MCQ Questions for Class 12 Engineering Chemistry Amines Quiz 10

Which of the following statements is correct?

0%
Aliphatic amines are stronger bases than ammonia.
0%
Aromatic amines are stronger bases than ammonia.
0%
The alkyl group in alkyl ammonium ion more stabilizes the ion relative to the amine.
0%
The aryl group in aryl ammonium ion less stabilizes the ion relative to the amine.

Explanation

(A)

$+I$ effect of alkyl groups in aliphatic amines make them more basic than ammonia.

(B) Aromatic amines are weaker bases than ammonia because the lone pair of

$N$ is in delocalization and not freely available to act as base.

$+I$ or electron releasing effect.

(D) The aryl group in aryl ammonium ion less stabilizes the ion relative to the amine due to the delocalization of positive charge on electron negative

$N$ .

Hence, A, C, D are correct statements.

$C$ and

$D$ in the given sequence are :

0%
Benzoic acid $+$ aniline
0%
Phthalic acid $+$ ethylamine
0%
Phthalic acid $+$ aniline
0%
Benzoic acid $+$ ethylamine

The correct set of products obtained in the following reactions

$RCN\xrightarrow [ ]{ reduction } (A)$ ,

$RCN \xrightarrow [ (ii){ H }_{ 2 }O ]{ (i){ CH }_{ 3 }MgBr } (B)$ ,

$RNC \xrightarrow [ ]{ hydrolysis } (C)$ ,

$R{NH}_{2}\xrightarrow [ ]{ H{NO}_{2} } (D)$
The answer is:

0%
$(A): {2}^{o}$ Amine; $(B)$ : Methyl ketone; $(C):{1}^{o}$ Amine; $(D)$ Alcohol
0%
$(A): {1}^{o}$ Amine; $(B)$ : Methyl ketone; $(C):{1}^{o}$ Amine; $(D)$ Alcohol
0%
$(A): {2}^{o}$ Amine; $(B)$ : Methyl ketone; $(C):{2}^{o}$ Amine; $(D)$ Acid
0%
$(A): {2}^{o}$ Amine; $(B)$ : Methyl ketone; $(C):{2}^{o}$ Amine; $(D)$ Aldehyde

Explanation

The compound (A) is a primary amine:

$\displaystyle RCN\xrightarrow [ ]{ reduction } \underset {(A)}{RNH_2}$

The compound (B) is a methyl ketone:

$\displaystyle RCN \xrightarrow [ (ii){ H }_{ 2 }O ]{ (i){ CH }_{ 3 }MgBr } \underset {(B)}{RCOCH_3}$

The compound (C) is a primary amine:

$\displaystyle RNC \xrightarrow [ ]{ hydrolysis } \underset{ (C) }{RNH_2}$

The compound (D) is an alcohol:

$\displaystyle R{NH}_{2}\xrightarrow [ ]{ H{NO}_{2} } \underset { (D)}{ROH}$

${CH}_{3}{CH}_{2}{NH}_{2}$ is soluble in:

0%
dilute $HCl$
0%
$Cu{SO}_{4}$ solution
0%
$Ag{NO}_{3}$
0%
none of these

Explanation

$CH_3CH_2NH_2$ being polar and basic in character is soluble in all - dilute

$HCl$ ,

$CuSO_4$ solution and

$AgNO_3$ .

The compound formed is an energetically stable, yellow azo dye (the azo group is

$-N=N-$ ). The stability is due to extensive delocalisation of electrons via the nitrogen-nitrogen double bonds.

The azo dye obtained on reacting 4-aminophenol with nitrous acid (in dilute hydrochloric acid) below

${10}^{o}C$ and coupling the resulting diazonium salt with phenol is:

Arrange the following resonating structure according to their contribution towards resonance hybrid?
(a)

$CH_2=\overset {\oplus}{N}=\overset {\ominus}{N}$ (b)

$\overset {\ominus}{C}H_2-N=\overset {\oplus}{N}:$ (c)

$\overset {\oplus}{C}H_2-\underset {. .}{N}=\overset {\ominus}{N}$ (d)

$\overset {\ominus}{C}H_2-\overset {\oplus}{N}\equiv \overset {. .}{N}$

0%
a > d > c > b
0%
b > a > c > d
0%
a > c > b > d
0%
d > a > b > c

Correct order of basicity of following molecule :

0%
I > II > III > IV
0%
II > I > III > IV
0%
II > III > I > IV
0%
II > III > IV > I

Consider the given ions.
The reactivities of these ions in a azo-coupling reaction (under similar conditions) will be such that :

0%
$(I)< (IV)< (II)< (III)$
0%
$(I)< (III)< (IV)< (II)$
0%
$(III)< (I)< (IV)< (II)$
0%
$(III)< (I)< (II)<(IV)$

Explanation

Aromatic diazonium ions act as electrophiles in coupling reactions with activated aromatics such as anilines or phenols. The substitution normally occurs at the para position.

The reactivities of given ions in an azo-coupling reaction will be

$I<III<IV<II$ as

$NO_2$ makes the compound strong electrophile. While

$CH_3, O-CH_3 \ and \ -NMe_2$ has electron releasing effects which successively make the compound weaker electrophile.

Amongst the following the most basic compound is-

Explanation

$\text { The basic strength of the compounds highly } \\$

$\text { depends on the }+I \text { effect of the groups. }$

$\mathrm{NH}_{2}$ is an

$e^{-}$ donating group. Hence it increases the basic strength of the compound.

$_{2}$ and

$C H O$ are

$e^{-}$ withdrawing groups.

Hence it decreases the basic strength of the group.

The basic character of substituted anilines (I to IV) are such that :

0%
I < II < IV < III
0%
I < II < III < IV
0%
II < I < III < IV
0%
III < IV < I < II

Explanation

The basic character of substituted anilines (I to IV) order is:

$\displaystyle III < IV < I < II$ .

When electron-withdrawing groups such as nitro and chloro are present, the basic strength of aniline is reduced.

When electron releasing groups such as methyl and methoxy are present, the basic strength of aniline is increased.

Option D is correct.

Among the following the weakest base is:

0%
$C_6H_5CH_2NH_2$
0%
$C_6H_5CH_2NHCH_3$
0%
$O_2N-CH_2NH_2$
0%
$CH_3NHCHO$

Explanation

The weakest base is

$\displaystyle CH_3NHCHO$ .

The lone pair of electrons cannot be easily donated to suitable Lewis acid as the nitro group is electron-withdrawing group.

As amides are not basic in nature and given option D is amide structure and other compounds are amines.

Option D is correct.

Which of the following is/are correct statement/statements ?

0%
Guanidine $\begin{bmatrix}NH_2-\underset{\underset{NH}{||}}{C}-NH_2\end{bmatrix}$ is more basic than pyridine because conjugate acid of guanidine has three equal contributing resonating structure.
0%
Diethylamine is weaker base than triethylamine in aqueous medium.
0%
Ortho-methyl aniline is weaker base than para-methyl aniline.
0%
$2,\,4,\,6$ -Trinitro- $N,\,N$ -dimethyl aniline is stronger base than $2,\,4,\,6$ -Trinitro aniline.

The correct order of increasing basic nature of the following base is :

0%
2 < 5 < 1 < 3 < 4
0%
5 < 2 < 1 < 3 < 4
0%
2 < 5 < 1 < 4 < 3
0%
5 < 2 < 1 < 4 < 3

Explanation

In structure 2, electron withdrawing group

$NO_2$ present in para position.

Also in structure 5, electron withdrawing group

$NO_2$ is present in meta position.

At para position

$NO_2$ shows more withdrawing effect than at meta position due to

$-R$ effect.

Hence, structure 2 is least basic.

Next basic structure is aniline i.e., structure 1.

Next comes the structure 3 and 4 in which methyl and methoxy groups are present which are electron-donating groups.

So the correct answer is A

Arrange the following in the increasing order of the

$\displaystyle{pK}_b$ values:

$C_2H_5NH_2$
ii)

$\,{ C }_{ 6 }{ H }_{ 5 }{ NHCH }_{ 3 }$
iii)

$\,({ C }_{ 2}{ H }_{ 5 }{ ) }_{ 2 }NH$
iv)

${ C }_{ 6 }{ H }_{ 5 }{ NH }_{ 2 }$

0%
i < ii < iii < iv
0%
iii < i < ii < iv
0%
i < iii < ii < iv
0%
iv < ii < i < iii

Explanation

$pK_b$ value gives a measure of the basicity of a compound. Higher the value of

$K_b$ of a base, stronger is the base. In aniline,

$C_6H_5NH_2$ , lone pair on nitrogen atom are withdrawn into the benzene ring. Benzene ring is an electron withdrawing group and hence, the lone pair is not easily available on nitrogen atom, so it cannot attract a proton easily.

N-Methylaniline,

$C_6H_5NHCH_3$ has benzene ring attached to it as well as a methyl group attached to the nitrogen atom. The benzene ring withdraws lone pair from nitrogen atom . Methyl group attached to the nitrogen atom is an electron donating group and it donates electron to nitrogen atom increasing electron density .

Thus the

$K_b$ value for N-Methylaniline is higher than aniline and it is a stronger base than aniline.

$C_2H_5NH_2$ , ethyl groups are electron donating group. It donates electrons to nitrogen atom increasing the electron density over nitrogen atom. Thus Ethylamine is a stronger base than N-Methylaniline.

In diethylamine,

$(C_2H_5)_2NH$ , two Ethyl groups attached to nitrogen atom are electron donating group. increasing the electron density on nitrogen atom. hence, diethylamine is a stronger base than ethylamine because of the presence of two electron donating groups.

Thus the correct option is D.

Arrange the following in the decreasing order of the

$\displaystyle{pK}_b$ values:

$i)\,\displaystyle { C}_{ 2 }{ H }_{ 5 }{ NH }_{ 2 }$

$ii) \,{ C }_{ 6 }{ H }_{ 5 }{ NHCH }_{ 3 }$

$iii) \,({ C }_{ 2}{ H }_{ 5 }{ ) }_{ 2 }NH$

$iv) { C }_{ 6 }{ H }_{ 5 }{ NH }_{ 2 }$

0%
$i > ii > iii > iv$
0%
$iii > i > ii > iv$
0%
$i > iii > ii > iv$
0%
$ii > i > iii > iv$

Explanation

$+I$ effect of alkyl group increases electron density on

$-N$ of amine and increases basic strength. Thus

$(C_2H_5)_2NH$ is the strongest base. In aniline, the lone pair of electrons of

$N$ is in delocalization with ring which makes it less available and least basic. Methyl substitution increases basic strength due to its inductive effect. Therefore the order of base strength or

$pK_b$ is

$iii>i>ii>iv$

Iso-propyl amine is a:

0%
primary amine
0%
secondary amine
0%
teritary amine
0%
quaternary amine

Explanation

The structure of isopropyl amine is

$CH_3-\underset{\underset{\displaystyle NH_2}{|}}{C}H-CH_3$
It is primary amine.

Identify the compounds from the following which form primary amines under suitable reduction conditions.
1.

$C_2H_5NC$

$C_2H_6$

$C_2H_5CONH_2$

$C_6H_5NO_2$

0%
$1, 4$
0%
$3, 4$
0%
$1, 3, 4$
0%
$2, 3, 4$

Explanation

$C_2H_5CONH_2 \xrightarrow{LiAlH_4}C_2H_5CH_2NH_2$

$C_6H_5NO_2 \xrightarrow{Sn/HCl} C_6H_5NH_2$

Within the list shown below, the correct pair of structures of alanine in pH ranges 2-4 and 9-11 is:
I.

$H_3N^+ - CH(CH_3)CO_2H$
II.

$H_2N - CH(CH_3)CO_2^-$
III.

$H_3N^+ - CH(CH_3)CO_2^-$
IV.

$H_2N -CH(CH_3)CO_2H$

0%
I and II
0%
I and III
0%
II and III
0%
III and IV

Explanation

The correct pair of structures of alanine in pH ranges 2-4 and 9-11 is
I.

$\displaystyle H_3N^+ - CH(CH_3)CO_2H$
II.

$\displaystyle H_2N - CH(CH_3)CO_2^-$
In pH range 2-4, the medium is acidic and the amino group is protonated.
In pH range 9-11, the medium is basic and carboxylic group is deprotonated.

Predict respectively

$'X'$ and

$'Y'$ in the following reactions

$Ar-N{ H }_{ 2 }\xrightarrow { \quad \quad X\quad \quad } Ar-\overset { + }{ N } \equiv N.{ Cl }^- \xrightarrow { \quad \quad Y\quad \quad }$

$Ar-Cl$

0%
$NaN{ O }_{ 3 }$ & ${ Cl }_{ 2 }$
0%
$NaN{ O }_{ 3 }-HCl$ & $HCl$
0%
$NaN{ O }_{ 2 }-HCl$ & ${ Cu }/{ HCl }$
0%
$NaN{ O }_{ 2 }-HCl$ & $NaN{ H }_{ 2 }$

Explanation

The balanced complete chemical equation is as follows:

$Ar-NH_2\xrightarrow{NaNO_2 - HCl} Ar-N^+_2-Cl^-$ , this reaction is the diazotisation reaction,

$Ar-N^+_2-Cl\xrightarrow{Cu/HCl} Ar-Cl + N_2$ , This is the Sandmeyer reaction.

The compound R is:

Explanation

Compound Q on treatment with

$Br_2/NaOH$ is nothing but Hoffmann Bromamide reaction which gives amine from amide.

Option A is correct.

A compound

$X$ has molecular formula

${C}_{7}{H}_{7}NO$ . On treatment with

${Br}_{2}$ and

$KOH$ ,

$X$ gives an amine

$Y$ . The later gives carbylamine test.

$Y$ upon diazotisation and coupling with phenol gives an azo dye. Thus,

$X$ is :

0%
${C}_{6}{H}_{5}N{O}_{2}$
0%
${C}_{6}{H}_{5}COON{H}_{4}$
0%
${C}_{6}{H}_{5}CON{H}_{2}$
0%
none of the above

Explanation

$C_7H_7NO$ , show the high degree of unsaturation, so it can be considered that there may be phenyl group, as phenyl group has 3 double bond.

2. As the compound

$C_7H_7NO$ , when reacts with

$Br_2/KOH$ , gives the amine Y. It shows that there must be amide group in

$C_7H_7NO$ compound, as this reaction is Hoffmann Bromamide synthesis.

3. As this amine is synthesized by Hoffmann Bromamide synthesis, it gives carbylamine test, so it means that amine must be primary amine.

4. Now as upon diazotisation this amine forms diazonium salt, also it gives the coupling reaction with phenol to give an azo dye, then it must be aromatic diazonium salt. If it were aliphatic then it would form alcohol by releasing

$N_2$ , and the coupling could not be possible.

So, all these steps show that this is

$C_6H_5CONH_2$ .
Reactions:
1.

$C_6H_5CONH_2 (X) + Br_2 + 4 KOH\longrightarrow C_6H_5NH_2 (Y) + K_2CO_3 + 2 KBr + 2H_2O$

$C_6H_5NH_2 (Y) + CHCl_3 + 3KOH +\xrightarrow{heat}C_6H_5-NC + 3 KCl + 3H_2O$ .

$C_6H_5NH_2 (Y) \xrightarrow{NaNO_2 + 2HCl (at\ 273-278 K)}C_6H_5-N_2^{+}Cl^- + 2NaCl + 2H_2O$ .

$C_6H_5N_2^{+}Cl^- \xrightarrow {C_6H_5OH /OH-} C_6H_5-N=N-C_6H_5-OH (azo\ dye)$ .

$CH_3CH_2CH_2CN+2H_2\xrightarrow {Pd}$

Which of the following compound is formed in the above reaction?

0%
$CH_3CH_2CH_2CH_2NH_2$
0%
$CH_3CH_2CH_2NH_2$
0%
$CH_3CH_2CH_2CH_2CH_2NH_2$
0%
None of the above

Explanation

Alkyl nitriles are reduced using a metal like palladium, platinum or nicke as a catalyst.
So,

$CH_3CH_2CH_2CN+2H_2\xrightarrow{Pd} CH_3CH_2CH_2CH_2NH_2$
Hence option

$A$ is correct.

In the reduction of nitriles to prepare primary amine:

0%
$LiAlH_4$ is used.
0%
$NaBH_4$ is used.
0%
$Sn/HCl$ is used.
0%
none of these.

Explanation

$LiAl{ H }_{ 4 }$ reduces nitriles to primary amines with ascent of amine series.

${ Sn }/{ HCl }$ reduce nitriles to immine hydrochloride.

${ NaB{ H }_{ 4 } }/{ { I }_{ 2 } }$ reduces nitriles to primary amines

Predict the product.

$CH_3-CH_2-CN + H_2/Ni \rightarrow$ ?

0%
$CH_3-CH_2-CH_2-NH_2$
0%
$CH_3-CH_2-NH_2$
0%
both a and b
0%
none of these

Explanation

$C{ H }_{ 3 }-C{ H }_{ 2 }-C\equiv N+{ { H }_{ 2 } }/{ Ni }\longrightarrow C{ H }_{ 3 }-C{ H }_{ 2 }-C{ H }_{ 2 }-N{ H }_{ 2 }$

The reduction of nitrile gives ascent of amine series indicating:

0%
preparation of amines containing two carbon atom more than starting amine
0%
preparation of amines containing one carbon atom more than starting amine
0%
preparation of amines containing three carbon atom more than starting amine
0%
preparation of amines containing four carbon atom more than starting amine

The compound T is:

0%
Glycine
0%
Alanine
0%
Valine
0%
Serine

Which of the following represents the correct order of basic strength?

0%
N,N-Dimethylaniline $>$ N-methylaniline $>$ aniline
0%
N,N-Dimethylaniline $>$ aniline $>$ N-methylaniline
0%
aniline $>$ N-methylaniline $>$ N,N-Dimethyl aniline
0%
N-methylaniline $>$ N,N-Dimethylaniline $>$ aniline

Explanation

The following represents the correct order of basic strength
N,N-Dimethylaniline

$>$ N-methylaniline

$>$ aniline
When hydrogen atoms of the amino group of arylamines are replaced by electron donating alkyl groups, the basic character of the resulting amine increases.

In the given set of reactions, the IUPAC name of product

$Y$ is:

$2-\text{Bromopropane} \xrightarrow [ alc.\ KOH ]{ AgCN }\ \ X\xrightarrow [ ]{ LiAl{ H }_{ 4 } }\ \ Y\quad$

0%
N-methylpropanamine
0%
N-isopropylmethanamine
0%
butan-2-amine
0%
N-methylpropan-2-amine

Point out the correct decreasing order of

$pK_b$ values of following amines:

$C_2H_5NH_2$ ,

$C_6H_5NHCH_3,$

$(C_2H_5)_2NH$ and

$C_6H_5NH_2$

0%
$(C_2H_5)_2NH > C_2H_5NH_2 > C_6H_5NHCH_3 > C_6H_5NH_2$
0%
$(C_2H_5)_2NH > C_6H_5NHCH_3 > C_6H_5NH_2 > C_2H_5NH_2$
0%
$C_6H_5NH_2 > C_6H_5NHCH_3 > C_2H_5NH_2 > (C_2H_5)_2NH$
0%
$C_2H_5NH_2 > (C_2H_5)_2NH > C_6H_5NHCH_3 > C_6H_5NH_2$

Explanation

The lesser the value of $pK_b$ the weaker the base.
Electron withdrawing groups like benzene decreases the basicity and increases the

$pK_b$ of amines and the alkyl groups (electron releasing groups) increases the basicity and decreases the

$pK_b$ value
So the correct order of

$pK_b$ is

$C_6H_5NH_2 > C_6H_5NHCH_3 > C_2H_5-NH_2 > (C_2H_5)_2NH$
Hence option C is correct.

Acetonitrile on reduction gives:

0%
ethanamine
0%
propanamine
0%
methanamine
0%
None of the above

Among the following substituted pyridines, the most basic compound is:

The correct basicity order of the following compound is :

0%
$I > III > II > IV$
0%
$III > I > II > IV$
0%
$II > I > III > IV$
0%
$I > II > III > IV$

Which of the following is the correct order of the solubility in water?

0%
propane $<$ ethylamine $<$ ethyl alcohol
0%
propane $<$ ethyl alcohol $<$ ethylamine
0%
ethyl alcohol $<$ propane $<$ ethylamine
0%
ethyl alcohol $<$ ethylamine $<$ propane

Explanation

Correct Answer: Option A

Explanation:

We know, like dissolves like. So, the polar compounds are soluble in the polar solvents. And the non-polar solvents are soluble in non-polar solvents.
Water is a polar solvent.
The order of polarity of the given compounds is:

$propane < ethylamine < ethyl\ alcohol$

So, the order of solubility in the water is:

$propane < ethylamine < ethyl\ alcohol$

Benzylamine is a stronger base than aniline because:

0%
the lone pair of electrons on the nitrogen atom in benzylamine is delocalised
0%
the lone pair of electrons on the nitrogen atom in aniline is delocalised
0%
the lone pair of electrons on the nitrogen atom in aniline is not involved in resonance
0%
benzylamine has a higher molecular mass than aniline

Explanation

Benzylamine is a stronger base than aniline because the lone pair of electrons on the nitrogen atom in aniline is delocalised over benzene ring.

Hence, due to resonance in aniline, the lone pair of electron on nitrogen is less available for donation to suitable acid.

However in case of benzylamine (

$C_6H_5-CH_2-NH_2$ ), the N atom is attached to saturated C atom.

Hence, no resonance is possible and the lone pair of electron can be easily donated.

Aniline is reacted with bromine water and the resulting product is treated with an aqueous solution of sodium nitrite in presence of dilute hydrochloric acid. The compound so formed is reacted with a tetrafluoroborate which is subsequently heated. The final product is:

0%
$p-bromoaniline$
0%
$p-bromofluorobenzene$
0%
$1, 3, 5-tribromobenzene$
0%
$2, 4, 6 - tribromofluorobenzene$

Gabriel synthesis is used for synthesis of:

0%
primary amines
0%
secondary amines
0%
aldehydes
0%
acids

Gabriel phthalimide synthesis is used in the preparation of primary amine from phthalimide which of the following reagent is not used during the process?

0%
HCl
0%
NaOH
0%
Alkyl Halides
0%
KOH

Which one of the following will be most basic?

0%
Aniline
0%
P-methoxyaniline
0%
P-nitroaniline
0%
Benzylamine

The correct order of increasing basic nature for the bases

$NH_3, CH_3NH_2$ and

$(CH_3)_2NH$ in aqueous solutions.

0%
$NH_3 < CH_3NH_2 <(CH_3)_2NH$
0%
$CH_3NH_2 < NH_3 < CH_3)_2NH$
0%
$CH_3NH_2 < (CH_3)_2NH < NH_3$
0%
$(CH_3)_2NH < NH_3 < CH_3NH_2$

Explanation

In aqueous solution, the correct order of increasing basic nature for the basicity is:

$\displaystyle NH_3 < CH_3NH_2 <(CH_3)_2NH$ .

It is due to combination of the inductive effect, solvation effect, steric effect etc.

Which of the following is an example of carbylamine reaction?

0%
$2PhNH_2+CS_2\xrightarrow[\begin{matrix}\Delta\\(ii)\,conc.HCl\end{matrix}]{(i)KOH}Ph-NCS$
0%
$R-NH_2+HNO_2\xrightarrow{0-5^oC} R- OH+N_2+HCl$
0%
$Ph-NH_2+CHCl_3+3KOH(alc.)\xrightarrow{\Delta} Ph-NC+3KCl+3H_2O$
0%
$Ph_2-NH+HONO\rightarrow Ph_2-N-N=O+H_2O$

Explanation

(a)

$2Ph-NH_2+S=C=S\xrightarrow[\begin{matrix}\Delta\\(ii)\,conc.HCl\end{matrix}]{(i) 2KOH}\underset{\text{Hotmann mustard oil reaction}}{Ph-N=C=S+Ph\overset{\oplus}{N}H_3Cl^-}$

(b)

$R-NH_2+HNO_2\xrightarrow{0-5^oC} \underset{\text{Diazotisation reaction}}{R- OH+N_2+HCl}$

(c)

$Ph-NH_2+CHCl_3+3KOH(alc.)\xrightarrow{\Delta} \underset{\text{Carbylamine reaction}}{Ph-NC+3KCl+3H_2O}$

(d)

$Ph_2-NH+HONO\rightarrow \underset{\text{Lieber mann's nitroso reaction}}{Ph_2-N-N=O+H_2O}$

Option C is correct.

Pick the correct statement among the following.

0%
Sodium lauryl sulphate forms an insoluble scum with hard water.
0%
Sodium dodecyl benzene sulphonate used in tooth paste is a cationic detergent.
0%
Cetyl-trimethyl ammonium bromide is a popular cationic detergent used in hair conditioner.
0%
Non-ionic detergents is formed when polyethylene glycol reacts with adipic acid.

The order of basicity among the following compounds is:

0%
$I > IV > III > II$
0%
$II > I > IV > III$
0%
$IV > I > II > III$
0%
$IV > II > III > I$

Explanation

The order of basicity among the following compounds is

$\displaystyle IV>I>II>III$

In IV, the extent of delocalization is maximum. When a lone pair of electrons on one of the

$N$ atom is donated to a suitable acid, still some delocalization is possible.

However, in case of I, when a lone pair of electrons on one of the

$N$ atom is donated to a suitable acid, further delocalization is not possible.

So, option C is correct.

Amines behave as:

0%
Lewis acids
0%
Lewis base
0%
aprotic acid
0%
neutral compound

Explanation

Amine is an organic compound derived from ammonia by replacement of one or more hydrogen atoms by organic groups. Ex: Aniline

( $C_6H_5NH_2$ ). Because of the presence of lone pair amines acts as Lewis bases.

Hence, option $B$ is correct

Between any two of the following molecules, hydrogen bonding is not possible:

0%
two primary amine molecules
0%
two secondary amine molecules
0%
two tertiary amine molecules
0%
two ammonia molecules

Explanation

Between two tertiary amine molecules, hydrogen bonding is not possible.
The hydrogen bonding is possible when

$\displaystyle H$ atom is attached to electronegative

$\displaystyle N, O$ or

$\displaystyle F$ atom. In tertiary amines, no

$\displaystyle H$ atom is attached to

$\displaystyle N$ atom. Hence, hydrogen bonding is not possible.

Gabriel's phthalimide synthesis can be used to prepare:

0%
ethanamine
0%
n-methylmethanamine
0%
benzeneamine
0%
$N, N-$ dimethylmethanamine
0%
p-toluidine

Which one of the following is used as a test for aliphatic primary amines ?

0%
Tollen's test
0%
Fehling's lest
0%
Isocyanide test
0%
Azo dye test
0%
Phthalein fusion test.

Explanation

Isocyanide test (carbylamine test) is used to confirm primary amines.

$R-NH_2 + CHCl_3 + 3KOH \rightarrow \underset{Isocyanide }{R-NC} + 3KCl + 3H_2O$

Which one of the following can be prepared by Gabriel phthalimide synthesis?

0%
Aniline
0%
$o$ -toluidine
0%
Benzylamine
0%
N-methylethanamine
0%
4-bromoaniline

$CH_3CH_2Cl \xrightarrow {NaCN}X \xrightarrow{Ni/H_2}Y \xrightarrow{acetic\, anhydride}Z$

In the above reaction sequence,

$Z$ is:

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$CH_3CH_2CH_2NHCOCH_3$
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$CH_3CH_2CH_2NH_2$
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$CH_3CH_2CH_2CONHCH_3$
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$CH_3CH_2CH_2CONHCOCH_3$

Explanation

$CH_3CH_2Cl \overset{NaCN}{\rightarrow} CH_3- \underset{X}{CH_2}-CN \overset{Ni/H_2}{\rightarrow} CH_3-CH_2 CH_2NH_2\xrightarrow[anhydride]{Acetic} CH_3-CH_2 -\underset{Z}{CH_2}-NHCOCH_3$

$C{ H }_{ 3 }Br+KCN\left( alc. \right) \longrightarrow X\xrightarrow [ \quad Na+{ C }_{ 2 }{ H }_{ 5 }OH\quad ]{ \quad Reduction\quad } Y$
What is

$Y$ in the series?

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$C{ H }_{ 3 }CN$
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${ C }_{ 2 }{ H }_{ 5 }CN$
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${ C }_{ 2 }{ H }_{ 5 }N{ H }_{ 2 }$
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$C{ H }_{ 3 }N{ H }_{ 2 }$

Explanation

$C{ H }_{ 3 }Br+KCN\left( alc. \right) \longrightarrow \underset {X}{C{ H }_{ 3 }CN}$

$\underset { Y }{ C{ H }_{ 3 }C{ H }_{ 2 }N{ H }_{ 2 } } \xleftarrow [ Reduction ]{ \quad { Na }/{ { C }_{ 2 }{ H }_{ 5 }OH }\quad }$

Which of the following are P and S in the series of reactions?

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$C_2H_5MgX$ and $(C_2H_5)_2NCH_3$
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$CH_3MgX$ and N-methyl- $2$ -butanamine
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$CH_3MgX$ and N-ethyl- $2$ -butanamine
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$C_2H_5MgX$ and $(CH_3)_2NC_2H_5$

CBSE Questions for Class 12 Engineering Chemistry Amines Quiz 10 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

CBSE Questions for Class 12 Engineering Chemistry Amines Quiz 10 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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