MCQ Questions for Class 12 Engineering Chemistry Amines Quiz 4

Which of the following compound is amphoteric in nature?

0%
0%
$CH_3COOH$
0%
0%
$CH_3CONH_2$

Explanation

Due to presence of lp of electrons over

$N-$ atom of amide and having resonance structure, amides are amphoteric in nature.

Which of the following is more basic than aniline ?

0%
Diphenylamine
0%
Triphenylamine
0%
p-nitroaniline
0%
Benzylamine

Explanation

Benzyl amine,

$C_6H_5CH_2\ddot{N}H_2$ is more basic that aniline because be group

$C_6H_5CH_2$ electron donating group due to

$+I$ effect. So, It is able to Increase the electron of N of

$-NH_2$ group. Thus, due to higher electron density, rate of donation of free pair of electron is increased i.e., basic character is higher. Phenyl and nitro groups are electron attractive group. So, they are able to decrease the electron density of

$N$ of

$\ddot{N}H_2$ group. Hence, they are less basic with aniline.

The strongest base among the following is :

0%
$C_6 H_5 NH_2$
0%
$(C_6 H_5 )_2 NH$
0%
$NH_3$
0%
$(C_2 H_5)_2 NH$

Explanation

$(C_2 H_5)_2 NH(2^o$ amine) is the strongest base. Basic nature of amines is due to the presence of lone pair of electrons on nitrogen atom which is available for the bond formation with Lewis acid. Due to the

$+I$ effect,

$2^o$ amine is better base than

$1^o$ amine and

$NH_3.$ In case of aromatic amines, the lone pair on nitrogen atom involved in resonance, therefore, not available for bond formation, so aromatic amines are less basic.

Compare boiling point of isomeric alkyl amines.

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

Explanation

Among isomeric amines, the primary and secondary amines have highest boiling point due to their large tendency to form hydrogen bonds whereas tertiary amines have a lowest boiling point because of their inability to form hydrogen bond due to the absence of hydrogen atom.

Therefore, the order of boiling point of isomeric amine is as follows

${ 1 }^{ o }>{ 2 }^{ o }>{ 3 }^{ o }$ amines

$C_3H_9N$ reacts with Hinsberg reagent and the product is insoluble in alkali but soluble in ether. This nitrogen containing compound is:

0%
primary amine
0%
secondary amine
0%
tertiary compound
0%
nitrite compound

Explanation

The compound with molecular formula

$C_3H_6N$ is a secondary amina. When it reacts with Hinsberg reagent, the product obtained is soluble in ether.

In case of substituted aniline the group which decreases the basic strength is :

0%
$- OCH_3$
0%
$-CH_3$
0%
$-NH_2$
0%
$-C_6H_5$

Explanation

In case of substituted aniline the group which decreases the basic strength is phenyl group

$\displaystyle (C_6H_5)$ . It withdraws the electron density from the ring through resonance.

$P$ is

0%
0%
0%
0%
None of these

Which compound is more basic ?

0%
0%
0%
0%
all of them are equally basic

Explanation

Basically increases with increase in electron releasing group. So (Refer image) is more due to high

$+H$ effect ( hyperconjugation effect )
1349897_698738_ans_8365d0289760469ba2593d00eb59ef1c.PNG

The basic character of amines is due to:

0%
tetrahedral structure
0%
presence of nitrogen atom
0%
lone pair of electrons on nitrogen atom
0%
high electronegativity of nitrogen

When ammonium cyanate is heated for a long time, the product is:

0%
Nitrous oxide
0%
Ammonium cyanide
0%
Nitrogen and water
0%
Biuret

Explanation

When ammonium cyanate is heated urea is first formed

$\underset { Ammonium\\ Cyanate }{ { NH }_{ 4 }CNO } \xrightarrow [ ]{ Heat } \underset { urea }{ { NH }_{ 2 }CO{ NH }_{ 2 } } \quad \quad$

Urea on further heating gives biuret

${ NH }_{ 2 }CO{ NH }_{ 2 }\xrightarrow [ ]{ Heat } \underset { Biuret }{ { NH }_{ 2 }CONHCO{ NH }_{ 2 } } +{ NH }_{ 3 }$

Choose the false statement.

0%
$-NH_2$ is strongly activating while $-NHCOCH_3$ is activating
0%
$-NO_2$ is strongly deactivating while $-CHO$ is deactivating
0%
$-NH\underset{\underset{\displaystyle O}{||}}{C}R$ and $-CH=O$ are m-directing
0%
$-CF_3$ and $-\underset{\underset{\displaystyle O}{||}}{C}R$ are m-directing

Compound

$C$ in the above reaction is:

Which nitrogen is maximum basic?

0%
$1$
0%
$2$
0%
$3$
0%
All are equally basic

Explanation

The more basic nitrogen is "2" because the nitrogen at '1' is involved in resonance with the benzene ring and nitrogen at '3' is in resonance with carbonyl carbon and is slightly

$sp^2$ .

But nitrogen at '2' position is

$sp^3$ hybridized and hence is more basic.

Hence the correct option is B.

Half-life period of a radioactive element is 100 yr. How long will it take for its 93.75% decay ?

0%
400 yrs
0%
300 yrs
0%
200 yrs
0%
193 yrs

Explanation

Given : Half-life period of a radioactive element is 100 yr. How long will it take for its 93.75% decay ?

Solution :

Since it is decayed by 93.75 % so left is 6,25%

That is

$\frac {1}{16}$ of total element

$\frac{1}{2^n}$ =

$\frac{1}{16}$ where n is number of half life.

n=4

To decay 93.75% time needed is

$4\times half -life$ =

$4 \times 100$ = 400 yr

The Correct Opt = A

What is the molarity of 0.2 N Na2 CO3 solution ?

0%
0.1M
0%
0M
0%
0.4M
0%
0.2M

Explanation

Given: Normality = 0.2N

Solution: Molarity refers to the number of moles dissolved per litre of the solution.

As we know that,

$Normality= molarity\times n$ (1)

where n is the valency factor.

For

$Na_{2}CO_{3}$ n=2.

From equation 1 we get,

$molarity= \frac{Normality}{n}$

Substituting the values we get,

$molarity= \frac{0.2}{2}$

molarity = 0.1M

Hence, the correct optionis (A).

In the above reaction, the product

$B$ is:

0%
0%
0%
0%
none of the above

Which of following statements is/are the not correct?

0%
$II$ is more soluble than $I$ in ${H}_{2}O$
0%
$II$ is more soluble than $III$ in ${H}_{2}O$
0%
$II$ is less soluble than $III$ in ${H}_{2}O$
0%
$III$ is more soluble than $I$ in ${H}_{2}O$

Explanation

Among the given compounds, compound (I) is very less soluble in

$H_2O$ because of large hydrophobic part, i.e. Carbon ring presence.

Compound (II) is less soluble than compound (III) because there is more H-bonding in case of compound (III).

So, statement (B) is not correct.

What is the name for red azo dye?

0%
p-hydroxy azo benzene
0%
$\beta$ - nathyl azo benzene
0%
p-amino azo benzene
0%
p- N, N dimethyl amino azo benzene

The correct statement among the following is:

0%
boiling point of amine is more than that of alcohol of nearly same molecular mass
0%
the solubility of amines in water increases with increase in molecular mass
0%
the order of boiling point of isomeric amines is primary > secondary > tertiary
0%
none of these

A definite amount of solid

${ NH }_{ 4 }HS$ is placed in a flask already containing ammonia gas at a certain temperature and

$0.50atm$ pressure.

${ NH }_{ 4 }HS$ decomposes to give

${ NH }_{ 3 }$ and

${ H }_{ 2 }S$ and at equilibrium total pressure in flask is

$0.84atm$ . The equilibrium constant for the reaction is:

0%
$0.30$
0%
$0.18$
0%
$0.17$
0%
$0.11$

The following sequence of reaction on

$"A"$ gives.

The correct order of basic strength of the following compounds is :

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

Explanation

Higher the electron density on electron donating group higher is the basic strength.

In compound

$3$ the electron density is highest because of

$+\ I$ effect of

$\text {methyl}$ groups attached to nitrogen. Hence it has the highest basicity.

In compound

$1$ there are two amine groups attached to the carbon in which one of the amine group is always available for donation. Hence it has the second highest basicity.

In compound

$2$ again the ethyl group increases the basic strength of amine due to its

$+\ I$ effect but lower than that of as in compound

$3$ .

The compound

$4$ has the least basic strength because the carbonyl group attached to amine decreases the basicity by withdrawing the lone pair from Nitrogen through resonance effect.

The correct order of basic strength is-

$3 > 1 > 2 > 4$

The correct order of increasing basicity of the given conjugate bases (R =

$CH_{3}$ ) is:

0%
$RCO\bar{O} < HC \equiv \bar{C} < \bar{R} < \bar{N}H_{2}$
0%
$\bar{R} < HC \equiv \bar{C} < RCO\bar{O} < \bar{N}H_{2}$
0%
$RCO\bar{O} < NH_{2} < HC \equiv \bar{C} < \bar{R}$
0%
$RCO\bar{O} < HC \equiv \bar{C} < \bar{N}H_{2} < \bar{R}$

Explanation

The order of acidity can be explained on the basis of the acids of the given conjugate base

$.$ Stronger is the acid

$,$ weaker is the conjugate base

$.$ Since

$,$

$RCOO$ is the stronger acid amongst all

$,$

$RCO{O}^{-}$ is the weakest base

$.$ Due to

$sp$ hybridised carbon

$,$ acetylene is also acidic and hence a weak base but stronger than

$RCO{O}^{-}.$ As

${sp}^{3}$ carbon is less electronegetive than

${sp}^{3}$ nitrogen

$,$

${R}^{-}$ is more basic than

$N{H}_{2}.$

What is the correct order of basicity values of the following compounds?

0%
0%
$NH_3\,>\,CH_3NH_2\,>\,CH_2\,=\,CH-NH_2\,>\,NH_2-CH_2-CH_2-N^+R_3$
0%
0%
None

Explanation

$1.$

$NH_3\longrightarrow pk_a=34.5$

$2.$

$CH_3NH_4Br\longrightarrow pk_a=3.4$

$3.$

$CH_3NH_2\longrightarrow pk_a=10.6$

$4.$

$\begin{matrix} Me \\ Me \end{matrix}>\overset { + }{ S } - { CH }_{ 2 }-{ NH }_{ 2 } \rightarrow { pk }_{ a }=\sim 2$

Order is

$4>2>3>1$

What would be the product for the following reaction ?

0%
Ph-Cl
0%
$Ph-N_2$
0%
0%

Explanation

REF.Image.

$Na\, NO_{2} + HCl \rightarrow Nacl+ HNO_{2}$

$HO -N = O + H^{+} \Rightarrow H_{2}O^{\oplus }-N =0$

$H_{2}O^{\oplus}-N = 0 \rightarrow H_{2}O+N^{\oplus} =0$

1155283_877087_ans_5eb65e9b33b64dc58727dc6a66933f4d.png

The correct order of basicity of the following compounds is?

$\underset {(3)}{(CH_{3})_{4}NH}$

$\underset {(4)}{CH_{3}CONH_{2}}$

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

Explanation

$\begin{array}{l} \text { (4) lone pair is on resonance. } \\ \text { Hence, basisty order is } \\ \qquad 1>2>3>4 \end{array}$
2005258_877302_ans_3591d1dd86274f4287b1eab8748d8449.PNG

2005258_877302_ans_3591d1dd86274f4287b1eab8748d8449.PNG

Which of the following is not a primary amine?

0%
tert - Butylamine
0%
sec - Butylamine
0%
Isobutylamine
0%
Dimethylamine

Explanation

Dimethylamine

$[(CH_3)_2 NH]$ is a secondary amine as the central Nitrogen atom is substituted by two methyl groups and the nitrogen has one hydrogen. Tert-butyl amine, secondary butyl amine, isobutylamine are all the isomers of n-butyl amine, varying in the position of the

$N-$ group in the alkyl chain.

Secondary amines can be prepared by:

0%
reduction of nitro compounds
0%
reduction of amides
0%
reduction of isonitriles
0%
reduction of nitriles

Explanation

Isonitriles on reduction with lithium aluminium hydride

$(LiAlH_4)$ or catalytic hydrogenation (

$H_2/Ni$ ) produce secondary amines.

$R-NC \xrightarrow{LiAlH_4, H_2/Ni} R-NH-CH_2$

$C_3H_9N$ cannot represent:

0%
$1^o$ amine
0%
$2^o$ amine
0%
$3^o$ amine
0%
quaternary ammonium salt

Explanation

${C}_{3}{H}_{9}N$ can not represent a quaternary salt

$.$ Because it has only three carbon atom

$.$

Among the following which statement (s) is/are correct?

0%
Both N of pyrimidine are same basic strength.
0%
In imidazole protonation take places on N-3.
0%
In purine only one lone pair of N is delocalized.
0%
Pyrimidine, imidazole and purine all are aromatic.

Explanation

Pyrimidine has nitrogen atoms at 1 and 3 positions.

In purine only one lone pair of

$N$ is delocalized. But lone pair electron does not take place in delocalization and imidazole also.

Which one of the following is aromatic amine?

0%
0%
0%
0%
Both $A$ and $B$

Explanation

(A) Aniline is aromatic as it obeys Huckel's rule of aromaticity. It is aromatic.

(B) This is also followed Huckel's rule of aromaticity

$(4n+2)\pi e^-$

$=6\pi e^-(n=1)$

It is also planar, cyclic and conjugated system.

So both A and B are aromatic amines.

While option C is aliphatic amine since the

$NH_2$ group attached to aliphatic carbon.

Option D is correct.

Identify

$X,\ Y$ and

$Z$ in the given reaction.

$CH_2 \, =\, CH_2 \xrightarrow [CCl_4]{Br_2} \,X\, \xrightarrow[(2 \, moles)]{NaCN} \, Y \, \xrightarrow {LiAlH_4} \, Z$

0%
$X \,=\, CH_2BrCH_2Br \, ,\, Y \, =\, CH_3CH_2CH_2CN$
0%
$X \, =\, CH_2BrCH_2Br \, , \, Y \, =\, CH_3CH_2CN$
0%
$X\, =\, CH_3CH_2Br \, ,\, Y\, = \, CH_3CH_2CN$
0%
$X \, =\, CH_2BrCH_2Br \, ,\, Y \,=\, NCCH_2CH_2CN$

Explanation

$CH_2 = CH_2 \xrightarrow [CCl_4]{Br_2} \underset{(X)}{BrCH_2 - CH_2Br} \xrightarrow {NaCN} \underset {(Y)}{NCCH_2CH_2CN} \xrightarrow {LiAlH_4} \underset{(Z)} {\underset {(1,4-Diaminobutane)}{H_2NCH_2CH_2CH_2CH_2NH_2}}$

Tertiary amines have lowest boiling points amongst isomeric amines because:

0%
they have highest molecular mass
0%
they do not form hydrogen bonds
0%
they are more polar in nature
0%
they are most basic in nature

Which of the following explode on heating?

0%
Azides
0%
Sulphates
0%
Chlorides
0%
Phosphides

Gabriel phthalimide reaction is used for the preparation of __________ amines.

0%
primary aromatic
0%
secondary
0%
primary aliphatic
0%
tertiary

The correct order of boiling points of the following isomeric amines is:

$C_4H_9NH_2,(C_2H_5)_2NH,\, C_2H_5N(CH_3)_2$

0%
$C_2H_5N(CH_3)_2 \, >\, (C_2H_5)_2NH \, >\, C_4H_9NH_2$
0%
$(C_2H_5)_2NH\, >\, C_2H_5N(CH_3)_2 \,>\, C_4H_9NH_2$
0%
$C_4H_9NH_2 \, >\, (C_2H_5)_2NH \, >\, C_2H_5N(CH_3)_2$
0%
$(C_2H_5)_2 NH \, >\, C_4H_9NH_2 \, >\, C_2H_5N(CH_3)_2$

Explanation

Order of Boiling points of amines:

Primary amines (

$RNH_2$ ) > Secondary amines (

$R_2NH$ ) > Tertiary amines (

$R^{'}NR_2$ ).

Intermolecular hydrogen bonding is more in primary amines than that in secondary amines and nil in tertiary amines as there are no hydrogens attached to the nitrogen atom.

$C_4H_9NH_2 > (C_2H_5)_2NH > C_2H_5N(CH_3)_2$

Which of the following Diazonium salt is stable at room temperature?

Which of the following compounds will not undergo azo coupling reaction with benzene diazonium chloride?

0%
Aniline
0%
Phenol
0%
Anisole
0%
Nitrobenzene

Explanation

Diazonium cation is a weak electrophile and hence it reacts with electron-rich compounds containing electron-donating groups such as

$-OH$ ,

$-NH_2$ and

$OCH_3$ groups and not with compounds containing electron-withdrawing groups such as

$-NO_2,$ etc.

Hence, option D is correct.

Which of the following is the weakest Bronsted base?

0%
0%
0%
0%
$CH_3NH_2$

Explanation

Aniline is the weakest Bronsted base due to delocalisation of lone pair of electron of

$N-$ atom into benzene ring.

The correct decreasing order of basic strength of the following species is:

$H_2O, \, NH_3, \, OH^-, \, NH_2^-$

0%
$NH_2^- \, > \, OH^- \,>\, NH_3 \, > \, H_2O$
0%
$OH^- \, >\, NH_2^- \,> \, H_2O \, > \, NH_3$
0%
$NH_3 \, > \, H_2O \, > \, NH_2^- \,> \, OH^-$
0%
$H_2O \, >\, NH_3 \, > \, OH^- \, > \, NH_2^-$

Explanation

$NH_3$ is more basic than

$H_2O$ , therefore

$NH^-_2$ (Conjugate base of weak acid

$NH_3$ ) is a stronger base than

$OH^-$ .Thus decreasing order of basic strength is

$NH^-_2$ >

$OH^-$ >

$NH_3$ >

$H_2O$ ,

Correct energy profile for amine inversion and hybridization of nitrogen in transition state is:

Explanation

Amine is

$sp^3$ hybridised and the transition state in

$Sp^2$ hybridised as the lone pair enters the nucleus leaving behind a vacant orbital. And after sometime, the lone pair popes out of the either side, since the inverted amine is an enantiomer, the

${\text {energy remains the same}}$ .
Option D is the correct curve.
1082173_988124_ans_69c782d57968412384695189dbf8a8f9.png

In order to prepare a 1 amine from an alkyl halide with simultaneous addition of one

$CH_2$ group in the carbon chain, the reagent used as source of nitrogen is:

0%
Sodium amide $NaNH_2$
0%
Sodium azide $NaN_3$
0%
Potassium cyanide $KCN$
0%
Potassium phthalimide $C_6H_4(CO)_2N^-K^+$

Explanation

$4R-X \xrightarrow{KCN} R-CN \xrightarrow {Na\ / C_2H_5OH} R-CH_2NH_2$

Basic strength of different alkyl amines depends upon:

0%
$+I$ effect
0%
steric effect
0%
solvation effect
0%
all of these

Explanation

The basic strength of different alkyl amines depends on many factors like:

$+ I$ effect is the polarization of a sigma bond due to electron donating effect of adjacent groups or atoms.

2) In water, the ammonium salts of primary and secondary amines undergo solvation effects due to hydrogen bonding to a much greater degree than ammonium salts of tertiary amines.

3) steric effect is the effect faced by incoming group or hydrogen by the already present bulky

$-R$ groups on the Nitrogen atom.

Which of the following has highest

$pK_b$ value?

0%
$(CH_3)_3CNH_2$
0%
$NH_3$
0%
$(CH_3)_2NH$
0%
$CH_3NH_2$

Explanation

Higher the basicity, lower is the

$pK_b$ value.

Since

$NH_3$ is the weakest base, hence it has highest

$pK_b$ value.

Option B is correct.

Among the following amines, the strongest Bronsted base is:

0%
0%
$NH_3$
0%
0%

Explanation

Option

$D$ is the strongest Bronsted base as there is no delocalisation of lone pair of electron of

$N$ atom which is not possible in aniline and in pyrrole.

The correct increasing order of basic strength for the following compounds is:

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

Explanation

Electron donating

$-CH_3$ group increases the basicity while electron withdrawing

$NO_2$ group decreases the basicity of amines.

Option D is correct.

Which of the following cyclic amine has lowest

$\Delta G$ for inversion?

Explanation

More is the steric hindrance the N of amine, lesser is the

$\Delta G$ for inversion because it has the least tendency to invert)

Refer to Image

Most steric hindrance is in C. Therefore it has lowest

$\Delta G$ for inversion.

1082455_989425_ans_f311672c35dc44fba56b4e0b152d44ad.png

Which (isomeric) amine has lowest boiling point?

0%
$1^o$ amine
0%
$2^$ amine
0%
$3^o$ amine
0%
cannot predict

Explanation

$3^o$ amine, there is no hydrogen directly attached to nitrogen, due to which hydrogen bonding is not possible in them. Due to this absence of hydrogen bonding, less energy is required for boiling leading to lower boiling point.

In sets a - d, only one of the set is incorrect regarding basic strength. Select it:

Explanation

The basic character of amines is because of the lone pair of nitrogen. More the availability of lone pair, more will be the basic strength.

In compound as shown in IMAGE

$01$ , the lone pair of nitrogen is involved in making the ring aromatic and hence is less available due to which this compound will be least basic.

The compound as shown in IMAGE

$02$ will be the most basic due to presence of

$2$ nitrogen i.e.,

$2$ lone pairs instead of

$1$ increase the basic strength.

The compound as shown in IMAGE

$03$ will be intermediate between theses compounds.

This order is not followed in option

$C$

1056220_989357_ans_9a9f087cc08b49b393bf9ed3e1ab3efb.PNG

Its basic strength is

$10^{10}$ more than 1-dimethyl amino naphthalene due to:

0%
resonance
0%
steric inhibitation of resonance
0%
ortho effect
0%
hyperconjugation

Explanation

For participation in resonance, the

$p-orbital$ of

$N$ should be perpendicular to the plane of the ring.
Presence of two bulky methyl groups does not allow the

$p-orbital$ to get perpendicular to the plane and hence the lone pair of

$N$ does not involve in resonance.
This increases the lone pair availability, thus increasing the basic strength.
Hence,

$1,8-bis(dimethylamino)$ napthalene is more basic than

$1-dimethyl amino$ napthalene due to steric inhibition of resonance.

CBSE Questions for Class 12 Engineering Chemistry Amines Quiz 4 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

CBSE Questions for Class 12 Engineering Chemistry Amines Quiz 4 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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