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**Answer** (c)

$S_N 1$ reaction proceeds through formation of carbocation. Hence, more stable be the carbocation more reactivity towards $S   _{N} 1$ mechanism.

|   |Alkyl <br> halides |  | Intermediate ||
| :--- | :--- | :--- | :--- | :--- |
| (a)| $CH _3 Br$ | $\longrightarrow$ | $CH _3^{\oplus}$ |  |
| (b) |$C _6 H _5 Br$ | $\longrightarrow$ | $C _6 H _5^{\oplus}$ |  |
| (c) |$C _6 H _5 CH _2 Br$ | $\longrightarrow$ | $C _6 H _5-CH _2^{\oplus}$ (more stable) |  |
| (d) |$C _2 H _5 Br$ | $\longrightarrow$  | $C _2 H _5^{\oplus}$ |  |

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Hence, the reaction will proceed through $S   _{N} 1$ mechanism when, $C _6 H _5 CH _2 Br$ is the substrate. because on ionisation it gives a resonance stabilised carbocation $(C _6 H _5 CH _2)$.

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**Answer** (b)

The best reagent for converting 2-phenylpropanamide into 1-phenylethanamine is by $NaOH / Br _2$ and chemical transformation can be done as

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

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**Ques.** The correct increasing order of basic strength for the following compounds is...... .

(a) II $<$ III $<$ I

(b) III $<$ I $<$ II

(d) II $<$ I $<$ III

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**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.

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**Answer** (c)

Nitration of benzene using a mixture of conc. $H _2 SO _4$ and conc. $HNO _3$ proceeds as

This reaction is known as electrophilic substitution reaction.

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**Answer** (b)

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

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**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., $-OH _1-NH _2$ and $-OCH _3$ groups and not with compounds containing electron withdrawing group, i.e., $NO _2$ etc.

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**Answer** (c)

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

Oxygen has more electronegative than $N$. So, $O-H$ 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.

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**Answer** (d)

Aniline is a weaker base than $NH _3$ due to delocalization of lone pair of electrons of the $N$-atom over the benzene ring. pyrrole is not more basic because the lone pair of electrons on the $N$-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.

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**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.

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**Answer** (b)

$1^{\circ}$ and $2^{\circ}$ amines have higher boiling points due to intermolecular $H$-bonding but less votatile than $3^{\circ}$ amines and hydrocarbons of comparable molecular mass. Further, because of polar $C-N$ 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.

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**Ques.** Which of the following methods of preparation of amines will give same number of carbon atoms in the chain of amines as in the reactant?

(a) Reaction of nitrite with $LiAlH _4$

(b) Reaction of amide with $LiAlH _4$ followed by treatment with water

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

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**Answer** $(a, b, c)$

Aliphatic and arylalkyl primary amines can be easily prepared by the reduction of the corrsponding nitriles with $LiAlH _4$.

$
R - \underset{\text{ Alkynitrile }}{C} \equiv N \text{ or } Ar - \underset{\text{ Arynitrile }}{C} \equiv N \to LiAlH _4 \underset{1^{\circ} \text{ amine }}{R CH _2 NH _2 \text{ or }} ArCH _2 NH _2
$

Heating alkyl halide with Primary, secondary and tertiary amine can be prepared by reduction of $LiAlH _4$ followed by treatment with water.

$
R-\underset{1^{\circ} \text{ amide }}{CONH _2} \xrightarrow[\text{ (ii) } H _2 O]{\text{ (i) } LiAlH _4 / \text{ ether }} R-CH _2-NH _2
$

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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.

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**Answer** $(a, b)$

$N$-acetylaniline on reaction with $Br _2$ in presence of acetic acid produces p-bromo $N$-acetyl aniline (major) and o-bromo-N acetyl aniline (minor) as follows

The $N$-acetyl group is a ortho, para directing group.

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

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**Ques.** Under which of the following reaction conditions, aniline gives p-nitro derivative as the major product?

(a) Acetyl chloride/pyridine followed by reaction with conc. $H _2 SO _4+$ conc. $HNO _3$

(b) Acetic anhydride/pyridine followed by conc. $H _2 SO _4+$ conc. $HNO _3$

(d) Reaction with conc. $HNO _3+$ conc. $H _2 SO _4$

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**Answer** $(a, b)$

Aniline or reaction with acetyl chloride or acetic anhydride in the presence of pyridine produces $N$-acetyl aniline which is a ortho, para directing group which on further reaction with nitrating mixture (conc. $HNO _3+$ conc. $H _2 SO _4$ ) produces $p$-nitroaniline preferentially as shown below.

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**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.

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**Answer**

Acylation of $-NH _2$ 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.

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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., $pH^{-1}-4-5$

$
\underset{\text{ Aniline }}{C _6 H _5 NH _2}+H^{+} \to \underset{\substack{\text{ Anilinium ion } \\\\\\ \text{ (coupling do not occur) }}}{C _6 H _5-\stackrel{+}{N} H _3}
$

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**Answer**

Aniline on reaction with $Br _2$ in non-polar solvent $CS _2$ produces $2,4,6$ tribromo aniline.

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

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**Answer**

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$N, N$-dimthyl benzenamine

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

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**Answer**

$Z(C _3 H _9 N)$ is an aliphatic amine. On reaction with $C _6 H _5 SO _2 Cl$ (Hinsberg's reagent), it gives a product insoluble in alkali. It means that the product does not have a replaceable $H$-atom attached to the $N$ - atom. So, compound $Z$ is a secondary amine (ethyl methyl amine).

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**Answer**

Primary amines show carbylamine reaction in which two $H$-atoms attached to $N$-atoms of $NH _2$ are replaced by one $C$-atom. On catalytic reduction, isocyanide (formed) produces secondary amine and not tertiary or quaternary salts.

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**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.

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**Ques.** Match the reactions given in Column I with the statements given in Column II.

| |Column I | |Column II   |
| :--- | :--- | :--- | :--- |
| A. | Ammonolysis | 1. | Amine with lesser number of carbon atoms |
| B. | Gabriel phthalimide synthesis | 2. | Detection test for primary amines. |
| C. | Hofmann bromamide reaction | 3. | Reaction of phthalimide with $KOH$ and $R-X$ |
| D. | Carbylamine reaction | 4. | Reaction of alkylhalides with $NH _3$ |

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#### Assertion and Reason

In the following questions a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct answer out of the following choices.
 
(a) Both assertion and reason are wrong.

(b) Both assertion and reason are correct statements but reason is not correct explanation of assertion.

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

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

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**Ques.** A hydrocarbon ' $A$ ' $(C _4 H _8)$ on reaction with $HCl$ gives a compound 'B', $(C _4 H _9 Cl)$, which on reaction with $1 mol$ of $NH _3$ gives compound ' $C$ ', $(C _4 H    _{11} N)$. On reacting with $NaNO _2$ and $HCl$ followed by treatment with water, compound ' $C$ ' yields an optically active alcohol, ' $D$ '. Ozonolysis of ' $A$ ' gives 2 mols of acetaldehyde. Identify compounds ' $A$ ' to ' $D$ '. Explain the reactions involved.

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**Answer**

(i) Addition of $HCl$ to compound ' $A$ ' shows that compound ' $A$ ' is alkene. Compound ' $B$ ' is $C _4 H _9 Cl$.

(ii) Compound ' $B$ ' reacts with $NH _2$, it forms amine ' $C$ '.

$
\underset{[A]}{C _4 H _8} \xrightarrow{HCl} \underset{[B]}{C _4 H _9 Cl} \xrightarrow{NH _3} \underset{[C]}{C _4} H    _{11} N \text{ or } C _4 H _9 NH _2
$

(iii) ' $C$ ' gives diazonium salt with $NaNO _2 / HCl$, which yields an optically active alcohol. So, ' $C$ ' is aliphatic amine.

(iv) ' $A$ ' on ozonolysis produces 2 moles of $CH _3 CHO$. So, ' $A$ ' is $CH _3-CH=CH-CH _3$ (But-2-ene).

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**Explanation**

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.

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**Ques.** A colourless substance ' $A$ ' $(C_6 H_7 N)$ is sparingly soluble in water and gives a water soluble compound ' $B$ ' on treating with mineral acid. On reacting with $CHCl _3$ and alcoholic potash ' $A$ ' produces an obnoxious smell due to the formation of compound ' $C$ '. Reaction of ' $A$ ' with benzenesulphonyl chloride gives compound ' $D$ ' which is soluble in alkali. With $NaNO _2$ and $HCl$, ' $A$ ' forms compound ' $E$ ' which reacts with phenol in alkaline medium to give an orange dye ' $F$ '. Identify compounds ' $A$ ' to ' $F$ '.

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