The correct set of products obtained in the following reactions $$RCN\xrightarrow [ ]{ reduction } (A)$$, $$RCN \xrightarrow [ (ii){ H }_{ 2 }O ]{ (i){ CH }_{ 3 }MgBr } (B)$$, $$RNC \xrightarrow [ ]{ hydrolysis } (C)$$, $$R{NH}_{2}\xrightarrow [ ]{ H{NO}_{2} } (D)$$ The answer is:

$$(A): {1}^{o}$$ Amine; $$(B)$$: Methyl ketone; $$(C):{1}^{o}$$ Amine; $$(D)$$ Alcohol

$$(A): {2}^{o}$$ Amine; $$(B)$$: Methyl ketone; $$(C):{1}^{o}$$ Amine; $$(D)$$ Alcohol

$$(A): {2}^{o}$$ Amine; $$(B)$$: Methyl ketone; $$(C):{2}^{o}$$ Amine; $$(D)$$ Acid

$$(A): {2}^{o}$$ Amine; $$(B)$$: Methyl ketone; $$(C):{2}^{o}$$ Amine; $$(D)$$ Aldehyde

MCQ Questions for Class 12 Medical Chemistry Organic Compounds Containing Nitrogen Quiz 10

Identify the most basic compound from the following.

0%
Primary amines
0%
Secondary amines
0%
Aromatic amines
0%
Aliphatic amines

Compound

(F)

$(F)$ is:

Explanation

D.U in

(A) = \frac{(2 n_{C} + 2) - n_{H}}{2} = \frac{(8 \times 2 + 2) - 10}{2} = 4^{o}

$(A)=\cfrac { \left( 2{ n }_{ C }+2 \right) -{ n }_{ H } }{ 2 } =\cfrac { \left( 8\times 2+2 \right) -10 }{ 2 } ={ 4 }^{ o }$
4 D.U and

C : H = 1 : 1

$C:H=1:1$ suggest that

(A)

$(A)$ contains benzene ring with two extra C atoms (i.e., two

(M e)

$(Me)$ groups]. Since compound

(A)

$(A)$ is steam volatile and on nittration gives two nitro-derivatives, so

(A)

$(A)$ is ortho-xylene

Identify the most basic amine among the above compounds :

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

Compound

(D)

$(D)$ is:

Explanation

D.U in

(A) = \frac{(2 n_{C} + 2) - n_{H}}{2} = \frac{(8 \times 2 + 2) - 10}{2} = 4^{o}

$(A)=\cfrac { \left( 2{ n }_{ C }+2 \right) -{ n }_{ H } }{ 2 } =\cfrac { \left( 8\times 2+2 \right) -10 }{ 2 } ={ 4 }^{ o }$
4 D.U and

C : H = 1 : 1

$C:H=1:1$ suggest that

(A)

$(A)$ contains benzene ring with two extra C atoms (i.e., two

(M e)

$(Me)$ groups]. Since compound

(A)

$(A)$ is steam volatile and on nittration gives two nitro-derivatives, so

(A)

$(A)$ is ortho-xylene

The following reaction is:

0%
benzidine rearrangement
0%
pinacol-pinacolone rearrangement
0%
fries rearrangement
0%
benzil-benzilic acid arrangement

Which statements are correct?

0%
Phenol and aniline give coupling reaction with diazonium salt.
0%
Phenol couples with diazonium salt in mild basic conditions ( $\displaystyle pH = 8-10$ ).
0%
Aniline couples with diazonium salt in mild acidic condition ( $\displaystyle pH = 4-6$ ).
0%
Both phenol and aniline couple with diazonium salt in neutral condition ( $pH = 7$ ).

Explanation

The condensation of diazonium salts wih aromatic amines in mildly acidic solution or with phenols in mildly alkaline solution to form coloured substances, called azo dyes, is known as coupling.
The statements A, B and C are correct.
(A) Phenol and aniline give coupling reaction with diazonium salt.
(B) Phenol couples with diazonium salt in mild basic conditions (

p H = 8 - 10

$\displaystyle pH = 8-10$ ).
(C) Aniline couples with diazonium salt in mild acidic condition (

p H = 4 - 6

$\displaystyle pH = 4-6$ ).

The decreasing order of the rate of bromination of the following compounds is:

I . P h \oplus N {M e}_{3}

$I. Ph\overset { \oplus }{ N{ Me }_{ 3 } }$

I I . P h {C H}_{3} \oplus N {M e}_{3}

$II. Ph{CH}_{3}\overset { \oplus }{ N{ Me }_{ 3 } }$

I I I . P h M e

$III. PhMe$

I V . P h N {M e}_{2}

$IV. PhN{Me}_{2}$

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

Explanation

Stronger the activating substituent (i.e EDG) faster is the rate of (bromination) reaction.

Order of

¯ ¯ ¯ e

$\overline { e }$ donating substituent

- \cdot \cdot N {M e}_{2} > - M e > - {C H}_{3} \oplus N {M e}_{3} > - \oplus N {M e}_{3}

$-\overset { \cdot \cdot }{ N } {Me}_{2}>-Me>-{CH}_{3}\overset { \oplus }{ N{ Me }_{ 3 } } >-\overset { \oplus }{ N{ Me }_{ 3 } }$

Hence, the rate of bromination is as given in (B)

Gabriel synthesis is used for the preparation of:

0%
$\displaystyle { 1 }^{ o }$ amine
0%
$\displaystyle { 2 }^{ o }$ amine
0%
$\displaystyle { 3 }^{ o }$ amine
0%
all of the above

Most basic compound in gaseous phase is :

0%
${NH}_{3}$
0%
${CH}_{3}{CH}_{2}{NH}_{2}$
0%
$({CH}_{3}{CH}_{2})_2NH$
0%
${({CH}_{3}{CH}_{2})}_{3}N$

Explanation

In gaseous phase (in case of amines)

$\text { In gaseous phase (in case of amines) }$

Stability of

2^{\circ}

$2^{\circ}$ carbocation

> 3^{\circ} > 1^{\circ} > {N H}_{3}

$>3^{\circ}>1^{\circ}>\mathrm{NH}_{3}$

Hence

{({C H}_{3} {C H}_{2})}_{2} N H

$\left(\mathrm{CH}_{3} \mathrm{CH}_{2}\right)_{2} \mathrm{NH}$ is the most stable.

Benzenediazonium chloride on reaction with aniline in a weakly basic medium gives :

0%
diphenyl ether
0%
p-aminoazobenzene
0%
chlorobenzene
0%
benzene

Explanation

Benzenediazonium chloride on reaction with aniline in a weakly basic medium gives p-aminoazobenzene.

C_{6} H_{5} - N = N - C l Benzene diazonium chloride + C_{6} H_{5} - N H_{2} Aniline O H^{-} - -- \to C_{6} H_{5} - N = N - C_{6} H_{4} - N H_{2} p-aminoazobenzene

$\displaystyle \underset {\text {Benzene diazonium chloride}}{C_6H_5-N=N-Cl}+\underset {\text {Aniline}}{C_6H_5-NH_2} \xrightarrow {OH^-} \underset {\text {p-aminoazobenzene}}{C_6H_5-N=N-C_6H_4-NH_2}$

Which of the following compounds has the lowest boiling point?

0%
2-propanamine
0%
ethylmethylamine
0%
1-propanamine
0%
N,N-dimethylmethanamine

Which of the following statement(s) is/are correct?

0%
Primary amines show intermolecular hydrogen bonding.
0%
Secondary amines intermolecular hydrogen bonding.
0%
Tertiary amines show intermolecular hydrogen bonding.
0%
Amines have lower boiling points as compared to those of alcohols and carboxylic acid of comparable molar masses.

Explanation

Primary and secondary amines have hydrogen atoms bonded to a nitrogen atom and are therefore capable of hydrogen bonding between

H

$H$ of

R_{2} N H o r R N H_{2}

$R_2NH \ or \ RNH_2$ with

N

$N$ of other amine molecule.

Therefore, A and B are correct.

Tertiary amines can not show intermolecular hydrogen bonding as there is no covalently bonded

H

$H$ with

N

$N$ .

Thus, C is incorrect.

Amines have lower boiling points as compared to those of alcohols and carboxylic acid of comparable molar masses because hydrogen bonding in amines is not as strong as alcohol molecules (which have hydrogen atoms bonded to an oxygen atom, which is more electronegative than nitrogen). This is the reason for their higher boiling points, such that they can form hydrogen bonds with each other as well as van der Waals dispersion forces and dipole-dipole interactions.

Thus, D is correct.

Which of the following statements is correct?

0%
Aliphatic amines are stronger bases than ammonia.
0%
Aromatic amines are stronger bases than ammonia.
0%
The alkyl group in alkyl ammonium ion more stabilizes the ion relative to the amine.
0%
The aryl group in aryl ammonium ion less stabilizes the ion relative to the amine.

Explanation

(A)

+ I

$+I$ effect of alkyl groups in aliphatic amines make them more basic than ammonia.

(B) Aromatic amines are weaker bases than ammonia because the lone pair of

N

$N$ is in delocalization and not freely available to act as base.

+ I

$+I$ or electron releasing effect.

(D) The aryl group in aryl ammonium ion less stabilizes the ion relative to the amine due to the delocalization of positive charge on electron negative

N

$N$ .

Hence, A, C, D are correct statements.

C

$C$ and

D

$D$ in the given sequence are :

0%
Benzoic acid $+$ aniline
0%
Phthalic acid $+$ ethylamine
0%
Phthalic acid $+$ aniline
0%
Benzoic acid $+$ ethylamine

Gabriel's phthalimide synthesis can be used to prepare:

0%
ethanamine
0%
n-methylmethanamine
0%
benzeneamine
0%
$N, N-$ dimethylmethanamine
0%
p-toluidine

Which one of the following is used as a test for aliphatic primary amines ?

0%
Tollen's test
0%
Fehling's lest
0%
Isocyanide test
0%
Azo dye test
0%
Phthalein fusion test.

Explanation

Isocyanide test (carbylamine test) is used to confirm primary amines.

R - N H_{2} + C H C l_{3} + 3 K O H \to R - N C I s o c y a n i d e + 3 K C l + 3 H_{2} O

$R-NH_2 + CHCl_3 + 3KOH \rightarrow \underset{Isocyanide }{R-NC} + 3KCl + 3H_2O$

The correct set of products obtained in the following reactions

R C N r e d u c t i o n - ---- \to (A)

$RCN\xrightarrow [ ]{ reduction } (A)$ ,

R C N (i) {C H}_{3} M g B r - ------- \to (i i) H_{2} O (B)

$RCN \xrightarrow [ (ii){ H }_{ 2 }O ]{ (i){ CH }_{ 3 }MgBr } (B)$ ,

R N C h y d r o l y s i s - ----- \to (C)

$RNC \xrightarrow [ ]{ hydrolysis } (C)$ ,

R {N H}_{2} H {N O}_{2} - --- \to (D)

$R{NH}_{2}\xrightarrow [ ]{ H{NO}_{2} } (D)$
The answer is:

0%
$(A): {2}^{o}$ Amine; $(B)$ : Methyl ketone; $(C):{1}^{o}$ Amine; $(D)$ Alcohol
0%
$(A): {1}^{o}$ Amine; $(B)$ : Methyl ketone; $(C):{1}^{o}$ Amine; $(D)$ Alcohol
0%
$(A): {2}^{o}$ Amine; $(B)$ : Methyl ketone; $(C):{2}^{o}$ Amine; $(D)$ Acid
0%
$(A): {2}^{o}$ Amine; $(B)$ : Methyl ketone; $(C):{2}^{o}$ Amine; $(D)$ Aldehyde

Explanation

The compound (A) is a primary amine:

R C N r e d u c t i o n - ---- \to R N H_{2} (A)

$\displaystyle RCN\xrightarrow [ ]{ reduction } \underset {(A)}{RNH_2}$

The compound (B) is a methyl ketone:

R C N (i) {C H}_{3} M g B r - ------- \to (i i) H_{2} O R C O C H_{3} (B)

$\displaystyle RCN \xrightarrow [ (ii){ H }_{ 2 }O ]{ (i){ CH }_{ 3 }MgBr } \underset {(B)}{RCOCH_3}$

The compound (C) is a primary amine:

R N C h y d r o l y s i s - ----- \to R N H_{2} (C)

$\displaystyle RNC \xrightarrow [ ]{ hydrolysis } \underset{ (C) }{RNH_2}$

The compound (D) is an alcohol:

R {N H}_{2} H {N O}_{2} - --- \to R O H (D)

$\displaystyle R{NH}_{2}\xrightarrow [ ]{ H{NO}_{2} } \underset { (D)}{ROH}$

Among the following the weakest base is:

0%
$C_6H_5CH_2NH_2$
0%
$C_6H_5CH_2NHCH_3$
0%
$O_2N-CH_2NH_2$
0%
$CH_3NHCHO$

Explanation

The weakest base is

C H_{3} N H C H O

$\displaystyle CH_3NHCHO$ .

The lone pair of electrons cannot be easily donated to suitable Lewis acid as the nitro group is electron-withdrawing group.

As amides are not basic in nature and given option D is amide structure and other compounds are amines.

Option D is correct.

{C H}_{3} {C H}_{2} {N H}_{2}

${CH}_{3}{CH}_{2}{NH}_{2}$ is soluble in:

0%
dilute $HCl$
0%
$Cu{SO}_{4}$ solution
0%
$Ag{NO}_{3}$
0%
none of these

Explanation

C H_{3} C H_{2} N H_{2}

$CH_3CH_2NH_2$ being polar and basic in character is soluble in all - dilute

H C l

$HCl$ ,

C u S O_{4}

$CuSO_4$ solution and

A g N O_{3}

$AgNO_3$ .

Arrange the following resonating structure according to their contribution towards resonance hybrid?
(a)

C H_{2} = \oplus N = ⊖ N

$CH_2=\overset {\oplus}{N}=\overset {\ominus}{N}$ (b)

⊖ C H_{2} - N = \oplus N :

$\overset {\ominus}{C}H_2-N=\overset {\oplus}{N}:$ (c)

\oplus C H_{2} - N . . = ⊖ N

$\overset {\oplus}{C}H_2-\underset {. .}{N}=\overset {\ominus}{N}$ (d)

⊖ C H_{2} - \oplus N \equiv . . N

$\overset {\ominus}{C}H_2-\overset {\oplus}{N}\equiv \overset {. .}{N}$

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

Which of the following orders regarding the basic strength of substituted aniline is correct?

0%
p-methylaniline> p-chloroaniline > p-aminoacetophenone
0%
p-methylaniline> p-aminoacetophenone> p-chloroaniline
0%
p-aminoacetophenone> p-methylaniline> p-chloroaniline
0%
p-aminoacetophenone> p-chloroaniline> p-methylaniline

Explanation

p - methylaniline will be the most basic because the methyl group is an electron releasing group that increases the electron density.

Halogen groups are weak

e^{-}

$e^{-}$ withdrawing groups.

{C H}_{3} C O

$\mathrm{CH}_{3} \mathrm{CO}$ group is an

e^{-}

$e^{-}$ withdrawing group which decreases the electron density, hence least basic.

Consider the given ions.
The reactivities of these ions in a azo-coupling reaction (under similar conditions) will be such that :

0%
$(I)< (IV)< (II)< (III)$
0%
$(I)< (III)< (IV)< (II)$
0%
$(III)< (I)< (IV)< (II)$
0%
$(III)< (I)< (II)<(IV)$

Explanation

Aromatic diazonium ions act as electrophiles in coupling reactions with activated aromatics such as anilines or phenols. The substitution normally occurs at the para position.

The reactivities of given ions in an azo-coupling reaction will be

I < I I I < I V < I I

$I<III<IV<II$ as

N O_{2}

$NO_2$ makes the compound strong electrophile. While

C H_{3}, O - C H_{3} a n d - N M e_{2}

$CH_3, O-CH_3 \ and \ -NMe_2$ has electron releasing effects which successively make the compound weaker electrophile.

The correct order of increasing basic nature of the following base is :

0%
2 < 5 < 1 < 3 < 4
0%
5 < 2 < 1 < 3 < 4
0%
2 < 5 < 1 < 4 < 3
0%
5 < 2 < 1 < 4 < 3

Explanation

In structure 2, electron withdrawing group

N O_{2}

$NO_2$ present in para position.

Also in structure 5, electron withdrawing group

N O_{2}

$NO_2$ is present in meta position.

At para position

N O_{2}

$NO_2$ shows more withdrawing effect than at meta position due to

- R

$-R$ effect.

Hence, structure 2 is least basic.

Next basic structure is aniline i.e., structure 1.

Next comes the structure 3 and 4 in which methyl and methoxy groups are present which are electron-donating groups.

So the correct answer is A

Arrange in decreasing order of basic strength :

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

Explanation

Aniline itself act as lewis base as the nitrogen of

- N H_{2}

$-NH_2$ can donate electron pairs. Its basic strength increases by presence of

a l k y l

$alkyl$ group which shows

+ I

$+I$ effect. Addition of

- C N o r - N O_{2}

$-CN \ or \ -NO_2$ decrease its basic strength as they both show

- M

$-M$ effect.

- M

$-M$ effect of nitro group is higher than nitrile group. Therfore the order of strength is

I V > I > I I I > I I

$IV>I>III>II$

Which of the following is/are correct statement/statements ?

0%
Guanidine $\begin{bmatrix}NH_2-\underset{\underset{NH}{||}}{C}-NH_2\end{bmatrix}$ is more basic than pyridine because conjugate acid of guanidine has three equal contributing resonating structure.
0%
Diethylamine is weaker base than triethylamine in aqueous medium.
0%
Ortho-methyl aniline is weaker base than para-methyl aniline.
0%
$2,\,4,\,6$ -Trinitro- $N,\,N$ -dimethyl aniline is stronger base than $2,\,4,\,6$ -Trinitro aniline.

Arrange the following in the increasing order of the

$\displaystyle{pK}_b$ values:

$C_2H_5NH_2$
ii)

$\,{ C }_{ 6 }{ H }_{ 5 }{ NHCH }_{ 3 }$
iii)

$\,({ C }_{ 2}{ H }_{ 5 }{ ) }_{ 2 }NH$
iv)

${ C }_{ 6 }{ H }_{ 5 }{ NH }_{ 2 }$

0%
i < ii < iii < iv
0%
iii < i < ii < iv
0%
i < iii < ii < iv
0%
iv < ii < i < iii

Explanation

$pK_b$ value gives a measure of the basicity of a compound. Higher the value of

$K_b$ of a base, stronger is the base. In aniline,

$C_6H_5NH_2$ , lone pair on nitrogen atom are withdrawn into the benzene ring. Benzene ring is an electron withdrawing group and hence, the lone pair is not easily available on nitrogen atom, so it cannot attract a proton easily.

N-Methylaniline,

$C_6H_5NHCH_3$ has benzene ring attached to it as well as a methyl group attached to the nitrogen atom. The benzene ring withdraws lone pair from nitrogen atom . Methyl group attached to the nitrogen atom is an electron donating group and it donates electron to nitrogen atom increasing electron density .

Thus the

$K_b$ value for N-Methylaniline is higher than aniline and it is a stronger base than aniline.

$C_2H_5NH_2$ , ethyl groups are electron donating group. It donates electrons to nitrogen atom increasing the electron density over nitrogen atom. Thus Ethylamine is a stronger base than N-Methylaniline.

In diethylamine,

$(C_2H_5)_2NH$ , two Ethyl groups attached to nitrogen atom are electron donating group. increasing the electron density on nitrogen atom. hence, diethylamine is a stronger base than ethylamine because of the presence of two electron donating groups.

Thus the correct option is D.

Arrange the following in the decreasing order of the

$\displaystyle{pK}_b$ values:

$i)\,\displaystyle { C}_{ 2 }{ H }_{ 5 }{ NH }_{ 2 }$

$ii) \,{ C }_{ 6 }{ H }_{ 5 }{ NHCH }_{ 3 }$

$iii) \,({ C }_{ 2}{ H }_{ 5 }{ ) }_{ 2 }NH$

$iv) { C }_{ 6 }{ H }_{ 5 }{ NH }_{ 2 }$

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

Explanation

$+I$ effect of alkyl group increases electron density on

$-N$ of amine and increases basic strength. Thus

$(C_2H_5)_2NH$ is the strongest base. In aniline, the lone pair of electrons of

$N$ is in delocalization with ring which makes it less available and least basic. Methyl substitution increases basic strength due to its inductive effect. Therefore the order of base strength or

$pK_b$ is

$iii>i>ii>iv$

The given imine can be prepared from which of the following

0%
0%
0%
0%
all of these

Select the correct statements regarding the final product (Y) :

0%
It can be identified with neutral $\displaystyle FeCl_{3}$ solution
0%
On treatment with Zn-dust it yields a product having degree of unsaturation = 4
0%
It undergoes coupling reaction with benzene diazonium chloride at low temperature (0 - $\displaystyle 5^{\circ}C$ ) in slightly basic medium
0%
It liberates $\displaystyle H_{2}$ gas with Na

Explanation

The product

$Y$ of reaction is

$Phenol (C_6H_5OH)$ .

Phenol can be identified with neutral

$FeCl_3$ solution as it gives violate coloured solution.

On treatment with Zn-dust it yields a product Benzene

$(C_6H_6)$ which is having degree of unsaturation

$= 4 \ (\Rightarrow \dfrac {2C+2−H}{2} = \dfrac {12+2-6}{2}=4)$

It undergoes coupling reaction with benzene diazonium chloride at low temperature

$(0 - 5^∘C)$ in slightly basic medium.

It liberates

$H_2$ gas with

$Na$ .

$C_6H_5OH + Na \rightarrow C_6H_5O^-Na^+ + H_2$

Iso-propyl amine is a:

0%
primary amine
0%
secondary amine
0%
teritary amine
0%
quaternary amine

Explanation

The structure of isopropyl amine is

$CH_3-\underset{\underset{\displaystyle NH_2}{|}}{C}H-CH_3$
It is primary amine.

On heating benzyl amine with chloroform and ethanolic

$KOH$ , product obtained is:

0%
benzyl alcohol
0%
benzaldehyde
0%
benzonitrile
0%
benzyl isocyanide

Identify the compounds from the following which form primary amines under suitable reduction conditions.
1.

$C_2H_5NC$

$C_2H_6$

$C_2H_5CONH_2$

$C_6H_5NO_2$

0%
$1, 4$
0%
$3, 4$
0%
$1, 3, 4$
0%
$2, 3, 4$

Explanation

$C_2H_5CONH_2 \xrightarrow{LiAlH_4}C_2H_5CH_2NH_2$

$C_6H_5NO_2 \xrightarrow{Sn/HCl} C_6H_5NH_2$

Within the list shown below, the correct pair of structures of alanine in pH ranges 2-4 and 9-11 is:
I.

$H_3N^+ - CH(CH_3)CO_2H$
II.

$H_2N - CH(CH_3)CO_2^-$
III.

$H_3N^+ - CH(CH_3)CO_2^-$
IV.

$H_2N -CH(CH_3)CO_2H$

0%
I and II
0%
I and III
0%
II and III
0%
III and IV

Explanation

The correct pair of structures of alanine in pH ranges 2-4 and 9-11 is
I.

$\displaystyle H_3N^+ - CH(CH_3)CO_2H$
II.

$\displaystyle H_2N - CH(CH_3)CO_2^-$
In pH range 2-4, the medium is acidic and the amino group is protonated.
In pH range 9-11, the medium is basic and carboxylic group is deprotonated.

Predict respectively

$'X'$ and

$'Y'$ in the following reactions

$Ar-N{ H }_{ 2 }\xrightarrow { \quad \quad X\quad \quad } Ar-\overset { + }{ N } \equiv N.{ Cl }^- \xrightarrow { \quad \quad Y\quad \quad }$

$Ar-Cl$

0%
$NaN{ O }_{ 3 }$ & ${ Cl }_{ 2 }$
0%
$NaN{ O }_{ 3 }-HCl$ & $HCl$
0%
$NaN{ O }_{ 2 }-HCl$ & ${ Cu }/{ HCl }$
0%
$NaN{ O }_{ 2 }-HCl$ & $NaN{ H }_{ 2 }$

Explanation

The balanced complete chemical equation is as follows:

$Ar-NH_2\xrightarrow{NaNO_2 - HCl} Ar-N^+_2-Cl^-$ , this reaction is the diazotisation reaction,

$Ar-N^+_2-Cl\xrightarrow{Cu/HCl} Ar-Cl + N_2$ , This is the Sandmeyer reaction.

The compound R is:

Explanation

Compound Q on treatment with

$Br_2/NaOH$ is nothing but Hoffmann Bromamide reaction which gives amine from amide.

Option A is correct.

The compound T is:

0%
Glycine
0%
Alanine
0%
Valine
0%
Serine

$CH_3CH_2CH_2CN+2H_2\xrightarrow {Pd}$

Which of the following compound is formed in the above reaction?

0%
$CH_3CH_2CH_2CH_2NH_2$
0%
$CH_3CH_2CH_2NH_2$
0%
$CH_3CH_2CH_2CH_2CH_2NH_2$
0%
None of the above

Explanation

Alkyl nitriles are reduced using a metal like palladium, platinum or nicke as a catalyst.
So,

$CH_3CH_2CH_2CN+2H_2\xrightarrow{Pd} CH_3CH_2CH_2CH_2NH_2$
Hence option

$A$ is correct.

In the reduction of nitriles to prepare primary amine:

0%
$LiAlH_4$ is used.
0%
$NaBH_4$ is used.
0%
$Sn/HCl$ is used.
0%
none of these.

Explanation

$LiAl{ H }_{ 4 }$ reduces nitriles to primary amines with ascent of amine series.

${ Sn }/{ HCl }$ reduce nitriles to immine hydrochloride.

${ NaB{ H }_{ 4 } }/{ { I }_{ 2 } }$ reduces nitriles to primary amines

Predict the product.

$CH_3-CH_2-CN + H_2/Ni \rightarrow$ ?

0%
$CH_3-CH_2-CH_2-NH_2$
0%
$CH_3-CH_2-NH_2$
0%
both a and b
0%
none of these

Explanation

$C{ H }_{ 3 }-C{ H }_{ 2 }-C\equiv N+{ { H }_{ 2 } }/{ Ni }\longrightarrow C{ H }_{ 3 }-C{ H }_{ 2 }-C{ H }_{ 2 }-N{ H }_{ 2 }$

The reduction of nitrile gives ascent of amine series indicating:

0%
preparation of amines containing two carbon atom more than starting amine
0%
preparation of amines containing one carbon atom more than starting amine
0%
preparation of amines containing three carbon atom more than starting amine
0%
preparation of amines containing four carbon atom more than starting amine

Which of the following is the correct order of the solubility in water?

0%
propane $<$ ethylamine $<$ ethyl alcohol
0%
propane $<$ ethyl alcohol $<$ ethylamine
0%
ethyl alcohol $<$ propane $<$ ethylamine
0%
ethyl alcohol $<$ ethylamine $<$ propane

Explanation

Correct Answer: Option A

Explanation:

We know, like dissolves like. So, the polar compounds are soluble in the polar solvents. And the non-polar solvents are soluble in non-polar solvents.
Water is a polar solvent.
The order of polarity of the given compounds is:

$propane < ethylamine < ethyl\ alcohol$

So, the order of solubility in the water is:

$propane < ethylamine < ethyl\ alcohol$

Which of the following represents the correct order of basic strength?

0%
N,N-Dimethylaniline $>$ N-methylaniline $>$ aniline
0%
N,N-Dimethylaniline $>$ aniline $>$ N-methylaniline
0%
aniline $>$ N-methylaniline $>$ N,N-Dimethyl aniline
0%
N-methylaniline $>$ N,N-Dimethylaniline $>$ aniline

Explanation

The following represents the correct order of basic strength
N,N-Dimethylaniline

$>$ N-methylaniline

$>$ aniline
When hydrogen atoms of the amino group of arylamines are replaced by electron donating alkyl groups, the basic character of the resulting amine increases.

In the given set of reactions, the IUPAC name of product

$Y$ is:

$2-\text{Bromopropane} \xrightarrow [ alc.\ KOH ]{ AgCN }\ \ X\xrightarrow [ ]{ LiAl{ H }_{ 4 } }\ \ Y\quad$

0%
N-methylpropanamine
0%
N-isopropylmethanamine
0%
butan-2-amine
0%
N-methylpropan-2-amine

Point out the correct decreasing order of

$pK_b$ values of following amines:

$C_2H_5NH_2$ ,

$C_6H_5NHCH_3,$

$(C_2H_5)_2NH$ and

$C_6H_5NH_2$

0%
$(C_2H_5)_2NH > C_2H_5NH_2 > C_6H_5NHCH_3 > C_6H_5NH_2$
0%
$(C_2H_5)_2NH > C_6H_5NHCH_3 > C_6H_5NH_2 > C_2H_5NH_2$
0%
$C_6H_5NH_2 > C_6H_5NHCH_3 > C_2H_5NH_2 > (C_2H_5)_2NH$
0%
$C_2H_5NH_2 > (C_2H_5)_2NH > C_6H_5NHCH_3 > C_6H_5NH_2$

Explanation

The lesser the value of $pK_b$ the weaker the base.
Electron withdrawing groups like benzene decreases the basicity and increases the

$pK_b$ of amines and the alkyl groups (electron releasing groups) increases the basicity and decreases the

$pK_b$ value
So the correct order of

$pK_b$ is

$C_6H_5NH_2 > C_6H_5NHCH_3 > C_2H_5-NH_2 > (C_2H_5)_2NH$
Hence option C is correct.

Which one of the following can be prepared by Gabriel phthalimide synthesis?

0%
Aniline
0%
$o$ -toluidine
0%
Benzylamine
0%
N-methylethanamine
0%
4-bromoaniline

The correct order of basicity of the following compounds is:

0%
$1 < 2 < 3 < 4$
0%
$1 < 2 < 4 < 3$
0%
$2 < 1 < 3 < 4$
0%
$4 < 3 < 2 < 1$

Explanation

The correct order of basicity is:

$2 < 1 < 3 < 4$

The basic strength order depends on:
(i) Accumulation of negative charge on

$N$ (double bonded) by another

$-NH_2$ group, thus intensifying the donor ability of

$N$ .

(ii) The higher donor ability of

$sp^3$ hybrid

$N$ as compared to

$sp^2\ N$ .

Option C is correct.

Acetonitrile on reduction gives:

0%
ethanamine
0%
propanamine
0%
methanamine
0%
None of the above

Among the following substituted pyridines, the most basic compound is:

The correct basicity order of the following compound is :

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

Benzylamine is a stronger base than aniline because:

0%
the lone pair of electrons on the nitrogen atom in benzylamine is delocalised
0%
the lone pair of electrons on the nitrogen atom in aniline is delocalised
0%
the lone pair of electrons on the nitrogen atom in aniline is not involved in resonance
0%
benzylamine has a higher molecular mass than aniline

Explanation

Benzylamine is a stronger base than aniline because the lone pair of electrons on the nitrogen atom in aniline is delocalised over benzene ring.

Hence, due to resonance in aniline, the lone pair of electron on nitrogen is less available for donation to suitable acid.

However in case of benzylamine (

$C_6H_5-CH_2-NH_2$ ), the N atom is attached to saturated C atom.

Hence, no resonance is possible and the lone pair of electron can be easily donated.

CBSE Questions for Class 12 Medical Chemistry Organic Compounds Containing Nitrogen Quiz 10 - 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 10 - MCQExams.com

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Practice Class 12 Medical Chemistry Quiz Questions and Answers

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