MCQ Questions for Class 12 Medical Chemistry Organic Compounds Containing Nitrogen Quiz 6

Which is most basic in aqueous solution?

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

Explanation

${ CH }_{ 3 }\rightarrow \begin{matrix} N \\ \uparrow \\ { CH }_{ 3 } \end{matrix}\leftarrow { CH }_{ 3 }$

$+I$ effects of Three methyl group

Here, it is 3 amine and three methyl group attached with N-atom. The inductive effect of three methyl group and lypez conjugation effect of

$9-\alpha -H$ atom, So this is more stable and basic in nature.

Which of the following is most basic ?

Explanation

Here, lone pair of electrons are available on the

$N-$ atom and it can easily donate electrons and

$pK_b$ value is also low.
1174962_1058202_ans_92c85bd3b89048b2b90c9b57ad44b137.png

Which of the following diazonium salt is relatively stable at

$0-{ 5^0 }^{ }C$ ?

0%
${ CH }_{ 3 }-N\equiv N{ \} }^{ \oplus }C{ l }^{ - }$
0%
${ \left( { CH }_{ 3 } \right) }_{ 2 }CH-N\equiv N{ \} }^{ \oplus }{ Cl }^{ - }$
0%
${ C }_{ 6 }{ H }_{ 5 }-N\equiv N{ \} }^{ \oplus }{ Cl }^{ - }$
0%
${ \left( { CH }_{ 3 } \right) }_{ 3 }C-N\equiv N{ \} }^{ \oplus }{ Cl }^{ - }$

Explanation

The

$+I$ effect which stabilizes the

${ N }^{ + }$ ion is in the following order

${ C }_{ 6 }{ H }_{ 5 }-<{ CH }_{ 3 }-<{ \left( { CH }_{ 3 } \right) }_{ 2 }-<{ \left( { CH }_{ 3 } \right) }_{ 3 }$

$\therefore$ The

${ \left\{ { \left( { CH }_{ 3 } \right) }_{ 3 }C-N\equiv N \right\} }^{ + }{ Cl }^{ - }$ is most stable

Select the basic strength order of following molecules.

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

Which of the following conditions is most suitable for the above reaction?

0%
strongly acidic
0%
strongly alkaline
0%
Slightly alkaine
0%
neutral

Explanation

In this reaction, primary aromatic amine is treated with nitrous acid (

${ HNO }_{ 2 }$ ).

${ HNO }_{ 2 }$ can be obtained by dissolving solid

${ NaNO }_{ 2 }$ in aqueous acidic solution.

Which of the following will accept

$H^{+}$ from

$NH_{4}^{+}$ ion?

0%
0%
0%
0%
$CH_{3}-CH_{2}-NH_{2}$

$H-COOH > Ph-COOH > CH_{3}-COOH$ (acidic nature)

$CF_{3}-COOH > CCI_{3}-COOH > CH_{2}-COOH$ (Acidic nature)

$N- methylmethanamine > methanamine > N, N- dimethylmethanamine$ (Basic nature in aqueous phase)

$N,N- diethlethanamine > N-ethylethanamine > Ethanamine$ (Basic nature in aqueous phase)Then the correct order is:

0%
$i,ii,iv$
0%
$ii,iii,iv$
0%
$i,ii,iii$
0%
$i,ii,iii,iv$

Explanation

In aqueous phase, the basicity of amines depends upon:

$(i)$ Inductive effect: which fovours tertiary amine to be more basic due to greater

$+I$ effect of

$3$ alkyl group directly attached to nitrogen.

$(ii)$ Solution Effect: Which favours primary amine to be more basic due to greater solution of its cation.

$(iii)$ Steric Factor: Which cause tertiary amine to be less basic due to steric hinderance .

Decreasing order of basicity of the three isomers of nitro aniline is:

0%
$p-nitroaniline> o-nitroaniline> m-nitroaniline$
0%
$p-nitroaniline> m-nitroaniline> o-nitroaniline$
0%
$m-nitroaniline> p-nitroaniline> o-nitroaniline$
0%
$m-nitroaniline> o-nitroaniline> p-nitroaniline$

Explanation

Nitro group is an electron-withdrawing group. It mainly withdraws electrons from the ortho and para positions. Therefore para isomer is less basic than meta isomer.

Ortho substituted anilines are generally weaker bases than aniline irrespective of the electron releasing or electron-withdrawing nature of the substituent.

Hence decreasing order of basicity of the isomers of nitroaniline:

$m-nitroaniline>p-nitroaniline>o-nitroaniline$

The order of basic strength of the given basic nitrogen atoms is :

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

Explanation

Since

$(II)$ is more basic then

$(III)$ due

$(II)$ nitrogen get negative charge due resonance hence it is more basic.

There for correct order is

$II$ >

$III$ >

$I$ >

$IV$

What is the major organic product of the following reaction?

${ CH }_{ 3 }{ CH }_{ 2 }{ NHCH }_{ 2 }{ CH }_{ 3 }\longrightarrow$ ?

The correct sequence regarding base strength of aliphatic amines in aqueous solution is:

0%
${R}_{3}N> {R}_{2}NH> R{NH}_{2}> {NH}_{3}$
0%
${R}_{2}NH> R{NH}_{2}> {R}_{3}N> {NH}_{3}$
0%
${R}_{2}NH> {R}_{3}N> R{NH}_{2}> {NH}_{3}$
0%
$R{NH}_{2}> {R}_{2}NH> {R}_{3}N> {NH}_{3}$

Explanation

Since the bascities of amine depend upon the no of alkyl group directly attached to the nitrogen directly due to

$+I$ effect to alkyl group increased the bascities of amine.

The correct order of basic strength of the compounds?

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

Explanation

the strength of basic nature of compounds depends on the ability to easily donate the lone pair of electrons.

$(i)$ lone pair is easily approachable so it has more basicity . in

$(ii)$ the lone pair is in different plane. while in

$(iii)$ the lone pair is in conjugation . hence order of basicity is

$I>II>III$

The product 'Y' in the following reaction sequence is:

Which of the following Statements is/are correct?

0%
$(SiH_3)_3N$ is weaker base than $(CH_3)_3N$ .
0%
Amongst $Cl-Cl$ , $N-F$ , $C-F$ And $O-F$ bond is most polar
0%
$OF_2$ has higher dipole moment than $H_2O$
0%
An alchohol is less volatile than other having same molecular formula on account of intermolecular hydrogen bonding

Explanation

$A.$

${(SiH_3)}_3N$ is a weaker base than

${(CH_3)}_3N$ .

${(CH_3)}_3N$ the lone pair is concentrated on the

$N-atom$ . But in case of Trisillyl amine, the lone pair on nitrogen gets delocalised on the

$3$ silicon atoms bonded to it.

Since, base strength is a measure of tendency to donate lone pair, trimethylamine is stranger base.

$D.$ In ethers there is no intermolecular hydrogen bond, where as in alcohol, intermolecular hydrogen bond is present which makes them less volatile and have high boiling point.

This reaction is example of:

0%
Intermolecular $C-N$ coupling
0%
Intramolecular $C-N$ coupling
0%
Intermolecular $N-N$ coupling
0%
Intramolecular $N-N$ coupling

Explanation

So this is example of

$N-N$ intra molecular coupling.
1157664_1072180_ans_f5a1882553b84c97bbf74d8d773533c9.png

$sp^3-$ hybridised nitrogen is present in:

0%
0%
0%
$CH_2 = CH - \overset{\oplus}{N}H_3$
0%

Explanation

$Sp^{2}$ hybridized Nitrogen

$S p^{2}$ hybridized Nitrogen due to participation

of lone pair into ring.

$C H_{2}=C H-NH_3+ \quad\left(S p^{3}\right.$ hybridised nitrogen

$)$

D) (sp

$^{2}$ hybridised) as nitrogen is in conjugation with double bond.

So, option C) is correct.

The increasing order of basicity of the given compounds is:

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

Explanation

Order of base nature depends on electron donation tendency.

In compound (b) nitrogen is

$sp^{ 2 }$ hybridized so least basic among all given compound

Compound (C) is a very strong nitrogeneous organic base as a lone pair of one nitrogen delocalize in resonance and make another nitrogen negatively charged and conjugate acid have two equivalent resonating structure.

Thus it is most basic in given compound.

Compound (d) is secondary amine more basic than compound (a) which is a primary amine.

Therefore, increasing order of basicity is

(b) < (a) < (d) < (c)

The correct order of increasing basic nature for the bases

$NH_3,CH_3NH_2$ and

$(CH_3)_2NH$ is?

0%
$(CH_3)_2NH < NH_3 < CH_3NH_2$
0%
$NH_3 < CH_3 NH_2 <(CH3)_2NH$
0%
$CH_3NH_2 <(CH_3)_2NH < NH_3$
0%
$CH_3NH_2 < NH_3 < (CH_3)_2NH$

Explanation

$(H_3C)_2NH$ ,

$CH_3NH_2$ ,

$NH_3$
Among the above molecules

$(CH_3)_2 NH$ is most basic as two electron donating group are attached to it which increase its basicity is

$NH_3 < CH_3NH_2 < (CH_3)_2NH$

The end product (Z) of the following reaction is?

0%
A cyanide
0%
A carboxylic acis
0%
An amine
0%
An arene

By heating ammonium chloride with two equivalents of formaldehyde it forms:

0%
dimethylamine
0%
ethylamine
0%
methylamine
0%
ammonium formate

Explanation

$6HCHO + 4N{H}_{4}Cl \rightarrow ({C{H}_{2}}_{6}){N}_{4} + 4HCl + 6{H}_{2}O$

So Methylamine is formed

Find product in the following reaction is Y. What is Y?

0%
benzamide
0%
benzophenone
0%
benzoic acid
0%
benzaldehyde

$CH_3Cl \xrightarrow{AgCN} A$ (major)

$\xrightarrow{H-2O^+} C + D$ here '

$C$ ' is formic acid '

$D$ is:'

0%
$2^o$ amine
0%
Carboxylic acid
0%
Alkyl Cyanide
0%
$1^o$ amine

Explanation

$\text { Hence correct Answer - D }$
2007123_1093890_ans_fb16420de45843dab84520727da33945.PNG

Reductive amination of

$A$ forms:

The reactivities of these ions in azo-coupling reactions will be in the order:

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

Explanation

In azo-coupling, the diazonium ion should be more electrophilic for more reactivity. Presence of electron-withdrawing groups increase the reactivity.

Considering these factors, the order of reactivity should be:

$1<2<3<4$

In the above reaction sequence (D) is?

Suitable explanation for the order of basic character,

$(CH_{3})_{3}N < (CH_{3})_{2}NH$ , is:

0%
steric hindrance by bulky methyl group
0%
higher volatility of $3^{\circ}$ amine
0%
stability of conjugate acid due to solvation
0%
all of these

Explanation

${ ({ CH }_{ 3 }) }_{ 3 }$ N is less basic due to steric hindrance by bulky methyl group prevent any acid to take a lone pair of nitrogen.

With which one of the following substrate do diazonium salt reacts to form a colourful compound?

In the reaction

$Me_{3}N \xrightarrow{BrCN} A \xrightarrow[ii)\Delta]{i)H_{2}O} B$ .
B is

0%
$MeNH_{2}$
0%
$Me_{3}NO$
0%
$Me_{2}NH$
0%
$Me_{2}NCOOH$

Explanation

$Me_3N\xrightarrow {BrCN}{}\underset{A}{Me_2N}-CN$

$Me_2N-Cn\xrightarrow[\triangle ]{H_2O}Me_2NCOOH$

The reagents/conditions K, L and M are respectively:

0%
$Ba(OH)_{2}$ , $KCN$ and $CHCl_{3} / NaOH$ heat, $H_{3}O^{+}$
0%
$H_{2}O$ (boil), $CO_{2} / KOH$ and $Ac_{2}O - AcONa$ , heat
0%
Steam ; $CHCl_{3} / NaOH$ , heat, $H_{3}O^{+}$ and $AC_{2}O - AcONa$ , heat
0%
$Cu_{2}Cl_{2} / HCl$ , $NaOH$ / high P and $CO_{2}|NaOH$ , heat

Explanation

$\text { Honre option } A \text { is correct }$

1998580_1109164_ans_bb41a8eec35a4338b8dc69d047a84257.PNG

The correct order of boiling points of the following amines

${ C }_{ 4 }{ H }_{ 9 }{ NH }_{ 2 },\left( { C }_{ 2 }{ H }_{ 5 } \right) _{ 2 }NH,{ C }_{ 2 }{ H }_{ 5 }N\left( { CH }_{ 3 } \right) _{ 2 }$ is :

0%
${ C }_{ 2 }{ H }_{ 5 }N\left( { CH }_{ 3 } \right) _{ 2 }>\left( { C }_{ 2 }{ H }_{ 5 } \right) _{ 2 }NH>{ C }_{ 4 }{ H }_{ 9 }{ NH }_{ 2 }$
0%
$\left( { C }_{ 2 }{ H }_{ 5 } \right) _{ 2 }NH>{ C }_{ 2 }{ H }_{ 5 }N\left( { CH }_{ 3 } \right) _{ 2 }>{ C }_{ 4 }{ H }_{ 9 }{ NH }_{ 2 }$
0%
${ C }_{ 4 }{ H }_{ 9 }{ NH }_{ 2} >\left( { C }_{ 2 }{ H }_{ 5 } \right) _{ 2 }NH>{ C }_{ 2 }{ H }_{ 5 }N\left( { CH }_{ 3 } \right) _{ 2 }$
0%
$\left( { C }_{ 2 }{ H }_{ 5 } \right) _{ 2 }NH>{ C }_{ 4 }{ H }_{ 9 }{ NH }_{ 2 }>{ C }_{ 2 }{ H }_{ 5 }N\left( { CH }_{ 3 } \right) _{ 2 }$

Explanation

Because

${ C }_{ 4 }{ H }_{ 9 }{ NH }_{ 2 }$ is primary amine and can form hydrogen bonding more than secondary amine

$({ { C }_{ 2 }{ H }_{ 5 } })_{ 2 }NH$ and tertiary amine

${ C }_{ 2 }{ H }_{ 5 }N{ ({ CH }_{ 3 }) }_{ 2 }$ .More hydrogen bonding leads to strong bonding between the molecules, hence increases the boiling point.

Which of the following compounds is an imine?

Explanation

The structure of imine is the given figure.

The nitrogen atom can be attached to hydrogen or an organic group.

Hence compound

$B$ is an imine.

1220829_1111560_ans_342deff7e8b94133a2ab57c57db13096.png

${ C }_{ 3 }{ H }_{ 9 }N$ represents

0%
primary amine
0%
secondary amine
0%
tertiary amine
0%
all of these

The correct order of basicity of the above compounds is:

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

Explanation

(Refer to Image)

The order of basicity is follows

$I >III >II >IV$

$\rightarrow$ in

$IV$ electron withdrawing is attached to amine. So, it is least basic.

1220003_1115396_ans_93814c82f5664738a0ec25321023c497.png

Basic nature order of the given compounds is
(a)

$CH_3 - CH = \ddot{N}H$
(b)

$CH_2= CH - CH = CH -NH_2$
(c)

$CH_3 -NH - CH_3$

0%
c>a>b
0%
a>b>c
0%
a>c>b
0%
c=a=b

Explanation

$C H_{2}=C H-C H=C H-N H_{2}(b)$

In this compound lone pair on

$-N H_{2}$ group

is involued in resonance. so givening lone

Pair is difficult. So this compound

has less basicity. than other two.

$\mathrm{CH}_{3}\mathrm{CH}=\ddot{\mathrm{N}}\mathrm{H}(\mathrm{O})$

$\begin{array}{l} \text { Nitrogen in }^{\prime}=\ddot{N} H^{\prime} \text { group is more electro. } \\ \text { negitiven than }-N H-\text { is } C H_{3}-N H-C H_{3}(C) \\ \text { due to double bond. } \\ \text { More electronegitive Nitrogen in ' }=\text { NH 'grap } \\ \text { cannot tolarate positive charge of it (when } \\ \text { lone pair is given it get positive charge) } \end{array}$

$\begin{array}{l} \text { Compared to Nitrogen in }-\mathrm{NH} \text { - group } \\ \text { So } \mathrm{CH}_{3}-\mathrm{NH}-\mathrm{Ct}_{3} \text { is more basic than } \\ \mathrm{CH}_{3}-\mathrm{CH}=\mathrm{NH} \text { (a) } \\ \text { Hence Basisity order is } c>a>b \\ \text { so option } A \text { is correct. } \end{array}$

Arrange these compounds 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

$iv>i>iii>ii$

Electron - donating substituents increase the electron density of N, thereby increasing the Lewis basicity. Amongst

$NO_2$ and

$CN,NO_2$ is slightly resonance stabilization.

Hence, the answer is

$iv>i>iii>ii.$

Which of the following is the strongest base?

Explanation

Aniline is the strongest base here. This is because,

$NH_2$ is stabilized by conjugation and resonance stabilised . More stability implies stronger the base.

The decreasing order of the basic character at

$a, b, c, d$ in the compound:

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

Explanation

$d>c>b>a.$
Due to ease of delocalisation of electron pair is least in

$(d)$ Nitrogen. For

$(a), (b), (c)$ Nitrogen atoms , basic character order is governed by electronic effects of rings and

$-CH_3$ group

Arrange the following in order of decreasing basic strength.

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

Explanation

$IV>I>II>III$

Which of the following is weak base than aniline ?

Explanation

An electron withdrawing groups decrease the availability of a lone pair of electrons on

$N-$ atom of aniline thereby decreasing its basicity.

Hence it is less basic than aniline.

1184080_1129241_ans_0e9ab19047b148cdb8190e5aedba700b.png

What is strong base?

An organic compound (A)

$C_9H_{13}N$ dissolves in dil. HCI and releases

$N_2$ with

$HNO_2$ giving an optically active alcohol. Alcohol on oxidation gives dicarboxylic acid, which on heating form anhydride. The organic compound 'A' is :-

The relative basics strenghts of

$NH_{3},CH_{3}NH_{2}$ and

$NF_{3}$ are in the order:

0%
$CH_{3}NH_{2} > NH_{3}>NF_3$
0%
$NF_{3} > CH_{3}NH_{2} > NH_{3}$
0%
$NH_{3} > CH_{3}NH_{2} > NF_{3}$
0%
$NF_{3} > CH_{3}NH_{2} = NH_{3}$

Explanation

$CH_3-$ group is electron donating group.

$F$ is electron withdrawing group.

$CH_3NH_2$ is more basic.

$NH_3$ is more basic than

$NF_3$ because

$H$ is not a withdrawing group.

$NF_3$ is less basic.

$\therefore$ Basic nature order=

$CH_3NH_2 > NH_3 >NF_3$

Acetic anhydride with ammonia gives the product:

0%
$CH_{3}CONH_{2}$
0%
$CH_{3}CONHCH_{3}$
0%
$CH_{3}CN$
0%
$CH_{3}COONH_{4}$

What will be the final product of the given reaction?

Explanation

Diazonium chloride on treating with

$HF/{ BF }_{ 3 }$ yields benzene diazonium fluoroborate. Benzene diazonium fluoroborate when heated undergoes decomposition reaction to give fluorobenzene.

Option B is correct.

1185398_1128733_ans_cf7d7e0df2fd46f8b94b367eb094d61c.png

which compound is Prepare propan -I- amine from?

0%
Butaniteuie
0%
I-Nitropropane
0%
Propanamide
0%
Butanamide

The major product of the reaction is:

Explanation

(Refer to Image)

In presence of

$NaNO_2$ , diazotisation reaction takes place and

$NH_2$ group leaves as

$-N_2$ . Then methyl group shift takes place, which makes the carboction more positive than former. Then oxygen stabilise the carbocation by

$+M$ effect and ketone formation takes place.

1163455_1117541_ans_62fce0ff7a5247f8b2e3868fbe152647.JPG

Adrenaline and ephedrine contain:

0%
${1}^{o}$ amino group
0%
${2}^{o}$ amino group
0%
${3}^{o}$ amino group
0%
quaternary amino group

Explanation

Thus Adrenaline and ephedrine contain

$2^0$ amino group.
1380602_1133505_ans_b5091b095b4342fca190460eb0262236.png

Sanger's reagent is used for the identification of:

0%
N-thermal of peptide chain
0%
C-terminal of a peptide chain
0%
side chain of amino acids
0%
number of amino acids in peptide chain

Explanation

1-Fluoro-2,4-dinitrobenzene. Undergoes nucleophilic aromatic substitution (NAS) with the N-terminal amino group of a peptide or protein, resulting in a chemical tag on the terminal amino acid residue. After hydrolysis into individual amino acids, the peptide's or protein's terminal amino acid can be identified by the specific wavelength of light absorbed (i.e., colorimetric analysis).

So option

$A$ is correct

Among the following which is not correct?

0%
P methyl benzyl carbonyl is more stable than ethyl benzyl carbonyl
0%
Aniline is more basic than pyridine
0%
Cyclopentadienyl anion is more stable than cyclopentadienyl Carbo cation
0%
K of Ethene 1 ol is more than that of propene 2 en 1 ol

${C}_{6}{H}_{5}-{NO}_{2}\xrightarrow [ ]{ Fe/HCl } A\xrightarrow [ H{ NO }_{ 3 } ]{ 273K } B\xrightarrow [ ]{ { C }_{ 6 }{ H }_{ 5 }OH } C$ ,

$C$ is

CBSE Questions for Class 12 Medical Chemistry Organic Compounds Containing Nitrogen Quiz 6 - 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 6 - MCQExams.com

0:0:1

Practice Class 12 Medical Chemistry Quiz Questions and Answers

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