MCQ Questions for Class 12 Medical Chemistry Organic Compounds Containing Nitrogen Quiz 2

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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 of the following effects of

$-NO_{2}$ group operates on

$-NH_{2}$ group in this molecule?

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Only $+M$ effect
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Only $-M$ effect
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Both $-l$ and $-M$ effect
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Only $-l$ effect

Explanation

Due to steric hindrance, the given species only show

$-I$ effect of

$-NO_{2}$ on

$-NH_{2}$ .

Which one of the following will be most basic?

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Aniline
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P-methoxyaniline
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P-nitroaniline
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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
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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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$H_5C_2-OH$
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0%
None

Benzylamine is a stronger base than________________

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Aniline
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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
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N-ethylethanamine
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N-methylmethanamine
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ethanamine
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phenylmethanamine

Which of the following is a

${ 1 }^o$ amine?

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

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

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ortho director
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only meta director
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only para director
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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:

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

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

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:

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

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

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

Amines are the derivatives of:

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acids
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ammonia
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alkanes
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esters

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

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aliphatic ${ 1 }^{ 0 }$ - amine
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aromatic ${ 1 }^{ 0 }$ - amine
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aliphatic ${ 2 }^{ 0 }$ - amine
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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).

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

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

Explanation

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

$NH_2CONH_2$ .

Which of the following amines form unstable diazonium salt?

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

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

$B$ ,

$C$ ,

$D$

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

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Inductive effect
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Steric hindrance
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Positive mesomeric effect
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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.

Correct gradation of basic character :

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

Explanation

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

An

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

Aniline when treated with

$NaNO_{ 2 }$ and

$HCl$ at

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

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chloroaniline
0%
benzene diazonium chloride
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chlorobenzene
0%
dichlorobenzene

Explanation

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

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

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

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supress the concentration of free aniline available for coupling
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supress hydrolysis of phenol
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ensure a stoichiometric amount of nitrous acid
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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 structural formula of methylamino methane is:

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

Explanation

Methylamino methane has 2 methane groups attached to

$N$ directly.

Hence option

$C$ is correct

What is the product

$A$ ?

The correct order of basic strength of given aniline is:

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o - nitroaniline > p - nitroaniline > m - nitroaniline
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m - nitroaniline > p - nitroaniline > o - nitroaniline
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m - nitroaniline > o - nitroaniline > p - nitroaniline
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o - nitroaniline > m - nitroaniline > p - nitroaniline

The correct decreasing order of strength of the bases

$OH^{-}$ ,

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

$HC\equiv C^{-}$ and

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

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

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 :

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

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

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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.

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

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

$NaOH$ .

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

An organic compound

$(C_{3}H_{9}N)$ (A), when treated with nitrous acid, gave an alcohol and

$N_{2}$ gas was evolved.
The compound (A) on warming with

$CHCl_{3}$ and alcoholic caustic potash gave (C), which on reduction gave isopropylmethylamine.
Predict the structure of (A).

Which one of the following is a primary amine?

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

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.

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.

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

Which of the following compounds is most basic?

Explanation

Basicily

$\propto + I, +m$ group (election denating) also line pairs of nitrogen should not be delocalised.

Which one of the following on reduction with
lithium aluminium hydride yield a secondary
amine:

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Methyl Cyanide
0%
Nitroethane
0%
Methylisocyanide
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Acetamide

Which of the following statements about primary amines is false?

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Alkyl amines are stronger bases than aryl amines
0%
Alkyl amines react with nitrous acid to produce alcohols
0%
Aryl amines react with nitrous acid to produce phenols
0%
Alkyl amines are stronger bases than ammonia

Explanation

Primary amines are those amines whose one hydrogen is substituted by some other groups like

$C{ H }_{ 3 }$ ,

$C{H}_{3}C{H}_{2}$ etc or those amines which have one

$C-N$ bond present in them. Example:

${ H }_{ 3 }C-N{ H }_{ 2 }\quad ,\quad { H }_{ 3 }C-C{ H }_{ 2 }-N{ H }_{ 2 }$

Aryl amines (if it is primary) on reaction with nitrous acid gives diazonium salt.

Which of the following has maximum

$pK_{b}$ value?

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

Explanation

The

$pK_{b}$ values of ethanamine, N-methylaniline, N-ethylethaneamine and aniline are 3.29, 9.30, 3.00 and 9.38 respectively.
Thus among the given amines, aniline is the weakest base and has maximum

$pK_{b}$ value.
Aliphatic amines ethanamine and N-methylaniline are more basic than aniline because in aniline, the lone pair on nitrogen is in resonance with benzene.
N-methylaniline is more basic than aniline because in N-methylaniline, the methyl group (an electron releasing group) increases the electron density on nitrogen.

Which of the following is most basic?

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Diphenylamine
0%
Triphenylamine
0%
p-Nitroaniline
0%
Benzylamine

Explanation

Benzylamine

$C_{6}H_{5}CH_{2}NH_{2}$ is more basic because benzyl group is electron donating group due to +I effect. So it is able to increase electron density of N of -NH

$_{2}$ group.

Thus due to higher electron density rate of donation of a free pair of electron is increased I.e basic character is higher While phenyl and nitro group are electron withdrawing group so they are able to decrease the electron density of N of

$-NH_{2}$ group. Hence they are less basic

Aromatic amines are insoluble in water because:

0%
Due to the larger hydrocarbon part which tends to retard the formation of H-bonds.
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Due to the larger hydrocarbon part which tends to retard the formation of H-Nbonds.
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Due to the larger hydrocarbon part which tends to retard the formation of H-Cbonds.
0%
none of these

Consider the Hofmann ammonolysis reaction:

$R = {C}_{2}{H}_{5}$ , the correct order of basic character of the above amines in aqueous medium is :

0%
$2 > 3 > 1 > N{H}_{3}$
0%
$2 > 1 > 3 > N{H}_{3}$
0%
$3 > 2 > 1 > N{H}_{3}$
0%
$N{H}_{3} > 1 > 2 > 3$

Explanation

$R = {C}_{2}{H}_{5}$ , the correct order of basic character of the above amines in aqueous medium is

$2 > 3 > 1 > N{H}_{3}$ . The factors that determine basicity are the +I effect of alkyl groups and the solvation /hydration in aqueous medium.

Give the decreasing order of basic strength of the following:
a.

$N{H}_{3}$
b.

$N{H}_{2}OH$
c.

$N{H}_{2}-N{H}_{2}$

0%
$c > b > a$
0%
$a > b > c$
0%
$c > a > b$
0%
$a > c > b$

Explanation

The decreasing order of basic strength is

$c > a > b$ or

$N{H}_{2}-N{H}_{2} > N{H}_{3} > N{H}_{2}OH$ .

$N{H}_{2}-N{H}_{2}$ is most basic as it has two lone pairs of electrons.
The basic strength of ammonia is lower than the basic strength of hydrazine as ammonia contains only one lone pair.
Hydroxyl amine is least basic due to

$-I$ effect of

$-OH$ group which decreases the ease with which lone pair of an electron can be donated.

Consider the Hofmann ammonolysis reaction:

$R = {Me}_{3}C-$ (

$t-butyl$ ), the correct order of basic character of the above amines in aqueous medium is :

0%
$N{H}_{3} > 1 > 2 > 3$
0%
$3 > 2 > 1 > N{H}_{3}$
0%
$1 > 2 > N{H}_{3} > 3$
0%
$1 > N{H}_{3} > 2 > 3$

Explanation

$R = {Me}_{3}C-$ (t-butyl), the correct order of basic character of the above amines in aqueous medium is

$N{H}_{3} > 1 > 2 > 3$
The factors that determine basicity are the +I effect of alkyl groups and the solvation /hydration in aqueous medium.

Indicate the following compound as

$1^o, 2^o$ and

$3^o$ .

$Et_3N$

0%
$1^o$
0%
$2^o$
0%
$3^o$
0%
Non of the above

Explanation

$C_2H_5 -\underset{\underset {\displaystyle {C_2H_5}}|}{N}-C_2H_5$

Triethylamine, is tertiary amines, as the three hydrogen atoms are replaced by ethyl substituents.

Option C is correct.

Consider the Hofmann ammonolysis reaction:

$R = {Me}_{2}CH-$ (isopropyl), the correct order of basic character of the above amines in aqueous medium is :

0%
$3 > 2 > 1 > N{H}_{3}$
0%
$N{H}_{3} > 1 > 2 > 3$
0%
$1 > 2 > N{H}_{3} > 3$
0%
$1 > N{H}_{3} > 2 > 3$

Explanation

$R = {Me}_{2}CH-$ (isopropyl), the correct order of basic character of the above amines in aqueous medium is

$1 > N{H}_{3} > 2 > 3$ .
The factors that determine basicity are the +I effect of alkyl groups and the salvation /hydration in aqueous medium.

Mendius reaction involves the:

0%
Reduction of aldehydes to give alcohols
0%
Reduction of nitriles with sodium and ethanol
0%
Oxidation of nitriles
0%
Hydrolysis of cyanides

Explanation

$\displaystyle R-C\equiv N\xrightarrow [ Mendius\quad reduction ]{ Na+EtOH } { RCH }_{ 2 }{ NH }_{ 2 }$

Which of the following statements are correct?

0%
In gas phase, the basic character of amine is 3 > 2 > 1. Due to the +1effect of (R-), the availability of $\displaystyle LP$ $\displaystyle \overset { - }{ e } 's$ on N increases.
0%
In aqueous medium, the basic character of amines is $\displaystyle { Me }_{ 2 }NH>{ Me }_{ 3 }N>{ MeNH }_{ 2 }>{ NH }_{ 3 }$ .
0%
In aqueous medium, the addition of protons increases crowding and thus strains setup, which being the highest in 3 amine decreases its basic character.
0%
In aqueous medium, the ammonium ions in solution are stabilised not only by alkyl groups but also by H-bond donation to the solvent.

Explanation

$(a,c,d)$
b. The order is:

$\displaystyle { Me }_{ 2 }NH>{ MeNH }_{ 2 }>{ { Me }_{ 3 }N>NH }_{ 3 }({ 2 }^{ o }>{ 1 }^{ o }>{ 3 }^{ o }>{ NH }_{ 3 })$ . Statements

$(a), (c),$ and

$(d)$ are correct.

(A) is subjected to reduction with

$\displaystyle Zn-{ Hg }/{ HCl }$ and the product formed is N-methylmethanamine. (A) can be:

0%
Ethane nitrile
0%
Nitroethane
0%
Carbylaminoethane
0%
Carbylaminomethane

Explanation

Reduction with

$Zn-Hg/HCl$ is Clemmensen reduction

which converts

$R-C\equiv N\rightarrow RCH_2NH_2$ and

$R-\overset{\oplus}{N}\equiv \overset{\ominus}{C}\rightarrow RNHCH_3$

Carbylamino methane is

$\left ( {Me-\overset{\oplus}{N}\equiv \overset{\ominus}{C}} \right )$

$\therefore Me-\overset{\oplus}{N}\equiv\overset{\ominus}{C}\xrightarrow {Zn-Hg/HCl}\underset{N-Methylmethanamine}{Me-NH-CH_3}$

Option D is correct.

Which of the following cannot be prepared by Gabriel phthalimide synthesis?

0%
Aliphatic primary amines
0%
Aromatic primary amines
0%
Cyclic primary amines
0%
None of the above

CBSE Questions for Class 12 Medical Chemistry Organic Compounds Containing Nitrogen Quiz 2 - MCQExams.com

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Practice Class 12 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 12 Medical Chemistry Organic Compounds Containing Nitrogen Quiz 2 - MCQExams.com

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Practice Class 12 Medical Chemistry Quiz Questions and Answers

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