MCQ Questions for Class 12 Medical Chemistry Organic Compounds Containing Nitrogen Quiz 4

$C_6H_5CN + H_2/Ni \rightarrow$ ?

0%
$C_6H_5-CH_2-CH_2-NH_2$
0%
$C_6H_5-CH_2-NH_2$
0%
$C_6H_6-CH_2-NH_2$
0%
$C_6H_5-NH_2$

Explanation

Reduction of cyanide occur to

$1^0$ amine in presence of

$H_2/Ni$ as reducing agent.
1158100_552636_ans_89e61f1dd2114a80a8d15de87e2db555.png

Identify the highest soluble amine from the following.

0%
Primary amine
0%
Secondary amine.
0%
Tertiary amine
0%
None of these

Explanation

The highest soluble amine would be primary amine, because it can form more H-bonds with water since

$1^o$ amine have

$-NH_2$ (2 Hydrogen) for H-bonding which decreases as we move on to secondary or tertiary amine.

The boiling point of amines are higher than that of hydrocarbon because:

0%
they form inter molecular H-bonds
0%
they are polar
0%
they form ionic bonds
0%
both A and B

In the preparation of primary amine by reduction of nitriles sodium amalgam can be used as reducing agent in presence of:

0%
ethyl alcohol
0%
methyl alcohol
0%
both a and b
0%
none of these

Explanation

RCN

$\xrightarrow [ Ethyl\quad alcohol ]{ Na-Hg }$

$RCH_2NH_2$

Reduction of cyanide in

$Na-Hg$ to amine (

$1^0$ ) occur in presence of ethyl alcohol.

Which of the following is the correct order of the boiling points?

0%
Propane $<$ Ethylamine $<$ Ethyl alcohol
0%
Ethylamine $<$ Propane $<$ Ethyl alcohol
0%
Ethylamine $<$ Ethyl alcohol $<$ Propane
0%
Propane $<$ Ethyl alcohol $<$ Ethylamine

Explanation

The following is the correct order of the boiling points.
propane

$<$ ethylamine

$<$ ethyl alcohol
In general, amines have higher boiling points than alkanes but lower boiling points than alcohols.

The most basic compound in aqueous solution is:

0%
${ NH }_{ 3 }$
0%
${ CH }_{ 3 }{ NH }_{ 2 }$
0%
$HN{ \left( { CH }_{ 3 } \right) }_{ 2 }$
0%
$N{ \left( { CH }_{ 3 } \right) }_{ 3 }$

Explanation

The combined effect of the pushing effect of the alkyl group (+I effect), steric hindrance and the solvation of amines causes the basicity order to be:

$NH_3< tertiary\ amine < primary \ amine< secondary\ amine$ .

$HN(CH_3)_2$ is secondary amine. So it is most basic.

Based on the concept of the hydration of protonated amines, which of the following represents correct order of decreasing basic strength of amines?

0%
primary amine $>$ secondary amine $>$ tertiary amine
0%
tertiary amine $>$ primary amine $>$ secondary amine
0%
tertiary amine $>$ secondary amine $>$ primary amine
0%
primary amine $>$ tertiary amine $>$ secondary amine

Explanation

Based on the concept of the hydration of protonated amines, the following represents the correct order of decreasing basic strength of amines
primary amine

$>$ secondary amine

$>$ tertiary amine
This is the same order in which the degree of hydrogen bond formation, the extent of hydration and the stabilisation of protonated amines decrease.

Based on the combined effects of inductive effect and solvation, which of the following represents correct order of decreasing basic strength of amines?

0%
Secondary amine $>$ Tertiary amine $>$ Primary amine
0%
Primary amine $>$ Secondary amine $>$ Tertiary amine
0%
Tertiary amine $>$ Secondary amine $>$ Primary amine
0%
Primary amine $>$ Tertiary amine $>$ Secondary amine

Explanation

Based on the combined effects of inductive effect and solvation, the following represents correct order of decreasing basic strength of amines
secondary amine

$>$ tertiary amine

$>$ primary amine
This is the order we observe in the aqueous solution. Secondary amines are strongest bases.

Which of the following represents the correct order of basic strength?.

0%
aniline $>$ diphenylaniline $>$ triphenylaniline
0%
triphenylaniline $>$ diphenylaniline $>$ aniline
0%
triphenylaniline $>$ aniline $>$ diphenylaniline
0%
aniline $>$ triphenylaniline $>$ diphenylaniline

Explanation

The following represents the correct order of basic strength
aniline

$>$ diphenylaniline

$>$ triphenylaniline
When hydrogen atoms of the amino group of arylamines are replaced by electron withdrawing phenyl groups, the basic character of the resulting amine decreases.

What makes diazonium ion one of the best leaving group of organic chemistry?

0%
Half filled electronic configuration of nitrogen atom
0%
High electronegativity of nitrogen atom
0%
Very high stability of dinitrogen molecule in the atmosphere
0%
Weaker $C-N$ bond

Which of the following is a true statement regarding the effect of substituent on the basic character of aromatic amines?

0%
Electron donating groups like $-OCH_3$ , $-CH_3$ increases the basic strength, while electron withdrawing groups like $-NO_2$ , $-SO_3$ , $-COOH$ etc. decrease the basic strength.
0%
Electron donating groups like $-OCH_3$ , $-CH_3$ decreases the basic strength, while electron withdrawing groups like $-NO_2$ , $-SO_3$ , $-COOH$ etc. increase the basic strength.
0%
Electron donating groups like $-OCH_3$ , $-CN$ increases the basic strength, while electron withdrawing groups like $-NO_2$ , $-SO_3$ , $-COOH$ etc. decrease the basic strength.
0%
Electron donating groups like $-OCH_3$ , $-CH_3$ increases the basic strength, while electron withdrawing groups like $-NO_2$ , $-SO_3$ , $-COOH$ , $-NH_2$ etc. decrease the basic strength.

Explanation

Electron donating groups like

$-OCH_3$ ,

$-CH_3$ ,

$-NH_3$ increases the basic strength, while electron withdrawing groups like

$-NO_2$ ,

$-SO_3$ ,

$-COOH$ ,

$-CN$ etc. decrease the basic strength.

The electron donating groups stabilize the cation and electron withdrawing groups destabilize the cation.

Which of the following is true statement?

0%
Every o-substituted aniline (electron releasing or electron withdrawing) is less basic than aniline.
0%
Every o-substituted aniline (electron releasing or electron withdrawing) is more basic than aniline.
0%
o-substituted aniline (electron releasing) is less basic than aniline and

o-substituted aniline (electron withdrawing) is more basic than aniline.
0%
o-substituted aniline (electron releasing) is more basic than aniline and

o-substituted aniline (electron withdrawing) is less basic than aniline.

In aqueous solution ______ amines are unexpectedly less basic than the others.

0%
Primary
0%
Secondary
0%
Primary and Secondary
0%
Tertiary

Which of the following represents the decreasing order of basic strength of amines in the gas phase?

0%
Tertiary amine $>$ Secondary amine $>$ Primary amine
0%
Tertiary amine $>$ Primary amine $>$ Secondary amine
0%
Primary amine $>$ Secondary amine $>$ Tertiary amine
0%
Secondary amine $>$ Tertiary amine $>$ Primary amine

Explanation

The following represents the decreasing order of basic strength of amines in the gas phase.

tertiary amine

$>$ secondary amine

$>$ primary amine

This is due +I (electron releasing) effect of alkyl groups. With an increase in the number of alkyl groups, the electron density on N increases and the lone pair of electrons can be easily donated.

Hence, the correct option is

$A$

The stability of benzene diazonium salts is because of:

0%
inductive effect
0%
resonance
0%
hyperconjugation
0%
mesomeric effect

Which of the following amine gives diazonium salt on reaction with

$HNO_2?$

0%
$(CH_3)_2NH$
0%
$(CH_3)_3N$
0%
$C_6H_5NH_2$
0%
$CH_3NH_2$

Explanation

$C_6H_5NH_2$ reacts with

$HONO$ to forms diazonium salts, the reaction are as follows

Refer tO image 01

$NaNO_2+HCl\longrightarrow NaCl+HONO$

Refer to Image 02

1083205_592861_ans_6395b376b3254c4aae0c0303b9e0feaf.png

This compound can be classified as:

0%
primary amine
0%
secondary amine
0%
tertiary amine
0%
none of these

Explanation

The

$-{ NH }_{ 2 }$ group in a compound makes it a primary amine.

In coupling reactions, diazonium ion acts as:

0%
nucleophile
0%
electrophile
0%
solvent
0%
none of the above

The conversion of primary aromatic amines into diazonium salts is known as:

0%
diazotisation
0%
displacement
0%
acylation
0%
alkylation

Explanation

Diazonium ions are prepared by diazotization, a procedure in which a primary aromatic amine (

$ArNH_2$ ) is treated with a source of nitrous acid (

$HNO_2$ ). Typically this involves adding sodium nitrite (

$NaNO_2$ ) to an aqueous acidic solution containing the amine.

Diazonium salts are used at a very lower temperature because:

0%
they are poisonous at room temperature
0%
they are highly unstable at room temperature
0%
they are highly volatile at room temperature
0%
none of these

Explanation

Diazonium salt is used at very low temperature because of the fact that at the higher temperature they are unstable, as at higher temperature

$N_2$ can leave out.

The molecule which does not exhibit strong hydrogen bonding is

$:$

0%
Methyl amine
0%
Diethyl ether
0%
Acetic acid
0%
Glucose

Explanation

The molecule which does not exhibit strong hydrogen bonding is

$\displaystyle diethyl \: ether$

$\displaystyle CH_3-O-CH_3$ .
Hydrogen bonding is possible when

$\displaystyle H$ atom is attached to an electronegative

$\displaystyle N, O$ or

$\displaystyle F$ atom.
Hence, the option (B) is the correct answer.

Which one of the following is strongly basic?

0%
Dimethyl amine
0%
Methyl amine
0%
Ammonia
0%
Aniline

Explanation

Dimethyl amine

$\displaystyle (CH_3)_2-NH$ is strongly basic due to electron releasing +I effect of two methyl groups that increase the electron density on N atom so that N can easily donate lone pair of electrons. Also, the dimethyl ammonium ion is stabilized due to the dispersal of positive charge by the +I effect of two methyl groups.

Aniline is least basic as the lone pair of electrons on N atom is in resonance with benzene ring and is less available for donation.

Correct gradation of basic character is:

0%
$NH_3 > CH_3NH_2 > NF_3$
0%
$CH_3NH_2 > NH_3 > NF_3$
0%
$NF_3 > CH_3NH_2 > NH_3$
0%
$CH_3NH_2 > NF_3 > NH_3$

Explanation

Correct gradation of basic character is

$\displaystyle CH_3NH_2 > NH_3 > NF_3$ . Due to

$\displaystyle +I$ effect of methyl group, the electron density on

$\displaystyle N$ increases.

Hence,

$\displaystyle CH_3-NH_2$ is most basic as it can easily donate lone pair of electrons on N.

Due to

$\displaystyle -I$ effect of F atom, the electron density on

$\displaystyle N$ decreases. Hence,

$\displaystyle NF_3$ is most basic as it can easily donate lone pair of electrons on N.

The product X and Y in the following reaction sequence are:

Explanation

$\displaystyle Nitrobenzene$ on reduction with

$\displaystyle Sn/HCl$ gives

$\displaystyle aniline$ .

Diazotization of

$\displaystyle aniline$ with

$\displaystyle NaNO_2 / HCl$ followed by heating with

$\displaystyle CuBr$ gives

$\displaystyle bromobenzene$ .

Option B is correct.

$C_3H_9N$ 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.

The source of nitrogen in Gabriel synthesis of amines is:

0%
sodium azide $Na{ N }_{ 3 }$
0%
potassium phthalimide ${ C }_{ 6 }{ H }_{ 4 }{ \left( CO \right) }_{ 2 }{ N }^{ - }{ K }^{ + }$
0%
sodium cyanide
0%
potassium nitrate

Select the basic strength order of following molecule.

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

Which one of the following on reduction with lithiium aluminium hydride yield a secondary amine?

0%
Nitroethane
0%
Methyl isocyanide
0%
Acetamide
0%
Methyl cyanide

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.

What is the name of red azo dye ?

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

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

Explanation

Alcohol undergo acetylation in presence of

$CH_3 - \overset{O}{\overset{||}{C}} - Cl$
1155198_692275_ans_2d3090b9e1eb427e87d948c6811eaf8e.png

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

The final compound is :

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

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.

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).

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

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.

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%
$0.11$

Compound

$C$ in the above reaction is:

The following sequence of reaction on

$"A"$ gives.

CBSE Questions for Class 12 Medical Chemistry Organic Compounds Containing Nitrogen Quiz 4 - 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 4 - MCQExams.com

0:0:1

Practice Class 12 Medical Chemistry Quiz Questions and Answers

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