MCQ Questions for Class 12 Engineering Chemistry Amines Quiz 2

Which of the following statements about primary amines is false?

0%
Alkyl amines are stronger bases than aryl amines
0%
Alkyl amines react with nitrous acid to produce alcohols
0%
Aryl amines react with nitrous acid to produce phenols
0%
Alkyl amines are stronger bases than ammonia

Explanation

Primary amines are those amines whose one hydrogen is substituted by some other groups like

$C{ H }_{ 3 }$ ,

$C{H}_{3}C{H}_{2}$ etc or those amines which have one

$C-N$ bond present in them. Example:

${ H }_{ 3 }C-N{ H }_{ 2 }\quad ,\quad { H }_{ 3 }C-C{ H }_{ 2 }-N{ H }_{ 2 }$

Aryl amines (if it is primary) on reaction with nitrous acid gives diazonium salt.

Which of the following has maximum

$pK_{b}$ value?

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

Explanation

The

$pK_{b}$ values of ethanamine, N-methylaniline, N-ethylethaneamine and aniline are 3.29, 9.30, 3.00 and 9.38 respectively.
Thus among the given amines, aniline is the weakest base and has maximum

$pK_{b}$ value.
Aliphatic amines ethanamine and N-methylaniline are more basic than aniline because in aniline, the lone pair on nitrogen is in resonance with benzene.
N-methylaniline is more basic than aniline because in N-methylaniline, the methyl group (an electron releasing group) increases the electron density on nitrogen.

What is the decreasing order of strengths of the following bases?

$\overset { \circleddash }{ O } H, N{ H }_{ 2 }^{ \circleddash }, HC\equiv { C }^{ \circleddash }$ and

$C{ H }_{ 3 } - C{ H }_{ 2 }^{ \circleddash }$

0%
$C{ H }_{ 3 } -C{ H }_{ 2 }^{ \circleddash } > N{ H }_{ 2 }^{ \circleddash } > H - C\equiv { C }^{ \circleddash } > \overset { \circleddash }{ O } H$
0%
$H -C\equiv { C }^{ \circleddash } > C{ H }_{ 3 } -C{ H }_{ 2 }^{ \circleddash } > N{ H }_{ 2 }^{ \circleddash }>\overset { \circleddash }{ O } H$
0%
$\overset { \circleddash }{ O } H > N{ H }_{ 2 }^{ \circleddash } > H - C\equiv { C }^{ \circleddash } > C{ H }_{ 3 } -C{ H }_{ 2 }^{ \circleddash }$
0%
$N{ H }_{ 2 }^{ \circleddash } > H -C\equiv { C }^{ \circleddash } > \overset { \circleddash }{ O } H > C{ H }_{ 3 } -C{ H }_{ 2 }^{ \circleddash }$

Explanation

The conjugate base of strong acid is weaker and conjugate base of weak acid is stronger. So, decreasing order of basic strength:

${ CH }_{ 3 }-{ CH }_{ 2 }^{ - }$ >

${ NH }_{ 2 }^{ - }$ >

${ CH }\equiv { C }^{ - }$ >

${ OH }^{ - }$

What is the product

$A$ ?

An organic compound

$(C_{3}H_{9}N)$ (A), when treated with nitrous acid, gave an alcohol and

$N_{2}$ gas was evolved.
The compound (A) on warming with

$CHCl_{3}$ and alcoholic caustic potash gave (C), which on reduction gave isopropylmethylamine.
Predict the structure of (A).

Give the decreasing order of basic strength of the following:
a.

$N{H}_{3}$
b.

$N{H}_{2}OH$
c.

$N{H}_{2}-N{H}_{2}$

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

Explanation

The decreasing order of basic strength is

$c > a > b$ or

$N{H}_{2}-N{H}_{2} > N{H}_{3} > N{H}_{2}OH$ .

$N{H}_{2}-N{H}_{2}$ is most basic as it has two lone pairs of electrons.
The basic strength of ammonia is lower than the basic strength of hydrazine as ammonia contains only one lone pair.
Hydroxyl amine is least basic due to

$-I$ effect of

$-OH$ group which decreases the ease with which lone pair of an electron can be donated.

Which one of the following on reduction with
lithium aluminium hydride yield a secondary
amine:

0%
Methyl Cyanide
0%
Nitroethane
0%
Methylisocyanide
0%
Acetamide

Indicate the following compound as

$1^o, 2^o$ and

$3^o$ .

$Et_3N$

0%
$1^o$
0%
$2^o$
0%
$3^o$
0%
Non of the above

Explanation

$C_2H_5 -\underset{\underset {\displaystyle {C_2H_5}}|}{N}-C_2H_5$

Triethylamine, is tertiary amines, as the three hydrogen atoms are replaced by ethyl substituents.

Option C is correct.

Which of the following is most basic?

0%
Diphenylamine
0%
Triphenylamine
0%
p-Nitroaniline
0%
Benzylamine

Explanation

Benzylamine

$C_{6}H_{5}CH_{2}NH_{2}$ is more basic because benzyl group is electron donating group due to +I effect. So it is able to increase electron density of N of -NH

$_{2}$ group.

Thus due to higher electron density rate of donation of a free pair of electron is increased I.e basic character is higher While phenyl and nitro group are electron withdrawing group so they are able to decrease the electron density of N of

$-NH_{2}$ group. Hence they are less basic

Consider the Hofmann ammonolysis reaction:

$R = {C}_{2}{H}_{5}$ , the correct order of basic character of the above amines in aqueous medium is :

0%
$2 > 3 > 1 > N{H}_{3}$
0%
$2 > 1 > 3 > N{H}_{3}$
0%
$3 > 2 > 1 > N{H}_{3}$
0%
$N{H}_{3} > 1 > 2 > 3$

Explanation

$R = {C}_{2}{H}_{5}$ , the correct order of basic character of the above amines in aqueous medium is

$2 > 3 > 1 > N{H}_{3}$ . The factors that determine basicity are the +I effect of alkyl groups and the solvation /hydration in aqueous medium.

Which of the following cannot be prepared by Gabriel phthalimide synthesis?

0%
Aliphatic primary amines
0%
Aromatic primary amines
0%
Cyclic primary amines
0%
None of the above

Consider the Hofmann ammonolysis reaction:

$R = {Me}_{3}C-$ (

$t-butyl$ ), the correct order of basic character of the above amines in aqueous medium is :

0%
$N{H}_{3} > 1 > 2 > 3$
0%
$3 > 2 > 1 > N{H}_{3}$
0%
$1 > 2 > N{H}_{3} > 3$
0%
$1 > N{H}_{3} > 2 > 3$

Explanation

$R = {Me}_{3}C-$ (t-butyl), the correct order of basic character of the above amines in aqueous medium is

$N{H}_{3} > 1 > 2 > 3$
The factors that determine basicity are the +I effect of alkyl groups and the solvation /hydration in aqueous medium.

Consider the Hofmann ammonolysis reaction:

For any value of

$R$ , the correct order of basic character of the above amines in gaseous phase of nonpolar solvent is :

0%
$3 > 2 > N{H}_{3} > 1$
0%
$3 > 2 > 1 > N{H}_{3}$
0%
$2 > 3 > N{H}_{3} > 1$
0%
$1 > 2 > 3 > N{H}_{3}$

Explanation

For any value of

$R$ , the correct order of basic character of the above amines in gaseous phase of nonpolar solvent is

$3 > 2 > 1 > N{H}_{3}$
This is because in gaseous or nonpolar solvent, the basic order is

$3 > 2 > 1 > N{H}_{3}$ .

Consider the Hofmann ammonolysis reaction:

$R = {Me}_{2}CH-$ (isopropyl), the correct order of basic character of the above amines in aqueous medium is :

0%
$3 > 2 > 1 > N{H}_{3}$
0%
$N{H}_{3} > 1 > 2 > 3$
0%
$1 > 2 > N{H}_{3} > 3$
0%
$1 > N{H}_{3} > 2 > 3$

Explanation

$R = {Me}_{2}CH-$ (isopropyl), the correct order of basic character of the above amines in aqueous medium is

$1 > N{H}_{3} > 2 > 3$ .
The factors that determine basicity are the +I effect of alkyl groups and the salvation /hydration in aqueous medium.

Which of the following statements are correct?

0%
The extent of H-bonding is greater in $1^0$ than $2^0$ and $3^0$ amines.
0%
The boiling points of isomeric amines are in the order: $1^0 > 2^0 > 3^0$ .
0%
The boiling points of
0%
The boiling points of $\displaystyle (I){ C }_{ 4 }{ H }_{ 9 }{ NH }_{ 2 }>(II){ Me }_{ 2 }N-Et>(III)({ C }_{ 2 }{ H }_{ 5 }{ ) }_{ 2 }NH$ .

Explanation

The order of boiling points of amines is:

$1^0 > 2^0 > 3^0$ .

1 amine forms more H-bonding with

$\displaystyle 2H$ than

$2^0$ amine

$\displaystyle (1 H)$ and than

$3^0$ (no hydrogen ) amine.

Options A, B and C are correct

The conjugate base of

$\displaystyle ({ CH }_{ 3 }{ ) }_{ 2 }{ NH }_{ 2 }^{ \oplus }$ is:

0%
$\displaystyle ({ CH }_{ 3 }{ ) }_{ 3 }N$
0%
$\displaystyle ({ CH }_{ 3 }{ ) }_{ 2 }NH$
0%
$\displaystyle ({ CH }_{ 3 }{ ) }_{ 2 }{ N }^{ \ominus }$
0%
$\displaystyle ({ CH }_{ 3 }{ ) }_{ 2 }{ N }^{ \oplus }$

Mendius reaction involves the:

0%
Reduction of aldehydes to give alcohols
0%
Reduction of nitriles with sodium and ethanol
0%
Oxidation of nitriles
0%
Hydrolysis of cyanides

Explanation

$\displaystyle R-C\equiv N\xrightarrow [ Mendius\quad reduction ]{ Na+EtOH } { RCH }_{ 2 }{ NH }_{ 2 }$

(A) is subjected to reduction with

$\displaystyle Zn-{ Hg }/{ HCl }$ and the product formed is N-methylmethanamine. (A) can be:

0%
Ethane nitrile
0%
Nitroethane
0%
Carbylaminoethane
0%
Carbylaminomethane

Explanation

Reduction with

$Zn-Hg/HCl$ is Clemmensen reduction

which converts

$R-C\equiv N\rightarrow RCH_2NH_2$ and

$R-\overset{\oplus}{N}\equiv \overset{\ominus}{C}\rightarrow RNHCH_3$

Carbylamino methane is

$\left ( {Me-\overset{\oplus}{N}\equiv \overset{\ominus}{C}} \right )$

$\therefore Me-\overset{\oplus}{N}\equiv\overset{\ominus}{C}\xrightarrow {Zn-Hg/HCl}\underset{N-Methylmethanamine}{Me-NH-CH_3}$

Option D is correct.

Which of the following statements is correct?

0%
Methyl amine is slightly acidic.
0%
Methyl amine is less basic than ammonia.
0%
Methyl amine is less basic than dimethyl amine.
0%
Methyl amine is less basic than aniline.

Explanation

Dimethyl amine (

$\displaystyle { Me }_{ 2 }NH$ ) is more basic than

$\displaystyle { MeNH }_{ 2 }$ , due to (+I) effect of two (Me) groups.

Diazo coupling is useful to prepare some:

0%
pesticides
0%
dyes
0%
proteins
0%
vitamins

Among the following, the strongest base is:

0%
$\displaystyle { C }_{ 6 }{ H }_{ 5 }{ NH }_{ 2 }$
0%
$\displaystyle p-{ NO }_{ 2 }{ C }_{ 6 }{ H }_{ 4 }{ NH }_{ 2 }$
0%
$\displaystyle m-{ NO }_{ 2 }{ C }_{ 6 }{ H }_{ 4 }{ NH }_{ 2 }$
0%
$\displaystyle { C }_{ 6 }{ H }_{ 5 }{ CH }_{ 2 }{ NH }_{ 2 }$

Explanation

(D) is an aliphatic amine; so it is stronger base than aromatic amine.

Moreover, EWG

$( -{ NO }_{ 2 } )$ at o, p and m-positions decreases the basic character.

The basicity order is:

$\displaystyle (D)>(A)>(C)>(B)$ .

Option D is correct.

Which one of the following is the strongest base in aqueous medium :

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

Explanation

$+I$ effect of

$C_2H_5$ group increases electron density on

$N$ and make it stronger base. So the order of basic strength would be

$3^o>2^o>1^o>NH_3$ but in aqueous medium presence of hydrogen bonding and bulkier groups affect the basicity. In the

$(C_2H_5)_3N$ , alkyl groups hinder the attack of proton on

$N$ . Therefore it become less basic. And due the additive effects of steric hindrance and induction

$(C_2H_5)_2NH$ is strongest base in aquous medium.

Which of the following amines will form stable diazonium salt at 273-283 K?

0%
$\displaystyle { C }_{ 2 }{ H }_{ 5 }{ NH }_{ 2 }$
0%
$\displaystyle { C }_{ 6 }{ H }_{ 5 }{ NH }_{ 2 }$
0%
$\displaystyle { C }_{ 6 }{ H }_{ 5 }{ CH }_{ 2 }{ NH }_{ 2 }$
0%
$\displaystyle { C }_{ 6 }{ H }_{ 5 }{ N\left( { CH }_{ 3 } \right) }_{ 2 }$

Which of the above is more acidic?

0%
I
0%
II
0%
Both have similar acidity
0%
None of the above

Arrange the following in correct activating order towards EAS.

$-\underset{I}{NHCOCH_3} \quad - O - \underset{II}{\overset{O}{\overset{||}{C}}} - CH_3 \quad - \underset{III}{\overset{\!\!\!\!\!+}{NR_3}}\quad - \underset{IV}{NH_2}$

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

Explanation

In Electrophilic aromatic substitution reaction, the electrons donating group is favoured over electron withdrawing group.

$-NH_2 > \overset{+}{-NR_3} > - O - \overset{O}{\overset{||}{C}} - CH_3 > - NHCOCH_3$

This is the required order of electrophilic aromatic substitution.

Which one is more basic,

$CH_3NH_2 \,\, or \,\, CF_3NH_2$ ?

0%
$CH_3NH_2$
0%
$CF_3NH_2$
0%
Both have similar basicity
0%
None of the above

Explanation

Between

${ H }_{ 3 }C\ddot { N } H_{ 2 },{ F }_{ 3 }C\ddot { N } { H }_{ 2 },{ H }_{ 3 }\ddot { N } { H }_{ 2 }$ is more basic than

$F_3C\ddot NH_2$ because

$-CH_3$ group is an electron donating group so shows posetive inductive effect and donates electron density to nitrogen which makes lone pair on nitrogen more destabilished and are available for reaction .but

$CF_3$ group is an electron density from nitrogen as it shows negative inductive effect and so lone pairs on nitrogen are not available for reactions.

Phenyl cyanide on reduction with

$Na/{C}_{2}{H}_{5}OH$ yields :

0%
${C}_{6}{H}_{5}{CH}_{2}{NH}_{2}$
0%
${C}_{6}{H}_{5}NH{CH}_{3}$
0%
0%
${C}_{6}{H}_{5}{NH}_{2}$

Explanation

Phenyl cyanide (benzonitrile) on reduction with sodium ethanol produces benzyl amine.

$\displaystyle -C \equiv N$ group is reduced to

$\displaystyle CH_2-NH_2$ group.

$\displaystyle Ph-C \equiv N \xrightarrow {Na/ ethanol} Ph-CH_2-NH_2$

Benzene diazonium chloride on reaction with phenol in weakly basic medium gives:

0%
diphenyl ether
0%
p-hydroxy azobenzene
0%
chlorobenzene
0%
benzene

Explanation

Benzene diazonium chloride on reaction with phenol in weakly basic medium gives p-hydroxy azobenzene.

$\displaystyle \underset {\text {Benzene diazonium chloride}}{C_6H_5-N\equiv N^+Cl^-}+\underset {\text {phenol}}{C_6H_5-OH} \xrightarrow {OH^-} \underset {\text {p-hydroxyazobenzene}}{C_6H_5-N=N-C_6H_4-OH}$

Option B is correct.

Tertiary butyl amine is a:

0%
${1}^{o}$ amine
0%
${2}^{o}$ amine
0%
${3}^{o}$ amine
0%
quaternary salt

Explanation

Tertiary butyl amine is a primary amine as it contains

$R-NH_2$ group.

If only one hydrogen of ammonia is replaced by one alkyl group then it is primary amine.

Option A is correct.

The following amine is:

0%
strongly acidic
0%
weakly acidic
0%
strongly basic
0%
weakly basic

Explanation

Explanation:

According to lewis's theory, the element which donates lone pair of electrons is basic.
When electrons of lone pair are get involved in resonance, the basicity decreases. because the electrons of the lone pair are not free to donate.

In the above image, the lone pairs of $N$ of $NMe_2$ are not involved in resonance due to steric hindrance.
Hence, it is more basic.

Hence, the correct answer is option $C$ .

How many primary amines can be formulated by

${C}_{3}{H}_{9}N$ and how many

${1}^{o}$ hydrogen are associated with carbon atoms of each compound?

0%
Two primary amines $[3,6]$
0%
One primary amine $[3]$
0%
Three primary amines $[3,6,6]$
0%
Two primary amines $[5,6]$

Explanation

Two primary amines, n propyl amine and iso propyl amine can be formulated with molecular formula

$C_3H_9N$
In n propyl amine, five primary hydrogens are attached with carbon atoms. In isopropyl amine, six primary hydrogen atoms are attached with carbon atoms.

Methyl cyanide on reaction with sodium and

${C}_{2}{H}_{5}OH$ forms:

0%
ethyl amine
0%
ethanoic acid
0%
ethanimine
0%
none of these

Explanation

Methyl cyanide on reaction with sodium and ethanol forms ethylamine.

$CH_3-C \equiv N +4[H] \xrightarrow {Na/ ethanol} CH_3-CH_2-NH_2$
In this reaction, the nitrile group is hydrogenated to form a primary amine. This is a useful method to increase the length of a carbon chain in preparing a primary amine which contains one more carbon atom than the alkyl halide from which alkyl cyanide is obtained.

Basic strength order:

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

Explanation

The decreasing order of basic strength is

$I > III> II$ .
Thus I is most basic and II is least basic. The lone pair on nitrogen atom of

$-NH_2$ group is in resonance with benzene ring. Due to this resonance, the electron density is maximum in ortho and para positions. Due to this, the basicity is maximum in I. The electron density is minimum in meta position. Hence, the basicity is minium in II. In the structure III, there is possibility of H bonding between ring nitrogen and the amino group. Hence, III is less basic than I.

Pyridine is less basic than triethylamine because :

0%
Pyridine has aromatic character
0%
Nitrogen in pyridine is $sp^2$ hybridised
0%
Pyridine is a cyclic system
0%
In pyridine, lone pair of nitrogen is delocalised

Explanation

Pyridine, is much less basic than typical aliphatic amines, but for a very different reason: the unshared pair is in an

$sp^2$ , which is much lower in energy than the electron pair of aliphatic amines, which is in an

$sp^3$ . Therefore, pyridine is less easily protonated than typical aliphatic amines such as triethylamine .

Arrange the following in increasing order of pH value

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

Write the basicity order of the following.

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

Explanation

1,2,3,4 is more basic as it contains two electron donating

$CH_3$ groups which destabilise the lone pair on nitrogen by donating more electron density due to + I effect and so lone pair are available easily 2 is having only one electron donating

$H_3C$ group so is less basic than 3 and 4 is least basic as it contains an electron .co group which stabilizes lone pair on nitrogen and therefore are not available for donation

How many primary amines are possible for the formula

$C_4H_{11}N$ ?

0%
One
0%
Two
0%
Three
0%
Four

Which of the following group will decrease the basicity of aniline when attached to its para position?

0%
$-OH$
0%
$-OC{H}_{3}$
0%
$-C{H}_{3}$
0%
$-N{O}_{2}$

Which of the following is more basic than aniline?

0%
p-nitroaniline
0%
Benzyl amine
0%
Diphenyl amine
0%
Triphenyl amine

Explanation

Although both are primary amines, the nitrogen lone pair of electrons are involved in resonance with the aromatic benzene and are less available to accept a proton whereas benzylamine is an aliphatic amine with a larger

$K_b$ than aniline.

So, Option B is correct.

Acetonitrile on reduction gives:

0%
propanamine
0%
methanamine
0%
ethanamine
0%
none of these

Explanation

Cyanides on reduction give

$1$ amine.
1142090_460287_ans_0411db8b614f4750a94a4bfaca9714cc.jpg

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

Which of the following method(s) does work for the preparation of secondary amine?

0%
$R_3N \xrightarrow[\begin{matrix} (ii) H_3O^+\\(iii) \Delta\end{matrix}]{(i) Br-CN}?$
0%
$R-\overset{O}{\overset{||}{C}}-NH - R' \xrightarrow{LiAlH_2}?$
0%
$\displaystyle RNC\xrightarrow[\Delta ]{Nl/H_{2}}?$
0%
All of these

Explanation

(A) Tertiary amines react with cyanogen bromide to form a dialkylcyanamide which, on hydrolysis, gives a secondary amine.

$R_3N + BrCN \rightarrow [R_3NCN]^+Br^- \rightarrow RBr + R_2NCN \overset {H^+}{\rightarrow} R_2NH$

(B) Secondary amides

$(RCONHR)$ react with lithium aluminium hydride (LiAlH4) to form secondary amines

$(RCH_2NHR')$

$R\overset{\overset{O}{||}}{C}NHR + H-Al^-H_3 \rightarrow R\overset {\overset{O^-}{|}}{C}=NR \rightarrow RCH_2NHR$

$RNC \overset {Ni / H_2}{\rightarrow} RNHCH_3$

Arrange the following in the decreasing order of boiling point:

${ C }_{ 2 }{ H }_{ 5}OH$

ii)

$(CH_3)_2NH$

iii)

$C_2H_5NH_2$

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

Explanation

All the given compounds have intermolecular hydrogen bonding due to which their boiling points are higher.

$C_2H_5OH$ possesses the strongest hydrogen bonding because the

$O$ is more electron negative than

$N$ in other compounds. Therefore it has the highest boiling point.

$(CH_3)_2NH$ and

$C_2H_5NH_2$ , the latter has a longer alkyl chain which makes hydrogen bonding weaker.

Hence the order of boiling points is

$i>ii>iii$

Arrange the following in the increasing order of boiling point:

(i) $C_2H_5OH$

(ii) ${(CH_3)_2}NH$

(iii) $C_2H_5NH_2$

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

Explanation

All the given compounds have intermolecular hydrogen bonding due to which their boiling points are higher.

$C_2H_5OH$ possesses strongest hydrogen bonding because the

$O$ is more electron negative than

$N$ in other compounds. Therefore it has the highest boiling point. In

$(CH_3)_2NH$ and

$C_2H_5NH_2$ , the latter has longer alkyl chain which makes hydrogen bonding weaker. Hence the order of boiling points is

$iii<ii<i$

Arrange the following in the decreasing order of solubility in water :

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

ii)

$({ C }_{ 2 }{ H }_{5 }{ ) }_{ 2 }NH$

iii)

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

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

Explanation

Ethylamine when added to water forms intermolecular H−bonds with water. And therefore it is most soluble in water.

Diethylamine having a bigger alkyl chain forms weaker intermolecular hydrogen bonds.

But aniline does not form H−bond with water to a very large extent due to the presence of a large hydrophobic

$−C_6H_5$ group. Hence, aniline has the least solubility in water.

Hence the order of solubility is

$iii>ii>i$

Which of the following is the most stable diazonium salt?

0%
$C_6H_5CH_2N_2^+X^-$
0%
$CH_3N_2^+X^-$
0%
$CH_3CH_2N_2^+X^-$
0%
$C_6H_5N_2^+X^-$

Explanation

Aliphatic amines do not give diazo compounds. The most stable diazonium salt is

$C_6H_5 N_2^+ X^-$ . Diazonium salt containing aryl group directly linked to the nitrogen atom is most stable due to resonance stabilization between the benzene nucleus and N-atom.

Arrange the following in the increasing order of solubility in water:

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

ii)

$({ C }_{ 2 }{ H }_{5 }{ ) }_{ 2 }NH$

iii)

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

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

Explanation

The decreasing order of solubility in water; is iii)

${ C }_{ 2 }{ H }_{ 5}{ NH }_{ 2 }$ > ii)

$({ C }_{ 2 }{ H }_{5 }{ ) }_{ 2 }NH$ > i)

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

The extent of

$H$ bond formation in iii)

${ C }_{ 2 }{ H }_{ 5}{ NH }_{ 2 }$ is maximum as it has two

$H$ attached to

$N.$ Hence, it has highest solubility.

$\displaystyle { C}_{ 6 }{ H }_{ 5 }{ NH }_{ 2 }$ solubility is least due to large hydrophobic part.

So, the correct option is

$A.$

How many primary amines are possible with the formula

$\displaystyle { C }_{ 4 }{ H }_{ 11 }N$ ?

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

Explanation

The possible primary amine with the formula

$\displaystyle { C }_{ 4 }{ H }_{ 11 }N$ are

(i)

$\displaystyle { CH }_{ 3 }{ CH }_{ 2 }{ CH }_{ 2 }{ CH }_{ 2 }{ NH }_{ 2 }$

(ii)

$\displaystyle { CH }_{ 3 }\overset { \overset { { CH }_{ 3 } }{ | } }{ CH } -{ CH }_{ 2 }-{ NH }_{ 2 }$

(iii)

$\displaystyle { CH }_{ 3 }-\overset { \overset { { CH }_{ 3 } }{ | } }{ \underset { \underset { { NH }_{ 2 } }{ | } }{ C } } -{ CH }_{ 3 }$

(iv)

$\displaystyle { CH }_{ 3 }-{ CH }_{ 2 }-\underset { \underset { { CH }_{ 3 } }{ | } }{ CH } -{ NH }_{ 2 }$

Gabriel phthalimide reaction is used for the preparation of :

0%
Primary aromatic amines
0%
Secondary amines
0%
Primary aliphatic amines
0%
Tertiary amines

Explanation

$Explanation:$

Gabriel phthalimide reaction is used to prepare primary amines from phthalimide the reaction steps are shown as below;

$Step\ 1:$ Potassium Phthalimide when treated with a primary alkyl halide gives N-alkylphthalimide.

$Step\ 2:$ N-alkylphthalimide with aqueous sodium hydroxide gives primary amine and sodium phthalate.

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Therefore, aliphatic primary amine is the correct answer.

$Final\ Answer:$ Option

$C$ .

Methyl amine reacts with methyl iodide. For completion of reaction, how many moles of methyl iodide are required?

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

Explanation

$\therefore$ 3 moles of

$C{H}_{3}I$ (methyl iodide) are required for reaction with methyl amine.

The correct answer is 3.

1056223_460299_ans_81226510599f4361933f1a5b351087c4.jpg

CBSE Questions for Class 12 Engineering Chemistry Amines Quiz 2 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

CBSE Questions for Class 12 Engineering Chemistry Amines Quiz 2 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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