Question 11 Mark
Give the structure of A, B and C in the reaction :
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Question 21 Mark
Give the structure of A, B and C in the reaction :
Question 31 Mark
Give the structure of A, B and C in the reaction :
Question 41 Mark
Give the structure of A, B and C in the reaction :
Question 51 Mark
Give the structure of A, B and C in the reaction :
Question 61 Mark
Give the structure of A, B and C in the reaction :
Question 71 Mark
Accomplish the conversion :
Benzamide to toluene
Benzamide to toluene
Question 81 Mark
Accomplish the conversion :
Aniline to p-bromo aniline
Aniline to p-bromo aniline
Question 91 Mark
Accomplish the conversion :
Chlorobenzene to p-chloroamine
Chlorobenzene to p-chloroamine
Question 101 Mark
Accomplish the conversion :
Benzyl chloride to 9-phenyl ethanomine
Benzyl chloride to 9-phenyl ethanomine
Question 111 Mark
Accomplish the conversion :
Aniline to benzyl alcohol.
Aniline to benzyl alcohol.
Question 121 Mark
Accomplish the conversion :
Aniline to 2,4,5-tribromo fluoro beznene
Aniline to 2,4,5-tribromo fluoro beznene
Question 131 Mark
Accomplish the conversion :
Benzoic acid to aniline
Benzoic acid to aniline
Question 141 Mark
Accomplish the conversion :
Benzene to m-bromo phenol
Benzene to m-bromo phenol
Question 151 Mark
Accomplish the conversion :
Nitrobenezene to benzoic acid
Nitrobenezene to benzoic acid
Question 161 Mark
Write short notes :
Hoffmann's bromamide reaction
Hoffmann's bromamide reaction
Answer
View full question & answer→Hoffmann Bromamide Reaction : In this reaction a amide is treated in an aqueous or ethanoic solution of NaOH or KOH , with bromine then primary amine is formed.
Question 171 Mark
How will you convert :
Propanoic acid into ethanoic acid?
Propanoic acid into ethanoic acid?
Question 181 Mark
How will you convert :
Nitromethane into dimethyl amine
Nitromethane into dimethyl amine
Question 191 Mark
How will you convert :
Methanamine into ethanamine
Methanamine into ethanamine
Question 201 Mark
How will you convert :
Ethanoic acid into propanoic acid
Ethanoic acid into propanoic acid
Question 211 Mark
How will you convert :
Ethanamine into methanamine
Ethanamine into methanamine
Question 221 Mark
How will you convert :
Methanol to ethanoic acid
Methanol to ethanoic acid
Question 231 Mark
How will you convert :
Hexane nitrile into 1 -amino pentane
Hexane nitrile into 1 -amino pentane
Question 241 Mark
How will you convert :
Ethanoic acid into methanamine
Ethanoic acid into methanamine
Question 251 Mark
Arrange the following :
In increasing order of solubility in water :
$C_6 H_5 NH_2,\left(C_2 H_5\right)_2 NH, C_2 H_5 NH_2$
View full question & answer→In increasing order of solubility in water :
$C_6 H_5 NH_2,\left(C_2 H_5\right)_2 NH, C_2 H_5 NH_2$
Question 261 Mark
Arrange the following :
In increasing order of boiling point :
$C_2 H_5 OH,\left(CH_3\right)_2 NH, C_2 H_5 NH_2$
View full question & answer→In increasing order of boiling point :
$C_2 H_5 OH,\left(CH_3\right)_2 NH, C_2 H_5 NH_2$
Question 271 Mark
Arrange the following :
In decreasing order of basic strength in gas phase
$C_2 H_5 NH_2,\left(C_2 H_5\right)_2 NH,\left(C_2 H_5\right)_3 N \text { and } NH_3$
In decreasing order of basic strength in gas phase
$C_2 H_5 NH_2,\left(C_2 H_5\right)_2 NH,\left(C_2 H_5\right)_3 N \text { and } NH_3$
Answer
View full question & answer→$\left(C_2 H_5\right)_3 N>\left(C_2 H_5\right)_2 NH>C_2 H_5 NH_2>NH_3$
(Decreasing order of basic strength in gas phase)
(Decreasing order of basic strength in gas phase)
Question 281 Mark
Arrange the following :
In incresing order of basic strength :
(a) Aniline, p-nitroaniline and p-toluidine
(b) $C _6 H _5 NH _2, C _6 H _5 NHCH _3, C _6 H _5 CH _2 NH _2$
In incresing order of basic strength :
(a) Aniline, p-nitroaniline and p-toluidine
(b) $C _6 H _5 NH _2, C _6 H _5 NHCH _3, C _6 H _5 CH _2 NH _2$
Answer
(b) $C _6 H _5 NH _2< C _6 H _5 NHCH _3< C _6 H _5 CH _2 NH _2$
View full question & answer→(b) $C _6 H _5 NH _2< C _6 H _5 NHCH _3< C _6 H _5 CH _2 NH _2$
Question 291 Mark
Arrange the following :
In increasing order of basic strength :
$C _6 H _5 NH _2, C _6 H _5 N\left( CH _3\right)_2,\left( C _2 H _5\right)_2 NH$ and $CH _3 NH _2$
In increasing order of basic strength :
$C _6 H _5 NH _2, C _6 H _5 N\left( CH _3\right)_2,\left( C _2 H _5\right)_2 NH$ and $CH _3 NH _2$
Answer
View full question & answer→$C _6 H _5 NH _2< C _6 H _5 N\left( CH _3\right)_2< CH _3- NH _2<$ $\left( C _2 H _5\right)_2 H$ (Increasing order of basic strength)
Question 301 Mark
Arrange the following :
In decreasing order of the $pK _{ b }$ values :
$C _2 H _5 NH _2, C _6 H _5 NHCH _3,\left( C _2 H _5\right)_2 NH$ and $C _6 H _5 NH _2$
In decreasing order of the $pK _{ b }$ values :
$C _2 H _5 NH _2, C _6 H _5 NHCH _3,\left( C _2 H _5\right)_2 NH$ and $C _6 H _5 NH _2$
Answer
View full question & answer→$C _6 H _5 NH _2> C _6 H _5 NHCH _3> C _2 H _5 NH _2>$ $\left( C _2 H _5\right)_2 NH$ (Decreasing order of $pK _{ b }$ value means increasing order of basic strength)
Question 311 Mark
Account for the following :
Gabriel phthalimide synthesis is preferred for synthesising primary amines.
Gabriel phthalimide synthesis is preferred for synthesising primary amines.
Answer
View full question & answer→In Garbriel phthalimide synthesis $RNH _2$ is formed from R-X in which pure primary amine is formed and no other side products are formed. Because phthallic acid obtained is used again while in other reactions mixture of products is formed. So Gabriel phthallimide synthesis is preferred for synthesising primary amines.
Question 321 Mark
Account for the following :
Diazonium salts of aromatic amines are more stable than those of aliphatic amines.
Diazonium salts of aromatic amines are more stable than those of aliphatic amines.
Answer
View full question & answer→Diazonium salts of aromatic amines are more stable than diazonium salts of aliphatic amines because due to resonance they attains stability. Resonating structures of $C _6 H _5 N_2^{+}$are as under :
Arene diazonium salts are stable for some time in solution at low temperature (273-278 K).
Arene diazonium salts are stable for some time in solution at low temperature (273-278 K).
Question 331 Mark
Account for the following :
Aniline does not undergo Friedel-Crafts reaction.
Aniline does not undergo Friedel-Crafts reaction.
Answer
View full question & answer→Aniline does not show Friedel-Craft reaction because in this reaction $AlCl _3$ is used as catalyst which is Lewis acid which form salt with aniline (Lewis base). Due to salt formation nitrogen of aniline attains + ve charge which is strong deactivating group so reactivity of this decreases.
Question 341 Mark
Account for the following :
Ethyl amine is soluble in water whereas aniline is not.
Ethyl amine is soluble in water whereas aniline is not.
Answer
View full question & answer→Ethyl amine $\left( C _2 H _5- NH _2\right)$ forms hydrogen bond with water while aniline does not have tendency to form hydrogen bond with water due to large size non-polar $C _6 H _5-$ group so ethyl amine is soluble in water whereas aniline is not.
Question 351 Mark
Give one chemical test to distinguish between the pair of compound :
Aniline and N-methyl aniline
Aniline and N-methyl aniline
Answer
View full question & answer→Aniline $\left(1^{\circ}\right)$. Gives carbyl amine test with $CHCl _3$ and alkali while N -methyl aniline $\left(2^{\circ}\right)$ do not give carbyl amine test.
Question 361 Mark
Give one chemical test to distinguish between the pair of compound :
Aniline and benzyl amine
Aniline and benzyl amine
Answer
View full question & answer→Aniline form azo dye by the reaction with benzene diazonium chloride $\left( C _6 H _5 \stackrel{+}{ N }_2 \overline{ C } l \right)$ while in benzyl amine this does not occurs.
Question 371 Mark
Give one chemical test to distinguish between the pair of compound :
Ethyl amine and aniline
Ethyl amine and aniline
Answer
View full question & answer→Ethyl amine does not form azo dye by the reaction with benzene diazonium chloride while aniline forms azo dye (yellow) by the reaction with benzene diazonium chloride.
Question 381 Mark
Give one chemical test to distinguish between the pair of compound :
Secondary and tertiary amines
Secondary and tertiary amines
Answer
View full question & answer→Secondary amine $\left( R _2 NH \right)$ reacts with Hinsberg reagent in a product formed insoluble in alkali while tertiary amine $\left( R _3 N\right)$ does not react with Hinsberg reagent.
Question 391 Mark
Give one chemical test to distinguish between the pair of compound :
Methylamine and dimethyl amine
Methylamine and dimethyl amine
Answer
View full question & answer→Methylamine $CH _3- NH _2\left(1^{\circ}\right)$ reacts $\left( C _6 H _5 SO _2 Cl \right)$ and product formed is soluble in alkali while product formed by the reaction of dimethyl amine $CH _3- NH -$ $CH _3\left(2^{\circ}\right)$ with Hinsberg reagent (benzene sulphonyl chloride) is insoluble in alkali.
Question 401 Mark
Write IUPAC name of the compound and classify them into primary, secondary and tertiary amine :
Answer
View full question & answer→3-bromoaniline or 3-bromobenzenamine $\left(1^{\circ}\right)$
Question 411 Mark
Write IUPAC name of the compound and classify them into primary, secondary and tertiary amine :
$\left( CH _3 CH _2\right)_2 NCH _3$
$\left( CH _3 CH _2\right)_2 NCH _3$
Answer
View full question & answer→N-ethyl-N-methyl ethanamine $\left(3^{\circ}\right)$
Question 421 Mark
Write IUPAC name of the compound and classify them into primary, secondary and tertiary amine :
$C _6 H _5 NHCH _3$
$C _6 H _5 NHCH _3$
Answer
View full question & answer→N -methyl benzeneamine or N -methylaniline $\left(2^{\circ}\right)$
Question 431 Mark
Write IUPAC name of the compound and classify them into primary, secondary and tertiary amine :
$\left( CH _3\right)_3 CNH _2$
$\left( CH _3\right)_3 CNH _2$
Answer
View full question & answer→2-methyl propan-2-amine $\left(1^{\circ}\right)$
Question 441 Mark
Write IUPAC name of the compound and classify them into primary, secondary and tertiary amine :
$CH _3 NHCH \left( CH _3\right)_2$
$CH _3 NHCH \left( CH _3\right)_2$
Answer
View full question & answer→N -methyl propan-2-amine $\left(2^{\circ}\right)$
Question 451 Mark
Write IUPAC name of the compound and classify them into primary, secondary and tertiary amine :
$CH _3\left( CH _2\right)_2 NH _2$
$CH _3\left( CH _2\right)_2 NH _2$
Answer
View full question & answer→Propan-1-amine $\left(1^{\circ}\right)$
Question 461 Mark
Write IUPAC name of the compound and classify them into primary, secondary and tertiary amine :
$\left( CH _3\right)_2 CHNH _2$
$\left( CH _3\right)_2 CHNH _2$
Answer
View full question & answer→Propan-2-amine $\left(1^{\circ}\right)$
Question 471 Mark
Give plausible explanation of :
Why are amines less acidic than alcohols of comparable molecular masses?
Why are amines less acidic than alcohols of comparable molecular masses?
Answer
View full question & answer→Amines are less acidic than alcohols of comparable molecular masses because in amines polarity of $N - H$ bond is less than $- O - H$ bond of alcohols due to more electronegativity of oxygen than nitrogen. So amines have less tendency to denote $H ^{+}$in comparison alcohols.
Question 481 Mark
Why cannot aromatic primary amines be prepared by Gabriel phthalimide synthesis ?
Answer
View full question & answer→Aromatic primary amines cannot be prepared by Gabriel phthalimide synthesis because in aryl halide double bond character is attained in carbon halogen bond due to resonance ( + M effect) so bond becomes strong due to which aryl halide do not undergo nucleophilic substitution with the anion formed by phthalimide.
Question 491 Mark
Complete the reaction :
$C _6 H _5 N_2 Cl \xrightarrow[\text { (ii) } NaNO _2 / Cu , \Delta]{\text { (i) } HBF _4}$
View full question & answer→$C _6 H _5 N_2 Cl \xrightarrow[\text { (ii) } NaNO _2 / Cu , \Delta]{\text { (i) } HBF _4}$
Question 501 Mark
Complete the reaction :
$C _6 H _5 NH _2+\left( CH _3 CO \right)_2 O \rightarrow$
View full question & answer→$C _6 H _5 NH _2+\left( CH _3 CO \right)_2 O \rightarrow$