MCQ Questions for Class 12 Medical Chemistry Organic Compounds Containing Nitrogen Quiz 7

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

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

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

Which of the following will be least basic?

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

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

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

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

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

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 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

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

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

In the following componds the order of basicity is:

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

Explanation

A specie with a higher ability to donate electrons is more basic.

In III, electrons of Nitrogen are in resonance. So, it has the least tendency to donate electrons. Hence, is the least basic among the all given.

$\mathrm{sp^3}$ hybridized Nitrogen has more tendency to donate electrons than

$\mathrm{sp^2}$ hybridized Nitrogen.

In Iv, the compound has a lesser tendency to donate electrons than that in I because of steric hindrance.

So, the order of basicity is:

$\mathrm{I>IV>II>III}$

Hence, Option "A" is the correct answer.

Which one of the following is the weakest base?

0%
$(C_2H_5)_3 N$
0%
$(C_2H_5)_2 NH$
0%
$C_2H_5NH_2$
0%
$NH_3$

Explanation

The

$+I$ effect increases with increase in the alkyl group.Therefore the basic strength will be the highest in

$(C_2H_5)3N$ and least in

$NH_3$ . Therefore the decreasing order of basic strength in gas phase will be

$(C_2H_5)3N > (C_2H_5)2NH > C_2H_5NH_2 > NH_3$ .

Among the following which one is most basic?

0%
$NH_3$
0%
$CH_3 NH_2$
0%
$CH_3 CH_2NH_2$
0%
$C_6H_5NH_2$

Explanation

The basicity of the among compounds is due to the presence of lone pair on Nitrogen.

$CH_3CH_2NH_2$ is more basic than

$NH_3$ because of electron-donating group in

$CH_3CH_2NH_2$ increases the electron density on nitrogen atom.

$CH_3NH_2$ is less basic than

$CH_3CH_2NH_2$ because of ethyl group present in ethylamine which increases electron density on N in ethylamine and attracts

$H^+$ better than methylamine.

$C_6H_5NH_2$ is less basic than ethylamine because in

$C_6H_5NH_2$ the lone pair of electron on nitrogen is less available for protonation. It is delocalised into the Benzene ring by resonance. In ethylamine no such delocalisation takes place.

Hence, among them

$CH_3CH_2NH_2$ is most basic.

The strongest base among the following is:

0%
Amide ion
0%
Hydroxide ion
0%
Trimethylamine
0%
Ammonia
0%
Aniline

Explanation

Lesser the stability of the species ion, more basic be the given species. As

$NH_2^-$ (amide ion) is the least stable species ion. Thus, it is the most basic species.

Hence, option A is the correct answer.

Which of the following is the strong base?

N-Ethylphthalimide

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

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

In the following compounds, piperidine (I) pyridine(II) morpholine (III) and pyrrole (IV) the order of basicity is:

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

Explanation

Pyrrole is least basic because the lone pair of $N$ is involved in the resonance and thus they are not available for formation of a new bond with proton.

Pyridine is more basic than pyrrole as in pyridine the lone pair of $N$ is not involved resonance.

Piperidine is most basic because the lone pair is present in $sp^3$ hybrid orbital of $N$ while In pyridine it is present in $sp^2$ hybrid orbital. Greater the $s$ character of orbital more strongly the electrons will be bonded with the atom and less will be the basicity.

morpholine is less basic than piperidine but more basic than pyridine. In morphiline the electron withdrawing tendency of ring $O$ makes the lone pair of $N$ to be less available for the bonding with proton.

$Piperidine>\ morpholine>\ pyridine>\ pyrrole$

$\mathbf{I > III > II > IV}$

$\mathbf{Hence\ the\ correct\ answer\ is\ option\ (D)}$

This set contains the questions with single correct answer.

0%
primary amine
0%
secondary amine
0%
tertiary amine
0%
quaternary salt

The compound

0%
Alkyne, $3^{\circ} amine$
0%
Alkene, $2^{\circ} amine$
0%
Alkene, $3^{\circ} amine$
0%
Alkyne, $2^{\circ} amine$

Strongest base is:

Which nitrogen is prtonated redily in the guanidine?

0%
1
0%
2
0%
3
0%
None of these

Explanation

Nitrogen which is attached with double bond i.e. 1 nitrogen is readily protonated because double-bonded nitrogen is

$sp^2$ hybridized and on protonating double-bonded nitrogen enables effective resonance stabilization. The positive charge can be delocalized across the molecule

$\mathbf{Hence\ the\ correct\ answer\ is\ option (A)}$

1557564_1717920_ans_fc69a88ed15f4ffb9e60fc6b6ffc385f.jpg

The basicity of aniline is less than that of cyclohexylamine. This is due to:

0%
-R effect of $- NH_2 \ group$
0%
-I effect of $- NH_2 \ group$
0%
+ R effect of $NH_2 \ group$
0%
Hyperconjugation effect

Explanation

The basicity of aniline is less than that of cyclohexylamine because of + R effect of

$NH_2 \ group$ in aniline.

Hence, Option "C" is the correct answer.

Gabriel phthalimide synthesis is used in the preparation of:

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

${ (CH }_{ 3 })_{ 2 }{NH}$ is a stronger base than

${ (CH }_{ 3 })_{ 3 }{N}$ .

0%
True
0%
False

Which of the following may be prepared by Gabriel phthalimide synthesis ?

0%
Aliphatic amines
0%
Aromatic amines
0%
Aliphatic amides
0%
Aromatic amides

Which of the following is insoluble in dil. HCI ?

0%
Aniline
0%
Triphenylamine
0%
Ethylamine
0%
Dimethylamine

Amines are more basic than:

0%
alcohols
0%
ethers
0%
esters
0%
all of these

Which of the following is the most basic amine ?

0%
$CH_3-NH_2$
0%
$ClCH_2NH_2$
0%
$Cl_2CH-NH_2$
0%
$CCl_3-NH_2$

Explanation

$CCl_3−NH_2$ this is most basic amine

Aliphatic amines are insoluble in water.

0%
True
0%
False

Which among the following is amphoteric ?

0%
$CH_3-NH_2$
0%
$(CH_3)_2NH$
0%
$(CH_3)_3N$
0%
$CH_3CONH_2$

Explanation

here

$CH_3CONH_2$ is amphoteric

Arrange following amines in the decreasing order of their basicity:

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

A solution of ethylamine:

0%
turns blue litmus red
0%
turns red litmus blue
0%
does not affect the litmus
0%
bleaches the litmus

Which of the following has the highest

$pK_b$ value ?

CBSE Questions for Class 12 Medical Chemistry Organic Compounds Containing Nitrogen Quiz 7 - 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 7 - MCQExams.com

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Practice Class 12 Medical Chemistry Quiz Questions and Answers

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