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

Which is most volatile?

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

Explanation

$CH_3-OH$ is more volatile. Volatility is due to the tendency to evaporate. The tendency of volatility of

$CH_3-OH$ is low. Dipole attraction and London force occurs in the molecules.

The correct order of the

$pKa$ values for the conjugate acids of the heterocyclic compound is:

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

Arrange the compounds in decreasing order of basicity

Ammonia, Triethylamine, Aniline, Ethylamine, Diethylamine

0%
Diethylamine > Triethylamine > Ethylamine > Ammonia > Aniline
0%
Diethylamine > Ethylamine >Triethylamine> Ammonia > Aniline
0%
Triethylamine > Diethylamine > Ethylamine> Ammonia > Aniline
0%
Aniline > Ethylamine> Diethylamine >Triethylamine> Ammonia

Which of the following aryl amine undergoes diazotization most readily?

Explanation

Diazotization reaction is an electrophilic

substitution reaction.

so, the benzene ring attached with more

The mesomeric effect group(+M) will readily

undergo a diazotization reaction.

so option

$c$ is correct.

1998752_1114361_ans_bb6472670e3b489583577bc3c6890ead.PNG

Which of the following amines cannot be prepared by path-1?

The correct order of basicites of the following compounds is:-

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

$CH_3NH_2 \quad \underrightarrow{{(CH_3COOCH_3)}, { \Delta}} \quad X$

0%
X is more basic than $CH_3NH_2$
0%
X is less basic than $CH_3NH_2$
0%
X is a tertiary amine
0%
None

The correct decreasing order of

$p{ K }_{ b }$ :

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

Explanation

We know that more is the

$pK_b$ value ,less basic will be the given compound.

Since chlorine exerts

$-I$ effect , hence it is least basic and has more

$pK_b$ value.

Similarly , on basis on electronic effects, the answer is option D.

Which of the following compound(s) will not dissolve in an alkali solution after it has undergone reaction with Hinsberg's reagent?

0%
Diethyl amine
0%
Isopropyl amine
0%
N-methyl ehtanamine
0%
N,N-dimethyl aniline

Explanation

Hinsberg reagent is Benzenesulphonyl chloride $({C_6}{H_5}S{O_2}Cl)$ . It reacts with primary and secondary amines to form sulphonamides.

It reacts with primary amine to form acidic compound which is soluble in alkali because hydrogen (H) attached to nitrogen(N) in sulphonamide is strongly acidic due to presence of stronf electron withdrawing sulphonyl group.

Now, out of the options given in the question, only isopropyl amine is a primary amine. Dimethyl amine and N-methyl ethaneamine are secondary amines and N,N-dimethylaniline is tertiary amine.

Hence,option (b) is correct.

Given compound is a:

0%
Primary amine
0%
Secondary amine
0%
Tertiary amine
0%
Ammonium salt

Which of the following cannot exist in room temperature?

Diazomethane was photolysed to generate singlet carbine which was treated with cis-2-butene. The main product(s) of the reaction is/are:

0%
cis-1, 2-dimethylcyclopropane
0%
trans-1, 2-dimethylcyclopropane
0%
mixture of $(A)$ and $(B)$
0%
Ethylcyclopropane

Explanation

Solution:- (A) cis-

$1,2$ -dimethylcyclopropane

$\underset{\text{Diazomethane}}{C{H}_{2}-\overset{+}{N}={N}^{-}} \xrightarrow[-N \equiv N]{h \nu} \underset{\text{Singlet carbene} \left( \text{Methylene} \right)}{:C{H}_{2}}$

The addition of methylene would yield trans-

$1,2$ -dimethylcyclopropane from trans-

$2$ -butene and cis-

$1,2$ -dimethylcyclopropane from cis-

$2$ -butene.

A 55- kDa protien was acid hydrolysed to obtain a misture of amino acids. How many amino acids could be present in the solution?

0%
550
0%
500
0%
1000
0%
1100

Which of the following arylamine is most reactive towards diazotisation?

Explanation

Diazotisation is an electrophilic substitution reaction. Electrophile

$(NO)^+$ attacks the

$-NH_2$ group to form diazo compound. Electron donating groups increase the reactivity of aryl amines towards electrophilic substitution reaction while electron withdrawing groups decrease the reactivity.

So, compounds C and D having electron withdrawing groups

$CN$ and

$Cl$ at a para position will be less reactive towards diazotization. Both

$CH_3$ and

$OCH_3$ will increase its reactivity, but

$OCH_3$ being strong electron donating group when present at ortho and para positions will increase its reactivity more effectively. So, B will be most reactive.

The most basic compound among the following is

0%
Benzylamine
0%
Aniline
0%
Acetanilide
0%
p-nitroaniline

Number of isomeric primary amines obtained from

$C_4H_{11}N$ are :

0%
3
0%
4
0%
5
0%
6

Explanation

The possible isomers are

$CH_{3}CH_{2}CH_{2}CH_{2}NH_{2}$

$(CH_{3})_{2}-CH-CH_{2}NH_{2}$

$(CH_{3})_{3}-NH_{2}$

$CH_{3}-CH(NH_{2})CH_{2}-CH_{3}$

Strongest base is

0%
$CH_2$ = $CH -CH_2-NH_2$
0%
$CH_2$ = $CH- CH_2 - NH_2$
0%
$CH_3CH_2CH_2NH_2$
0%

Explanation

Hint: Basic strength is directly proportional to the electron pair intensity on the nitrogen atom.

Correct Answer: Option C

Explanation of correct option:

Amines are basic in nature as they have lone pair of electrons on nitrogen. Therefore, they have a strong tendency to donate this lone pair of electrons to electron acceptors.

Here, in the case of option C

It is a primary amine and the presence of the alkyl group in alkyl amines increases the electron density on the nitrogen atom by virtue of the +I effect. Hence it is most basic.

Explanation of incorrect options:

Options A and B: The double-bonded carbon is more electronegative, so it shows -I effect. Hence electron richness on

$N$ decreases. Hence Both A andB are less basic.

Option D: As we know Aryl amines are less basic than alkyl-substituted amines because some electron density provided by the nitrogen atom is distributed throughout the aromatic ring. Hence it is also less basic.

Amongest the following the most basic compound is:

0%
p-nitro aniline
0%
Acetanilide
0%
Aniline
0%
Benzylamine

Explanation

Basicity in these compounds is due to the presence lone pair of electrons on

$N$ atom of

$NH_2$ group. In p-nitroaniline, acetanilide and aniline the

$NH_2$ group is attached directly to the benzene ring and the lone pair of electrons on

$N$ atom will get involved in resonance in the benzene ring. As a result its ability to donate electrons to the incoming group will decrease, thereby decreasing its basicity. While in benzylamine,

$NH_2$ group is not attached directly to the benzene ring and its lone pair of electron is available for donation. Therefore, it is most basic among these compounds.
1124803_1164434_ans_454635d4a0574a0d9b5e68a49efb7bd6.jpg

1124803_1164434_ans_454635d4a0574a0d9b5e68a49efb7bd6.jpg

Which of the following order is correct ?

0%
0%
0%
0%
All of the above

Phenyl acetonitrile reacts with sodium and alcohol to give

0%
$Ph-CH_2\ CH_2NH_2$
0%
$Ph\ CH_2-NH_2$
0%
$CH_3-CH_2-NH-Ph$
0%
$Ph-NH_2$

$-NH_2$ group in aniline is:

0%
m- directing and activating
0%
m- directing and deactivating
0%
o,p- directing and deactivating
0%
o,p- directing and activating

Find the B in the following reaction

Which of the following amines cannot be prepared by Gabriel Phthalimide reaction?

0%
0%
0%
0%
$CH_3-\underset{NH_2}{\underset{|\,\,\,\,\,\,\,}{NH_2}}-CH_3$

Which of the following will be most stable diazonium salt

$RN_2^{+}X^{-}$ ?

0%
$CH_3N_2^{+}X^{-}$
0%
$C_6H_5N_2^{+}X^{-}$
0%
$CH_3 CH_2 N_2^{+}X^{-}$
0%
$C_6H_5 CH_2 N_2^{+}X^{-}$

Which of the following is correct about Guanidine base?

0%
Resonance possible
0%
Odd number of p- electrons
0%
Localized lone pair of $e^{-}$
0%
Delocalized lone pair of $e^{-}$

The order of basicity of amines in gaseous state is :-

0%
$3^{o} > 2^{o} > NH_{3} > 1^{o}$
0%
$1^{o} > 2^{o} > 3^{o} > NH_{3}$
0%
$NH_{3} > 1^{o} > 2^{o} > 3^{o}$
0%
$3^{o} > 2^{o} > 1^{o} > NH_{3}$

Which of the following order regarding basicity of amines is correct?

Factors responsible for base strength comparison of amines in aqueous solution ?

0%
Availability of lone pair on nitrogen
0%
Lone pair donating power of nitrogen
0%
Stability of conjugate acid formed by solvation through H-bonding.
0%
All of the above

Explanation

The basicity of amines depend upon its ability to donate the lone pair of electrons on nitrogen.

Order of basicity of amines in aqueous solution is,

$NH_3<Primary \ amines(RNH_2)<Tertiary \ amines(R_3N)<Secondary \ amines(R_2NH)$

This is due to the combined effect of +I effect of alkyl group, steric hindrance caused by alkyl groups and solvation of amines through H-bonding.

The power to donate lone pair of electrons increases as the number of alkyl groups increase. Therefore, making

$R_3N$ most basic, followed by

$R_2NH$ and

$RNH_2$ .

But as the number of alkyl groups increase, the steric hindrance caused by them also increases. Thus, decreasing the electron donating power of amines. Hence, making

$RNH_2$ most basic followed by

$R_2NH$ and

$R_3N$ .

Solvation is the formation of protonated amines when they are dissolved in water. As the number of H-atoms on nitrogen increases the possibility of H-bonding also increases, providing greater stability to the amine. Thus, primary amine

$(RNH_2)$ is more stable than secondary amine

$(R_2NH)$ , which is more stable than tertiary amine

$(R_3N)$ .

The combined effect of these three factors give the observed order of basicity of amines in aqueous solution.

Minimum C-atom required for

$2^\circ$ amine to show chain isomerism are:

0%
$4$
0%
$6$
0%
$3$
0%
$5$

The correct order of increasing basic nature for the bases

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

$(CH_{3})_{2}NH$ is :

0%
$NH_{3}<CH_{3}NH_{2}$ < $(CH_{3})_{2}NH$
0%
$(CH_3)_2NH<CH_3NH_2<NH_3$
0%
$CH_3NH_2<NH_3<(CH_3)_2NH$
0%
$NH_3 < (CH_3)_2NH<CH_3NH_2$

$CH_3NH_2+CHCl_3+KOH \rightarrow$ Nitrogen containing compound +

$KCl +H_2O$ . Nitrogen containing compound is:

0%
$CH_3CN$
0%
$CH_3NHCH_3$
0%
$CH_3-\bar{N}\equiv\overset{+}{C}$
0%
$CH_3-\overset{+}{{N}}\equiv {C}:^-$

Explanation

The reaction of primary amine (

$R-NH_2$ ) with chloroform

$(CHCl_3)$ and a base

$(KOH)$ is known as carbylamine reaction. The reaction yields isocyanide as the product, along with salt and water.

$CH_3NH_2+CHCl_3+KOH\rightarrow CH_3N^+\overline{=}C^-+KCl+H_2O$

What is the decreasing order of basicity of pri-, sec-, and tert-, ethyl amines and

$NH_3$ ?

0%
$NH_3$ > $C_2H_5NH_2$ > $(C_2H_5)_2NH$ > $(C_2H_5)_3N$
0%
$(C_2H_5)_3N$ > $(C_2H_5)_2NH$ > $C_2H_5NH_2$ > $NH_3$
0%
$(C_2H_5)_2NH$ > $C_2H_5NH_2$ > $(C_2H_5)_3N$ > $NH_3$
0%
$(C_2H_5)_2NH$ > $C_2H_5NH_2$ > $NH_3$ > $(C_2H_5)_3N$

$[Cr(H_2O)_6]^{3+}$ typically absorbs at around 574 nm. It is allowed to react with ammonia to form a new complex

$[Cr(NH_3)_6]^{3+}$ that should have absorption at

0%
800 nm
0%
462 nm
0%
620 nm
0%
574 nm

Which of the following compound has highest boiling point ?

Which of the following are soluble in aq.

$NaOH$ solution ?

Which nitrogen atom is most basic in the above given compound?

0%
$3$
0%
$4$
0%
$5$
0%
$2$

What will be the final product in the following reaction sequence?

$CH_3CH_2CN\xrightarrow {H^{+}/H_2O } A \xrightarrow [ \Delta ]{ { NH }_{ 3 } } B \xrightarrow { NaOBr }$

0%
$CH_3CH_2CONH_2$
0%
$CH_3CH_2COBr$
0%
$CH_3CH_2NH_2$
0%
$CH_3CH_2CH_2NH_2$

$H_3C-CH_2-NH_2$
3)

$(CH_3)_2NH$
Write basicity order of the above

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

$H_3C-CH_2-NH_2$
3)

$(CH_3)_2NH$
Write basicity order of the above

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

Among the following which amine can be prepared by Gabriel phthalimide method?

0%
$C_6H_5NH_2$
0%
$CH_3 - NH - CH_2CH_3$
0%
0%

Which of the following is an Incorrect statement about benzenediazonium chloride?

0%
It is prepared by the reaction of aniline with nitrous acid at 273-278 K
0%
It is a colourless crystalline solid
0%
Alkyl diazonium salts are more stable than arenediazonium salts
0%
Benzene diazonium fluoroborate is water insoluble & stable at room temperature

$1^0, 2^0$ and

$3^0$ amine can be distinguished by?

0%
Hinsberg reaction
0%
Hoffmann reaction
0%
Reaction with Nitrous acid
0%
All of these

Which of the following is most basic:

0%
$CH_{3}CH_{2}NH_{2}$
0%
$CH_{3}CH_{2}NHCH_{3}$
0%
0%
$CH_{3}CONH_{2}$

Which of the following compounds has the most basic nitrogen?

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

Which of the following is most basic in nature?

Compare basic strength:

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

Arrange basicity of given compounds in decreasing order :
(1)

$CH_3-CH_2-NH_2$
(2)

$CH_2= CH-NH_2$
(3)

$CH\equiv C-NH_2$
(4)

$C_6H_5-NH_2$

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

Arrange the following in the decreasing order of their

$pK_{a}$ value:
I .

$CH_{2} = CHCH_{2}NH_{2}$
II .

$CH_{3}CH_{2}CH_{2}NH_{2}$
III .

$CH \equiv CCH_{2}NH_{2}$

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

Which of the following will not react with sodiumnitrite and

$HCl$ ?

Arrange the following compounds in decreasing order of basicity:

0%
P > Q > R > S
0%
Q > P > R > S
0%
R > S > Q > P
0%
Q > P > S > R

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

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Practice Class 12 Medical Chemistry Quiz Questions and Answers

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