Organic Compounds Containing Nitrogen — Page 5 | NEET Chemistry

Q41

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}

$

Q42

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.

Q43

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.

Q44

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

A.

Propanamine N–Methylethanamine B.

Hexan–2–one Hexan–3–one C.

Ethanamide Hydroxyethanimine D. o–Nitrophenol p–nitrophenol

Q45

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

Q46

Identify correct statements : (A) Primary amines do not give diazonium salts when treated with

\mathrm{NaNO}_2

in acidic condition. (B) Aliphatic and aromatic primary amines on heating with

\mathrm{CHCl}_3

and ethanolic KOH form carbylamines. (C) Secondary and tertiary amines also give carbylamine test. (D) Benzenesulfonyl chloride is known as Hinsberg's reagent. (E) Tertiary amines reacts with benzenesulfonyl chloride very easily. Choose the correct answer from the options given below :

A (B) and (C) only

B (D) and (E) only

C (A) and (B) only

D (B) and (D) only

Correct Answer

Option D

Solution

(A) $\mathrm{R}-\mathrm{NH}_2 \xrightarrow[\mathrm{HCl}]{\mathrm{NaNO}_2} \mathrm{R}-\mathrm{N}_2^{\oplus} \mathrm{Cl}^{\ominus}$ (B) (C) Only primary amine gives carbyl amine test (D) (E) Tertiary amine do not react with $\mathrm{Ph}-\mathrm{SO}_2 \mathrm{Cl}$ So correct options are (B) and (D) only

Q47

Match with $,$ \Delta$$

List - I		List - II
(B)	Clemenson reduction	(II)	$\mathrm{CHCl_3,NaOH/H_3O}$ $^ \oplus$
(C)	Cannizaro reaction	(III)	$\mathrm{Br_2,NaOH}$
(D)	Reimer-Tiemann Reaction	(IV)	$\mathrm{Zn-Hg/HCl}$

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

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

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

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

Correct Answer

Option A

Solution

.tg .tg Reaction Reagents (A) $\begin{aligned} & \text{ Hoffmann } \\ & \text{ degradation }\end{aligned}$ $\longrightarrow$ $\mathrm{Br}_{2}, \mathrm{NaOH}$ (B) $\begin{aligned} & \text{ Clemenson } \\ & \text{ reduction }\end{aligned}$ $\longrightarrow$ $\mathrm{Zn}-\mathrm{Hg} / \mathrm{HCl}$ (C) $\begin{aligned} & \text{ Cannizaro- } \\ & \text{ reaction }\end{aligned}$ $\longrightarrow$ Conc.

$\mathrm{KOH}, \Delta$ (D) $\begin{aligned} & \text{ Reimer-Tiemann } \\ & \text{ reaction }\end{aligned}$ $\longrightarrow$ $\begin{aligned} & \mathrm{CHCl}_{3}, \\ & \mathrm{NaOH} / \mathrm{H}_{3} \mathrm{O}^{+}\end{aligned}$ $\therefore$ Correct match is : (A)-III, (B) -IV, (C) -I, (D)-II

Q48

The gas leaked from a storage tank of the Union Carbide plant in Bhopal gas tragedy was:

A Methylamine

B Ammonia

C Phosgene

D Methyl isocyanate

Correct Answer

Option D

Solution

Methyl isocyanate

C{H_3} - N = C = O

Q49

The correct order of basic nature in aqueous solution for the bases

\mathrm{NH}_3, \mathrm{H}_2 \mathrm{~N}-\mathrm{NH}_2, \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{NH}_2,\left(\mathrm{CH}_3 \mathrm{CH}_2\right)_2 \mathrm{NH}

and

\left(\mathrm{CH}_3 \mathrm{CH}_2\right)_3 \mathrm{~N}

is :

A

\mathrm{NH}_3<\mathrm{H}_2 \mathrm{~N}-\mathrm{NH}_2<\left(\mathrm{CH}_3 \mathrm{CH}_2\right)_3 \mathrm{~N}<\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{NH}_2<\left(\mathrm{CH}_3 \mathrm{CH}_2\right)_2 \mathrm{NH}

B

\mathrm{H}_2 \mathrm{~N}-\mathrm{NH}_2<\mathrm{NH}_3<\left(\mathrm{CH}_3 \mathrm{CH}_2\right)_3 \mathrm{~N}<\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{NH}_2<\left(\mathrm{CH}_3 \mathrm{CH}_2\right)_2 \mathrm{NH}

C

\mathrm{NH}_2-\mathrm{NH}_2<\mathrm{NH}_3<\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{NH}_2<\left(\mathrm{CH}_3 \mathrm{CH}_2\right)_3 \mathrm{~N}<\left(\mathrm{CH}_3 \mathrm{CH}_2\right)_2 \mathrm{NH}

D

\mathrm{NH}_3<\mathrm{H}_2 \mathrm{~N}-\mathrm{NH}_2<\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{NH}_2<\left(\mathrm{CH}_3 \mathrm{CH}_2\right)_2 \mathrm{NH}<\left(\mathrm{CH}_3 \mathrm{CH}_2\right)_3 \mathrm{~N}

Correct Answer

Option C

Solution

Basic strength of amine depends on hydrogen bonding and electronic inductive effect.

$\mathrm{NH}(\mathrm{Et})_2>\mathrm{N}(\mathrm{Et})_3>\mathrm{NH}_2 \mathrm{Et}>\stackrel{\bullet\bullet}{\mathrm{NH}_3}>\mathrm{NH}_2-\mathrm{NH}_2$

Q50

Which one of the following is the strongest base in aqueous solution?

A Trimethylamine

B Aniline

C Dimethylamine

D Methylamine

Correct Answer

Option C

Solution

NOTE : Aromatic amines are less basic than aliphatic amines. Among aliphatic amines the order of basicity is

{2^ \circ } > {1^ \circ } > {3^ \circ }.

The electron density is decreased in

{3^ \circ }

amine due to crowing of alkyl group over

N

atom which makes the approach and bonding by a proton relatively difficult.

Therefore the basicity decreases.

Further Phenyl group show

- {\rm I}

effect, thus decreases the electron density on nitrogen atom and hence the basicity. $\therefore$ dimethylamine (

{2^ \circ }

aliphatic amine) is strongest base among gtiven choices.

$\therefore$ The correct order of basic strength is Dimethyl amine

>

Methyl amine

>

Trimethyl-amine

>

Aniline.