MCQ Questions for Class 12 Engineering Chemistry Amines Quiz 9

Which of the following is correct formula for BDC?

0%
${ C }_{ 6 }{ H }_{ 4 }{ N }_{ 2 }Cl$
0%
${ C }_{ 6 }{ H }_{ 5 }{ N }_{ 2 }$
0%
${ C }_{ 6 }{ H }_{ 5 }{ N }_{ 2 }Cl$
0%
${ C }_{ 6 }{ H }_{ 4 }{ N }_{ 2 }$

Explanation

${ C }_{ 6 }{ H }_{ 5 }{ N }_{ 2 }Cl$ or benzene diazonium chloride is stable for a short time only. It slowly decomposes even at low temperatures. It is used immediately after preparation.

In the reaction series:

${ C }_{ 6 }{ H }_{ 5 }Cl\xrightarrow [Cu _{ 2 }O,\: 200^oC ]{NH_3} X; X\xrightarrow [ { 0-5 }^{ 0 }C ]{ { HNO }_{ 2 } } Z; X+Z\rightarrow A$ , the number of

$\sigma$ and

$\pi$ bonds in

$A$ are:

0%
$25$ $\sigma$ and $6$ $\pi$
0%
$25 \sigma$ and $7 \pi$
0%
$27\sigma$ and $7\pi$
0%
$27\sigma$ and $6\pi$

Explanation

Nucleophilic substitution of chloro-benzene with ammonia gives aniline.

${ C }_{ 6 }{ H }_{ 5 }Cl\xrightarrow [ { Cu }_{ 2 }O,\: { 200 }^{ 0 }C ]{ { NH }_{ 3 } } \underset {X} {C_6H_5NH_2} +HCl$

Diazotization of aniline gives benzene diazonium chloride.

$\underset {X} {C_6H_5NH_2} \xrightarrow [ { 0-5 }^{ 0 }C ]{ { HNO }_{ 2 } } \underset {Z} {C_6H_5N_2Cl}$

Azo coupling of aniline with benzenediazonium chloride gives p-aminoazobenzene.

$\underset {X} {C_6H_5NH_2}+ \underset {Z} {C_6H_5N_2Cl}\rightarrow \underset {A} {C_6H_5-N=N-C_6H_4-NH_2}$

The number of

$\sigma$ and

$\pi$ bonds in

$C_6H_5-N=N-C_6H_4-NH_2$ are

$27$

$\sigma$ and

$7$

$\pi$ .

Hence, the correct option is C.

Reaction at

${ 0-5 }^{ 0 }C$ between aniline,

${ NaNO }_{ 2 }$ and HCl is known as____________

0%
Suphonation of benzene
0%
Acylation of aniline
0%
Alkylation of aniline
0%
Diazotisation

Explanation

This is the diazotization reaction.

The reaction goes as follows :

$PhNH_{2} + NaNO_{2} + HCl \rightarrow PhN_{2}^{+} Cl^{-}$

The benzendiazonium salt thus formed acts as a synthetic intermediate and can be used to synthesize a wide variety of substituted benzene compounds.

For example,

$PhN_{2}^{+} Cl^{-} + CuCN \rightarrow PhCN$ (cyanobenzene)

$PhN_{2}^{+} Cl^{-} + CuI (KI) \rightarrow PhI$ (iodobenzene)

Option D is correct.

Among the following which is more basic :

Explanation

Hint: If $N$ lone pair is more available to donation more will be the basicity.

Explanation:

According to lewis's theory, the species which donate, its lone pair of electrons are basic in nature.
The electron releasing groups increases the electron density in the ring and hence lone pair of electrons in $NH_2$ can be easily donated.
Electron withdrawing groups decrease the ring's electron density, and the $NH_2$ group doesn't donate its lone pairs easily.
Aniline is a basic compound. As the electron-donating group on aniline increases, basicity increases, and basicity decreases when electron-withdrawing groups increase.
From the given options, $NO_2$ and $COOH$ are electron-withdrawing groups, while $CH_3O$ is an electron releasing group. Therefore, the basicity of the compound is in order of $A>C>B>D$ .
Hence, option $A$ is the most basic.

Correct Option: $A$

Statement I: Aniline on reaction with

$\mathrm{N}\mathrm{a}\mathrm{N}\mathrm{O}_{2}/\mathrm{H}\mathrm{C}l$ at

$0^{\mathrm{o}}\mathrm{C}$ followed by coupling with

$\beta$ -naphthol gives a dark blue coloured precipitate.
Statement II: The colour of the compound formed in the reaction of aniline with

$\mathrm{N}\mathrm{a}\mathrm{N}\mathrm{O}_{2}/\mathrm{H}\mathrm{C}l$ at

$0^{\mathrm{o}}\mathrm{C}$ followed by coupling with

$\beta$ -naphthol.

0%
Statement I and statement II are true and statement II is the correct explanation of statement I
0%
Statement I and statement II are true and statement II is not the correct explanation of statement I
0%
Statement I is true and the statement II is false
0%
Statement I is false and the statement II is true

Explanation

Statement I: Aniline on reaction with

$NaNO_2/HCl$ at

$0^oC$ followed by coupling with β-naphthol gives scarlet red coloured dye. Aniline on reaction with

$NaNO_2/HCl$ forms a diazonium salt. This diazonium salt is then coupled with β-naphthol.
Statement II: The colour obtained is the colour of the compound formed in the reaction of aniline with

$NaNO_2/HCl$ at

$0^oC$ followed by coupling with β-naphthol.

Hence, statement I is false and the statement II is true.

A compound

$A$ has molecular formula

$C_{7} H_{7} NO$ . On treatment with

$Br_{2}$ and

$KOH$ ,

$A$ gives an amine

$B$ which gives carbylamine test.

$B$ upon diazotization and coupling with phenol gives an azo dye.

$A$ can be:

0%
$C_{6}H_{5}CONHCOCH_{3}$
0%
$C_{6}H_{5} CONH_{2}$
0%
$C_{6}H_{5} NO_{2}$
0%
$o,\:m$ or $p$ - $C_{6}H_{4}(NH_{2})CHO$

Explanation

According to the reactions given for compound

$B$ ,

$B$ should be aniline as it gives carbylamine test and also couples with phenol to form dye.

The action of

$Br_2$ and

$KOH$ is Hoffmann bromamide reaction in which aniline is formed.

Therefore,

$A$ should be an amide.

Thus,

$A$ is

$C_6H_5CONH_2$ .

Option B is correct.

The starting reagents needed to make the azo compound shown below is:

Which of the following orders with regards to 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

Which of the following forms unstable diazonium ion when treated with

$NaNO_{2}$ in aqueous

$HCl$ ?

0%
p-nitrotoluene
0%
ethylamine
0%
N,N-dimethyl aniline
0%
All of the above

Explanation

Aromatic diazonium salts are stable if kept cold

$(0-5^oC)$ and in solution, but they often decompose violently when isolated. Their stability (compared to alkane diazonium salts) is due in part to resonance stabilization of the diazonium ion and in part to the very high energy of the aryl cation that would result from the loss of

$N_2$ (g).
Therefore except ethylamine all other given compounds forms stable diazonium ion when treated with

$NaNO_2$ in aqueous

$HCl$ .

Hence, the correct option is B.

The correct order of increasing basic nature for the bases

$NH_{3} ,CH_{3} NH_{2} and (CH_{3})_{2}NH$ in gas phase is :

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

Amongst the following the most basic compound is :

0%
p - nitroaniline
0%
acetanilide
0%
aniline
0%
benzylamine

Explanation

in p-nitroaniline, acetanilide and aniline, the lone pair of nitrogen is in conjugation with the aromatic system (benzene ring). So, it is delocalized and is not freely available for donation. So, A,B and C are less basic.
In benzyl amine (

$PhCH_{2}NH_{2}$ ), the lone pair of nitrogen is not in conjugation with the benzene ring and it is free for donation to an electrophile. So, D is the most basic compound.

The major product expected, when Phthalmide is treated with NaOH, is :

Which of the following order for basic strength of amines is correct?

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

The increasing order of basic strength for the following compounds is:
(I)

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

$CH_{3}-NH-CH_{3}$
(III)

$CH_{3}-NH-Ph$
(IV)

$CH_{3}-NH-CHO$

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

Explanation

Aliphatic amines are stronger bases than ammonia and aromatic amines are weaker bases than

$N{ H }_{ 3 }$ and

$3°>2°>1°\rightarrow$ basic strength in a gaseous phase.

Thus, the increasing order of basic strength from the options can be

$IV<III<II>I$

$II,III,IV$ are secondary amines.

$III$ is aromatic, hence is less basic than

$II, IV$ . The lone pair of amine resonates with the

$-CHO$ bond, hence it cannot be freely available, thus it is less basic.

Option C is correct.

The product

$U$ is:

Explanation

Benzoic acid reacts with thionyl chloride to form benzoyl chloride. Reaction with dinitrogen pentoxide introduces nitro group. Reaction with lithium dimethyl cuprate converts acid chloride to methyl ketone. The carbonyl group is protected with ethylene glycol, nitro group is hydrogenated and the protecting group is removed to form

$1-(4-aminophenyl)$

$ethanone$ . This is followed by diazotization and coupling with phenol to obtain the end product

$U$ .

Which compound does not give positive test in Lassaigne's test for nitrogen?

0%
Urea
0%
Hydrazine
0%
Azobenzene
0%
Phenyl hydrazine

Explanation

In Lassaigne's test, the organic compound is fused with sodium.

Carbon and nitrogen present in the organic compound gives sodium cyanide (

$NaCN$ ).

This is further used in the test for nitrogen.
Hydrazine

$(H_2N-NH_2)$ do not contain carbon. It cannot form

$NaCN$ on fusion with sodium.
Hence, hydrazine cannot give Lassaigne's test for nitrogen.

The correct order of basicity of the above compounds is given.
Identify the correct statement(s).

0%
In III, lone pair of electrons is involved in delocalisation but not in II.
0%
In IV, lone pair of electrons is involved in delocalisation but not in III.
0%
I is more basic than II because six membered ring is more stable than five membered ring.
0%
In III, lone pair of electrons is present in sp orbital but in I, II and IV lone pair of electrons are present in $sp^{2}$ orbital.

Aniline when diazotized in cold and then treated with dimethylaniline gives a coloured product. Its structure would be:

Explanation

Aniline when diazotized in cold I.e 0-5

$^{0}$ C give benzene diazonium chloride . When this compound coupled with dimethyl aniline it give coloured compound I.e p-(N,N-dimerhyle) amino azobenzene I.e azo dye

$3HC\equiv CH \xrightarrow {polymerisation} X \xrightarrow[conc.\ HNO_{3}]{conc.\ H_{2}SO_{4}} Y \xrightarrow {Sn/HCl}Z\xrightarrow[NaNO_{2},HCl]{(0-5^{0}C)} V\xrightarrow[Phenol]{dil\ NaOH}W$
Product W is :

Which of the following is weakest base?

0%
0%
0%
$H_2N-CH_2-NO_2$
0%
$CH_3NH-CH=O$

Which compound is soluble in water :

0%
$\displaystyle \left [ \left ( CH_{3} \right )_{2}NH_{2}\right ]^{+}Cl^{-}$
0%
$\displaystyle \left [ CH_{3}NH_{3} \right ]^{+}Cl^{-}$
0%
$\displaystyle \left [ \left ( CH_{3} \right )_{3}NH\right ]^{+}Cl^{-}$
0%
$\displaystyle PhNO_{2}$

Explanation

Nitro benzene is insoluble in water.
Mono, di, tri and tetra alkyl ammonium halides are soluble in water.
Thus

$\displaystyle \left [ \left ( CH_{3} \right )_{2}NH_{2}\right ]^{+}Cl^{-}, \displaystyle \left [ CH_{3}NH_{3} \right ]^{+}Cl^{-} and \displaystyle \left [ \left ( CH_{3} \right )_{3}NH\right ]^{+}Cl^{-}$ are soluble in water.

What is the principle product when aniline is treated with

$NaNO_2$ and

$HCl$ at

$0-5^{\circ}C$ , and this mixture is added to p-toludine (p-methyl aniline)?

The following product(s) will be obtained from the above reaction:

$\displaystyle -$ Ethyl pthalimide on hydrolysis gives:

0%
methyl alcohol
0%
ethyl amine
0%
dimethyl amine
0%
diethyl amine

Which compound will liberate

$\displaystyle CO_{2}$ from

$\displaystyle NaHCO_{3}$ solution :

0%
$\displaystyle CH_{3}CO\:NH_{2}$
0%
$\displaystyle CH_{3}NH_{2}$
0%
$\displaystyle \left ( CH_{3} \right )_{4}N^{+}OH^{-}$
0%
$\displaystyle CH_{3}N^{+}H_{3}Cl^{-}$

Explanation

$\displaystyle CH_{3}N^{+}H_{3}Cl^{-}$ will react with

$\displaystyle NaHCO_{3}$ to liberate

$\displaystyle CO_{2}$ .

$\displaystyle CH_{3}N^{+}H_{3}Cl^{-} + \displaystyle NaHCO_{3} \rightarrow CH_{3}NH_2 + NaCl + H_2O + \displaystyle CO_{2}$

Which of the following diazonium salt is relatively stable of

$0-5^\circ C$ - :

0%
$CH_{3}CH_2N\equiv N\left. \right \} ^\oplus Cl^{-}$
0%
$CH_{3}-C(CH_{3})_{2}-N\equiv N\left. \right \} ^\oplus Cl^{-}$
0%
$CH_{3}-N\equiv N\left. \right \} ^\oplus Cl^{-}$
0%
$(CH_{3})_{3}C-N\equiv N\left. \right \} ^\oplus Cl^{-}$

Explanation

The diazonium salt

$CH_{3}-N\equiv N\left. \right \} ^\oplus Cl^{-}$ obtained by treatment of methylamine with nitrous acid is relatively stable at

$0-5^\circ C$ .

Which of the following diazonium salt is relatively stable at

$\displaystyle 0-5^{\circ}$ C?

0%
$\displaystyle CH_{3}-N\equiv N^{\oplus }Cl^{-}$
0%
$\displaystyle CH_{3}CH\left ( CH_{3} \right )-N\equiv N ^{\oplus }Cl^{-}$
0%
$\displaystyle C_{6}H_{5}-N\equiv N ^{\oplus }Cl^{-}$
0%
$\displaystyle \left ( CH_{3} \right )_{3}C-N\equiv N ^{\oplus }Cl^{-}$

Explanation

Diazonium salts obtained from aliphatic primary amines are unstable. Those obtained from aromatic primary amines are relatively stable.
Benzene diazonium chloride

$(\displaystyle C_{6}H_{5}-N\equiv N ^{\oplus }Cl^{-})$ is obtained from aromatic primary amine. Hence, it is relatively stable.

Of the following statements :

$(P)\:C_{6}H_{5}N=CH-C_{6}H_{5}$ is a Schiff's base.

$(Q)\:$ A dye is obtained by the reaction of aniline and

$C_{6}H_{5}N^+ \equiv NCl$ .

$(R)\:C_{6}H_{5}CH_{2}NH_{2}$ on treatment with

$[NaNO_{2}+HCl]$ gives diazonium salt.

$(S)\:p-$ Toluidine on treatment with

$[HNO_{2}+HCl]$ gives diazonium salt.

0%
only $(P)\:and\:(Q)$ are correct
0%
only $(P)\:and\:(R)$ are correct
0%
only $(R)\:and\:(S)$ are correct
0%
$(P),\:(Q)\:$ and $\:(S)$ are correct

Explanation

A Schiff base is a substituted imine

$(R_2C=N-R')$ . It is used as antioxidant. Thus

$(P)\:C_{6}H_{5}N=CH-C_{6}H_{5}$ is a Schiff's base. Thus statment

$(P)$ is correct.

Aniline reacts with

$C_{6}H_{5}N^+\equiv NCl$ to form higly coloured azo dye. This reaction is known as azo coupling. Thus the statement

$(Q)$ is correct.

$\:C_{6}H_{5}CH_{2}NH_{2}$ is a primary aliphatic amine. When it is treated with

$[NaNO_{2}+HCl]$ , it gives unstable diazonium salt which decomposes to form benzyl alcohol. Hence, statement

$(R)$ is incorrect.

p-Toluidne is an aromatic primary amine. Hence, it on treatment with

$[NaNO_{2}+HCl]$ gives diazonium salt. Hence, statement

$(S)$ is correct.

Option D is correct.

Arrange the following in the increasing order of their basicities.
I. p - Toluidine
II. N, N - Dimethyl - p - toluidine
III. p - Nitroaniline
IV. Aniline

0%
III < IV < I < II
0%
II < IV < I < III
0%
III < II < I < IV
0%
I < IV < III < II

Explanation

N,N-Dimethyl p-toluidine is most basic whereas p-nitroaniline is least basic.
II +I effect of two Me groups of N atom and Me group at p position (+I and hyperconjugative. effects) are responsible for highest basicity.
I +I and H.C. effects of (Me) group at p position makes p-toluidine more basic than aniline.
IV (Standard) (Aniline)
III -I and -R effect of (-

$N{O}_{2}$ ) group at p position are responsible for least basicity.

Which one(s) is/are true?

0%
(I) and (III) are modest Bronsted bases whereas (II) is not
0%
In (III) ${N}^{a}$ is more basic than ${N}^{b}$
0%
When (II) is protonated in the presence of a strong acid, protonation occurs at $C - 2$
0%
All the nitrogen present in (I), (II), and (III) is $s{p}^{2}$ hybridised

Explanation

(I) and (III) are modest Bronsted bases whereas (II) is not
This is because, (II) is not protonated, hence the statement (A) is true.
In (II),

${N}^{b}$ is more basic due to the presence of a lone pair of electrons.
Hence, the statement (B) is wrong.
Statement (C): When (II) is protonated in the presence of a strong acid, protonation occurs at

$C - 2$ .
This is clear from the resonance structures as shown below.
Statement (C) is true.
Statement (D): Due to resonance all the

$N$ in I, II, and III is

$s{p}^{2}$ hybridized and hence the statement is true.

Arrange the following in their decreasing order of acidity.

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

Explanation

The decreasing order of acidity is

$III > IV > I > II$ .
In (III)

$+I$ effect of two

$Me$ groups makes it most basic
In (I) electron-withdrawing effect of the carbonyl group makes it less basic than ammonia.
In (II) electron-withdrawing effect of two (

$C = O$ ) groups makes it least basic.

Which of the following is the product of the reaction shown above?

0%
$(CH_3)_2CHCH_2CH_2NHNH_2$
0%
$[(CH_3)_2CHCH_2CH_2]_2NH$
0%
$(CH_3)_2CHCH_2CH_2NH_2$
0%
$(CH_3)_2CHCH_2CH_2CONH_2$

The decreasing order of basic characters of the following is :

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

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

0%
(I) and (II) are aromatic and have equal basic strength
0%
(I) is aromatic and (II) is anti-aromatic, but (II) is stronger base than (I)
0%
The order of basicity of the above compounds is (IV) > (III) > (II) > (I)
0%
The conjugate acid of (IV) is more stabilised than the conjugate acid of (II)

Explanation

In IV, lone pair of two N makes it more basic and does not delocalize in the benzene ring.
In III, no delocalisation of lone pair of electrons on

$\ddot { N } { H }_{ 2 }$ .
In II, no delocalization of lone pair of electrons on N.
In I, delocalization of lone pair of electrons on N via resonance.
Hence, the order of basic character is: IV > III > II > I.
Also The conjugate acid of (IV) is more stabilized than the conjugate acid of (II).
This is due to resonance in II which is absent in IV.
(I) and (II) are aromatic (follow Huckel's Rule) and have different basic strengths.
Thus, the options A and B are incorrect whereas options C and D are correct.

Identify the most basic amine among the above compounds :

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

Alkyl isocyanides (

$R N^{\bigoplus} \equiv C^{\ominus}$ ) are reduced to 2

$^{\circ}$ amines(

$R-NH-CH_3$ ) with:

0%
$LAH$
0%
$NaBH_4$
0%
$HI + P$
0%
$H_2 / Pt$

Explanation

Alkyl isocyanides

$(RN≡C)$ are reduced to

$2^o$ amines with sodium and alcohol. Alkyl isocyanides (RN≡C) are reduced to

$2 ^o$ amines with

$H_2 /Pt$ .

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 (

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

$\displaystyle pH = 4-6$ ).

Compound

$(E)$ is:

Explanation

D.U in

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

$C:H=1:1$ suggest that

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

$(Me)$ groups]. Since compound

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

$(A)$ is ortho-xylene

Compound

$(A)$ is:

0%
Ethylbenzene
0%
$p-Xylene$
0%
$m-Xylene$
0%
$o-Xylene$

Explanation

D.U in

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

$C:H=1:1$ suggest that

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

$(Me)$ groups]. Since compound

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

$(A)$ is ortho-xylene

Compound

$(F)$ is:

Explanation

D.U in

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

$C:H=1:1$ suggest that

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

$(Me)$ groups]. Since compound

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

$(A)$ is ortho-xylene

Compound

$(D)$ is:

Explanation

D.U in

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

$C:H=1:1$ suggest that

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

$(Me)$ groups]. Since compound

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

$(A)$ is ortho-xylene

The following reaction is:

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

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

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

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

$III. PhMe$

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

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

$\overline { e }$ donating substituent

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

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

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

$e^{-}$ withdrawing groups.

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

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

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

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

Stability of

$2^{\circ}$ carbocation

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

Hence

$\left(\mathrm{CH}_{3} \mathrm{CH}_{2}\right)_{2} \mathrm{NH}$ 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.

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

Which of the following compounds has the lowest boiling point?

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

Among the following, the strongest base is :

0%
$C_6H_5NH_2$
0%
$p-NO_2C_6H_4NH_2$
0%
$m-NO_2C_6H_4NH_2$
0%
$C_6H_5CH_2NH_2$

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

$H$ of

$R_2NH \ or \ RNH_2$ with

$N$ of other amine molecule.

Therefore, A and B are correct.

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

$H$ with

$N$ .

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.

CBSE Questions for Class 12 Engineering Chemistry Amines Quiz 9 - MCQExams.com

0:0:2

Practice Class 12 Engineering Chemistry Quiz Questions and Answers

CBSE Questions for Class 12 Engineering Chemistry Amines Quiz 9 - MCQExams.com

0:0:2

Practice Class 12 Engineering Chemistry Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.