Compounds Containing Nitrogen

$ \begin{aligned} &\text{In Hinsberg’s test, the sulphonamide formed from a }2^{\circ}\text{ (secondary) amine is alkali insoluble.} \\\\ &\text{This is because a }2^{\circ}\text{ amine has no N–H left in the sulphonamide, so it cannot form a soluble salt with alkali.} \\\\ &\text{So, we have to count the }2^{\circ}\text{ amines from the given list.} \\\\ &\begin{aligned} & \mathrm{Ph}-\mathrm{NH}-\mathrm{CH}_3, \ \mathrm{Me}-\mathrm{NH}-\mathrm{Me}, \\\\ & \mathrm{Ph}-\mathrm{NH}-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_3, \ \mathrm{Ph}-\mathrm{NH}-\mathrm{Ph} \end{aligned} \end{aligned} $

Statement I : False

$ \mathrm{R}-\mathrm{NC} \xrightarrow{\mathrm{H}_3 \mathrm{O}^{+}} \mathrm{R}-\mathrm{NH}_2+\mathrm{HCOOH} $

Statement II : True

$ \mathrm{R}-\mathrm{CN} \xrightarrow{\mathrm{H}_3 \mathrm{O}^{+}} \mathrm{RCOOH} \xrightarrow{\mathrm{SOCl}_2} \mathrm{RCOCl} $

$ \xrightarrow{\mathrm{NH}_3} \mathrm{RCONH}_2 \xrightarrow{\mathrm{NaOCl} / \mathrm{OH}^{-}} \mathrm{RNH}_2 $

$ \xrightarrow{\mathrm{CHCl}_3+\mathrm{KOH}} \mathrm{R}-\mathrm{NC} $

Reaction (1:1):

$ \ce{C6H5NH2 + (CH3CO)2O -> C6H5NHCOCH3 + CH3COOH} $

• Molar mass of aniline $\ce{C6H7N}$ $\approx 93\,\text{g mol}^{-1}$

$ n(\text{aniline})=\frac{9.3}{93}=0.10\ \text{mol} $

• Molar mass of acetanilide $\ce{C8H9NO}$ $\approx 135\,\text{g mol}^{-1}$

Theoretical mass:

$ m_{\text{theor}} = 0.10 \times 135 = 13.5\ \text{g} $

Percent yield:

$ \%\,\text{yield}=\frac{11}{13.5}\times 100 \approx 81.5\% $

Answer: Option D (81.5%)

Correct order of stability

EDG increases stability

EWG decreases stability

Statement-I

Statement-II

Both statement (I) and (II) are true.

$\begin{aligned} & \text { Mole }=\frac{137 \times 10^{-3}}{137}=0.001 \mathrm{~mole} \\ & \text { Mole of product }=0.001 \mathrm{~mole} \\ & \text { Mass of product }=0.001 \times 228 \mathrm{gm} \\ & =0.228 \mathrm{gm}=228 \mathrm{mg}\end{aligned}$

Number of Br atom in major product $(\mathrm{P})=5$

$ \mathrm{C}_6 \mathrm{H}_7 \mathrm{N} \Rightarrow \mathrm{DU}=4 $

DU (Degree of Unsaturation) equal to 4 tells us there is one benzene ring in the molecule.

With the formula $\mathrm{C}_6 \mathrm{H}_7 \mathrm{~N}$ and one benzene ring, the compound must be aniline, which is a benzene ring attached to an $\mathrm{NH_2}$ group.

Molar mass $=121$

A $\rightarrow$ Benzamide Shows positive Lassaigh's test.

$B \rightarrow$ Benzoic acid gives effervescence with aq. $\mathrm{NaHCO}_3$.

$\mathrm{C} \rightarrow$ Ester gives fruity smell.

Aromatic halide give nucleophilic substitution reaction at high temperature or in presence of $-\mathrm{I} /-$ M group rate of reaction high even at low temperature.

A-IV

B-III

C-II

D-I

Only $1^{\circ}$ or primary amines gives positive carbylamines test.

Option (A) and (C) are primary amine and given $\oplus \mathrm{ve}$ carbyl amine test

0.1 mole of red-azo dye $($ Molar Mass $=327 \mathrm{gm} / \mathrm{mol})$ will have 32.7 gm mass. Nearly 33 gm .

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

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

Azo dyes are organic compounds that contain an azo bond $(-N=N-)$

(1)

There is no azo bond $-N=N-$ formation in this case.

So, azo dye compound is not formed

(2)

Azo dye $(-N=N-$) is not formed in this reaction.

Benzonitrile reaction with H$_2$SO$_4$, gives benzoic acid product. $-CN$ becomes $-COOH$. Then, reaction with $PCl_5$ gives benzoyl chloride $(C_6H_5COCl)$. Its on reaction with aniline does not give $-N=N-$ type compound. So, azo dye compound is not formed.

(3)

Nitrobenzene on reaction with $\mathrm{Sn} / \mathrm{HCl}$ undergoes reduction reaction to form aniline. $-\mathrm{NO}_2$ is reduced to $-\mathrm{NH}_2$.

Aniline on reaction with $\mathrm{NaNO}_2 / \mathrm{HCl}$ undergoes diazotization reaction to form benzene diazonium chloride.

Nitrobenzene on reaction with So/HCl undergoes reduction reaction to form aniline. $-NO_2$ is reduced to $-NH_2$.

Aniline on reaction with $NaNO_2/HCl$ undergoes diazotization reaction to form benzene diazonium chloride.

$\mathrm{NaNO}_2+\mathrm{HCl} \longrightarrow \mathrm{NaCl}+\mathrm{HNO}_2$

Benzene diazonium chloride (Benzene diazonium salt) on reaction with $\beta$-naphthol and NaOH gives an azo compound.

Azo dye is formed in this reaction

(4)

Azo dye is not formed.

The reaction is an example of nucleophilic aromatic substitution reaction.

The ethoxide ion $\left(\mathrm{C}_2 \mathrm{H}_5 \mathrm{O}^{-}\right)$replaces a bromine atom on the aromatic ring.

The nitro group $\left(-\mathrm{NO}_2\right)$ activates the benzene ling towards nucleophilic aromatic substitution. The Br atom para to the $-\mathrm{NO}_2$ group is more susceptible to substitution due to resonance stabilisation of the intermediate carbanion. Therefore, the $\mathrm{C}_2 \mathrm{H}_5 \mathrm{O}^{-}$ion replaces the Br atom para to the nitrogroup.

Steam volatile compounds can be distilled using steam distillation. Steam volatility is influenced by the intermolecular forces present in the substance. Steam volatile compounds have lower boiling points and. weaker intermolecular forces.

A compound's. Steam volatility is directly related to hydrogen bonding.

The presence of intramolecular hychogen bonds make a molecule more volatile. A compound's intermolecular hydrogen bonding decrease its volatility.

So, molecules with intramolecular hydrogen bonds are more likely to be steam volatile:

(A)

Intramolecular hydrogen bonding is present in this molecule.

So, it is steam volatile

(B)

The $-\mathrm{NH}_2$ group and $-\mathrm{NO}_2$ group are at oitho positions (adjacent positions) and hence there is intramolecular hydrogen bonding.

The electronegative oxygen of $-\mathrm{NO}_2$ group and hydrogen of $-\mathrm{NH}_2$ group makes the intramolecular hydrogen bonding. Due to this, intramolecular hydrogen bonding, O-nitroaniline is considered as steam volatile.

(C)

The amino group and hydroxy group are at para positions and the possibility of intramolecular hydrogen bonding is less in this compound. Therefore, this compound is not considered as steam volatile.

(D)

The $-\mathrm{NH}_2$ group and group are fan away to form intramolecular hydrogen bonding. As a result, it is not considered as steam volatile.

The steam volatile compounds are

Correct answer option (3) (A) and (B) only.

(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

In Dumas method nitrogen present in organic compound is converted into $\mathrm{N}_2$ gas whose volumetric analysis gives the percentage of nitrogen atom in the organic compound.

B and D are $3^{\circ}$ amine which does not have replaceable H on N , So does not react.

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).