Match List I with List II. .tg .tg List I List II (A) Benzenesulphonyl chloride (I) Test for primary amines (B) Hoffmann bromamide reaction (II) Anti Saytzeff (C) Carbylamine reaction (III) Hinsb

(A) Benzene sulphonyl chloride is also known as Hinsberg reagent. (B) Hoffmann bromamide reaction involves conversion of amide to amine having one $\mathrm{C}$ atom less. This reaction involves isocyanate as intermediate. (C) Carbylamine reaction is a test given by all primary amines. (D) Hoffmann orientation refers to the addition of molecules to unsymmetrical alkenes according to anti Saytzeff's rule. Correct match is A-III, B-IV, C-I, D-II

Organic Compounds Containing Nitrogen

JEE Chemistry · 46 questions · Page 3 of 5 · Click an option or "Show Solution" to reveal answer

Q21

Reaction of propanamide with

\mathrm{Br_2/KOH(aq)}

produces :

A Propanenitrile

B Propylamine

C Ethylnitrile

D Ethylamine

Correct Answer

Option D

Solution

Propanamide

\mathrm{\overset{{B{r_2}/KOH}}\longrightarrow}

Ethylamine

Q22

Which of the following nitrogen containing compound does not give Lassaigne's test ?

A Glycene

B Phenyl hydrazine

C Urea

D Hydrazine

Correct Answer

Option D

Solution

Hydrazine

(\mathrm{N}_2 \mathrm{H}_4)

doesn't contain any carbon atom and hence doesn't give Lassaigne test.

Q23

Given below are two statements : Statement (I) : Kjeldahl method is applicable to estimate nitrogen in pyridine. Statement (II) : The nitrogen present in pyridine can easily be converted into ammonium sulphate in Kjeldahl method. In the light of the above statements, choose the correct answer from the options given below :

A Both Statement I and Statement II are true

B Both Statement I and Statement II are false

C Statement I is true but Statement II is false

D Statement I is false but Statement II is true

Correct Answer

Option B

Solution

In order to determine the correct option, let’s analyze both statements in the context of the Kjeldahl method.

Statement I: Kjeldahl method is applicable to estimate nitrogen in pyridine.

The Kjeldahl method is a widely used procedure for estimating the nitrogen content in organic compounds.

However, this method is generally not applicable to aromatic nitrogen compounds like pyridine because the nitrogen in these compounds is not easily converted to ammonium form.

Therefore, Statement I is false.

Statement II: The nitrogen present in pyridine can easily be converted into ammonium sulphate in the Kjeldahl method.

The nitrogen in pyridine is part of the aromatic ring structure, making it more stable and less reactive compared to nitrogen in aliphatic amines or amides.

This means it cannot be easily converted into ammonium compounds such as ammonium sulphate via the Kjeldahl method.

Therefore, Statement II is also false.

Thus, the correct option is: Option B Both Statement I and Statement II are false.

Q24

Hinsberg's reagent is :

A C6H5COCl

B C6H5SO2Cl

C SOCl2

D (COCl)2

Correct Answer

Option B

Solution

The correct answer is Option B: C6H5SO2Cl.

Hinsberg's reagent is benzenesulfonyl chloride (C6H5SO2Cl).

It is used to differentiate primary, secondary, and tertiary amines.

Here's why the other options are incorrect : Option A: C6H5COCl is benzoyl chloride, a reagent used in Friedel-Crafts acylation reactions.

Option C: SOCl2 is thionyl chloride, a reagent used to convert alcohols into alkyl chlorides.

Option D: (COCl)2 is oxalyl chloride, a reagent used for the conversion of carboxylic acids into acid chlorides.

Q25

Match with . .tg .tg

List - I		List - II
(A)	Benzenesulphonyl chloride	(I)	Test for primary amines
(B)	Hoffmann bromamide reaction	(II)	Anti Saytzeff
(C)	Carbylamine reaction	(III)	Hinsberg reagent
(D)	Hoffmann orientation	(IV)	Known reaction of Isocyanates.

A A-IV, B-III, C-II, D-I

B A-IV, B-II, C-I, D-III

C A-III, B-IV, C-I, D-II

D A-IV, B-III, C-I, D-II

Correct Answer

Option C

Solution

(A) Benzene sulphonyl chloride is also known as Hinsberg reagent.

(B) Hoffmann bromamide reaction involves conversion of amide to amine having one $\mathrm{C}$ atom less.

This reaction involves isocyanate as intermediate.

(D) Hoffmann orientation refers to the addition of molecules to unsymmetrical alkenes according to anti Saytzeff's rule.

Correct match is A-III, B-IV, C-I, D-II

Q26

Considering the basic strength of amines in aqueous solution, which one has the smallest pKb value?

A (CH3)3N

B C6H5NH2

C (CH3)2NH

D CH3NH2

Correct Answer

Option C

Solution

Arylamines are less basic than alkyl amines and even ammonia.

This is due to resonance.

In aryl amines the lone pair of electrons on

N

is partly shared with the ring and is thus less available for sharing with a portion.

In alkylamines, the electron releasing alkyl group increases the electron density on nitrogen atom and thus also increases the ability of amine for protonation.

Hence more the no. of alkyl groups higher should be the basicity of amine.

But a slight discrepancy occurs in case of trimethyl amines due to steric effect.

Hence the correct order is

{\left( {CH{}_3} \right)_2}NH > C{H_3}N{H_2} > \left( {C{H_3}} \right){}_3N > {C_6}{H_5}N{H_2}

Q27

Given below are two statements : Statement (I) : The

\mathrm{NH}_2

group in Aniline is ortho and para directing and a powerful activating group. Statement (II) : Aniline does not undergo Friedel-Craft's reaction (alkylation and acylation). In the light of the above statements, choose the most appropriate answer from the options given below :

A Both Statement I and Statement II are correct

B Both Statement I and Statement II are incorrect

C Statement I is correct but Statement II is incorrect

D Statement I is incorrect but Statement II is correct

Correct Answer

Option A

Solution

The given statements pertain to aniline, a primary amine where the amino group ( $\mathrm{NH}_2$ ) is directly attached to a benzene ring.

Let's analyze the statements: Statement (I): The $\mathrm{NH}_2$ group in Aniline is ortho and para directing and a powerful activating group.

This statement is correct.

The amino group ( $\mathrm{NH}_2$ ) in aniline is an electron-donating group due to the lone pair of electrons on the nitrogen atom.

It increases the electron density on the benzene ring, particularly at the ortho and para positions.

This in turn makes the ortho and para positions more reactive toward electrophilic aromatic substitution reactions.

Therefore, the amino group is considered to be an ortho and para director and is one of the most powerful activating groups in the context of electrophilic aromatic substitution reactions.

Statement (II): Aniline does not undergo Friedel-Craft's reaction (alkylation and acylation).

This statement is also correct.

Aniline does not undergo Friedel-Crafts alkylation or acylation reactions.

The reason for this is twofold: Firstly, the Lewis acid catalysts used in Friedel-Crafts reactions (such as AlCl $_3$ ) react with the amino group to form a complex that deactivates the benzene ring toward further reaction.

Secondly, the strong interaction between the Lewis acid and the lone pair on the nitrogen can lead to a salt formation rather than the desired alkylation or acylation of the benzene ring.

Moreover, the acidic conditions can lead to protonation of the amino group, converting it into an $\mathrm{NH}_3^+$ group, which is meta-directing and deactivating, further inhibiting the reaction.

Given this analysis, the correct option is: Option A: Both Statement I and Statement II are correct.

Q28

Given below are two statements : Statement (I) : Aminobenzene and aniline are same organic compounds. Statement (II) : Aminobenzene and aniline are different organic compounds. In the light of the above statements, choose the most appropriate answer from the options given below :

A Both Statement I and Statement II are correct

B Statement I is incorrect but Statement II is correct

C Statement I is correct but Statement II is incorrect

D Both Statement I and Statement II are incorrect

Correct Answer

Option C

Solution

Statement (I) says that aminobenzene and aniline are the same organic compounds.

Statement (II) says that aminobenzene and aniline are different organic compounds.

To evaluate these statements, let's examine what each name refers to.

Aminobenzene and aniline both refer to the same compound, which has the chemical formula

C_6H_5NH_2

. This molecule consists of a benzene ring with an amino group (

NH_2

) attached to it.

Aminobenzene is a name that describes the structure based on its functional groups - an amino group attached to a benzene ring.

Aniline is the common name for this compound and is recognized by IUPAC as an acceptable name.

Given this information: Statement I is correct because aminobenzene and aniline refer to the same compound, that is,

C_6H_5NH_2

Statement II is incorrect because it states that aminobenzene and aniline are different organic compounds, which is not true.

Therefore, the most appropriate answer from the given options is: Option C Statement I is correct but Statement II is incorrect.

Q29

Match List I with List II .tg .tg List I Isomeric pairs List II Type of isomers A. Propanamine and N-Methylethanamine I. Metamers B. Hexan-2-one and Hexan-3-one II. Positional isomers C. Ethanamide and Hydroxyethanimine III. Functional isomers D. o-nitrophenol and p-nitrophenol IV. Tautomers Choose the correct answer from the options given below :

A A-III, B-I, C-IV, D-II

B A-III, B-IV, C-I, D-II

C A-II, B-III, C-I, D-IV

D A-IV, B-III, C-I, D-II

Correct Answer

Option A

Solution

Propanamine N–Methylethanamine B.

Hexan–2–one Hexan–3–one C.

Ethanamide Hydroxyethanimine D. o–Nitrophenol p–nitrophenol

Q30

In the chemical reaction, CH3CH2NH2 + CHCl3 + 3KOH

\to

(A) + (B) + 3H2O, the compound (A) and (B) are respectively

A C2H5CN and 3KCl

B CH3CH2CONH2 and 3KCl

C C2H5NC and K2CO3

D C2H5NC and 3KCl

Correct Answer

Option D

Solution

This is carbylamine reaction.

C{H_3}C{H_2}N{H_2} + CHC{l_3} + 3KOH

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\overset{\,}\longrightarrow {C_2}{H_5}NC + 3KCl + 3{H_2}O

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