Question 11 Mark
Give a chemical test to distinguish between a primary and a secondary amine.
AnswerAromatic primary amine gives orange dye on treating with ice cold $\ce{NaNO_2 + HCl}$ followed by $\beta-$ naphthol whereas secondary amine does not give this test or describe carbyl amine test or Hinsberg test.
View full question & answer→Question 21 Mark
How will you convert?
Benzene into N, N-dimethylaniline.
Question 31 Mark
How will you convert:
Methanol to ethanoic acid.
AnswerConversion of methanol to Ethanoic acid.
$\text{CH}_3\text{OH} \xrightarrow[\text{P}/\text{I}_2]{\text{P}\text{I}_3}\text{CH}_3\text{I}\xrightarrow[\text{alc}]{\text{KCN}}\text{CH}_3\text{CN}\xrightarrow[(\text{dil HCl})]{\text{Hydrolysis}}\\\text{Methanol}\ \ \ \ \ \ \ \text{Methyl} \ \ \ \ \ \ \ \text{Methyl}\ \ \ \ \ \ \ \ \ \ \ \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{iodide}\ \ \ \ \ \ \ \ \ \text{iodide}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\text{COOH}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Ethanic acid}$
View full question & answer→Question 41 Mark
Give plausible explanation for each of the following:
Why are aliphatic amines stronger bases than aromatic amines?
AnswerIn 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.
View full question & answer→Question 51 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.
View full question & answer→Question 61 Mark
How will you convert:
Hexanenitrile into 1-aminopentane.
AnswerHexane nitrite into 1-amino pentane$\text{CH}_3(\text{CH}_2)\text{CN}\xrightarrow[(\text{Partial hydrolysis})]{\text{H}_2\text{O}/ \text{H}^+}\text{CH}_3(\text{CH}_2)_4\text{CONH}_2\xrightarrow[]{\text{Br}_2/\text{KOH}}\text{CH}_3(\text{CH}_2)_4\text{NH}_2\\\text{Hexane nitrate} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Hexanamide} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{I-amino Pentane}$
View full question & answer→Question 71 Mark
Convert
Aniline into 1, 3, 5 - tribromobenzene.
Question 81 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 $\ce{Na_2CO_3}$ solution $, N, N$ -dimethylaniline produces $N, N, N -$ trimethylanilinium carbonate.
View full question & answer→Question 91 Mark
Why cannot aromatic primary amines be prepared by Gabriel phthalimide synthesis?
AnswerGabrielphthalimide synthesis is used for the preparation of aliphatic primary amines. It involves nucleophilic substitution $\ce{(SN_2)}$ 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. View full question & answer→Question 101 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.
AnswerThe nitration of aniline is carried out using conc. $\ce{HNO_3}$ and $\ce{H_2SO_4}$.
However, in the presence of conc. $\ce{H_2SO_4},$ aniline forms aniline hydrogen sulphate in which the anilinium ion $, \ce{C_6H_5NH_3}^+$ is meta directing because the positive charge on the nitrogen attracts electrons from the benzene ring.
View full question & answer→Question 111 Mark
Arrange the following:
In increasing order of basic strength:
- Aniline $,p-$ nitroaniline and $p-$ toluidine.
- $\ce{C_6H_5NH_2, C_6H_5NHCH_3, C_6H_5CH_2NH_2}.$
Answer
- $P-$ nitro aniline $ < $ aniline $ < p-$toluidine.
- $\ce{C_6H_5NHCH_3 < C_6H_5NH_2 < C_6H_5CH_2NH_{2.}}$
View full question & answer→Question 121 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.
View full question & answer→Question 131 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.
View full question & answer→Question 141 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.
View full question & answer→Question 151 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.
View full question & answer→Question 161 Mark
Convert
3-Methylaniline into 3-nitrotoluene.
Question 171 Mark
Account for the following:
Ethylamine is soluble in water whereas aniline is not.
AnswerIn $\ce{C_2H_5NH_2},$ the $\ce{C_2H_5}$ group has a $+ I$ effect and increases the electron density on $N$ atom. This results in stronger intermolecular $H-$ bonding.
While in $\ce{C_6H_5NH_2}$ 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 $\ce{C_6H_5NH_2}$ is insoluble.
View full question & answer→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.
View full question & answer→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.
$\text{CH}_3\text{CH}_2\text{NH}_2+\text{CHCl}_3+3\text{KOH}(\text{alc})\xrightarrow[]{\text{warm}}\\\text{Ethyl amine}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\text{CH}_2\stackrel{{\rightarrow}}{=}\text{C}+3\text{KCl}+3\text{H}_2\text{O}\\ \ \ \ \ \ \ \ \ \ \ \text{Ethyl isocyanide}$
View full question & answer→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 remains
insoluble in aqueous KOH.
View full question & answer→Question 211 Mark
Accomplish the following conversion:
Benzoic acid to aniline.
Answer
Benzoic acid to aniline
View full question & answer→Question 221 Mark
Accomplish the following conversion:
Benzyl chloride to 2 - phenylethanamine.
Answer
Benzyl chloride to 2-phenyl ethanamine.
View full question & answer→Question 231 Mark
How will you convert?
Benzene into aniline.
Question 241 Mark
Write short notes on the following: Ammonolysis.
AnswerAmmonolysis: Alkyl halide reacts with ammonia to form primary amine.
The reaction of ammonia with alkyl halide is known as ammonolysis.
$\ce{CH_3Cl + NH_{3 }\rightarrow CH_3NH_2.HCl}$
$\ce{CH_3NH_2.HCl + NH_{3 }\rightarrow CH_3NH_2 + NH_4Cl}$
View full question & answer→Question 251 Mark
Arrange the following: In decreasing order of basic strength in gas phase:
$\ce{C_2H_5NH_2, (C_2H_5)_2NH, (C_2H_5)_3N}$ and $\ce{NH_3}$
Answer$\ce{(C_2H_5)_2NH > C_2H_5NH > (C_2H_5)_3N > NH_3}$
View full question & answer→Question 261 Mark
Complete the following reaction:
$\text{C}_6\text{H}_5\text{N}_2\text{Cl}+\text{C}_2\text{H}_5\text{OH}\rightarrow$
Answer$ \ \ \ \ \ \ \ \ \ \ \ \ \text{C}_6\text{H}_5\text{N}_2\text{Cl} \ \ + \ \ \ \text{C}_2\text{H}_5\text{OH}\rightarrow\text{C}_6\text{H}_6+\text{CH}_3\text{CHO}+\text{N}_2+\text{HCl}\\\text{Benzenediazonium} \ \ \ \ \ \ \ \text{Ethanol} \ \ \ \ \ \ \ \text{Benzene} \ \ \ \ \ \text{Ethanal}\\ \ \ \ \ \ \ \ \ \ \ \ \text{Cloride}$
View full question & answer→Question 271 Mark
How will you convert:
Ethanamine into methanamine.
AnswerConversion of Ethanamine into Methanamine.
$ \text{C}_2\text{H}_5\text{NH}_2\xrightarrow[]{\text{HONO}}\text{C}_2\text{H}_5\text{OH}\xrightarrow[\text{K}_2\text{Cr}_2\text{O}_7/ \text{H}_2\text{SO}_4]{\text{Oxidation}}\\\text{Ethanamine} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Ethyl} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\\ \ \ \ \ \ \text{alcohol}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\text{CHO}\xrightarrow[]{\text{Oxidation}}\text{CH}_3\text{COOH}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Acetaldahyde} \ \ \ \ \ \ \ \ \ \ \text{Acetic acid}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Big\downarrow{\text{NH}_3}\\\text{CH}_3\text{NH}_2\xleftarrow[]{\text{Br}_2/\text{KOH}}\text{CH}_3\text{CONH}_2\xleftarrow[]{\text{heat}}\text{CH}_3\text{COONH}_4\\\text{methamine} \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Acetamide} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Ammonium}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{acetate}$
View full question & answer→Question 281 Mark
Give plausible explanation for each of the following:
Why do primary amines have higher boiling point than tertiary amines?
AnswerAmines are polar compounds and can form H-bonds as proton acceptors. Only primary and secondary amines can form intermolecular H-bonding due to the presence of H atom at N atom. Tertiary amines cannot form intermolecular in-bonding. Primary amines are higher boiling due to the presence of a stronger intermolecular force of attraction due to H-bonding.
$\cdot\cdot\cdot\text{H} \ \ \ \ \ \ \ \ \ \text{R} \ \ \ \ \ \ \ \ \ \ \text{H}\cdot\cdot\cdot\\ \ \ \ \ \ \ \ | \ \ \ \ \ \ \ \ \ \ \ | \ \ \ \ \ \ \ \ \ \ \ |\\\text{RNH}\cdot\cdot\cdot\text{NH}\cdot\cdot\cdot \ \text{NR} \ \ \ \ \ \ \ \ \ \ \ \ \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ | \ \ \ \ \ \ \ \ \ \ \ \ |\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{H}\cdot\cdot\cdot \ \ \text{H}\cdot\cdot\cdot \ \$\text{Hydrogen bonding in primary})$
View full question & answer→Question 291 Mark
How will you convert:
Nitromethane into dimethylamine.
AnswerNitromethane into dimethylamine
$\text{CH}_3\text{NO}_2+6[\text{H}]\xrightarrow[]{\text{Sn}/\text{HCl}}\text{CH}_3\text{NH}_2\xrightarrow[-\text{HBr}]{\text{CH}_3\text{Br}}\\\text{Nitromethane} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Methyl}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{amine}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\text{CH}_3)_2\text{NH}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{dimethyl amine}$
View full question & answer→Question 301 Mark
Account for the following: $\ce{pK_b}$ of aniline is more than that of methylamine.
AnswerIt is because, in aniline, the $\ce{NH_2}$ 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 $, \ce{pk_b}$ of aniline is more than that of methylamine.
View full question & answer→Question 311 Mark
Give plausible explanation for each of the following: Why are amines less acidic than alcohols of comparable molecular masses?
AnswerOxygen in alcohol is more electro negative than nitrogen in amines. $\text{RC—H}$ bond in alcohol is more polar with ${δ+}$ charge on.
$RO^{δ-}H^{δ+}$ as compared to $\text{RN—H}$ bond in Alcohols can loose proton to some extent but are proton acceptors.
$\text{RO}-\text{H}\rightleftharpoons\text{RO}^-+\text{H}^+\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Unstable}$
$\text{R}\stackrel{{. .} }{{\text{N}}}\text{H}_2+\text{H}^+\rightarrow\text{R}\stackrel{{+} }{{\text{N}}}\text{H}_3\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Stable}$
View full question & answer→Question 321 Mark
Write short notes on the following:
Hofmann’s bromamide reaction.
AnswerHoffmann 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}$
View full question & answer→Question 331 Mark
Account for the following: Aniline does not undergo Friedel $-$ Crafts reaction.
AnswerA Friedel $-$ crafts reaction is carried out in the presence of $\ce{AlCl_3}$.
but $\ce{AlCl_3}$ is acidic in nature, while aniline is a strong base.
Thus, aniline reacts with $\ce{AlCl_3}$ 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.
View full question & answer→Question 341 Mark
Account for the following:
Gabriel phthalimide synthesis is preferred for synthesising primary amines.
AnswerGabriel 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.
View full question & answer→Question 351 Mark
Complete the following reaction:
$\text{C}_6\text{H}_5\text{NH}_2+(\text{CH}_3\text{CO})_2\text{O}\rightarrow$
Answer$\text{C}_6\text{H}_5\text{NH}_2+(\text{CH}_3\text{CO})_2\text{O}\rightarrow\text{C}_6\text{H}_5-\text{N}-\text{C}-\text{CH}_3+\text{CH}_3\text{COOH}\\\text{Aniline} \ \ \ \ \ \ \ \ \text{acitic anhydride} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ | \ \ \ \ \ \ ||\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{H} \ \ \ \ \ \text{O}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{N}-\text{Phenylethanamide}$
View full question & answer→Question 361 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$\text{C}_6\text{H}_5\text{N}_2\text{Cl}\xrightarrow[(\text{ii})\text{NaNO}_2/\text{Cu}.\Delta]{(\text{i})\text{HBF}_4}\text{C}_6\text{H}_5\text{NO}_2+\text{N}_2+\text{NaBF}_4\\\text{Benzenediazonium} \ \ \ \ \ \ \ \ \ \ \ \ \text{Nitrobenzene}\\ \ \ \ \ \ \ \text{chloride}$
View full question & answer→Question 371 Mark
What are the hydrolysis products of
- Sucrose
- lactose?
AnswerBoth 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. $\text{C}_{12}\text{H}_{22}\text{O}_{11}+\text{H}_2\text{O}\xrightarrow[]{\text{H}_3\text{O}^+} \ \ \ \ \ \ \ \ \text{C}_6\text{H}_{12}\text{O}_6+ \ \ \ \ \ \ \ \ \text{C}_6\text{H}_{12}\text{O}_6\\\text{Sucrose} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{D}-(+)-\text{Glucose} \ \ \ \ \text{D}-(-)-\text{Fructose}$ $\text{C}_{12}\text{H}_{22}\text{O}_{11}+\text{H}_2\text{O}\xrightarrow[]{\text{H}_3\text{O}^+} \ \ \ \ \ \ \ \ \text{C}_6\text{H}_{12}\text{O}_6+ \ \ \ \ \ \ \ \ \text{C}_6\text{H}_{12}\text{O}_6\\\text{Lactose} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{D}-(+)-\text{Glucose} \ \ \ \ \text{D}-(+)-\text{Galactose}$
View full question & answer→Question 381 Mark
Complete the following reaction:
$\text{C}_6\text{H}_5\text{NH}_2+\text{H}_2\text{SO}_4(\text{conc.})\rightarrow$
Answer$\text{C}_6\text{H}_5\text{NH}_2+\text{conc}.\text{H}_2\text{SO}_4\rightarrow\text{C}_6\text{H}_5\stackrel{{+}}{{\text{N}}}\text{H}_3\text{HS}\stackrel{{-}}{{\text{O}_4}}\\\text{Aniline} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Anilinium hydrogen sulphate}$
View full question & answer→Question 391 Mark
Arrange the following: In decreasing order of the $\ce{pK_b}$ values:
$\ce{C_2H_5NH_2, C_6H_5NHCH_3, (C_2H_5)_2NH}$ and $\ce{C_6H_5NH_2}$
Answer$\ce{(C_2H_5)_2NH > C_2H_5NH_2 > C_6H_5NHCH_3 > C_6H_5NH_2}$
View full question & answer→Question 401 Mark
Give one chemical test to distinguish between the following pairs of compounds.
Methylamine and dimethylamine
AnswerHeat 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}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $
View full question & answer→Question 411 Mark
Arrange the following: In increasing order of solubility in water:
$\ce{C_6H_5NH_2, (C_2H_5)_2NH, C_2H_5NH_{2.}}$
Answer$\ce{C_6H_5NH_2 < (C_2H_5)_2NH < C_2H_5NH_2}$.
View full question & answer→Question 421 Mark
Complete the following reaction:
$\text{C}_6\text{H}_5\text{NH}_2+\text{CHCl}_3+\text{alc.KOH}\rightarrow$
Answer$\text{C}_6\text{H}_5\text{NH}_2+\text{CHCl}_3+\text{alc.KOH}\xrightarrow[]{\text{Carbyamine} \text{ reaction}}3\text{H}_2\text{O}+3\text{KCl}+\text{C}_6\text{H}_5-\text{NC}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Phynyl}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{isocynide}$
View full question & answer→Question 431 Mark
Arrange the following: In increasing order of basic strength:
$\ce{C_6H_5NH_2, C_6H_5N(CH_3)_2, (C_2H_5)_2NH}$ and $\ce{CH_3NH_2}$
Answer$\ce{CH_3NH_2 > (C_2H_5)_2 NH > C_6H_5N(CH_3)_{2 }> C_6H_5 NH_2}$
View full question & answer→Question 441 Mark
How will you convert:
Ethanoic acid into propanoic acid.
AnswerConversion of Ethanoic acid into Propanoic acid.
$\text{CH}_3\text{COOH}\xrightarrow[\text{Reduction}]{\text{LiAlH}_4}\text{CH}_3\text{CH}_2\text{OH}\xrightarrow[]{\text{PI}_3}\\\text{Ethanoic} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Ethyl} \ \ \ \ \ \ \ \ \ \ \\ \ \text{acid} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{alcohol}\\\text{CH}_3\text{CH}_2\text{I}\xrightarrow[]{\text{KCN}}\text{CH}_3\text{CH}_2\text{CN}\xrightarrow[]{2\text{H}_2\text{O}/\text{H}^+}\\\text{Ethyl Iodide} \ \ \ \ \ \ \ \ \ \text{Ethyl cynide}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\text{CH}_2\text{COOH}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Propinic acid}$
View full question & answer→Question 451 Mark
Account for the following:
Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide.
AnswerDue to the $+I$ effect of $-\ce{CH_3}$ group, methylamine is more basic than water. therefore, in water methylamine produces $\ce{OH}^-$ ions by accepting $H^+$ ions form water.
Methyl amine is a base and dissolves in water to produce hydroxide ions.
$\ce{CH_3 NH_2 + H_2O \leftrightharpoons CH_3 NH_3^⊕ + OH^Θ}$
$\ce{FeCl_3}$ combines with $OH^-$ ions to give reddish brown precipitate of $\ce{Fe (OH)_3}$
$\ce{FeCl_3 + 3OH^{- }\rightarrow Fe(OH)_3}$
View full question & answer→Question 461 Mark
Arrange the following in increasing order of their basic strength:
$\ce{CH_3NH_2, (CH_3)_2NH, (CH_3)_3N, C_6H_5NH_2, C_6H_5CH_2NH_2}$
Answer$\ce{C_6H_5NH_2 < C_6H_5CH_2NH_2 < (CH_3)_3 N < CH_3NH_{2 } < (CH_3)_2NH}$
View full question & answer→Question 471 Mark
Give one chemical test to distinguish between the following pairs of compounds.
Aniline and N-methylaniline.
AnswerAniline 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}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $
View full question & answer→Question 481 Mark
How will you convert:
Ethanoic acid into methanamine.
AnswerEthanoic acid into methanamine.
$\text{CH}_3\text{COOH}+\text{HN}_3\xrightarrow[\text{Warm}]{\text{conc}.\ \text{H}_2 \text{SO}_4}\text{CH}_3\text{NH}_2+\text{CO}_2+\text{N}_2\\\text{Ethanoic} \ \ \ \ \ \ \ \text{Hydrazoic} \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Methanamine}\\\text{acid}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{acid}$
View full question & answer→Question 491 Mark
Arrange the following in increasing order of their basic strength: $\ce{C_2H_5NH2, C_6H_5NH_2, NH_3, C_6H_5CH_2NH_2}$ and $\ce{(C_2H_5)_2NH.}$
Answer$\ce{C_6H_5NH_2 < NH_3 < C_6H_5CH_2NH_2 < C_2H_5NH_2<(C_2H_5)_2NH}$
View full question & answer→Question 501 Mark
How will you convert? $\ce{Cl – (CH_2)_4 –Cl}$ 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}$
View full question & answer→Question 511 Mark
Write $\text{IUPAC}$ name of the following compounds and classify them into primary, secondary and tertiary amines.
$\ce{(CH_3CH_2)_2NCH_3}$
Answer$\ce{(CH_3CH_2)_2 NCH_3}$
$N -$ Ethyl $- N -$ methylethanamine $(3^\circ$ amine$)$
View full question & answer→Question 521 Mark
Write $\text{IUPAC}$ name of following and classify them into primary, secondary, tertiary amines:
$\ce{(CH_3)_2CHNH_2}$
Answer$\ce{(CH_3)_2CHNH}_2 : 1-$ Methylethanamine $(1^\circ$ amines$)$
View full question & answer→Question 531 Mark
Write $\ce{IUPAC}$ names of the following compounds and classify them into primary, secondary and tertiary amines.
$\ce{(CH3)_3CNH_2}$
Answer$\ce{(CH_3)_3CNH_2}$ Tert$-$butylamine $($primary amine$)$
View full question & answer→Question 541 Mark
Arrange the following: In increasing order of boiling point: $\ce{C_2H_5OH, (CH_3)_2NH, C_2H_5NH_2}$
AnswerIncreasing order of given: $\ce{(CH_3)_2NH < C_2H_5NH_2 < C_2H_5OH}$
View full question & answer→Question 551 Mark
Arrange the following in increasing order of their basic strength:
$\ce{C_2H_5NH_2, (C_2H_5)_2NH, (C_2H_5)_3N, C_6H_5NH_2}$
Answer$\ce{C_6H_5NH_2 < C_2H_5NH_2 < (C_2H_5)_3N < (C_2H_5)_2NH}$
View full question & answer→Question 561 Mark
Write $\ce{IUPAC}$ names of the following compounds and classify them into primary, secondary and tertiary amines.
$\ce{m–BrC_6H_4NH_2}$
Answer$\ce{m-Br C_6H_{4 }NH_2}. 3-$Isobromo aniline $($primary$)$
View full question & answer→Question 571 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$\text{C}_6\text{H}_5\text{N}_2\text{Cl}+\text{H}_3\text{PO}_2+\text{H}_2\text{O}\rightarrow\text{C}_6\text{H}_6+\text{N}_2+\text{H}_3\text{PO}_3+\text{HCl}\\\text{Benzenediazonium} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Benzene}\\\text{Cloride}$
View full question & answer→Question 581 Mark
Complete the following acid-base reaction and name the product:
$(\text{C}_2\text{H}_5)_3\text{N}+\text{HCl}\rightarrow$
Answer$(\text{C}_2\text{H}_5)_3\text{N}+\text{HCl}\rightarrow\Big[(\text{C}_2\text{H}_5)_3\stackrel{{+}}{{\text{N}}}\text{H}\Big]\text{Cl}^-\\\text{Triethylamine} \ \ \ \ \ \ \ \ \text{Triemethylammonium chloride}$
View full question & answer→Question 591 Mark
Complete the following acid-base reaction and name the product:
$\text{CH}_3\text{CH}_2\text{CH}_2\text{NH}_2+\text{HCl}\rightarrow$
Answer$\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{NH}_2\rightarrow\text{CH}_3\text{CH}_2\text{CH}_2\text{NH}_3 \ ^ +\text{Cl}^-\\\text{n-propylamine} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{n-propylammoniumchloride}$
View full question & answer→Question 601 Mark
Give one chemical test to distinguish between the following pairs of compounds. Aniline and benzylamine.
AnswerAniline and benzylamine: Add $\ce{NaNO_2}$ and $\ce{HCl}$ to each separately. Cool, it to $0-5^\circ C.$ Then add an alkaline solution of phenol. Orange azo dye is formed in aniline. Benzylamine $\ce{(C_6H_5CH_2NH_2)}$ does not form an azo dye.
View full question & answer→Question 611 Mark
Write $\ce{IUPAC}$ names of following and classify them into primary, secondary, tertiary amines: $\ce{CH_3NHCH(CH_3)_2}$
Answer$\ce{CH_3NHCH(CH_3)_2}$ Methyl isopropyl amine $($secondary$)$
View full question & answer→Question 621 Mark
Complete the following reaction:
$\text{C}_6\text{H}_5\text{N}_2\text{Cl}+\text{Br}_2(\text{aq})\rightarrow$
View full question & answer→Question 631 Mark
The conversion of primary aromatic amines into diazonium salts is known as ___________.
Question 641 Mark
Write the structure of n-methy-lethanamine.
Question 651 Mark
Rearrange the following in an increasing order of their basic strengths:$ \ce{C_6H_5NH_2, C_6H_5N(CH_3)_2, (C_6H_5)_2NH and CH_3NH_2.}$
Answer$\ce{(C_6H_5)_2NH < C_6H_5NH_2 < C_6H_5N(CH_3)_2 < CH_3NH_{2}}$
View full question & answer→Question 661 Mark
Arrange the following compounds in an increasing order of basic strengths in their aqueous solutions:$\ce{NH_3, CH_3NH_2, (CH_3)_3NH, (CH_3)_3N}.$
Answer$\ce{NH_3 < (CH_3)_3N < CH_3NH_2 < (CH_3)_2NH}.$
View full question & answer→Question 671 Mark
Why do amines react as nucleophiles?
AnswerDue to the presence of lone pair of electrons on the nitrogen atom.
View full question & answer→Question 681 Mark
Write the IUPAC name of the given compound.
Question 691 Mark
Write the IUPAC name of the given compound.
Question 701 Mark
Write the IUPAC name of the given compound.
Question 711 Mark
Write the IUPAC name of the given compound:
Answer2,4,6-Tribromoaniline/2,4,6-Tribromobenzenamine.
View full question & answer→Question 721 Mark
Write the IUPAC name of the given compound:
Answer2,4,6-Tribromoaniline/2,4,6-Tribromobenzenamine.
View full question & answer→Question 731 Mark
Write $\ce{IUPAC}$ name of the following compound: $\ce{(CH_3CH_2)_2NCH_3}$
Answer$N-$Ethyl$-N-$methylethanamine.
View full question & answer→Question 741 Mark
Write the IUPAC name of the given compound:
Answer2,4,6-Tribromoaniline/2,4,6-Tribromobenzenamine.
View full question & answer→Question 751 Mark
Arrange the following compounds in increasing order of solubility in water: $\ce{C_6H_5NH_2, (C_2H_5)_2NH, C_2H_5NH_{2.}}$
Answer$\ce{C_6H_5NH_2<(C_2H_5)_2NH2H_5NH_{2}}.$
View full question & answer→Question 761 Mark
Arrange the following in increasing order of their basic strength in aqueous solution: $\ce{CH_3.NH_{2,} (CH_3)_3N, (CH_3)_2NH}$
Answer$\ce{(CH_3)_3N3NH_2<(CH_3)_2NH.}$
View full question & answer→Question 771 Mark
Arrange the following in the decreasing order of their basic strength in aqueous solutions: $\ce{CH_3NH_2, (CH_3)_2NH, (CH_3)_3N}$ and $\ce{NH_3}.$
Answer$\ce{(CH_3)_2NH > CH_3NH_{2 }> (CH_3)_3N > NH_3}.$
View full question & answer→Question 781 Mark
Give the $\ce{IUPAC}$ name of $\ce{H_2N-CH_2–CH_2-CH=CH_2}.$
AnswerBut$-3-$en$-1-$amine.
View full question & answer→Question 791 Mark
Why is an alkylamine more basic than ammonia?
AnswerDue to electron donating nature of alkyl group or +I effect.
View full question & answer→Question 801 Mark
Write a chemical reaction in which the iodide ion replaces the diazonium group in a diazonium salt.
Answer$\ce{C_6H_5N_2^+Cl^- + KI \rightarrow C_6H_5I + KCl + N_2}.$
View full question & answer→Question 811 Mark
Why do nitro compounds have high boiling points in comparison with other compounds of same molecular mass?
AnswerBecause of polar nature, they are much more strongly associated.
View full question & answer→Question 821 Mark
Give a chemical test to distinguish between aniline and N-methyl aniline.
AnswerAdd chloroform and KOH and heat. Aniline forms pungent smelling isocyanide.
View full question & answer→Question 831 Mark
What is the role of $HNO_3$ in the nitrating mixture used for nitration of benzene?
Answer$HNO_3$ acts as a base in the nitrating mixture $\ce{(HNO_3 + H_2SO_4)}$ and provides the electrophile.$\text{HNO}_3+\text{H}_2\text{SO}_4\xrightarrow{\ \ \ \ \ \ \ \ }\text{NO}_2^++\text{HSO}_4^-+\text{H}_2\text{O}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Electrofile}$
View full question & answer→Question 841 Mark
Mention two important uses of sulphanilic acid.
AnswerSulphanilic acid is used in the manufacture of (i) dyes (ii) sulpha drugs.
View full question & answer→Question 851 Mark
Complete the following reaction:$\text{C}_6\text{H}_5\text{N}_2\text{Cl}+\text{C}_2\text{H}_5\text{OH}\rightarrow$
Answer$ \ \ \ \ \ \ \ \ \ \ \ \ \text{C}_6\text{H}_5\text{N}_2\text{Cl} \ \ + \ \ \ \text{C}_2\text{H}_5\text{OH}\rightarrow\text{C}_6\text{H}_6+\text{CH}_3\text{CHO}+\text{N}_2+\text{HCl}\\\text{Benzenediazonium} \ \ \ \ \ \ \ \text{Ethanol} \ \ \ \ \ \ \ \text{Benzene} \ \ \ \ \ \text{Ethanal}\\ \ \ \ \ \ \ \ \ \ \ \ \text{Cloride}$
View full question & answer→Question 861 Mark
Complete the following reaction:$\text{C}_6\text{H}_5\text{NH}_2+(\text{CH}_3\text{CO})_2\text{O}\rightarrow$
Answer$\text{C}_6\text{H}_5\text{NH}_2+(\text{CH}_3\text{CO})_2\text{O}\rightarrow\text{C}_6\text{H}_5-\text{N}-\text{C}-\text{CH}_3+\text{CH}_3\text{COOH}\\\text{Aniline} \ \ \ \ \ \ \ \ \text{acitic anhydride} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ | \ \ \ \ \ \ ||\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{H} \ \ \ \ \ \text{O}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{N}-\text{Phenylethanamide}$
View full question & answer→Question 871 Mark
What is the product when $\ce{C_6H_5CH_2NH_2}$ reacts with $\ce{HNO_2}$?
Answer$\text{C}_6\text{H}_5\text{CH}_2\text{NH}_2+\text{HONO}\xrightarrow{\ \text{HCl}\ \ }\text{C}_6\text{H}_5\text{CH}_2\text{OH}+\text{N}_2+\text{H}_2\text{O}\\\text{Benzyl amine}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Benzyl alohol}$
View full question & answer→Question 881 Mark
Give an example of a zwitter ion.
Answer
sulphanilic acid.
View full question & answer→Question 891 Mark
Give the structure of 'A' in the following reaction.
Question 901 Mark
Complete the following reaction:$\text{C}_6\text{H}_5\text{N}_2\text{Cl}+\text{Br}_2(\text{aq})\rightarrow$
View full question & answer→Question 911 Mark
Convert Aniline into p-nitroaniline.
Question 921 Mark
Why does acetylation of $-NH_2$ group of aniline reduce its activating effect?
AnswerThe activating effect of $-NH_2$ group can be controlled by protecting the $-NH_2$ group by acetylation with acetic anhydride, then carrying out the desired substitution followed by hydrolysis of the substituted amide to the substituted amine.
The lone part of electrons on nitrogen of acetanilide interacts with oxygen atom due to resonance as shown below:
View full question & answer→Question 931 Mark
Complete the following reaction:$\text{C}_6\text{H}_5\text{NH}_2+\text{CHCl}_3+\text{alc.KOH}\rightarrow$
Answer$\text{C}_6\text{H}_5\text{NH}_2+\text{CHCl}_3+\text{alc.KOH}\xrightarrow[]{\text{Carbyamine} \text{ reaction}}3\text{H}_2\text{O}+3\text{KCl}+\text{C}_6\text{H}_5-\text{NC}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Phynyl}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{isocynide}$
View full question & answer→Question 941 Mark
Why is $NH_2$ group of aniline acetylated before carrying out nitration?
AnswerDirection nitration of aniline yields carry oxidation products in addition to the nitro derivatives. Moreover, in the strongly acidic medium, aniline is protonated to form the anilinium ion which is meta directing. That is why besides the ortho and para derivatives, significant amount of meta derivative is also formed.
However, by protecting the $-NH_2$ group by acetylation reaction with acetic anhydride, the nitration reaction can be controlled and the $p-$nitro derivative can be obtained as the major product.
View full question & answer→Question 951 Mark
How will you convert 4-nitrotoluene to 2-bromobenzoic acid?
Question 961 Mark
Answer the following questions:
Write a chemical test to distinguish between aniline and methylamine.
Question 971 Mark
Why is aniline soluble in aqueous HCl?
Answer
Aniline forms the salt anilinium chloride which is water soluble.
View full question & answer→Question 981 Mark
For an amine $RNH_2,$ write the expression for $K_b$ to indicate its strength.
Answer$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \oplus$
$\text{RNH}_2+\text{H}_2\text{O}\rightleftharpoons\text{RNH}_3+\text{OH}^-\text{K}_\text{b}=\frac{\ \ \oplus
[\text{RNH}_3][\text{OH}^-]}{\text{[RNH}_2]}$
View full question & answer→Question 991 Mark
Predict the product of reaction of aniline with bromine in non$-$polar solvent such as $CS_2.$
Answer
In non$-$polar solvent $($such as $CS_2)$ the activating effect of $-NH_2$ group is reduced $($due to resonance$)$ and
hence, mono substitution occurs only at $o-$and $p-$positions. View full question & answer→Question 1001 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$\text{C}_6\text{H}_5\text{N}_2\text{Cl}\xrightarrow[(\text{ii})\text{NaNO}_2/\text{Cu}.\Delta]{(\text{i})\text{HBF}_4}\text{C}_6\text{H}_5\text{NO}_2+\text{N}_2+\text{NaBF}_4\\\text{Benzenediazonium} \ \ \ \ \ \ \ \ \ \ \ \ \text{Nitrobenzene}\\ \ \ \ \ \ \ \text{chloride}$
View full question & answer→Question 1011 Mark
Complete the following acid-base reaction and name the product:$\text{CH}_3\text{CH}_2\text{CH}_2\text{NH}_2+\text{HCl}\rightarrow$
Answer$\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{NH}_2\rightarrow\text{CH}_3\text{CH}_2\text{CH}_2\text{NH}_3 \ ^ +\text{Cl}^-\\\text{n-propylamine} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{n-propylammoniumchloride}$
View full question & answer→Question 1021 Mark
What type of bonding helps in stabilising the $\alpha$ - helix structure of protins.
AnswerIn the $\alpha$ - helix configuration, a polypeptide chain forms all possible hydrogen bonds by twisting into a right handed screw (helix) with -NH group of each amino acid residue hydrogen to the bonded $ \ \ \ \ \ \text{O}\\ \ \ \ \ \ ||\\ -\text{C}-$of an adjacent turn of the helix.
View full question & answer→Question 1031 Mark
Complete the following reaction:$\text{C}_6\text{H}_5\text{NH}_2+\text{H}_2\text{SO}_4(\text{conc.})\rightarrow$
Answer$\text{C}_6\text{H}_5\text{NH}_2+\text{conc}.\text{H}_2\text{SO}_4\rightarrow\text{C}_6\text{H}_5\stackrel{{+}}{{\text{N}}}\text{H}_3\text{HS}\stackrel{{-}}{{\text{O}_4}}\\\text{Aniline} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Anilinium hydrogen sulphate}$
View full question & answer→Question 1041 Mark
Complete the following acid-base reaction and name the product:$(\text{C}_2\text{H}_5)_3\text{N}+\text{HCl}\rightarrow$
Answer$(\text{C}_2\text{H}_5)_3\text{N}+\text{HCl}\rightarrow\Big[(\text{C}_2\text{H}_5)_3\stackrel{{+}}{{\text{N}}}\text{H}\Big]\text{Cl}^-\\\text{Triethylamine} \ \ \ \ \ \ \ \ \text{Triemethylammonium chloride}$
View full question & answer→Question 1051 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$\text{C}_6\text{H}_5\text{N}_2\text{Cl}+\text{H}_3\text{PO}_2+\text{H}_2\text{O}\rightarrow\text{C}_6\text{H}_6+\text{N}_2+\text{H}_3\text{PO}_3+\text{HCl}\\\text{Benzenediazonium} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Benzene}\\\text{Cloride}$
View full question & answer→Question 1061 Mark
AnswerEnzymes are proteins that catalyze biological reaction. They are very specific in nature and catalyze only a particular reaction for a particular substrate. Enzymes are usually named after a particular. For example, the enzyme used to catalyses the hydrolysis of maltose in glucose is named as maltase.$\text{C}_{12}\text{H}_{22}\text{O}_{11}\xrightarrow[]{\text{Maltase}}2\text{C}_6\text{H}_{12}\text{O}_6\\\text{maltose} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{glucose}$
View full question & answer→Question 1071 Mark
What is the best reagent to convert nitrile to primary amine?
AnswerReduction of nitriles with sodium alcohol or $\ce{LiAlH_4}$ gives primary amine.
$\text{R}-\text{CN}+4[\text{H}]\xrightarrow[\text{or LiAh}_4/\text{ether}]{\ \ \text{Na/C}_2\text{H}_5\text{OH}\ \ \ }\text{R}-\text{CH}_2\text{NH}_2\\\text{Alkyl nitrile}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Alkyl amine}(1^\circ)$
View full question & answer→Question 1081 Mark
Arrange the following compounds in the order of property indicated for each set:
Methylamine, dimethylamine, aniline, N-methylaniline (increasing order of their acid strength).
AnswerThe acid strength increases in the reverse order of their basic strength, i.e.,
Dimethylamine < methylamine < N-methylaniline < aniline.
View full question & answer→Question 1091 Mark
Explain why $\ce{MeNH_2}$ is stronger base than $\ce{MeOH}$?
AnswerNitrogen is less electronegative than oxygen.
therefore, lone pair of electrons on nitrogen is readily available for donation.
Hence, $\ce{MeNH_2}$ is more basic than $\ce{MeOH}.$
View full question & answer→Question 1101 Mark
What is the difference between a nucleoside and a nucleotide?
AnswerA nucleoside is formed by the attachment of a base to$^{\ce{I}\ '}$ position of sugar.
Nucleoside $=$ Sugar $+$ Base
On the other hand, all the three basic components of nucleic acids $($i.e., pentose sugar, phosphoric acid, and base$)$ are present in a nucleotide.
Nucleotide $=$ Sugar $+$ Base $+$ Phosphoric acid
View full question & answer→Question 1111 Mark
Suggest a route by which the following conversion can be accomplished.
Question 1121 Mark
Complete the following reaction.
Question 1131 Mark
Identify A and B in the following reaction.
Question 1141 Mark
Why do amines behave as nucleophiles?
AnswerDue to the presence of a lone pair of electrons on nitrogen atom, amines behave as nucleophiles.
View full question & answer→Question 1151 Mark
How will you carry out the following conversions?
toluene → p-toluidine.
Question 1161 Mark
Under what reaction conditions (acidic/ basic), the coupling reaction of aryldiazonium chloride with aniline is carried out?
AnswerThe azo products obtained have an extended conjugate system having both the aromatic rings joined through the –N=N- bond. These compounds are often coloured and are used as dyes. Benzene diazonium chloride reacts with phenol in which the phenol molecule at its para position is coupled with the diazonium salt to form p-hydroxy azobenzene. This type of reaction is known as coupling reaction.
View full question & answer→Question 1171 Mark
The two strands in DNA are not identical but are complementary. Explain.
AnswerIn the helical structure of DNA, the two strands are held together by hydrogen bonds between specific pairs of bases. Cytosine forms hydrogen bond with guanine, while adenine forms hydrogen bond with thymine. As a result, the two strands are complementary to each other.
View full question & answer→Question 1181 Mark
What is the role of pyridine in the acylation reaction of amines?
AnswerThe products obtained by acylation reaction are known as amides. The reaction is carried out in the presence of a base stronger than the amine, like pyridine, which removes HCl so formed and shifts the equilibrium to the right-hand side.
View full question & answer→Question 1191 Mark
What is the structure and $\ce{IUPAC}$ name of the compound, allyl amine?
Answer$\ce{CH_2-CH-CH_2-NH_2},$prop$-2-$en$-1-$amine.
View full question & answer→Question 1201 Mark
Write down the IUPAC name of:
AnswerN, N-Dimethyl aminobenzene.
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Arrange the following compounds in increasing order of dipole moment. $\ce{CH_3CH_2CH_3, CH_3CH_2NH_2, CH_3CH_2OH}.$
Answer$\ce{CH_3CH_2CH_3 < CH_3CH_2NH_2 < CH_3CH_2OH}$
Since, $O$ is more electronegative than $N,$ therefore, dipole moment of ethyl alcohol is higher than that of ethyl amine.
Propane however, has the least dipole moment since it is almost a non$-$polar molecule.
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What is Hinsberg reagent?
AnswerBenzenesulphonyl chloride $\ce{(C_6H_5SO_2Cl)},$ is also known as Hinsberg’s reagent. It reacts with primary and secondary amines to form sulphonamides.
Secondary and tertiary amines can be distinguished by allowing them to react with Hinsberg’s reagent $($benzenesulphonyl chloride $\ce{C_6H_5SO_2Cl)}.$
Secondary amines react with Hinsberg’s reagent to form a product that is insoluble in an alkali.
For example, $N, N-$diethylamine reacts with Hinsberg’s reagent to form $N, N-$diethylbenzenesulphonamide.
which is insoluble in an alkali. Tertiary amines.
however, do not react with Hinsberg’s reagent.
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Arrange the following compounds in the order of property indicated for each set:
p-toluidine, N, N-dimethyl-p-toluidine, p-nitroaniline, aniline (increasing basicity)
Answerp-nitroaniline < aniline < p-toluidine < N, N-dimethyl-p-toluidine.
View full question & answer→Question 1241 Mark
A compound $Z$ with molecular formula $\ce{C_3H_9N}$ reacts with $\ce{C_6H_5SO_2Cl}$ to give a solid, insoluble in alkali. Identify $Z.$
AnswerCompound $'Z\ '$ with molecular formula $\ce{C_3H_9N}$ is an aliphatic amine which on treatment with $\ce{C_6H_5SO_2Cl}$ gives a solid, insoluble in alkali.
Therefore, the product does not have any replaceable hydrogen on the nitrogen atom. In other words, the amines $(Z)$ must be secondary amine, i.e., $Z$ is ethyl methylamine $\ce{(C_2H_5NHCH_3)}.$
View full question & answer→Question 1251 Mark
Why is benzene diazonium chloride not stored and is used immediately after its preparation?
AnswerBenzene diazonium chloride is very unstable. Therefore, it cannot be stored and is used immediately after its preparation.
View full question & answer→Question 1261 Mark
Mention the chief use of quaternary ammonium salts derived from long chain amines.
AnswerThe quaternary ammonium salts derived from long chain aliphatic amines are used as detergents, such as,
$\ce{CH_3(CH_2)_{15} ^+N(CH^3)^3Br^–}.$
View full question & answer→Question 1271 Mark
Account for the following: Aniline gets coloured on standing in air for a long time.
AnswerDue to electron$-$donating effect $(+$R$-$effect$)$ of $-NH_2$ group, the electron density on the benzene ring increases. As a result, aniline is easily oxidised on standing in air for a long time to form coloured products.
View full question & answer→Question 1281 Mark
How will you carry out the following conversions?
p-toluidine diazonium chloride → p-toluicacid.
Question 1291 Mark
Write the type of hybridisation of orbitals of nitrogen present in amine.
AnswerThe nitrogen atom in amines is $s p^3$ hybridized.
View full question & answer→Question 1301 Mark
Explain why alkyl amine is more basic than ammonia?
AnswerAlkyl amines are more basic than ammonia due to the +I (inductive) effect of the alkyl group, which increases the electron density on the nitrogen atom.
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