1 Marks Question

Question 11 Mark

Give a chemical test to distinguish between a primary and a secondary amine.

Answer

Aromatic primary amine gives orange dye on treating with ice cold NaNO₂ + HCl followed by β-naphthol whereas secondary amine does not give this test/or describe carbyl amine test or Hinsberg test.

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Question 21 Mark

How will you convert:
Methanol to ethanoic acid.

Answer

Conversion of methanol to Ethanoic acid.
$\text{CH}_3\text{OH} \xrightarrow[\text{P}/\text{I}_2]{\text{P}\text{I}_3}\text{CH}_3\text{I}\xrightarrow[\text{alc}]{\text{KCN}}\text{CH}_3\text{CN}\xrightarrow[(\text{dil HCl})]{\text{Hydrolysis}}\\\text{Methanol}\ \ \ \ \ \ \ \text{Methyl} \ \ \ \ \ \ \ \text{Methyl}\ \ \ \ \ \ \ \ \ \ \ \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{iodide}\ \ \ \ \ \ \ \ \ \text{iodide}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\text{COOH}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Ethanic acid}$

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Question 31 Mark

Give plausible explanation for each of the following:
Why are aliphatic amines stronger bases than aromatic amines?

Answer

In aliphatic amines, alkyl groups are electron releasing and increases the electron pair availability at N atom while aryl group in aromatic amines are electron withdrawing decreasing the availability of electron pair at N atom.

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Question 41 Mark

Write short notes on the following:
Gabriel phthalimide synthesis.

Answer

Gabriel phthalimide synthesis: Primary amines can be prepared by this process. In this process, phthalimide is reacted with alcoholic KOH to get potassium phthalimide which reacts with an alkyl halide to form N-alkyl phthalimide which on basis hydrolysis gives primary amine and phthalic acid. Phthalic acid can be reused to get phthalimide.

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Question 51 Mark

How will you convert:
Hexanenitrile into 1-aminopentane.

Answer

Hexane nitrite into 1-amino pentane

$\text{CH}_3(\text{CH}_2)\text{CN}\xrightarrow[(\text{Partial hydrolysis})]{\text{H}_2\text{O}/ \text{H}^+}\text{CH}_3(\text{CH}_2)_4\text{CONH}_2\xrightarrow[]{\text{Br}_2/\text{KOH}}\text{CH}_3(\text{CH}_2)_4\text{NH}_2\\\text{Hexane nitrate} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Hexanamide} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{I-amino Pentane}$

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Question 61 Mark

Convert
Aniline into 1, 3, 5 - tribromobenzene.

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Question 71 Mark

Write reactions of the final alkylation product of aniline with excess of methyl iodide in the presence of sodium carbonate solution.

Answer

Aniline reacts with methyl iodide to produce N, N-dimethylaniline.

With excess methyl iodide, in the presence of Na₂CO₃ solution, N, N-dimethylaniline produces N, N, N-trimethylanilinium carbonate.

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Question 81 Mark

Why cannot aromatic primary amines be prepared by Gabriel phthalimide synthesis?

Answer

Gabrielphthalimide synthesis is used for the preparation of aliphatic primary amines. It involves nucleophilic substitution (SN₂) of alkyl halides by the anion formed by the phthalimide.

But aryl halides do not undergo nucleophilic substitution with the anion formed by the phthalimide.

Hence, aromatic primary amines cannot be prepared by this process.

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Question 91 Mark

Account for the following:
Although amino group is o– and p– directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m-nitroaniline.

Answer

The nitration of aniline is carried out using conc. HNO₃ and H₂SO₄. However, in the presence of conc. H₂SO₄, aniline forms aniline hydrogen sulphate in which the anilinium ion, C₆H₅NH₃⁺ is meta directing because the positive charge on the nitrogen attracts electrons from the benzene ring.

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Question 101 Mark

Arrange the following:
In increasing order of basic strength:

Aniline, p-nitroaniline and p-toluidine.
C₆H₅NH₂, C₆H₅NHCH₃, C₆H₅CH₂NH₂.

Answer

P-nitro aniline < aniline < p-toluidine.
C₆H₅NHCH₃ < C₆H₅NH₂ < C₆H₅CH₂NH_2.

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Question 111 Mark

Write short notes on the following:
Acetylation.

Answer

Acetylation: The process, in which acetyl group $ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{O} \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ||\\\ \ \ \ \Bigg(\text{CH}_3-\text{C}-\Bigg)$ is introduced, is called acety lation. It is done by reaction with acetyl chloride or acetic
anhydride.

Aliphatic and aromatic primary and secondary amines undergo acetylation reaction by nucleophilic substitution when treated with acid chlorides, anhydrides or esters.

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Question 121 Mark

Write short notes on the following:
Diazotisation.

Answer

Diazotization: It is a process of treating primary aromatic amine with nitrous acid at 273 - 278 K to get diazonium salts which are very useful compounds.

They are used to prepare benzene, phenol, halo-arenes, cyano benzene, dyes, etc.

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Question 131 Mark

Give one chemical test to distinguish between the following pairs of compounds.
Ethylamine and aniline.

Answer

Ethylamine and aniline: To the ice cold solution of both these compounds prepared in excess of dilute HCl, add ice cold solution of sodium nitrite in water and of β—Naphthol (2—Naphthol) prepared in diluted in sodium hydroxide solution.

Further, cool the reaction mixture in both the cases. The mixture which forms a brilliant orange dye (azodye) contains aniline while the one in which no dye is formed has ethylamine present in it.

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Question 141 Mark

Account for the following:
Diazonium salts of aromatic amines are more stable than those of aliphatic amines.

Answer

Primary aliphatic amines form a highly unstable alkyl diazonium salts. Primary aromatic amines form arene diazonium salts which are stable for a short time in solution at low temperature (273 - 278 K). The stability of diazonium ions of aromatic amines is explained on the basis of resonance.

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Question 151 Mark

Convert
3-Methylaniline into 3-nitrotoluene.

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Question 161 Mark

Account for the following:
Ethylamine is soluble in water whereas aniline is not.

Answer

In C₂H₅NH₂, the C₂H₅ group has a + I effect and increases the electron density on N atom. This results in stronger intermolecular H-bonding. While in C₆H₅NH₂ due to resonance, N acquires a +ve charge and electron density on N decreases. The tendency to form H-bonding diminishes. Hence, ethylamine is soluble while C₆H₅NH₂ is insoluble.

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Question 171 Mark

How will you convert?
Benzene into N, N-dimethylaniline.

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Question 181 Mark

Write short notes on the following:
Coupling reaction

Answer

Coupling reaction: When benzene diazonium chloride reacts with phenol in which the phenol molecule at its para position is coupled with coupled with the diazonium salt to form p-hydroxyazobenzene. This is known as coupling reaction.

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Question 191 Mark

Write short notes on the following:
Carbylamine reaction.

Answer

Carbylamine reaction: The carbylamines reaction test used for detection of primary amines. In this reaction, the analyte/ given compound is heated with alcoholic potassium hydroxide and chloroform. If a primary amine is present, the isocyanide(carbylamines) is formed which gives unpleasant smell.

$\text{CH}_3\text{CH}_2\text{NH}_2+\text{CHCl}_3+3\text{KOH}(\text{alc})\xrightarrow[]{\text{warm}}\\\text{Ethyl amine}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\text{CH}_2\stackrel{{\rightarrow}}{=}\text{C}+3\text{KCl}+3\text{H}_2\text{O}\\ \ \ \ \ \ \ \ \ \ \ \text{Ethyl isocyanide}$

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Question 201 Mark

Give one chemical test to distinguish between the following pairs of compounds.
Secondary and tertiary amines

Answer

Shake the given amines separately with Hinsberg’s reagent (benzene sulphonyl chloride) in the presence of an excess of aqueous KOH solution.

A secondary amine forms N, N—dialkyl benzene sulphonamide which remains insoluble in aqueous KOH and even after acidification with dilute HCl.

A tertiary amine does not react with benzene sulphonyl chloride and remainsinsoluble in aqueous KOH.

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Question 211 Mark

Accomplish the following conversion:
Benzoic acid to aniline.

Answer

Benzoic acid to aniline

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Question 221 Mark

Accomplish the following conversion:
Benzyl chloride to 2 - phenylethanamine.

Answer

Benzyl chloride to 2-phenyl ethanamine.

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Question 231 Mark

Account for the following:
pK_b of aniline is more than that of methylamine.

Answer

It is because, in aniline, the NH₂ group is attached directly to the benzene ring. It results in the unshared electron pair of the nitrogen atom to be in conjugation with the benzene ring and thus making it less available for protonation.
on another hand, in the case of methylamine (due to the +I effect of a methyl group), the electron density on the N-atom is increased. As a result, aniline is less basic than methylamine. Thus, pk_b of aniline is more than that of methylamine.

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Question 241 Mark

Give plausible explanation for each of the following:
Why are amines less acidic than alcohols of comparable molecular masses?

Answer

Oxygen in alcohol is more electro negative than nitrogen in amines. RC—H bond in alcohol is more polar with ^δ+ charge on.
RO^δ-H^δ+ as compared to RN—H bond in Alcohols can loose proton to some extent but are proton acceptors.
$\text{RO}-\text{H}\rightleftharpoons\text{RO}^-+\text{H}^+\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Unstable}$
$\text{R}\stackrel{{. .} }{{\text{N}}}\text{H}_2+\text{H}^+\rightarrow\text{R}\stackrel{{+} }{{\text{N}}}\text{H}_3\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Stable}$

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Question 251 Mark

Write short notes on the following:
Hofmann’s bromamide reaction.

Answer

Hoffmann bromamide degradation reaction: Hoffman develops a method for preparation of primary amines by treating an amide with bromine in an aqueous or ethanolic solution of sodium hydroxide.
In this reaction migration of an alkyl or aryl group takes place from carbonyl carbon of the amide to the nitrogen atom. The amine so formed contains one carbon less than that present in the amide.
$ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{O}\\\ \ \ \ \ \ \ \ \ \ \ \ \ ||\\\text{CH}_3-\text{C}-\text{NH}_2+\text{Br}_2+4\text{KOH}\rightarrow{\\\\\\\\\\\\}\text{CH}_3\text{NH}_2\\\text{Ethanamide} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Methanamine}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +\text{K}_2\text{CO}_3+2\text{KBr}+2\text{H}_2\text{O}$

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Question 261 Mark

Account for the following:
Aniline does not undergo Friedel-Crafts reaction.

Answer

A Friedel-crafts reaction is carried out in the presence of AlCl₃. but AlCl₃ is acidic in nature, while aniline is a strong base. Thus, aniline reacts with AlCl₃ to form a salt. Due to the positive charge on the N-atom, electrophilic substitution in the benzene ring is deactivated.Hence, aniline does not under go the Friedel-crafts reactions.
Aniline does not under Friedel-Craft reaction (alkylation and acetylation) due to the salt formation with aluminium chloride, the Lewis acid which is used as a catalyst. Due to this, the nitrogen of aniline acquires a positive charge and hence acts as a strong deactivating group for further reaction.

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Question 271 Mark

Account for the following:
Gabriel phthalimide synthesis is preferred for synthesising primary amines.

Answer

Gabriel phthalimide synthesis results in the formation of primary amine only. secondary or tertiary amines are not formed in this synthesis. thus, the pure primary amine can be obtained. Therefore, Gabriel phthalimide synthesis is preferred for synthesising primary amines.
Phthalimide is alkylated with alkyl or benzyl halide and then hydrolysed or hydrazinolysis to get pure primary amine. In this method, phthalic acid is produced which can be again converted into phthalimide and used over and over again.

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Question 281 Mark

Complete the following reaction:
$\text{C}_6\text{H}_5\text{NH}_2+(\text{CH}_3\text{CO})_2\text{O}\rightarrow$

Answer

$\text{C}_6\text{H}_5\text{NH}_2+(\text{CH}_3\text{CO})_2\text{O}\rightarrow\text{C}_6\text{H}_5-\text{N}-\text{C}-\text{CH}_3+\text{CH}_3\text{COOH}\\\text{Aniline} \ \ \ \ \ \ \ \ \text{acitic anhydride} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ | \ \ \ \ \ \ ||\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{H} \ \ \ \ \ \text{O}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{N}-\text{Phenylethanamide}$

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Question 291 Mark

Complete the following reaction:
$\text{C}_6\text{H}_5\text{N}_2\text{Cl}\xrightarrow[(\text{ii})\text{NaNO}_2/\text{Cu}.\Delta]{(\text{i})\text{HBF}_4}$

Answer

$\text{C}_6\text{H}_5\text{N}_2\text{Cl}\xrightarrow[(\text{ii})\text{NaNO}_2/\text{Cu}.\Delta]{(\text{i})\text{HBF}_4}\text{C}_6\text{H}_5\text{NO}_2+\text{N}_2+\text{NaBF}_4\\\text{Benzenediazonium} \ \ \ \ \ \ \ \ \ \ \ \ \text{Nitrobenzene}\\ \ \ \ \ \ \ \text{chloride}$

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Question 301 Mark

What are the hydrolysis products of

Sucrose
lactose?

Answer

Both sucrose and lactose are disaccharides. Sucrose on hydrolysis gives one molecule each of glucose and fructose but lactose on hydrolysis gives one molecule each of glucose and galactose.

$\text{C}_{12}\text{H}_{22}\text{O}_{11}+\text{H}_2\text{O}\xrightarrow[]{\text{H}_3\text{O}^+} \ \ \ \ \ \ \ \ \text{C}_6\text{H}_{12}\text{O}_6+ \ \ \ \ \ \ \ \ \text{C}_6\text{H}_{12}\text{O}_6\\\text{Sucrose} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{D}-(+)-\text{Glucose} \ \ \ \ \text{D}-(-)-\text{Fructose}$

$\text{C}_{12}\text{H}_{22}\text{O}_{11}+\text{H}_2\text{O}\xrightarrow[]{\text{H}_3\text{O}^+} \ \ \ \ \ \ \ \ \text{C}_6\text{H}_{12}\text{O}_6+ \ \ \ \ \ \ \ \ \text{C}_6\text{H}_{12}\text{O}_6\\\text{Lactose} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{D}-(+)-\text{Glucose} \ \ \ \ \text{D}-(+)-\text{Galactose}$

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Question 311 Mark

Complete the following reaction:
$\text{C}_6\text{H}_5\text{NH}_2+\text{H}_2\text{SO}_4(\text{conc.})\rightarrow$

Answer

$\text{C}_6\text{H}_5\text{NH}_2+\text{conc}.\text{H}_2\text{SO}_4\rightarrow\text{C}_6\text{H}_5\stackrel{{+}}{{\text{N}}}\text{H}_3\text{HS}\stackrel{{-}}{{\text{O}_4}}\\\text{Aniline} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Anilinium hydrogen sulphate}$

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Question 321 Mark

Arrange the following:

In decreasing order of the pK_b values:

C₂H₅NH₂, C₆H₅NHCH₃, (C₂H₅)₂NH and C₆H₅NH₂

Answer

(C₂H₅)₂NH > C₂H₅NH₂ > C₆H₅NHCH₃ > C₆H₅NH₂

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Question 331 Mark

Give one chemical test to distinguish between the following pairs of compounds.
Methylamine and dimethylamine

Answer

Heat both these amines with chloroform and alcohol KOH solution. The compound which gives unpleasant (offensive) smell is methylamine while the compound which does not give any smell is diethylamine.
$\text{CH}_3\text{NH}_2+\text{CHCl}_3+3\text{KOH}\xrightarrow[]{\text{Heat}}\text{CH}_3\text{N}\stackrel{{\rightarrow}}{{=}}\text{C}\\\text{Methyl amine}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Methyl}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{isocyanide}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\text{offencive smell})\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +3\text{KCl}+3\text{H}_2\text{o}$
$(\text{CH}_3)\text{NH}+\text{CHCl}_3+3\text{KOH}\xrightarrow[]{\text{Heat}}\text{NO smell}\\ \ {\text{Dimethylamine}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $

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Question 341 Mark

Arrange the following:
In increasing order of solubility in water:
C₆H₅NH₂, (C₂H₅)₂NH, C₂H₅NH_2.

Answer

C₆H₅NH₂ < (C₂H₅)₂NH < C₂H₅NH₂.

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Question 351 Mark

Complete the following reaction:
$\text{C}_6\text{H}_5\text{NH}_2+\text{CHCl}_3+\text{alc.KOH}\rightarrow$

Answer

$\text{C}_6\text{H}_5\text{NH}_2+\text{CHCl}_3+\text{alc.KOH}\xrightarrow[]{\text{Carbyamine} \text{ reaction}}3\text{H}_2\text{O}+3\text{KCl}+\text{C}_6\text{H}_5-\text{NC}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Phynyl}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{isocynide}$

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Question 361 Mark

Arrange the following:
In increasing order of basic strength:
C₆H₅NH₂, C₆H₅N(CH₃)₂, (C₂H₅)₂NH and CH₃NH₂

Answer

CH₃NH₂ > (C₂H₅)₂ NH > C₆H₅N(CH₃)₂> C₆H₅ NH₂

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Question 371 Mark

How will you convert:
Ethanoic acid into propanoic acid.

Answer

Conversion of Ethanoic acid into Propanoic acid.
$\text{CH}_3\text{COOH}\xrightarrow[\text{Reduction}]{\text{LiAlH}_4}\text{CH}_3\text{CH}_2\text{OH}\xrightarrow[]{\text{PI}_3}\\\text{Ethanoic} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Ethyl} \ \ \ \ \ \ \ \ \ \ \\ \ \text{acid} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{alcohol}\\\text{CH}_3\text{CH}_2\text{I}\xrightarrow[]{\text{KCN}}\text{CH}_3\text{CH}_2\text{CN}\xrightarrow[]{2\text{H}_2\text{O}/\text{H}^+}\\\text{Ethyl Iodide} \ \ \ \ \ \ \ \ \ \text{Ethyl cynide}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\text{CH}_2\text{COOH}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Propinic acid}$

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Question 381 Mark

Account for the following:
Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide.

Answer

Due to the +I effect of -CH₃ group, methylamine is more basic than water. therefore, in water methylamine produces OH^- ions by accepting H⁺ions form water.
Methyl amine is a base and dissolves in water to produce hydroxide ions.
CH₃ NH₂ + H₂O ⇌ CH₃ NH₃^⊕ + OH^Θ
FeCl₃ combines with OH^- ions to give reddish brown precipitate of Fe (OH)₃
FeCl₃ + 3OH^-→ Fe(OH)₃

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Question 391 Mark

Arrange the following in increasing order of their basic strength:
CH₃NH₂, (CH₃)₂NH, (CH₃)₃N, C₆H₅NH₂, C₆H₅CH₂NH₂

Answer

C₆H₅NH₂ < C₆H₅CH₂NH₂ < (CH₃)₃ N < CH₃NH₂< (CH₃)₂NH

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Question 401 Mark

Give one chemical test to distinguish between the following pairs of compounds.
Aniline and N-methylaniline.

Answer

Aniline and N—Methylaniline Carbylamine test: Heat both the compounds separately with chloroform and alcoholic KOH. The compound which gives an unpleasant or offensive smell in aniline while the compound which does not give any smell is N-Methylaniline.
$\text{C}_6\text{H}_5\text{NH}_2+\text{CHCl}_3+3\text{KOH}\xrightarrow[\ \ \ \ \ \ \ \ \ \ \ \ \ ]{\text{Heat}}\text{C}_6\text{H}_5\stackrel{{\rightarrow}}{{=}}\text{C}\\\text{Aniline}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\text{alc})\ \ \ \ \ \ \ \ \ \text{Phenyl isocyanide}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\text{Offencive smell})\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +3\text{KCl}+3\text{H}_2\text{O}$
$\text{C}_6\text{H}_5\text{NHCH}_3+\text{CHCl}_3+3\text{KOH}\xrightarrow[\ \ \ \ \ \ \ \ \ \ \ \ \ ]{\text{Heat}}\text{No smell}.\\\text{N-methyl}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\text{alc}) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\\ \ \ \ \ \text{aniline}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $

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Question 411 Mark

How will you convert:
Ethanoic acid into methanamine.

Answer

Ethanoic acid into methanamine.
$\text{CH}_3\text{COOH}+\text{HN}_3\xrightarrow[\text{Warm}]{\text{conc}.\ \text{H}_2 \text{SO}_4}\text{CH}_3\text{NH}_2+\text{CO}_2+\text{N}_2\\\text{Ethanoic} \ \ \ \ \ \ \ \text{Hydrazoic} \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Methanamine}\\\text{acid}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{acid}$

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Question 421 Mark

Arrange the following in increasing order of their basic strength:
C₂H₅NH2, C₆H₅NH₂, NH₃, C₆H₅CH₂NH₂ and (C₂H₅)₂NH.

Answer

C₆H₅NH₂ < NH₃ < C₆H₅CH₂NH₂ < C₂H₅NH₂<(C₂H₅)₂NH

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Question 431 Mark

How will you convert?
Cl – (CH₂)₄ –Cl into hexan-1, 6 - diamine?

Answer

$\text{Cl}-(\text{CH}_2)_4-\text{Cl}\xrightarrow[]{\text{Ethanolic NaCN}}\text{N}\equiv\text{C}-(\text{CH}_4)-\text{C}\equiv\text{N}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Bigg\downarrow{\text{H}_2/\text{Ni}}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{H}_2\text{N}-\text{CH}_2-(\text{CH}_2)_4-\text{CH}_2-\text{NH}_2\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Hexane- 1,6-diamine}$

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Question 441 Mark

Write IUPAC name of the following compounds and classify them into primary, secondary and tertiary amines.
(CH₃CH₂)₂NCH₃

Answer

(CH₃CH₂)₂ NCH₃
N - Ethyl - N - methylethanamine (3° amine)

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Question 451 Mark

Write IUPAC name of following and classify them into primary, secondary, tertiary amines:
(CH₃)₂CHNH₂

Answer

(CH₃)₂CHNH₂: 1-Methylethanamine (1° amines)

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Question 461 Mark

Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.
(CH3)₃CNH₂

Answer

(CH₃)₃CNH₂
Tert-butylamine (primary amine)

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Question 471 Mark

Arrange the following:
In increasing order of boiling point:
C₂H₅OH, (CH₃)₂NH, C₂H₅NH₂

Answer

Increasing order of given:
(CH₃)₂NH < C₂H₅NH₂ < C₂H₅OH

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Question 481 Mark

Arrange the following in increasing order of their basic strength:
C₂H₅NH₂, (C₂H₅)₂NH, (C₂H₅)₃N, C₆H₅NH₂

Answer

C₆H₅NH₂ < C₂H₅NH₂ < (C₂H₅)₃N < (C₂H₅)₂NH

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Question 491 Mark

Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.
m–BrC₆H₄NH₂

Answer

m—Br C₆H₄NH₂.
3-Isobromo aniline (primary)

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Question 501 Mark

Complete the following reaction:
$\text{C}_6\text{H}_5\text{N}_2\text{Cl}+\text{H}_3\text{PO}_2+\text{H}_2\text{O}\rightarrow$

Answer

$\text{C}_6\text{H}_5\text{N}_2\text{Cl}+\text{H}_3\text{PO}_2+\text{H}_2\text{O}\rightarrow\text{C}_6\text{H}_6+\text{N}_2+\text{H}_3\text{PO}_3+\text{HCl}\\\text{Benzenediazonium} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Benzene}\\\text{Cloride}$

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