MCQ Questions for Class 12 Engineering Chemistry Amines Quiz 6

Which of the N atoms is least basic in the given compound ?

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

Explanation

Electron pair of

$NH_{2}$ attached to benzene undergoes resonance i.e lone pair on Nitrogen is delocalised and hence less basic than other amine groups.

Which is most basic?

An organic compound (A)

$C_9H_{13}N$ dissolves in dil. HCI and releases

$N_2$ with

$HNO_2$ giving an optically active alcohol. Alcohol on oxidation gives dicarboxylic acid, which on heating form anhydride. The organic compound 'A' is :-

The correct order of basicity of the above compounds is:

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

Explanation

(Refer to Image)

The order of basicity is follows

$I >III >II >IV$

$\rightarrow$ in

$IV$ electron withdrawing is attached to amine. So, it is least basic.

1220003_1115396_ans_93814c82f5664738a0ec25321023c497.png

Basic nature order of the given compounds is
(a)

$CH_3 - CH = \ddot{N}H$
(b)

$CH_2= CH - CH = CH -NH_2$
(c)

$CH_3 -NH - CH_3$

0%
c>a>b
0%
a>b>c
0%
a>c>b
0%
c=a=b

Explanation

$C H_{2}=C H-C H=C H-N H_{2}(b)$

In this compound lone pair on

$-N H_{2}$ group

is involued in resonance. so givening lone

Pair is difficult. So this compound

has less basicity. than other two.

$\mathrm{CH}_{3}\mathrm{CH}=\ddot{\mathrm{N}}\mathrm{H}(\mathrm{O})$

$\begin{array}{l} \text { Nitrogen in }^{\prime}=\ddot{N} H^{\prime} \text { group is more electro. } \\ \text { negitiven than }-N H-\text { is } C H_{3}-N H-C H_{3}(C) \\ \text { due to double bond. } \\ \text { More electronegitive Nitrogen in ' }=\text { NH 'grap } \\ \text { cannot tolarate positive charge of it (when } \\ \text { lone pair is given it get positive charge) } \end{array}$

$\begin{array}{l} \text { Compared to Nitrogen in }-\mathrm{NH} \text { - group } \\ \text { So } \mathrm{CH}_{3}-\mathrm{NH}-\mathrm{Ct}_{3} \text { is more basic than } \\ \mathrm{CH}_{3}-\mathrm{CH}=\mathrm{NH} \text { (a) } \\ \text { Hence Basisity order is } c>a>b \\ \text { so option } A \text { is correct. } \end{array}$

Arrange these compounds in decreasing order of basic strength?

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

Explanation

$iv>i>iii>ii$

Electron - donating substituents increase the electron density of N, thereby increasing the Lewis basicity. Amongst

$NO_2$ and

$CN,NO_2$ is slightly resonance stabilization.

Hence, the answer is

$iv>i>iii>ii.$

Which of the following is the strongest base?

Explanation

Aniline is the strongest base here. This is because,

$NH_2$ is stabilized by conjugation and resonance stabilised . More stability implies stronger the base.

The decreasing order of the basic character at

$a, b, c, d$ in the compound:

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

Explanation

$d>c>b>a.$
Due to ease of delocalisation of electron pair is least in

$(d)$ Nitrogen. For

$(a), (b), (c)$ Nitrogen atoms , basic character order is governed by electronic effects of rings and

$-CH_3$ group

Arrange the following in order of decreasing basic strength.

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

Explanation

$IV>I>II>III$

The major product of the reaction is:

Explanation

(Refer to Image)

In presence of

$NaNO_2$ , diazotisation reaction takes place and

$NH_2$ group leaves as

$-N_2$ . Then methyl group shift takes place, which makes the carboction more positive than former. Then oxygen stabilise the carbocation by

$+M$ effect and ketone formation takes place.

1163455_1117541_ans_62fce0ff7a5247f8b2e3868fbe152647.JPG

Which of the following is weak base than aniline ?

Explanation

An electron withdrawing groups decrease the availability of a lone pair of electrons on

$N-$ atom of aniline thereby decreasing its basicity.

Hence it is less basic than aniline.

1184080_1129241_ans_0e9ab19047b148cdb8190e5aedba700b.png

Sanger's reagent is used for the identification of:

0%
N-thermal of peptide chain
0%
C-terminal of a peptide chain
0%
side chain of amino acids
0%
number of amino acids in peptide chain

Explanation

1-Fluoro-2,4-dinitrobenzene. Undergoes nucleophilic aromatic substitution (NAS) with the N-terminal amino group of a peptide or protein, resulting in a chemical tag on the terminal amino acid residue. After hydrolysis into individual amino acids, the peptide's or protein's terminal amino acid can be identified by the specific wavelength of light absorbed (i.e., colorimetric analysis).

So option

$A$ is correct

What is strong base?

The relative basics strenghts of

$NH_{3},CH_{3}NH_{2}$ and

$NF_{3}$ are in the order:

0%
$CH_{3}NH_{2} > NH_{3}>NF_3$
0%
$NF_{3} > CH_{3}NH_{2} > NH_{3}$
0%
$NH_{3} > CH_{3}NH_{2} > NF_{3}$
0%
$NF_{3} > CH_{3}NH_{2} = NH_{3}$

Explanation

$CH_3-$ group is electron donating group.

$F$ is electron withdrawing group.

$CH_3NH_2$ is more basic.

$NH_3$ is more basic than

$NF_3$ because

$H$ is not a withdrawing group.

$NF_3$ is less basic.

$\therefore$ Basic nature order=

$CH_3NH_2 > NH_3 >NF_3$

Acetic anhydride with ammonia gives the product:

0%
$CH_{3}CONH_{2}$
0%
$CH_{3}CONHCH_{3}$
0%
$CH_{3}CN$
0%
$CH_{3}COONH_{4}$

What will be the final product of the given reaction?

Explanation

Diazonium chloride on treating with

$HF/{ BF }_{ 3 }$ yields benzene diazonium fluoroborate. Benzene diazonium fluoroborate when heated undergoes decomposition reaction to give fluorobenzene.

Option B is correct.

1185398_1128733_ans_cf7d7e0df2fd46f8b94b367eb094d61c.png

which compound is Prepare propan -I- amine from?

0%
Butaniteuie
0%
I-Nitropropane
0%
Propanamide
0%
Butanamide

Adrenaline and ephedrine contain:

0%
${1}^{o}$ amino group
0%
${2}^{o}$ amino group
0%
${3}^{o}$ amino group
0%
quaternary amino group

Explanation

Thus Adrenaline and ephedrine contain

$2^0$ amino group.
1380602_1133505_ans_b5091b095b4342fca190460eb0262236.png

Among the following which is not correct?

0%
P methyl benzyl carbonyl is more stable than ethyl benzyl carbonyl
0%
Aniline is more basic than pyridine
0%
Cyclopentadienyl anion is more stable than cyclopentadienyl Carbo cation
0%
K of Ethene 1 ol is more than that of propene 2 en 1 ol

${C}_{6}{H}_{5}-{NO}_{2}\xrightarrow [ ]{ Fe/HCl } A\xrightarrow [ H{ NO }_{ 3 } ]{ 273K } B\xrightarrow [ ]{ { C }_{ 6 }{ H }_{ 5 }OH } C$ ,

$C$ is

In an amino acid, the carboxyl group ionises at

$p{K}_{{a}_{1}}=2.34$ and ammonium ion at

$p{K}_{{a}_{2}}=9.60$ . The isoelectric point of the amino acid is at

$pH$ :

0%
$5.97$
0%
$2.34$
0%
$9.60$
0%
$6.97$

Explanation

The isoelectronic point or isoionic point is the pH at which the amino acid does not migrate in an electric field.

This means it is the pH at which the amino acid is neutral, i.e. the zwitter ion form is dominant.

the isoelectronic point is halfway between the two given values, i.e.,

$pH=\cfrac{pK_{a_{1}}+pK_{a_{2}}}{2}=\cfrac{2.34+9.60}{2}=5.97$

Acid anhydride on reaction with primary amine give:

0%
Amide
0%
Imide
0%
Secondary amine
0%
Imine

Explanation

Acid anhydrides react with primary and secondary amines to form amides.

See the given figure.

Option

$A$ is the answer.

1174492_1136718_ans_f79516d9c24c4415ab9fb568a91806ba.png

Nitrobenzene when reduced with tin and hydrochloric acid, i.e., in acidic medium, the product formes is:

0%
Aniline
0%
Benzene
0%
Both a and b
0%
None of these

The structural feature which distinguishes proline from other natural

$\alpha$ -amino acids is that:

0%
it is optically inactive
0%
it contains aromatic group
0%
it contains two amino groups
0%
it is secondary amine

Isopropylamine

$\xrightarrow{KMnO_4} X \xrightarrow{H_3O^+}Y$ .

In the above sequence,

$X$ and

$Y$ are _________, ____________ respectively

0%
Aceraldimine, Ethanal
0%
Ethanal, Ketimine
0%
Ketimine, Acetone
0%
Acetone, Propan-2-ol

Explanation

Isopropylamine on treatment with KMnO

$_4$ produce ketimine first which gets hydrolysed to give acetone as the final product.

Hence, the correct option is

$\text{C}$

1475651_1143766_ans_d68b6cf1ea614908b54fc26e826bb907.png

Benzyl amine is............. basis than aniline while ethyl amine is..............basis than diethyl amine

0%
More, less
0%
Less, more
0%
Both
0%
None

Explanation

Benzyl amine is

$more$ basic than anniline

while ethyl amine is

$less$ basic than diethyl amine

The order of the basicity of the following compound is
(1) is refer to image
(2)

${ CH }_{ 3 }CH_{ 2 }{ NH }_{ 2 }$
(3)

${ ({ CH }_{ 3 }) }_{ 2 }NH$
(4)

${ CH }_{ 3 }CO{ NH }_{ 2 }$

0%
2>1>3>4
0%
1>3>2>4
0%
3>1>2>4
0%
1>2>3>4

Explanation

Amines are basic in nature as they have lone pair of electrons on Nitrogen the stability of conjugate acids formed is

dependent on the extent of

$H-$ bonding and greater the number of

$H-$ atom on

$N$ more stable is the conjugate acidHence, option

$A$ is correct answer.

1166245_1158764_ans_efd48ca9722948efa07719ba28a5a86c.png

The reaction :[

${ C }_{ 2 }{ H }_{ 5 }{ B }r+{ N }{ H }_{ 3 }$ ] is in fact an example of

0%
Ammonolysis only
0%
Nucleophilic substitution only
0%
Ammonolysis as well as nucleophilic substitution
0%
None

The increasing order of basicity of the following compound is:

Explanation

Due to resonance electron density increases at

$sp^2$ hybridisation

$NH$ electron density increase basically increase so this is most basic these are called as anidine here an

$sp^3$ hybridisation

$NH_2$ group is in resonance with

$sp^2$ hybridised

$NH$

1054205_1150640_ans_bfbce320fb4a4cb1a65e76b9df8c943c.PNG

Gabriel phthalimide synthesisis used in the preparation of:

0%
1 $^{o}$ amine
0%
2 $^{o}$ amine
0%
3 $^{o}$ amine
0%
4 $^{o}$ amine

Explanation

The chemical reaction used for transforming primary alkyl halides into primary amines is called Gabriel phthalimide synthesis. It uses potassium phthalimide. Therefore, Gabriel phthalimide synthesis is used in the preparation of

$1^o$ amide.

The increasing order of basicity of the following compounds is:

0%
$(d)<(b)<(a)<(c)$
0%
$(a)<(b)<(c)<(d)$
0%
$(b)<(a)<(c)<(d)$
0%
$(b)<(a)<(d)<(c)$

Amongst the following the strongest base is:

0%
benzylamine
0%
aniline
0%
m-nitroaniline
0%
p-nitroaniline

Explanation

Correct option is

$A$

Benzyl amine is strongest base than aniline because

The lone pair of electron on

$N$ -atom in aniline is delocalised & not involved in resonance

Which of the following will be least basic?

Identify the statement about the basic nature of amines.

0%
Alkylamines are weaker bases than ammonia
0%
Arylamines are stronger bases than alkyl amines
0%
Secondary aliphatic amines are stronger bases than
primary aliphatic amines
0%
Tertiary aliphatic amines are weaker bases than
arylamines

Amine that can't be prepared by Gabriel phthalimide synthesis is:

0%
Aniline
0%
Benzylamine
0%
Methyl amine
0%
Iso-butyl amine

Which product can not be formed in given reaction?

Which is most basic among the following?

0%
$C{H}_{3}N{H}_{2}$
0%
$C{H}_{3}C{H}_{2}N{H}_{2}$
0%
$N{H}_{3}$
0%
$(C{H}_{3})_{2}CHN{H}_{2}$

When methyl iodide treated with ammonia, the product obtained is:

0%
methyl amine
0%
dimethyl amine
0%
trimethyl amine
0%
all of these

Which of the following aromatic rings having greater electronics density than the given compound?

Explanation

The compound given in the problem contains both

$+I$ and

$+H$ group. Aniline contains

$-NH_2$ as a functional group which is a

$-I$ and

$+R$ group. Since

$+R$ effect is predominant when compared to

$+I$ and

$+H$ effect, it will have the greatest electron density

When

$CH_{3}CH_{2}CHCl_{2}$ is treated with

$NaNH_{2}$ the product formed is:

0%
$CH_{3}-CH=CH_{2}$
0%
$CH_{3}-C\equiv CH$
0%
0%

Explanation

Solution:- (B)

$C{H}_{3}-C \equiv CH$

When

$C{H}_{3}-C{H}_{2}-CH-{Cl}_{2}$ is treated with

$NaN{H}_{2}$ , the product formed is propyne.

$C{H}_{3}-C{H}_{2}-CH-{Cl}_{2} \xrightarrow[\Delta]{NaN{H}_{2}} \underset{\left( \text{Propyne} \right)}{C{H}_{3}-C \equiv CH}$

The product 'Y' in the following reaction sequence is:

the increasing order of diazotisation of the following compound is?

0%
$(a)<(d)<(b)<(c)$
0%
$(d)<(c)<(b)<(a)$
0%
$(a)<(b)<(c)<(d)$
0%
$(a)<(d)<(c)<(b)$

Explanation

SOLUTION:

Electron withdrawing substituents decrease electron density on the benzene ring. Electron donating substituents increase electron density on the benzene ring. They increase the stability of diazonium salts.

Aromatic diazonium salts are more stable than aliphatic diazonium salts. The stability of aryl diazonium salts is due to resonance. Electron donating substituents increase electron density on the benzene ring. They increase the stability of diazonium salts.

Electron withdrawing substituents decrease electron density on the benzene ring. They decrease the stability of diazonium salts.(

$-O-COCH_3$ ) group is electron-withdrawing and hence, (d) is less stable than (b).

Although

$-O-COCH_3$ is an electron-donating substituent, but it is present in the meta position. Hence, it will not have a significant effect on stability.

hence the correct option: A

The order of bascity is
(I)

$Ph-CONH_{2}$
(II)

$Ph- NH_{2}$
(III)

$Ph-CH_{2}-NH_{2}$
(IV)

$p-OCH_{3}Ph0-NH_{2}$

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

The most nucleophilic nitrogen is in :

The most basic

$N-$ atom is:

0%
$a$
0%
$b$
0%
$c$
0%
$d$

Explanation

Here the most basic nitrogen atom is

$\underset{(c)}{-NH}$ . This is because here nitrogen is attached to the aliphatic carbon atom.

The nitrogen in

$\underset{(b)}{-NH_2}$ is in conjugation with aromatic ring thus it does not have enough electron to act as a Lewis base.

Similarly, In nitrogen,

$a$ and

$d$ the nitrogen is present in the ring thus there is not enough electron available for donation.

Which of the following amines will react with cyclohexane to give enamine?

0%
$C{H_3}N{H_2}$
0%
${(C{H_3})_2}NH$
0%
0%

Image

N-Ethylphthalimide

$\longrightarrow$ Ethylamine
Reagent for the conversion of this reaction is:

0%
$H_2O$
0%
$NaBH_4$
0%
$NH_2-NH_2$
0%
$CaH_2$

Consider the following statements
(i) Out of two

$- NH_{2}$ groups in semicarbazide only one is involved in formation of semicarbazones.
(ii) Cyclohexane froms cyanohydrin in good yield but

$2, 2, 6-trimethylcyclohexanone$ does not.
(iii) The

$\alpha - hydrogens$ in aldehydes and ketones are acidic.

0%
i and iii
0%
ii and iii
0%
i and ii
0%
i, ii and iii

The correct order of

$K_b$ value of following is:

$(C_2H_5)_2\ddot{N}H, NH_3, \ddot{N}(C_2H_5)_3$

0%
$1 > 2 > 3$
0%
$1 > 3 > 2$
0%
$3 > 2 > 1$
0%
$3 > 1 > 2$

Explanation

$+I$ effect is maximum in secondary amine and minimum in primary amine.

Thus the correct order of

$K_b$ will be

$(C_2H_5)_2\ddot{N}H > \ddot{N}(C_2H_5)_3 > NH_3$

CBSE Questions for Class 12 Engineering Chemistry Amines Quiz 6 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

CBSE Questions for Class 12 Engineering Chemistry Amines Quiz 6 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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