MCQ Questions for Class 12 Medical Chemistry Organic Compounds Containing Nitrogen Quiz 11

Aniline is reacted with bromine water and the resulting product is treated with an aqueous solution of sodium nitrite in presence of dilute hydrochloric acid. The compound so formed is reacted with a tetrafluoroborate which is subsequently heated. The final product is:

0%
$p-bromoaniline$
0%
$p-bromofluorobenzene$
0%
$1, 3, 5-tribromobenzene$
0%
$2, 4, 6 - tribromofluorobenzene$

Gabriel synthesis is used for synthesis of:

0%
primary amines
0%
secondary amines
0%
aldehydes
0%
acids

Gabriel phthalimide synthesis is used in the preparation of primary amine from phthalimide which of the following reagent is not used during the process?

0%
HCl
0%
NaOH
0%
Alkyl Halides
0%
KOH

Which one of the following will be most basic?

0%
Aniline
0%
P-methoxyaniline
0%
P-nitroaniline
0%
Benzylamine

The correct order of increasing basic nature for the bases

$NH_3, CH_3NH_2$ and

$(CH_3)_2NH$ in aqueous solutions.

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

Explanation

In aqueous solution, the correct order of increasing basic nature for the basicity is:

$\displaystyle NH_3 < CH_3NH_2 <(CH_3)_2NH$ .

It is due to combination of the inductive effect, solvation effect, steric effect etc.

Which of the following is an example of carbylamine reaction?

0%
$2PhNH_2+CS_2\xrightarrow[\begin{matrix}\Delta\\(ii)\,conc.HCl\end{matrix}]{(i)KOH}Ph-NCS$
0%
$R-NH_2+HNO_2\xrightarrow{0-5^oC} R- OH+N_2+HCl$
0%
$Ph-NH_2+CHCl_3+3KOH(alc.)\xrightarrow{\Delta} Ph-NC+3KCl+3H_2O$
0%
$Ph_2-NH+HONO\rightarrow Ph_2-N-N=O+H_2O$

Explanation

(a)

$2Ph-NH_2+S=C=S\xrightarrow[\begin{matrix}\Delta\\(ii)\,conc.HCl\end{matrix}]{(i) 2KOH}\underset{\text{Hotmann mustard oil reaction}}{Ph-N=C=S+Ph\overset{\oplus}{N}H_3Cl^-}$

(b)

$R-NH_2+HNO_2\xrightarrow{0-5^oC} \underset{\text{Diazotisation reaction}}{R- OH+N_2+HCl}$

(c)

$Ph-NH_2+CHCl_3+3KOH(alc.)\xrightarrow{\Delta} \underset{\text{Carbylamine reaction}}{Ph-NC+3KCl+3H_2O}$

(d)

$Ph_2-NH+HONO\rightarrow \underset{\text{Lieber mann's nitroso reaction}}{Ph_2-N-N=O+H_2O}$

Option C is correct.

Arrange the following in the decreasing order of basic character.
I. p-toluidine
II. N, N-dimethyl-p-toluidine
III. p-nitroaniline
IV. Aniline

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

Explanation

1. The

$+I$ effect of 2 methyl group of N and hyper conjugation of methyl group at para position makes the

$N,N-dimethyl-p-toluidine$ more basic.

2. para toluidine shows the + I and hyperconjugation of one methyl group at position makes it less basic than N,N-dimethyl-p-toluidine.

3. paranitroaniline shows -R effect of

$NO_2$ group at para position, it is the least basic compound.

Hence, the order is

$II>I>IV>III$

Option C is correct.

Pick the correct statement among the following.

0%
Sodium lauryl sulphate forms an insoluble scum with hard water.
0%
Sodium dodecyl benzene sulphonate used in tooth paste is a cationic detergent.
0%
Cetyl-trimethyl ammonium bromide is a popular cationic detergent used in hair conditioner.
0%
Non-ionic detergents is formed when polyethylene glycol reacts with adipic acid.

Amines behave as:

0%
Lewis acids
0%
Lewis base
0%
aprotic acid
0%
neutral compound

Explanation

Amine is an organic compound derived from ammonia by replacement of one or more hydrogen atoms by organic groups. Ex: Aniline

( $C_6H_5NH_2$ ). Because of the presence of lone pair amines acts as Lewis bases.

Hence, option $B$ is correct

What is the final product?

Which is the strongest acid amongst the following compounds?

$CH_3CH_2Cl \xrightarrow {NaCN}X \xrightarrow{Ni/H_2}Y \xrightarrow{acetic\, anhydride}Z$

In the above reaction sequence,

$Z$ is:

0%
$CH_3CH_2CH_2NHCOCH_3$
0%
$CH_3CH_2CH_2NH_2$
0%
$CH_3CH_2CH_2CONHCH_3$
0%
$CH_3CH_2CH_2CONHCOCH_3$

Explanation

$CH_3CH_2Cl \overset{NaCN}{\rightarrow} CH_3- \underset{X}{CH_2}-CN \overset{Ni/H_2}{\rightarrow} CH_3-CH_2 CH_2NH_2\xrightarrow[anhydride]{Acetic} CH_3-CH_2 -\underset{Z}{CH_2}-NHCOCH_3$

The order of basicity among the following compounds is:

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

Explanation

The order of basicity among the following compounds is

$\displaystyle IV>I>II>III$

In IV, the extent of delocalization is maximum. When a lone pair of electrons on one of the

$N$ atom is donated to a suitable acid, still some delocalization is possible.

However, in case of I, when a lone pair of electrons on one of the

$N$ atom is donated to a suitable acid, further delocalization is not possible.

So, option C is correct.

$C{ H }_{ 3 }Br+KCN\left( alc. \right) \longrightarrow X\xrightarrow [ \quad Na+{ C }_{ 2 }{ H }_{ 5 }OH\quad ]{ \quad Reduction\quad } Y$
What is

$Y$ in the series?

0%
$C{ H }_{ 3 }CN$
0%
${ C }_{ 2 }{ H }_{ 5 }CN$
0%
${ C }_{ 2 }{ H }_{ 5 }N{ H }_{ 2 }$
0%
$C{ H }_{ 3 }N{ H }_{ 2 }$

Explanation

$C{ H }_{ 3 }Br+KCN\left( alc. \right) \longrightarrow \underset {X}{C{ H }_{ 3 }CN}$

$\underset { Y }{ C{ H }_{ 3 }C{ H }_{ 2 }N{ H }_{ 2 } } \xleftarrow [ Reduction ]{ \quad { Na }/{ { C }_{ 2 }{ H }_{ 5 }OH }\quad }$

Which of the following are P and S in the series of reactions?

0%
$C_2H_5MgX$ and $(C_2H_5)_2NCH_3$
0%
$CH_3MgX$ and N-methyl- $2$ -butanamine
0%
$CH_3MgX$ and N-ethyl- $2$ -butanamine
0%
$C_2H_5MgX$ and $(CH_3)_2NC_2H_5$

Among the following amines, which one has the highest

$pK_{b}$ value in aqueous solution?

0%
Methanamine
0%
N, N-dimethylaniline
0%
Ethanamine
0%
Benzenamine
0%
None of the above

Explanation

+Inductive effct increases the electron density on nitrogen thus makes it more basic while +R effect involves the lone pair of N in conjugation thereby reduces its basicity.

Therefore aniline are less basic than alkyl amines and secondary aniline are more basic than aniline.

Highest

$pK_b$ means least basic nature, therefore option D is correct.

Which of the following is a correct decreasing order of boiling points of the given isometric compounds:
Ethyldimethyl amine (I), n-butylamine (II), Diethylamine (III)

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

Explanation

Answer:

Primary amine has higher boiling point than secondary followed by tertiary amine, this is because higher boiling points of the primary amines is that they can form hydrogen bonds with each other as well as van der Waals dispersion forces and dipole-dipole interactions.

Tertiary amine has no hydrogen left for forming Hydrogen bonds among themselves so can not show intermolecular H-bonding. So intermolecular force will be weak and low boiling point of tertiary amines

$I=Ethyldimethyl\ amine=3^0$

$II=n-butylamine=1^0$

$III=diethylamine=2^0$

In primary amines only one hydrogen is substituted by an alkyl group, in secondary amines, two hydrogens are substituted by alkyl groups and in tertiary amines, all the three hydrogens are substituted by alkyl groups thereby resulting in the absence of intermolecular hydrogen bonding. Higher the hydrogen bonding difficult is to break the bonds thereby high boiling point.

Intermolecular hydrogen bonding is maximum in primary amines and absent in tertiary amines.

So order of boiling points will be $II>III>I$

The correct order of basic strength of the following are:

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

Explanation

As it is known that electron-releasing group increases the basic strength and electron withdrawing group decreases the basic strength of

$-NH_2$ group.
As in group is planked between two electron withdrawing groups, compound

$(1)$ is least basic among all. In compound

$(2)$ ,

$-NH-$ is attached with one electron withdrawing group

$-CO-$ and one electron releasing group -

$CH_3$ , it is a bit basic than compound

$(1)$ . In compound

$(4)$ -

$COCH_3$ , group is attached to aniline ring, not directly with -

$NH_2$ group.
Hence it is more basic than compound

$(2)$ . Compound

$(3)$ is most basic among all compounds since in it, -

$NH_2$ is attached to only one electron releasing group

$(-CH_2-)$ . Thus, the correct order of basic strength will be

$3 > 4 > 2 > 1$ .

$CH_2 = CHCH_2NH_2,$ II:

$CH_3CH_2CH_2NH_2,$ III:

$CH\equiv CCH_2NH_2$

What is the correct order of basicity of the above compounds?

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

Explanation

For the basicity of compounds, specially amines, the

$+I$ Effect on them is counted. Basicity is directly proportionality to the

$+I$ effect.

Among the given compounds, maximum

$+I$ effect is present in (II) as it is an alkyl group. Minimum

$+I$ effect is present in (III) as

$CH\equiv C-$ group is present.

So, (II) compound is most basic and (III) is least basic among the given compounds.

$II>I>III$

How many

$\pi$ electrons are present in the above conjugation compound?

0%
2
0%
4
0%
6
0%
None of these

Which one of the following is least basic in character? (Compare basicity at N-H position).

The order of basicity in increasing order is:

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

The increasing order of reactivity of these diazonium ions in azo-coupling reaction (under similar conditions) will be:

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

Explanation

Strong the electron withdrawing group, more will be the electrophilic nature.

Increasing order of electron withdrawing group is given by :

$NO_2 > -CH_3 > -OCH_3 > -(CH_3)_2N$

Therefore, correct order of reactivity is 1 < 3 < 4 < 2.

Hence, the correct option is B.

In the following reaction the major product is:

Decreasing order of basicity is?
(i)

$CH_3NH_2$
(ii)

$C_6H_5CH_2NH_2$
(iii)

$p-CH_3O-C_6H_4-NH_2$
(iv)

$m-CH_3O-C_6H_4-NH_2$

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

Explanation

Basicity depends upon the availability of the lone pair on the

$N$ atom. If the lone pair on

$N$ is more available then it makes it a better Lewis base and if the lone pair is less available it makes the amine a weak Lewis base. Tertiary amines are the most basic followed by secondary, primary and phenyl amines. Amides are still weaker than the phenyl amines and can be considered neutral.

Therefore, the correct answer will be option A)

$I > II > III > IV$

1996296_699664_ans_f34c803f0f3c4cebbb71e18c0e70f59d.jpg

The compunds X and Y are respectively

0%
$C{ H }_{ 3 }C{ H }_{ 2 }N{ H }_{ 2 },C{ H }_{ 2 }COCl$
0%
$C{ H }_{ 3 }CON{ H }_{ 2 },C{ H }_{ 3 }C{ H }_{ 2 }Cl$
0%
$C{ H }_{ 3 }C{ OOH }_{ 2 },C{ H }_{ 3 }C{ ONH }_{ 2 }$
0%
$C{ H }_{ 3 }CON{ H }_{ 2 },C{ H }_{ 3 }C{ OOH }$

Explanation

The solution is given to the image

$.$

1155882_877086_ans_1d1f32a692654ce1a28e62b041a1eb22.png

Which of the following amino acid has pH greater than

$7$ ?

0%
Glutamic acid
0%
Lysine
0%
Glycine
0%
Alanine

Explanation

Lysine is a basic amino acid. It has pH greater than

$7$ . Note: Basic solutions have pH greater than

$7$ .

The increasing order of diazotisation of the following compounds is:

0%
$(D) < (C) < (B) < (A)$
0%
$(A) < (D) < (B) < (C)$
0%
$(A) < (B) < (C) < (D)$
0%
$(A) < (D) < (C) < (B)$

Explanation

Aromatic diazonium salts are more stable than aliphatic diazonium salts. The stability of aryl diazonium salts is due to resonance. Electron donating substituents increase electron density on benzene ring. They increase the stability of diazonium salts.

Electron withdrawing substituents decrease electron density on benzene ring. They decrease the stability of diazonium salts.

$\displaystyle -COCH_3$ group is electron withdrawing and hence, (d) is less stable than (b).

Although

$\displaystyle -O-COCH_3$ is electron donating substituent, but it is present in meta position. Hence, it will not have significant effect on stability. The increasing order of diazotisation is

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

The increasing order of basicity of the above compounds is:

0%
$(b)<(a)<(d)<(c)$
0%
$(d)<(b)<(a)<(c)$
0%
$(a)<(b)<(c)<(d)$
0%
$(b)<(a)<(c)<(d)$

Explanation

$Basicity \propto \text {stability of conjugate acid}$

Since

$C$ is most basic due to resonance stabilization of its conjugate acid and

$b$ is least stable as positive charge is on

$sp^2$

$N$ .

Out of the given options, the answer is option

$A$ .

Which of the following compound is strongest base?

0%
${ CH }_{ 3 }CO{ NH }_{ 2 }$
0%
${ NH }_{ 2 }CO{ NH }_{ 2 }$
0%
$NH_2 - \overset{NH}{\overset{||}{C}} - NH_2$
0%
${ CH }_{ 3 }CONHCO{ CH }_{ 3 }$

Which order for amines is incorrect?

0%
Order of basicity in aq. medium.
0%
Order of boiling point.
0%
Order of basicity in gaseous state.
0%
The order of aqueous solubility.

Explanation

Order of basicity in aq. medium. for methyl amines.

$2>1>3$

In gaseous medium, the order of basicity is

$3>2>1$

B.P order is

$1>2>3$ . Hence option B is incorrect.

Increase in the size of hydrophobic alkyl part increases the molar mass of amines. This usually results in a decrease in its solubility in water. Hence the given order for solubility is correct.

The order of basic strength the above compound is:

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

The decreasing order of boiling points of ethyldimethylamine, n- butylamine and diethylamine is n - Butylamine > Diethylamine > Ethyldimethylamine. This trend of the boiling point can be explained as:

0%
boiling point increases with increase in molecular mass
0%
tertiary amines have highest boiling point due to highest basicity
0%
intermolecular hydrogen bonding is maximum in primary amines and absent in tertiary amines
0%
intramolecular hydrogen bonding is present in tertiary amines

Explanation

The decreasing order of boiling points:

n-butylamine(primary amine) > diethylamine(secondary amine) > ethyldimethylamine (tertiary amine).

$CH_3CH_2CH_2CH_2NH_2 > (C_2H_5)_2NH > C_2H_5NH(CH_3)_2$

Intermolecular hydrogen bonding is maximum in primary amines and absent in tertiary amines.

Choose the correct order of basic strength?

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

Explanation

The first step is refers to the image

$.$

The order

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

In case of compound

$(I),(II)$ and

$(III),$ compound

$(II)$ have one me group having

$+I$ effect at para position

$.$

$(III)$ have me group at the ortho position

$.$ So it pushed the electron towards

$N$ atom and increases the electron density on

$N$ atom

$,$ it is more basic than other two

$.$

$(I)$ have me group at meta position which does increase the density of electron on

$N$ -atom

$.$ So it is less basic than the other two compound

$.$

1021022_877181_ans_26dd863560784ef7832e462deec07b35.png

Reduction of

$CH_3CH_2NC$ with hydrogen in presence of

$Ni$ or

$Pt$ as catalyst gives:

0%
$CH_3CH_2NH_2$
0%
$CH_3CH_2NHCH_3$
0%
$CH_3CH_2NHCH_2CH_3$
0%
$(CH_3)_3N$

Explanation

Reduction of isonitriles

$(NC)$ in the presence of

$H_2/Ni$ or

$Pt$ gives a secondary amine (

$R - NH- R^{'}$ ).

$CH_3CH_2NC\xrightarrow{H_2/Ni\ or \ Pt} CH_3CH_2NHCH_3$

Option B is correct.

Most basic species amongst the following is:

Explanation

Cyclohexylamine is the most basic among all, as the nitrogen is outside the ring and exhibits

$sp^3$ hybridization, and the cyclic ring is also non resonating, there by the lone pair of electrons on the nitrogen can be easily available.

$\xrightarrow [ { H }^{ + } ]{ \Delta }$ (Major product )

0%
$CH_3 - \underset{O}{\underset{||}{C}} - NH - Ph$
0%
$PhCH_2 - \underset{O}{\underset{||}{C}} - NH_2$
0%
$Ph - \underset{O}{\underset{||}{C}} - NHCH_3$
0%
$PhCH_2CH_2NH_2$

Explanation

Here migration of group takes place. The group migrates with a negative charge to the electron deficient

$N$ . Out of

$-CH_{3}$ and

$-pH$ group, this is possible for

$-CH_{3}$ group. This is because , negative charge on

$-pH$ group is unstable.

1085199_1067472_ans_465356ad1a3b4ea08e2f552d5a130cc2.png

What is the order of basic strength of the following?

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

Explanation

In case of

$(1)$ and

$(2),(1)$ is more basic than

$(2),$ because

$(1)$ has lone pair on the

$N$ -atom in the non aromatic ring and ready to donate same as pyridine

$.$ Although pyride act as a base and will be protonated in presence of a strong acid

$,$ it is a weak base than it's non aromatic counterpast

$.$

On the other hand methyl cynide

$+I$ effect of methyl group pushed the electrons towards cyanide and it increasesthe density of electron on

$N$ -atom

$.$ Hence the order is

$1>2>3>4.$

The reaction is given to the image

$.$

1021568_878759_ans_dd97f69b5bad49319cc8ef24f84e0fc0.png

The correct basic strength order of following anions is :

0%
${CH_3 -{C}{H_2}}^- >{{N}{H_2}}^- > {CH_2 = {C}{H}}^- >{ CH \equiv {C}}^- > H{O}^- > {F}^-$
0%
${N}{H_2}^- > {C}{H_3} - {C}{H_2}^- > CH_2 = {C}{H}^- > CH \equiv {C}^- > {F}^- > H{O}^-$
0%
$CH_3 -{C}{H_2}^- > CH_2 = {C}{H}^- >{N}{H_2}^- > CH \equiv {C}^- > H{O}^- > {F}^-$
0%
${F}^- > H{O}^- > CH \equiv{C}^- > {C}{H_2} = {C}{H}^- >{N}{H_2} ^-> CH_3 -{C}{H_2}^-$

.The order of basicity of the above compounds is:

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

Explanation

We know that Phenyl (

$-C_6H_5$ ) is an electron withdrawing group. Higher the electron density on the phenyl group less likely is the compound to withdraw electron from the attached electron releasing group (here

$-{\overset {\bullet \bullet}{N}H}_2$ group).

Hence the correct order is-

$S>R>Q>P$

1532534_879703_ans_179d030fbf704f74902adcd23bb3a61a.png

Aniline when diazotised in cold and then treated with N, N - dimethylaniline gives a coloured product. The structure of this product is:

Explanation

Hence, option

$B$ is correct.

The best method for preparing primary amines from alkyl halides without changing the number of carbon atoms in the chain is:

0%
Hofmann bromamide reaction
0%
Gabriel phthalimide synthesis
0%
Sandmeyer reaction
0%
Reaction with $NH_3$

Product of this reaction is?

Arrange the following compounds in increasing order of basicity.

$CH_3NH_2, \, (CH_3)_2NH,\,NH_3, \, C_6H_5NH_2$

0%
$C_6H_5NH_2 \, <\, NH_3 \, <\,(CH_3)_2NH \,<\, CH_3NH_2$
0%
$CH_3NH_2 \, <\, (CH_3)_2NH\, <\, NH_3 \,<\,C_6H_5NH_2$
0%
$C_6H_5NH_2 \, <\, NH_3 \,<\, CH_3NH_2\, <\, (CH_3)_2NH$
0%
$(CH_3)_2NH \,<\, CH_3NH_2\,<\, NH_3 \,<\, C_6H_5NH_2$

Explanation

Aromatic amines are least basic as the lone pair of electrons on the nitrogen are not freely available due to the presence of the conjugated ring.

The secondary amines are most basic here as both the

$-R$ groups are electron-donating to the nitrogen, thereby the electron density increases on the nitrogen.

Hence, the order is

$C_6H_5NH_2 < NH_3 < CH_3NH_2 < (CH_3)_2NH$

Option C is correct.

The solubility of diethyl ether in water is comparable with that of:

0%
ethanamine
0%
chlorobutane
0%
1-butanol
0%
ethanol

Explanation

Diethyl ether is a common polar aprotic solvent. It has limited solubility in water and dissolves in water. In di-ethyl ether, the flexible chain of atoms sweeps out a volume that prevents hydrogen bonding between the ether oxygen atom and hydrogen atoms of water. But ethanamine soluble in water and form intermolecular

$H-$ bonds with water.

Identify

$Z$ in the sequence.

$C_6H_5NH_2 \xrightarrow [273K]{NaNO_2 + HCl} \, X\, \xrightarrow{CuCN} \, Y\, \xrightarrow [Boil]{H^+/H_2O} \, Z\,$

0%
$C_6H_5CN$
0%
$C_6H_5CONH_2$
0%
$C_6H_5COOH$
0%
$C_6H_5CH_2NH_2$

Explanation

$C_6H_5NH_2 \xrightarrow [273K]{NaNO_2+HCl} \underset{(X)}{C_6H_5\overset{+}{N_2}C \overset {-}{l}} \xrightarrow {CuCN} \underset{(Y)}{C_6H_5CN} \xrightarrow [boil]{H^+/H_2O} \underset{benzoic \, acid (Z)}{C_6H_5COOH}$

Compare the basic strength of compounds given:

0%
$(i) > (ii) > (iii)$
0%
$(ii) > (i) > (iii)$
0%
$(ii) > (iii) > (i)$
0%
$(iii) > (ii) > (i)$

Which of the following compound is the most basic?

The correct order of decreasing basicity of the above compounds is :

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

Explanation

The order of decreasing basicity of the given compounds is (C) III > IV > I > II.

As the

$-NH_2$ group has a lone pair of electrons on nitrogen to donate to electron deficient species.

Hence, its power to donate electrons it increased when an electron donation group (+R, +I) is attached and its donating power is decreased when an electron withdrawing group is attached at its para position.

Since

$NO_2$ is an electron withdrawing group hence it decreases the electron density over

$-NH_2$ group hence it decreases the basicity.

On the other hand

$-OCH_3$ (+M effect) group increase the electron density more than

$CH_3$ group (+I effect).

$CH_3-\overset{\,\,\,\,\,NH}{\overset{||}{\underset{(1)}{C}}}-NH_2$ ,

$CH_3-\underset{(2)}{CH_2}-NH_2$

$\underset{(3)}{(CH_3)_2}NH$ ,

$CH_3-\overset{O}{\overset{||}{\underset{(4)}{C}}}-NH_2$

The correct order of the basicities 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

The correct order is

$2<1<3<4$ .Since, more the number of

${ 2 }^{ o }-NH$ - group,more is the basic nature.

(i)Thus, 3 and 4 are more basic than 1 and 2

(ii)Between 3 and 4, 4 has one two

$-N{ H }_{ 2 }$ group , while 3 has only one

$-N{ H }_{ 2 }$

$\therefore$ 4 is more basic than 3.

(iii)Between 1 and 2,1 has one

$-N{ H }_{ 2 }$ group with no resonance,thus is more basic than 2.

CBSE Questions for Class 12 Medical Chemistry Organic Compounds Containing Nitrogen Quiz 11 - 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 11 - MCQExams.com

0:0:1

Practice Class 12 Medical Chemistry Quiz Questions and Answers

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