MCQ Questions for Class 12 Engineering Chemistry Amines Quiz 1

Which of the following is NOT a correct method of the preparation of benzylamine from cyanobenzene?

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(i) $HCl/ H_{2}O$ (ii) $NaBH_{4}$
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(i) $LiAlH_{4}$ (ii) $H_{3}O^{+}$
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(i) $SnCl_{2} + HCl(gas)$ (ii) $NaBH_{4}$
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$H_{2}/ Ni$

Considering the basic strength of amines in aqueous solution, which one has the smallest

$pK_{b}$ value?

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$(CH_{3})_{3}N$
0%
$C_{6}H_{5}NH_{2}$
0%
$(CH_{3})_{2}NH$
0%
$CH_{3}NH_{2}$

In the following reactions, the major product W is:

Consider the following four compounds I, II, III, and IV.
Choose the correct statement(s).

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The order of basicity is $II > I > III > IV$ .
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The magnitude of $pK_b$ difference between $I$ and $II$ is more than that between $III$ and $IV.$
0%
Resonance effect is more in $III$ than in $IV.$
0%
Steric effect makes compound $IV$ more basic than $III.$

Explanation

The correct basic strength order is

(A)

$K_0 : IV > II > I > III$ :

$\bullet$ IV is stronger base due to SIR effect

$\bullet$ III is the weakest base due to

$-M$ group of three nitro group present at Ortho and Para positions.

$\bullet$ II is stronger than I since III is tertiary and I primary aromatic amine

(B) IV is found to be

$40,000$ times more basic than III while I and II differ vary little in basic strength

(C, D) Due to SIR effect in IV both

$-NO_2$ and

$N(CH_3)_2$ will be out of plane, hence resonance effect is more in III than in IV

The correct statement regarding the basicity of arylamines is:

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arylamines are generally less basic than alkylamines because the nitrogen lone-pair electrons are delocalized by interaction with the aromatic ring $\pi$ electron system.
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arylamines are generally more basic than alkylamines because the nitrogen lone-pair electrons are not delocalized by interaction with the aromatic ring $\pi$ electron system.
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arylamines are generally more basic than alkylamines because of aryl group.
0%
arylamines are generally more basic than alkylamines, because the nitrogen atom in arylamines is sp-hybridized.

Explanation

Arylamines are generally less basic than alkylamines because the nitrogen lone-pair electrons are delocalized by interaction with the aromatic ring

$\pi$ electron system which makes electrons less available. While in alkyl amines, the

$+I$ electron releasing effect of the alkyl group makes it strongly basic. Thus, option A is correct.

Which of the following will be most stable diazonium salt

$RN_2^+X^-$ ?

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$CH_3N_2^+X^-$
0%
$C_6H_5N_2^+X^-$
0%
$CH_3CH_2N_2^+X^-$
0%
$C_6H_5CH_2N_2^+X^-$

Explanation

Among the given diazonium salts, benzene diazonium halide

$C_6H_5N_2^+X^-$ is most stable due to conjugation of the

$N \equiv N$ triple bond with benzene ring.

Which of the following compounds is most basic?

Explanation

$\bf{Hint:}$ Compound in which

$N$ lone pair is not involved in conjugation are more basic.

$\bf{Correct \ answer:}$ Option

$B$

$\bf{Explanation \ for \ correct \ option:}$

The basic strength of an a-amino group-containing compound depends on how easily

$N$ atom of

$-NH_{2}$ group share its lone pair electrons.

If these lone pairs are taking part in conjugation with a ring or an electron-withdrawing group attached with the compound then it becomes least basic because the lone pair is not available for sharing to act as a base.

In compound

$B$ , the lone pair of

$N$ atom of

$-NH_{2}$ group not involved in resonance with ring neither an electron-withdrawing group present there, hence, it is the most basic compound among all given compounds.

$\bf{Explanation \ for \ incorrect \ options:}$

In compound

$A$ , the lone pair of

$N$ atom of

$-NH_{2}$ group involved in resonance with the ring as well as an electron-withdrawing

$-NO_{2}$ group present there, hence, its basic strength reduced.

In compound

$C$ , the lone pair of

$N$ atom of

$-NH_{2}$ group involved in resonance with the ring as well as adjacent

$CO$ , hence, its basic strength is reduced. and it is the least basic of all.

In compound

$D$ , the lone pair of

$N$ atom of

$-NH_{2}$ group involved in resonance with the ring, hence, its basic strength is reduced.

The decreasing order of basicity is shown as below;

$B > D > A > C$

In the following reaction, the product (A) is:

Explanation

In the given coupling reaction, compound A is yellow dye represented by the structure given in option D.

It is an electrophilic substitution reaction.

The coupling reaction of aniline takes place at the para-position to

$NH_2$ group in the benzene nucleus gives azodye.

The major product of the following reaction is:

The correct order of the basic strength of methyl substitited amines in aqueous solution is :

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$(CH_3)_2NH> CH_3NH_2> (CH_3)_3N$
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$(CH_3)_3N> CH_3NH_2>(CH_3)_2NH$
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$(CH_3)_3N> (CH_3)_2NH>CH_3NH_2$
0%
$CH_3NH_2>(CH_3)_2NH> (CH_3)_3N$

Explanation

In aqueous solution, electron donating inductive effect, solvation effect (H-bonding) and steric hindrance all together affect basic strength of substituted amines.
Basic character:

$\underset{2^0}{(CH_3)_2NH}> \underset{1^0}{CH_3NH_2}>\underset{3^0}{(CH_3)_3N}$

In the given replacement reaction, the reaction will be most favourable if

$M$ happens to be:

0%
$K$
0%
$Rb$
0%
$Li$
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$Na$

Explanation

Reaction is faster with

$Rb$ because lattice of

$RbF$ is less than

$LiF, NaF, KF$ e energy

Which of the following reagents would not be a good choice for reducing an aryl nitro compound to an amine?

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${H}_{2}(excess)/Pt$
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$LiAl{H}_{4}$ in ether
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$Fe$ and $HCl$
0%
$Sn$ and $HCl$

Explanation

$LiAl{ H }_{ 4 }/ether$ reduces aryl nitro compounds to azo compounds

$2{C}_{6}{H}_{5}{NO}_{2} \xrightarrow [ ]{ LiAl{ H }_{ 4 } } {C}_{6}{H}_{5}N=N-{C}_{6}{H}_{5}$
nitrobenzene azobenzene

The correct decreasing order of

$pK_{b}$ is:

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$I > II > III > IV$
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$III > IV > III > I$
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$II > III > IV > I$
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$IV > II > I > III$

Explanation

Substituent with strong

$+R$ effect,

$+I$ effect and weaker

$-I$ effect increases the basicity.

Hence

$pK_{b}$ decreases.

Also, alkylamines are stronger bases than arylamines.

So, the order is option D.

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Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
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Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
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Assertion is correct but Reason is incorrect
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Both Assertion and Reason are incorrect

Which one of the following will be most basic?

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

Explanation

Aliphatic amines are more basic than aromatic amines. This is because lone pair of nitrogen takes part in resonance in case of aromatic amines and hence, are less available for donation. Thus, benzylamine

$(C_{6}H_{5}CH_{2}NH_{2})$ is most basic among the given.

Which of the following statements is not correct?

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Methylamine is more basic than $NH_{3}$ .
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Amines form hydrogen bonds.
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Ethylamine has higher boiling point than propane.
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Dimethylamine is less basic than methylamine.

Which amino acid in present only in bacteria and cyanobacteria ?

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Glycine
0%
Tyrosine
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Glutamic acid
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Diaminopimellic acid

Which of the following compound gives secondary amine on reduction?

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Alkyl nitrile
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Carbylamine
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Primary amine
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Secondary nitro compound

A mixture containing primary, secondary and tertiary amine is treated with diethyl oxalate. Choose the correct statement:

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The distillate of the mixture after treatment mainly contains $1^o$ amines.
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Tertiary amine do not react with diethyl oxalate
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This is Hinsberg method of separating $1^o ,2^o$ and $3^o$ amines.
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$3^∘$ amine is removed by filtration

Product D is

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$H_5C_2-NH_2$
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$H_5C_2-OH$
0%
0%
None

Benzylamine is a stronger base than________________

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Aniline
0%
Acetamide
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o-methylaniline
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All of these

The strongest base in aqueous solution among the following amines is:

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N-N-diethylethanamine
0%
N-ethylethanamine
0%
N-methylmethanamine
0%
ethanamine
0%
phenylmethanamine

A primary amine {B), on treatment with a solution of

$AuCl_{3}$ in conc.

$HCl$ yields a compound whose formula would be?

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$(BH)^{+}.AuCl_{4}^{-}$
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$(BH_{2})^{+}.AuCl_{4}^{-2}$
0%
$(BH)^{+}.HAuCl_{4}^{-}$
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$B.AuCl_{4}$

Explanation

$B+AuCl_3 \xrightarrow []{HCl}(BH)^+AuCl_4^-$ .

Which of the following is a

${ 1 }^o$ amine?

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$\left( { CH }_{ 3 } \right) _{ 3 }{ CNH }_{ 2 }$
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${ C }_{ 6 }{ H }_{ 5 }{ NHCH }_{ 3 }$
0%
${ CH }_{ 3 }NHCH\left( { CH }_{ 3 } \right) _{ 2 }$
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$\left( { CH }_{ 3 }{ CH }_{ 2 } \right) _{ 2 }NCH_{ 3 }$

Explanation

Hint: The amine in which $N$ is attached to only one $C$ atom is called $1^◦$ amine.

Explanation:

The structures of the given amines are-

$(CH_3)_3NH_2$

Tert-Butylamine at Best Price in India

Here, $N$ is attached to one $C$ atom. So it is a $1^◦$ amine.

$C_6H_5NHCH_3$

Write the IUPAC name of C6H5NHCH3 .

Here, $N$ is attached to two $C$ atoms. So it is a $2^◦$ amine.

$(C_2H_5)_2NCH_3$

Here, $N$ is attached to three $C$ atoms. So it is a $3^◦$ amine.

Here, $N$ is attached to two $C$ atoms. So it is a $2^◦$ amine.

Final Answer: Correct option is (A).

Which of the following is a secondary amine?

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${ H }_{ 2 }N-{ NH }_{ 2 }$
0%
${ CH }_{ 3 }-{ NH }_{ 2 }$
0%
$\left( { CH }_{ 3 } \right) _{ 2 }{ NC }_{ 6 }{ H }_{ 5 }$
0%
$C_{ 6 }{ H }_{ 5 }-NH-{ CH }_{ 3 }$

Explanation

$C_{6}H_{5}NHCH_{3}$ contains 2 alkyl group attached to the nitrogen atom
So, it is a secondary amine

Which of the following is a secondary amine?

0%
Dimethylamine
0%
Aniline
0%
Isobutylamine
0%
Secondary butyl amine

Explanation

Dimethylamine is

$(CH_{3})_{2}NH$
So clearly, 2 alkyl groups are attached to the nitrogen atom.
So, it is a secondary amine

The correct order of basic strength is:

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${ C }_{ 6 }{ H }_{ 5 }{ NH }_{ 2 }>{ CH }_{ 3 }{ NH }_{ 2 }>{ NH }_{ 3 }$
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${ CH }_{ 3 }{ NH }_{ 2 }>{ C }_{ 6 }{ H }_{ 5 }{ NH }_{ 2 }>{ NH }_{ 3 }$
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${ CH }_{ 3 }{ NH }_{ 2 }>{ NH }_{ 3 }>{ C }_{ 6 }{ H }_{ 5 }{ NH }_{ 2 }$
0%
${ NH }_{ 3 }={ CH }_{ 3 }{ NH }_{ 2 }={ C }_{ 6 }{ H }_{ 5 }{ NH }_{ 2 }$

Explanation

To determine the order of basicity of amines, we have to check for the availability of lone pair, means how easily the lone pair on the nitrogen atom is available for the attack of an electrophile.
A)

$NH_3$ :- it has lone pair which is available on it.
B)

$CH_3NH_2$ :- there is positive inductive effect of methyl group, so electron density on nitrogen is more than that of

$NH_3$ .
C)

$C_6H_5NH_2$ :- the lone pair of nitrogen is taking part in resonance with benzene ring, so it is least available to the attack of an electrophile compared to others.
So, the order is

${ CH }_{ 3 }{ NH }_{ 2 }>{ NH }_{ 3 }>{ C }_{ 6 }{ H }_{ 5 }{ NH }_{ 2 }$ .

Term tertiary, secondary and primary amine represents the:

0%
number of amino groups
0%
nature of C atom
0%
degree of substitution on nitrogen
0%
degree of uunsaturation

Which of the following is secondary amine?

0%
${ CH_3-NH }_{ 2 }$
0%
$CH_3-N-(C_2H_5)_2$
0%
$CH_3-NHCH_3$
0%
$CH_3-N(CH_3)_2$

Explanation

In secondary amine, 2 alkyl groups and one hydrogen atom are directly attached to N atom.

It is represented by general formula

$R_2NH$ .

$CH_3-NHCH_3$ represents

$R-NH-R$ .

Hence, option C is correct.

Amines are the derivatives of:

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acids
0%
ammonia
0%
alkanes
0%
esters

$o-Br-{ C }_{ 6 }{ H }_{ 4}-{ NH }_{ 2 }$ is :

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aliphatic ${ 1 }^{ 0 }$ - amine
0%
aromatic ${ 1 }^{ 0 }$ - amine
0%
aliphatic ${ 2 }^{ 0 }$ - amine
0%
aromatic ${ 2 }^{ 0 }$ - amine

Explanation

Hint: The amine in which $N$ is attached to only one $C$ atom is called $1^◦$ amine. The amine which has a phenyl group is called an aromatic amine.

Explanation:

It has a phenyl group so, it is an aromatic amine and here $N$ is attached to only one $C$ atom so, it is $1^◦$ amine.

So, it is aromatic $1^◦$ -amine.

Final Answer: Correct option (B).

Which of the following amines form unstable diazonium salt?

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${ C }_{ 6 }{ H }_{ 5 }{ NH }_{ 2 }$
0%
$\left( { C }_{ 2 }{ H }_{ 5 } \right) _{ 2 }NH$
0%
$\left( { C }_{ 2 }{ H }_{ 5 } \right) _{ 3 }N$
0%
${ CH }_{ 3 }{ NHC }_{ 2 }{ H }_{ 5 }$

Explanation

Only primary aromatic amines form stable arene diazonium salts.
The reaction is as follows:

$C_{6}H_{5}NH_{2} + NaNO_{2} + HCl (0-5^{\circ}C) \rightarrow PhN_{2}^{+}Cl^{-}$

Remain forms the unstable diazonium salt.

So, the correct options are

$B$ ,

$C$ ,

$D$

The structural formula of N-methylaminomethane is:

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$\left( { CH }_{ 3 } \right) _{ 2 }{ CHNH }_{ 2 }$
0%
$\left( { CH }_{ 3 } \right) _{ 3 }N$
0%
$\left( { CH }_{ 3 } \right) _{ 2 }NH$
0%
${ CH }_{ 3 }{ NH }_{ 2 }$

Explanation

N-methylaminomethane contains two methyl groups directly attached to nitrogen atom.

Hence, its structural formula is

$\left( { CH }_{ 3 } \right) _{ 2 }NH$ .

Option C is correct.

Which of the following is more appropriate reason for aniline being less basic than aliphatic amines?

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Inductive effect
0%
Steric hindrance
0%
Positive mesomeric effect
0%
Negative mesomeric effect

Explanation

$NH_{2}$ gives e- away from the group.
So e- density of N decrease and hence less available for Protonation.
So less Basicity.
In aliphatic amines, no mesomerism (resonance) takes place. Thus it has more electrons available for donation thus shows more basic property.

$-{ NH }_{ 2 }$ group is only:

0%
ortho director
0%
only meta director
0%
only para director
0%
both ortho, para director

Explanation

$-NH_{2}$ group is very strongly activating.
It activated both the ortho as well as para positions.

Note that all the positions - ortho, meta and para are activated.
But ortho and para are activated too strongly (due to +M effect) while the meta position is activated weakly (due to +I effect)

So, D is correct answer.

The structure given below represents:

0%
quarternary ammonium Salt.
0%
primary amine.
0%
secondary amine.
0%
tertiary amine.

Explanation

only 1 hydrogen atom of ammonia

$(NH_3$ ) has been replaced by the alkyl group.

So, clearly it is a primary amine.

Note that it does not matter what kind of alkyl group is present.

To classify as primary/secondary/tertiary, we only count the number of replaced hydrogens.

If 1 has been replaced -> Primary
2 replaced -> secondary
3 replaced -> tertiary

Option B is correct.

Aniline when treated with

$NaNO_{ 2 }$ and

$HCl$ at

$0- { 5 }^{ 0 }C$ gives ____________ as a product.

0%
chloroaniline
0%
benzene diazonium chloride
0%
chlorobenzene
0%
dichlorobenzene

Explanation

Aniline reacts with sodium nitrite and HCl in cold condition to form benzene diazonium chloride.

$\underset {aniline} {{ C }_{ 6 }{ H }_{ 5 }{ NH }_{ 2 }}\xrightarrow [ { NaNO }_{ 2 } ]{ HCl } \underset {benzene diazonium chloride} {[C_6H_5-N_2^+Cl^-]}$

The structural formula of methylamino methane is:

0%
$(CH_{3})_{2}CHNH_{2}$
0%
$(CH_{3})_{3}N$
0%
$(CH_{3})_{2}NH$
0%
$CH_{3}NH_{2}$

Explanation

Methylamino methane has 2 methane groups attached to

$N$ directly.

Hence option

$C$ is correct

${ C }_{ 3 }{ H }_{ 9 }N$ cannot represent:

0%
${ 1 }^{ 0 }$ amine
0%
${ 2 }^{ 0 }$ amine
0%
${ 3 }^{ 0 }$ amine
0%
quaternary salt

The correct order of basic strength of given aniline is:

0%
o - nitroaniline > p - nitroaniline > m - nitroaniline
0%
m - nitroaniline > p - nitroaniline > o - nitroaniline
0%
m - nitroaniline > o - nitroaniline > p - nitroaniline
0%
o - nitroaniline > m - nitroaniline > p - nitroaniline

When aniline in hydrochloric acid is treated with ice cold solution of sodium nitrite, it gives:

0%
benzene diazonium chloride
0%
chlorobenzene
0%
nitro Aniline
0%
chloro Aniline

Explanation

Aniline reacts with sodium nitrite and HCl in cold condition to form benzene diazonium chloride.

$\underset {\text {aniline}} {{ C }_{ 6 }{ H }_{ 5 }{ NH }_{ 2 }}\xrightarrow [ { NaNO }_{ 2 } ]{ HCl } \underset {\text {benzene diazonium chloride}} {[C_6H_5-N_2^+Cl^-]}$

Butanamine-1 structure is shown as:

Explanation

$CH_3CH_2CH_2CH_2NH_2$
It is a four carbon compound with an amino group on a terminal carbon atom. Hence, its name is 1 - butanamine

The first organic compound synthesized in the laboratory from an inorganic compound is:

0%
$NH_{4}NCO$
0%
$NH_{2}-CO-NH_{2}$
0%
$CH_{3}COOH$
0%
$CH_{4}$

Explanation

The first organic compound synthesized in the laboratory from an inorganic compound is urea which is

$NH_2CONH_2$ .

Which one of the following is a primary amine?

0%
0%
$CH_3-CH_2-CH_2-NH-CH_3$
0%
$CH_3-H_2C-\!\!\!\!\!\!\!\!\!\!\!\!\!\underset{H}{\underset{|}{\overset{\quad \quad \,\,CH_2CH_3}{\overset{|}{C}}}}\!\!\!\!\!\!\!\!\!\!\!\!\!-NH_2$
0%
$CH_3H_2C-\!\!\!\!\!\!\!\!\!\!\!\!\!\underset{H}{\underset{|}{\overset{\quad \quad \,\,CH_2CH_3}{\overset{|}{C}}}}\!\!\!\!\!\!\!\!\!\!\!\!\!-NH-CH_3$

Explanation

In a 1 degree amine, only 1 hydrogen atom of

$NH_3$ is replaced by alkyl group.

Primary amine have general formula

$R-NH_2$

So, A is correct because 1 of the hydrogens has been replaced by the aryl group.

Note that the kind of alkyl/aryl group does not matter in deciding the type of amine,what matters is the number of hydrogen replaced in

$NH_3$ .

Option A and C are correct.

Correct gradation of basic character :

0%
$NH_{3} > CH_{3}NH_{2} >NF_{3}$
0%
$CH_{3}NH_{2} >NH_{3} >NF_{3}$
0%
$NF_{3} >CH_{3}NH_{2} >NH_{3}$
0%
$CH_{3}NH_{2} >NF_{3} >NH_{3}$

Explanation

$\ e^{-}$ donating group (EDa) increase the basic character

An

$\ e^{-}$ withdrawing group (EWa) decrease the basic character.

Among the following isomeric

$C_{4} H_{11}N$ amines, one having the lowest boiling point is:

Assertion : Benzylamine is more basic than aniline.
Reason :

$NH_{2}$ group is electron releasing group.

0%
Assertion and Reason are true and Reason is the correct explanation of Assertion
0%
Assertion and Reason are true and Reason is not the correct explanation of Assertion
0%
Assertion is true Reason is false
0%
Assertion is false Reason is true

Explanation

Aliphatic amines are stronger bases as compared to aromatic amines.

This is because, in aromatic amines, the lone pair on nitrogen is involved in delocalization with aromatic nucleus.

Benzyl amine is an aliphatic amine whereas aniline is aromatic amine.

Hence, benzyl amine

$({ C }_{ 6 }{ H }_{ 5 }{ CH }_{ 2 }{ NH }_{ 2 })$ is more basic than aniline.

Option B is correct.

In the diazotization of aryl amines with sodium nitrite and hydrochloric acid, an excess of hydrochloric acid is used primarily to :

0%
supress the concentration of free aniline available for coupling
0%
supress hydrolysis of phenol
0%
ensure a stoichiometric amount of nitrous acid
0%
neutralise the base liberated

Explanation

If excess

$HCl$ is not used, the arene diazonium salt will react with free aniline to form azo compound.

Excess

$HCl$ forms hydrochloride salt of aniline. This prevents azo coupling.

Option A is correct.

The correct decreasing order of strength of the bases

$OH^{-}$ ,

${NH_{2}}^{-}$ ,

$HC\equiv C^{-}$ and

${CH_{3}-CH_{2}}^{-}$ is:

0%
${CH_{3}-CH_{2}}^{-} >{NH_{2}}^{-} > H-C \equiv C^{-} >OH^{-}$
0%
$H-C\equiv C^{-} >{CH_{3}-CH_{2}}^{-} >{NH_{2}}^{-} >OH^{-}$
0%
$OH^{-} > {NH_{2}}^{-} >H-C\equiv C^{-} >{CH_{3}-CH_{2}}^{-}$
0%
${NH_{2}}^{-} > H-C\equiv C^{-} >OH^{-} >{CH_{3}-CH_{2}}^{-}$

Explanation

Making conjugate acids of all:

$i)CH_{3}-CH_{3}\rightarrow CH_{3}-\bar{CH_{2}}+H^{+}$
Acid (Conjugate Base)
strong weak
ii)

$NH_{3}\rightarrow NH_{2}^{-}+H^{+}$
Acid (CB)
strong weak

$iii)H-C=C-H\rightarrow H-C=C+H^{+}$
(Acid) (CB)
strong weak Base
iv)

$H_{2}O\rightarrow H^{+} + OH^{-}$
Acid (CB)

$\therefore$ order of Basicity :

$CH_3-\bar{CH}_{2}>\bar{NH}_{2}>H-C\equiv C^{-}>OH^{-}$

Which of the following would not react with benzene sulphonyl chloride in aq.

$NaOH$ ?

0%
Aniline
0%
Methylamine
0%
$N,N$ -dimethyl aniline
0%
$N-$ methyl aniline

Explanation

N,N-dimethyl aniline does not contain

$H$ attached to nitrogen.
Hence, it would not react with benzene sulphonyl chloride in aqueous

$NaOH$ .

CBSE Questions for Class 12 Engineering Chemistry Amines Quiz 1 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

CBSE Questions for Class 12 Engineering Chemistry Amines Quiz 1 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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