^{\circ}$ amines also have higher boiling points due to intermolecular hydrogen bonding, making them less volatile compared to hydrocarbons.

• (d) III: This option is incorrect because $3^{\circ}$ amines, although they do not have hydrogen bonding, still have polar $\mathrm{C}-\mathrm{N}$ bonds leading to dipole-dipole interactions, making them less volatile compared to hydrocarbons.

Answer

$(a, b, c)$

Aliphatic and arylalkyl primary amines can be easily prepared by the reduction of the corrsponding nitriles with $\mathrm{LiAlH} _{4}$.

$$ R - \underset{\text { Alkynitrile }}{\mathrm{C}} \equiv \mathrm{N} \text { or } \mathrm{Ar} - \underset{\text { Arynitrile }}{\mathrm{C}} \equiv \mathrm{N} \rightarrow \mathrm{LiAlH} _{4} \underset{1^{\circ} \text { amine }}{R \mathrm{CH} _{2} \mathrm{NH} _{2} \text { or }} \mathrm{ArCH} _{2} \mathrm{NH} _{2} $$

Heating alkyl halide with Primary, secondary and tertiary amine can be prepared by reduction of $\mathrm{LiAlH} _{4}$ followed by treatment with water.

$$ R-\underset{1^{\circ} \text { amide }}{\mathrm{CONH} _{2}} \xrightarrow[\text { (ii) } \mathrm{H} _{2} \mathrm{O}]{\text { (i) } \mathrm{LiAlH} _{4} / \text { ether }} R-\mathrm{CH} _{2}-\mathrm{NH} _{2} $$

Heating alkyl halide with potassium salt of phthalimide followed by hydrolysis produces primary amine. This process is known as Gabriel phthalimide reaction. The number of carbon atoms in the chain of amines of product is same as reactant.

• Option (d): Treatment of amide with bromine in aqueous solution of sodium hydroxide.

  This method is known as the Hofmann rearrangement (or Hofmann degradation). In this reaction, an amide is converted to a primary amine with one fewer carbon atom than the original amide. Therefore, the number of carbon atoms in the chain of the amine product is not the same as in the reactant.

Answer

(c, d)

Sandmeyer’s reaction is used for preparation of chlorobenzene and bromobenzene.

lodobenzene and fluorobenzene can be prepared by direct reaction of diazonium salt with $\mathrm{KI}$ and $\mathrm{HBF} _{4} / \Delta$.

• Chlorobenzene: Chlorobenzene can be prepared by Sandmeyer’s reaction, where benzene diazonium chloride reacts with cuprous chloride (CuCl) to form chlorobenzene.

• Bromobenzene: Bromobenzene can be prepared by Sandmeyer’s reaction, where benzene diazonium chloride reacts with cuprous bromide (CuBr) to form bromobenzene.

Answer

$(a, b, c)$

Reduction of nitrobenzene by $\mathrm{Sn} / \mathrm{HCl}, \mathrm{Fe} / \mathrm{HCl}$ and $\mathrm{H} _{2}$ - $\mathrm{Pd}$ gives aniline as follows

• Option (d) $\mathrm{Sn} / \mathrm{NH} _{4} \mathrm{OH}$ is incorrect because $\mathrm{NH} _{4} \mathrm{OH}$ (ammonium hydroxide) is a weak base and does not provide the necessary acidic conditions required for the reduction of nitrobenzene to aniline. The reduction typically requires an acidic medium, which is provided by $\mathrm{HCl}$ in options (a) and (b).

Answer

( $a, b$ )

Carbylamine reaction Amine on reaction with a mixture of $\mathrm{CHuCl} _{3}$ and $\mathrm{KOH}$ produces alkyl isocyanate. $R-\mathrm{NH} _{2}+\mathrm{CHCl} _{3}+3 \mathrm{KOH} \longrightarrow \mathrm{RNC}+3 \mathrm{KCl}+3 \mathrm{H} _{2} \mathrm{O}$

Only $R \mathrm{NC}$ and $\mathrm{CHCl} _{3}$ are involved in carbylamine reaction. Hence, (a) and (b) are correct.

• Option (c) $\mathrm{COCl}_2$: Phosgene ($\mathrm{COCl}_2$) is not involved in the carbylamine test. The carbylamine reaction specifically requires chloroform ($\mathrm{CHCl}_3$) and a primary amine, not phosgene.

• Option (d) $\mathrm{NaNO}_2 + \mathrm{HCl}$: Sodium nitrite ($\mathrm{NaNO}_2$) and hydrochloric acid ($\mathrm{HCl}$) are used in the diazotization reaction, not in the carbylamine test. The carbylamine test involves chloroform ($\mathrm{CHCl}_3$) and a primary amine, not these reagents.

Answer

$(b, c)$

Benzene diazonium chloride can be converted into benzene using protic acid as follows

• (a) $\mathrm{SnCl} _{2} / \mathrm{HCl}$: This reagent is typically used for the reduction of nitro groups to amines or for the reduction of other functional groups, but it is not effective for converting benzenediazonium chloride to benzene.

• (d) $\mathrm{LiAlH} _{4}$: Lithium aluminium hydride is a strong reducing agent commonly used to reduce carbonyl compounds, nitro groups, and other functional groups. However, it is not suitable for the reduction of benzenediazonium chloride to benzene.

Answer

$(a, b)$

$\mathrm{N}$-acetylaniline on reaction with $\mathrm{Br} _{2}$ in presence of acetic acid produces p-bromo $\mathrm{N}$-acetyl aniline (major) and o-bromo-N acetyl aniline (minor) as follows

The $\mathrm{N}$-acetyl group is a ortho, para directing group.

Hence, (a) and (b) are correct.

• The $\mathrm{N}$-acetyl group is an ortho, para directing group, which means it directs incoming substituents to the ortho and para positions relative to itself. Therefore, any product that does not place the bromine atom at the ortho or para positions relative to the $\mathrm{N}$-acetyl group is incorrect.

• If an option suggests a meta-substituted product, it is incorrect because the $\mathrm{N}$-acetyl group does not direct substituents to the meta position.

• If an option suggests a product with substitution at a position that is not adjacent to the benzene ring (e.g., on the acetyl group itself), it is incorrect because the bromination reaction occurs on the aromatic ring.

• If an option suggests a product with multiple bromine atoms or bromination at positions other than ortho or para, it is incorrect because the reaction conditions specified (bromine in acetic acid) typically lead to mono-substitution at the ortho and para positions.

• If an option suggests a product with no bromine substitution, it is incorrect because the question specifies a bromination reaction.

Answer

$(a, b, c)$

Arenium ion involved in the bromination of aniline are as follows

• Option (d) is incorrect because it does not represent a resonance structure of the arenium ion formed during the bromination of aniline. The positive charge is not delocalized correctly in the aromatic ring.
• Option (e) is incorrect because it does not show the correct position of the positive charge in the arenium ion. The positive charge should be on the ortho or para positions relative to the amino group, not meta.
• Option (f) is incorrect because it does not depict a valid resonance structure of the arenium ion. The positive charge is not properly delocalized within the aromatic ring, and the structure does not follow the rules of resonance stabilization.

Answer

$(a, b)$

Isobutylamine and 2-phenylethyl amine are primary amine can be prepared easily by Gabriel phthalimide reaction.

Refer to answer of question 8.

• (c) N-methylbenzylamine: This compound is a secondary amine, not a primary amine. Gabriel synthesis is specifically used for the preparation of primary amines, and thus it cannot be used to prepare secondary amines like N-methylbenzylamine.

• (d) Aniline: Aniline is an aromatic amine. Gabriel synthesis is not suitable for the preparation of aromatic amines because the reaction conditions and intermediates are not compatible with the aromatic ring structure.

Answer

(a, c)

This is an example of nucleophilic subsitution reaction.

This is an example of elimination reaction.

• Option (b): The reaction shown in option (b) is incorrect because it does not follow the mechanism of nucleophilic substitution or elimination. The product formed is not consistent with the expected outcome of such reactions.

• Option (d): The reaction shown in option (d) is incorrect because it does not follow the expected mechanism of elimination or nucleophilic substitution. The product formed is not consistent with the typical products of these reactions.

Answer

$(a, b)$

Aniline or reaction with acetyl chloride or acetic anhydride in the presence of pyridine produces $\mathrm{N}$-acetyl aniline which is a ortho, para directing group which on further reaction with nitrating mixture (conc. $\mathrm{HNO} _{3}+$ conc. $\mathrm{H} _{2} \mathrm{SO} _{4}$ ) produces $p$-nitroaniline preferentially as shown below.

• Option (c): Dilute HCl followed by reaction with conc. H₂SO₄ and conc. HNO₃ is incorrect because the presence of HCl will protonate the aniline to form anilinium ion, which is a meta-directing group. This will lead to the formation of meta-nitroaniline as the major product rather than the para derivative.

• Option (d): Direct reaction with conc. HNO₃ and conc. H₂SO₄ is incorrect because aniline is highly reactive towards nitration and can lead to multiple nitration products, including ortho, meta, and para derivatives. Additionally, the strong acidic conditions can lead to the formation of anilinium ion, which is meta-directing, thus not favoring the para product specifically.

Answer

$(a, b)$

Bromination of acetanilide and coupling reaction of aryldiazonium salts is an example of electrophilic aromatic substitution reaction.

Coupling reaction of aryldiazonium salts

• (c) Diazotisation of aniline: This reaction involves the conversion of aniline into a diazonium salt using nitrous acid. It is not an electrophilic aromatic substitution reaction but rather a diazotization reaction, which is a type of substitution reaction where the amino group is replaced by a diazonium group.

• (d) Acylation of aniline: This reaction involves the introduction of an acyl group into the aniline molecule, typically using an acyl chloride or anhydride in the presence of a Lewis acid catalyst. This is an example of an electrophilic aromatic substitution reaction, but it specifically falls under the category of Friedel-Crafts acylation. However, the question may be interpreted to focus on more common examples of electrophilic aromatic substitution like halogenation and coupling reactions, which might be why it was not included in the correct options.

Answer

$\mathrm{HNO} _{3}$ acts as a base in the nitrating mixture and provide the electrophile, $\mathrm{NO} _{2}^{+}$on reaction with $\mathrm{H} _{2} \mathrm{SO} _{4}$ as follows

Answer

In order to check the activation of benzene ring by amino group, first it is acetylated with acetic anhydride or acetyl chloride in presence of pyridine to form acetanilide which can be further nitrated easily by nitrating mixture.

Answer

$\mathrm{C} _{6} \mathrm{H} _{5} \mathrm{CH} _{2} \mathrm{NH} _{2}$ on reaction with $\mathrm{HNO} _{2}$ produces $\mathrm{C} _{6} \mathrm{H} _{5} \mathrm{CH} _{2} \mathrm{~N} _{2}^{+} \mathrm{Cl}^{-}$as follows

Answer

Best reagent to convert nitrile to aniline is sodium/alcohol or $\mathrm{LiAlH} _{4}$.

Thinking Process

This problem is based on the concept of preparation of diazonium salt and its chemical properties.

Answer

Complete conversion can be shown as

Answer

Benzene sulphonyl chloride $\left(\mathrm{C} _{6} \mathrm{H} _{5} \mathrm{SO} _{2} \mathrm{Cl}\right)$ is known as Hinsberg reagent. It is used to distinguish between primary, secondary and tertiary amine.

Answer

Benzene diazonium chloride are highly unstable and stable for a very short time span in solution at low temperature. Due to its instability, it is used immediately after its preparation.

Answer

Acylation of $-\mathrm{NH} _{2}$ group of aniline reduces its activity due to resonance of lone pair of nitrogen towards the carbonyl group hence $0^{-}, p^{-}$directive influence of amino group get disturbed.

Answer

Basicity of $\mathrm{MeNH} _{2}$ and $\mathrm{MeOH}$ can be explained on the basis of electronegativity of $\mathrm{N}$ and $\mathrm{O}$ atom. $\mathrm{MeNH} _{2}$ is stronger base than $\mathrm{MeOH}$ because of low electronegativity value of $\mathrm{N}$, it is easy for nitrogen to loose its lone pair readily than compared to $\mathrm{MeOH}$.

Answer

Pyridine being a base, is used to remove the side product i.e., $\mathrm{HCl}$ from reaction mixture.

Answer

In strongly basic conditions, benzenediazonium chloride is converted into diazohydroxide and diazoate as both of which are not electrophilic and do not couple with aniline.

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$$ \mathrm{C} _{6} \mathrm{H} _{5} \stackrel{+}{\mathrm{N}} \equiv \mathrm{NC} \overline{\mathrm{I}}+\mathrm{OH} \mathrm{SO} _{2} \underset{\substack{\text { Diazohydroxide }}}{\mathrm{CH} _{5}-\mathrm{N}=\mathrm{N}-\mathrm{OH}} \stackrel{\mathrm{NaOH}}{\rightleftharpoons} \mathrm{C} _{6} \mathrm{H} _{5}-\mathrm{N}=\mathrm{N}-\overline{\mathrm{O}} \mathrm{Na}^{+} $$

Similarly, in highly acidic conditions, aniline gets converted into anilinium ion. From this, result aniline is no longer nucleophilic acid and hence will not couple with diazonium chloride. Hence, the reaction is carried out under mild conditions, i.e., $\mathrm{pH}^{-1}-4-5$

$$ \underset{\text { Aniline }}{\mathrm{C} _{6} \mathrm{H} _{5} \mathrm{NH} _{2}}+\mathrm{H}^{+} \rightarrow \underset{\substack{\text { Anilinium ion } \\ \text { (coupling do not occur) }}}{\mathrm{C} _{6} \mathrm{H} _{5}-\stackrel{+}{\mathrm{N}} \mathrm{H} _{3}} $$

Answer

Aniline on reaction with $\mathrm{Br} _{2}$ in non-polar solvent $\mathrm{CS} _{2}$ produces $2,4,6$ tribromo aniline.

Aniline has high reactivity towards bromine as it gives the triply substituted product.

Answer

Dipole moment of amine, alcohol and hydrocarbon can be explained on the basis of bond polarity of $\mathrm{C}-\mathrm{H}, \mathrm{N}-\mathrm{H}$ and $\mathrm{O}-\mathrm{H}$ bond. As the bond polarity increase, dipole moment increases $\mathrm{CH} _{3} \mathrm{CH} _{2} \mathrm{CH} _{3}<\mathrm{CH} _{3} \mathrm{CH} _{2} \mathrm{NH} _{2}<\mathrm{CH} _{3} \mathrm{CH} _{2} \mathrm{OH}$

Answer

Structural formula of allyl amine is as follows

$$ \underset{\text { Prop-2-ene -1-amine (IUPAC name) }}{\stackrel{3}{\mathrm{C}} \mathrm{H} _{2}=\stackrel{1}{\mathrm{C}} \mathrm{H}-\mathrm{CH} _{2}-\mathrm{NH} _{2}} $$

Answer

$\mathrm{N}, \mathrm{N}$-dimthyl benzenamine

During naming of $\mathrm{N}$-substituted amine, substituted group present at $\mathrm{N}$ are added as suffix on $\mathrm{N}$-alkyl in IUPAC nomenclature.

Thinking Process

This process is based on concept of Hinsberg test. Only amine containing replaceable $\mathrm{H}$ gives Hinsberg test.

Answer

$Z\left(\mathrm{C} _{3} \mathrm{H} _{9} \mathrm{~N}\right)$ is an aliphatic amine. On reaction with $\mathrm{C} _{6} \mathrm{H} _{5} \mathrm{SO} _{2} \mathrm{Cl}$ (Hinsberg’s reagent), it gives a product insoluble in alkali. It means that the product does not have a replaceable $\mathrm{H}$-atom attached to the $\mathrm{N}$ - atom. So, compound $\mathrm{Z}$ is a secondary amine (ethyl methyl amine).

Answer

Primary amines show carbylamine reaction in which two $\mathrm{H}$-atoms attached to $\mathrm{N}$-atoms of $\mathrm{NH} _{2}$ are replaced by one $\mathrm{C}$-atom. On catalytic reduction, isocyanide (formed) produces secondary amine and not tertiary or quaternary salts.

Answer

The reaction exhibits azo-coupling reaction of phenols. Benzene diazonium chloride reacts with phenol in such a manner that the para position of phenol is coupled with diazonium salt to form $p$-hydroxy azobenzene.

Answer

Aniline is soluble in aqueous $\mathrm{HCl}$ due to formation of ionic anilinium chloride.

Answer

Complete conversion can be performed as

Answer

Complete conversion can be performed as

(B)

Hence,

Answer

(i) Conversion of toluene to $p$-toluidine can be done as

(ii) Conversion of $p$-toluidine diazonium chloride to $p$-toluic acid can be done as

Answer

(i) Nitrobenzene can be converted into acetanilide as follows

(ii) Acetanilide can be converted into $p$ - nitroaniline as follows

Thinking Process

This problem is based upon conceptual mixing of electrophilicity of ring system and diazo-coupling reaction.

Answer

The above stated reaction is an example of electrophilic aromatic substitution. In alkaline medium, phenol generates phenoxide ion which is more electron rich than phenol and more reactive for electrophilic attack.

The electrophile in this reaction is aryldiazonium cation. As we know, stronger the electrophile faster is the reaction. $p$-nitrophenyldiazonium cation is a stronger electrophile than $p$-toluene diazonium cation.

So, nitrophenyl diazonium chloride couples preferentially with phenol.

Thinking Process

This problem includes conceptual mixing of bromination, nitration and Sandmeyer’s reaction. Follow the steps to approach towards given product.

Bromination of p-nitroaniline followed by diazotisation and Sandmeyer’s reaction

Answer

Complete conversion of above reaction can be shown as

3, 4, 5-tribromonitrobenzene

Answer

Conversion of benzene to $p$-nitroaniline can be done as

Answer

Conversion of aniline to $m$-bromo nitrobenzene can be completed as

Answer

(i) Conversion of aniline to 3,5-dibromonitrobenzene can be completed as

(ii) Conversion (A) given below is same as in part (i) given above after that reaction $(B)$ can be carried out.

Answer

A. $\rightarrow(4)$

B. $\rightarrow(3)$

C. $\rightarrow(1)$

D. $\rightarrow(2)$

Answer

A. $\rightarrow(2) \quad$

B. $\rightarrow(1) \quad$

C. $\rightarrow(4) \quad$

D. $\rightarrow(3)$

Answer

(c) Assertion is correct statement but reason is wrong statement.

Acylation of amine gives a monosubstituted product whereas alkylation of amine gives polysubstituted product because acylation in amine takes place at $\mathrm{N}$-atom and alkylation takes place at $o$ and $p$ position.

Answer

(a) Both assertion and reason are wrong.

Correct Assertion Hofmanns bromamide reaction is given by amide.

Correct Reason Amide on reaction with $\mathrm{NaOX}$ produces amine having one carbon less than amide.

Answer

(d) Both assertion and reason are correct and reason is the correct explanation of assertion.

$\mathrm{N}$-ethylbenzene is soluble in alkali because hydrogen attached to nitrogen in sulphonamide is strongly acidic and forms a salt during reaction between these two.

Answer

(d) Both assertion and reason are correct and reason is not the correct explanation of assertion.

$\mathrm{N}, \mathrm{N}$-diethylbenzene sulphonamide is insoluble in alkali due to absence of acidic $\mathrm{H}$ attached to nitrogen.

Answer

(d) Assertion and reason both are correct and reason is the correct explanation of assertion.

Only small amount of $\mathrm{HCl}$ is required in the reduction of nitro compounds with iron scrap and $\mathrm{HCl}$ in the presence of steam because $\mathrm{FeCl} _{2}$ formed gets hydrolysed to release $\mathrm{HCl}$ during the reaction.

Answer

(a) Both assertion and reason are wrong. Aryl $1^{\circ}$ amine can’t be prepared by Gabriel phthalimide reaction because aryl halide don’t undergo nucleophilic substitution with anion formed by phthalimide.

Answer

(d) Assertion and reason both are correct and reason is the correct explanation of assertion.

Acetanilide is less basic than aniline because acetylation of aniline results in decrease of electron density on nitrogen.

Thinking Process

This problem includes conceptual mixing of ozonolysis, optical activity, ammonolysis and diazotisation. Follow the steps to solve the problem

Analyse the overall reaction once then sequentially predict a molecule for each $A, B, C$ and $D$ on the basis of information provided in the question.

Fit every molecule in a flow chart made by using information provided in the question and reach to the correct compounds.

Answer

(i) Addition of $\mathrm{HCl}$ to compound ’ $A$ ’ shows that compound ’ $A$ ’ is alkene. Compound ’ $B$ ’ is $\mathrm{C} _{4} \mathrm{H} _{9} \mathrm{Cl}$.

(ii) Compound ’ $B$ ’ reacts with $\mathrm{NH} _{2}$, it forms amine ’ $C$ ‘.

$$ \underset{[A]}{\mathrm{C} _{4} \mathrm{H} _{8}} \xrightarrow{\mathrm{HCl}} \underset{[B]}{\mathrm{C} _{4} \mathrm{H} _{9} \mathrm{Cl}} \xrightarrow{\mathrm{NH} _{3}} \underset{[C]}{\mathrm{C} _{4}} \mathrm{H} _{11} \mathrm{~N} \text { or } \mathrm{C} _{4} \mathrm{H} _{9} \mathrm{NH} _{2} $$

(iii) ’ $C$ ’ gives diazonium salt with $\mathrm{NaNO} _{2} / \mathrm{HCl}$, which yields an optically active alcohol. So, ’ $C$ ’ is aliphatic amine.

(iv) ’ $A$ ’ on ozonolysis produces 2 moles of $\mathrm{CH} _{3} \mathrm{CHO}$. So, ’ $A$ ’ is $\mathrm{CH} _{3}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH} _{3}$ (But-2-ene).

Reactions

(i)

(ii)

(iii)

(iv)

Thinking Process

This problem is based on chemical properties of aniline and property and solubility of their derivatives.

Answer

Answer

Correct sequence can be represented as follows including all reagents.

Hence,