Answer

(b) The structure of given amines are as follows

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

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.

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).

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.

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 $\left(\mathrm{C} _{6} \mathrm{H} _{5} \mathrm{CH} _{2}\right)$.

Answer

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

Hence, option (b) is the correct choice.

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

Answer

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

Answer

(c)

Chemical transformation can be shown as

While other given set of reactants give primary amine only.

Answer

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

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.

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.

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.

Answer

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

Answer

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

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.

Answer

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

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.

Answer

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

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.

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

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.

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.

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.

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.

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.

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.

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 $\mathrm{HBF} _{4} / \Delta$.

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

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.

Answer

$(b, c)$

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

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.

Answer

$(a, b, c)$

Arenium ion involved in the bromination of aniline are as follows

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.

Answer

(a, c)

This is an example of nucleophilic subsitution reaction.

This is an example of elimination reaction.

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.

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

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,