Compounds Containing Nitrogen

$ \text { The complete reaction sequence is as follows } $

$ \text { The complete reaction sequence is as follows } $

Statement I

In the nitration of aniline, more amount of m-nitroaniline is formed than expected.

Normally, –NH₂ is an activating, ortho/para-directing group.

Hence, we expect mainly o- and p-nitroaniline.

However, nitration is carried out in strongly acidic medium (H₂SO₄/HNO₃), and under these conditions, aniline gets protonated to form the anilinium ion (C₆H₅NH₃⁺).

The –NH₃⁺ group is an electron-withdrawing, deactivating, meta-directing group.

So the nitration mixture now contains protonated aniline, which gives meta nitro product.

As a result, the actual nitration product mixture contains a higher proportion of m-nitroaniline than we would expect based on the activating nature of –NH₂ alone.

✅ Hence, Statement I is correct.

Statement II

In the presence of a strongly acidic medium, aniline is protonated to form anilinium ion, which is meta-directing.

As discussed above, this is true — in a strongly acidic medium, aniline is converted to the anilinium ion (C₆H₅NH₃⁺).

The anilinium ion is meta-directing because the positively charged nitrogen withdraws electrons from the ring, deactivating it, especially at ortho and para positions.

✅ Hence, Statement II is also correct.

✅ Final Answer

Option A: Both Statement I and Statement II are correct.

The reaction of benzene diazonium chloride with Cu and HCl is called as Gattermann reaction to form chlorobenzene.

The sequence of reagent required is, $\mathrm{NaNO}_2 / \mathrm{HCl} / 273-278 \mathrm{~K} ; \mathrm{CuCN} / \mathrm{KCN} ; \mathrm{H}_3 \mathrm{O}^{+}$ The reaction is,

$ \mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2+\mathrm{NaNO}_2+2 \mathrm{HCl} \xrightarrow{273-278 \mathrm{~K}} \mathrm{C}_6 \mathrm{H}_5 \mathrm{~N}_2^{+} \mathrm{Cl}^{-}+\mathrm{NaCl}+2 \mathrm{H}_2 \mathrm{O} $

$ \begin{aligned} & \mathrm{C}_6 \mathrm{H}_5 \mathrm{~N}_2^{+} \mathrm{Cl}^{-}+\mathrm{CuCN} \longrightarrow \mathrm{C}_6 \mathrm{H}_5 \mathrm{CN}+\mathrm{CuCl}+\mathrm{N}_2 \\ & \mathrm{C}_6 \mathrm{H}_5 \mathrm{CN}+2 \mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}+\mathrm{NH}_3 \end{aligned} $

The reaction mentioned is a Fitting reaction, The complete reaction is as follow,

$ \text { The complete reaction sequence is as follow } $

Total three amines with formula $\mathrm{C}_3 \mathrm{H}_9 \mathrm{~N}$ can react with benzene sulphonyl chloride.

(i) $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{NH}_2$ : Propylamine

(ii) $\left(\mathrm{CH}_3\right)_2 \mathrm{CHNH}_2$ : Isopropylamine

(iii) $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{NHCH}_3$ :

N -methylethylamine

$ \text { The complete reaction is as follows } $

Statement I:

$ \mathrm{CH_3NH_2} $ is more basic than $ \mathrm{NH_3} $ but $ \mathrm{C_6H_5NH_2} $ is less basic than $ \mathrm{NH_3} $.

Explanation:

• Methylamine (CH₃NH₂):

  The –CH₃ group is electron donating (+I effect), which increases the electron density on nitrogen, making it more basic than ammonia.

• Aniline (C₆H₅NH₂):

  The lone pair on nitrogen gets delocalized into the benzene ring through resonance, reducing its availability to accept a proton.

  So aniline is less basic than ammonia.

✅ Hence, Statement I is correct.

Statement II:

The order of basic strength of amines in aqueous phase follows the order

$ (\mathrm{C_2H_5})_3\mathrm{N} > (\mathrm{C_2H_5})_2\mathrm{NH} > \mathrm{C_2H_5NH_2} $.

Let's check this:

In the aqueous phase, basic strength is influenced by two factors:

1. +I effect (electron donating alkyl groups increase basicity)

2. Solvation effect (stabilization of the conjugate ammonium ion)

• In aqueous solution, the solvation effect is strong for primary > secondary > tertiary amines, because primary amines can form more hydrogen bonds.

• However, the +I effect increases as we move from primary → secondary → tertiary.

The combined effect in aqueous phase gives the observed order:

$ \textbf{Secondary amine} > \textbf{Primary amine} > \textbf{Tertiary amine} $

That is:

$ (\mathrm{C_2H_5})_2\mathrm{NH} > \mathrm{C_2H_5NH_2} > (\mathrm{C_2H_5})_3\mathrm{N} $

But the statement says the order is:

$ (\mathrm{C_2H_5})_3\mathrm{N} > (\mathrm{C_2H_5})_2\mathrm{NH} > \mathrm{C_2H_5NH_2} $

❌ This is incorrect in aqueous medium.

✅ Correct Answer:

Statement I is correct but Statement II is not correct.

✅ Option C

The complete reaction sequence is as follows

Let's analyze both statements one by one.

Statement (I): All the following compounds react with p-toluenesulfonyl chloride.

The compounds given are:

$\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2 \quad (\mathrm{Aniline})$

$\left(\mathrm{C}_6 \mathrm{H}_5\right)_2 \mathrm{NH} \quad (\mathrm{Diphenylamine})$

$\left(\mathrm{C}_6 \mathrm{H}_5\right)_3 \mathrm{N} \quad (\mathrm{Triphenylamine})$

P-toluenesulfonyl chloride (TsCl) reacts with amines to form sulfonamides. The general reaction is:

$\mathrm{RNH}_2 + \mathrm{TsCl} \rightarrow \mathrm{RNHTs} + \mathrm{HCl}$

For this reaction to occur, the amine must have at least one hydrogen attached to the nitrogen. Let's check the provided compounds:

1. $\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2$ (Aniline) - primary amine, has 2 hydrogens, so it reacts.

2. $\left(\mathrm{C}_6 \mathrm{H}_5\right)_2 \mathrm{NH}$ (Diphenylamine) - secondary amine, has 1 hydrogen, so it reacts.

3. $\left(\mathrm{C}_6 \mathrm{H}_5\right)_3 \mathrm{N}$ (Triphenylamine) - tertiary amine, has no hydrogen, so it does not react.

Hence, Statement I is false, since $\left(\mathrm{C}_6 \mathrm{H}_5\right)_3 \mathrm{N}$ does not react with p-toluenesulfonyl chloride.

Statement (II): Their products in the above reaction are soluble in aqueous $\mathrm{NaOH}$.

The products of the reaction of aniline and diphenylamine with p-toluenesulfonyl chloride (if it were to happen for the sake of analysis) would be sulfonamides:

$\mathrm{RNH}_2 + \mathrm{TsCl} \rightarrow \mathrm{RNHTs}$

Sulfonamides (RNHTs) are generally not soluble in aqueous NaOH because they are neutral compounds and NaOH is a strong base. Only aniline itself is soluble in aqueous NaOH due to its basic nature, but its product with TsCl is not.

Thus, Statement II is also false, as the products would not be soluble in aqueous NaOH.

Conclusion:

Both Statement I and Statement II are false. Therefore, the correct answer is:

Option C

Both Statement I and Statement II are false.

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.

Conversion of diazonium chloride to chlorobenzene with $\mathrm{Cu}_2 \mathrm{Cl}_2$ is Sandmeyer reaction and the fusion of chlorobenzene to sodium phenoxide in presence of $\mathrm{NaOH}$ at high temperature and high pressure is Dow's process.

Statement I is incorrect, and Statement II is correct.

The correct name for picric acid is 2,4,6-trinitrophenol, not 2,4,6-trinitrotoluene. Meanwhile, phenol-2,4-disulphonic acid, when treated with concentrated nitric acid ($\mathrm{HNO}_3$), results in the formation of picric acid.

Thus, the most appropriate option is :

Option A

Statement I is incorrect but Statement II is correct.

Hydrazine $(\mathrm{N}_2 \mathrm{H}_4)$ doesn't contain any carbon atom and hence doesn't give Lassaigne test.

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.

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.

This is known as Griess-Ilosvay test.

Since $-\ddot{\mathrm{N}}\mathrm{H}_2$ group is $\mathrm{o} / \mathrm{p}$ directing hence arenium ion will not be formed by attack at meta position i.e.

Hence Answer is (3).

Thus, $ X $ and $ Y $ are $ \left(\mathrm{NaOH}+\mathrm{Br}_{2}\right) $ and [(i) $ \mathrm{LiAlH}_{4} $

(ii) $ \mathrm{H}_{2} \mathrm{O} $ ]

The Wurtz-Fittig reaction is a chemical process that produces substituted aromatic compound from aryl halides, alkyl halide in presence of dry ether and sodium metal. The complete reaction is as follows.

The complete reaction is as follows.

Product $(Z)$ is an aldehyde.

Ethyl aldehyde is only aldehyde which gives iodoform test.

Hence, statements I and III are correct.

The lesser the $\mathrm{p} K_b$ value, more stronger will be the base.

Hence, the correct match is A-II, B-III, C-IV, D-I.

The reaction involved is as follows,

Thus, reagent/chemical $X$ and $Y$ are

DIBAL - $\mathrm{H}, \mathrm{H}_2 \mathrm{O}$ and

Statement given in B is correct which $A$ and $C$ are incorrect. The correct form of A and C are:

(A) Aniline can form stable diazonium salt at 273-283 K.

(C) Formula of Hinsberg reagent is $\mathrm{C}_6 \mathrm{H}_5 \mathrm{SO}_2 \mathrm{Cl}$.

Among the given options, $p$-methyl benzene nitrile can be prepared by