Compounds Containing Nitrogen

$\begin{aligned} & \text { Mol. } \mathrm{wt}=6 \times 12+(6 \times 1)+(2 \times 14)+(2 \times 16) \\ & =138 \\ & \% \mathrm{~N}=\frac{28}{138} \times 100=20.29 \%\end{aligned}$

Step 1: Hydrolysis – The compound first undergoes acid-catalysed hydrolysis in the presence of HCl. This breaks the bond between the carbon and nitrogen atoms, forming the corresponding amine intermediate.

Step 2: Role of aniline – Aniline is used both as a solvent and as a reactant. It helps in dissolving the intermediate and also participates in the next reaction.

Step 3: Rearrangement on heating – When the mixture is heated gently, a rearrangement occurs, producing a new compound that leads to the formation of the yellow product.

Step 4: Isolation of the product – After the reaction finishes, the mixture is cooled and acetic acid is added. Acetic acid neutralises extra acid and helps in the precipitation of the final yellow product.

Step 5: Calculation of nitrogen percentage

The molecular mass of the product is given as $197$, and the total mass of nitrogen atoms present is $42$.

Hence, the percentage of nitrogen is:

$ = \frac{42}{197} \times 100 = 21.30\% \approx 21\% $

Most basic compound (benzyl amine) has localized electron pairs (electrons are not involved in resonance). The basicity of aniline is reduced due to the resonance delocalisation of electrons. $-OCH_3$ group is also electron withdrawing (less than $-NO_2$) and hence electron density decrease and basicity decrease.

Mass of benzyl amine given = 1.0 g

Molar mass of benzyl amine = 107.15 g mol$^{-1}$

Moles of benzyl amine

$ = {{mass} \over {molar\,mass}}$

$ = {{1.0\,g} \over {107.15\,g\,mo{l^{ - 1}}}}$

$ = 0.0093327\,mol$

The stoichiometric ratio between both the reactants is 1 : 1.

Benzyl amine : HCl

$1:1$

So, moles of HCl = 0.0093327 mol

Mass of HCl consumed by benzyl amine is calculated as:

Mass = moles $\times$ molar mass

Molar mass of HCl = 36.5 g mol$^{-1}$

Mass = 0.0093327 mol $\times$ 36.5 g mol$^{-1}$

= 0.34064355 g

= 340.64355 mg

= 340.6 mg $\approx$ 341 mg

Molar mass of product $\left(\mathrm{C}_7 \mathrm{H}_7 \mathrm{Br}\right)(\mathrm{A})$ is $171 \mathrm{~g} \mathrm{~mol}^{-1}$

Molar mass of $A=154 \frac{\mathrm{gm}}{\mathrm{mole}}$

Compounds which can not undergo Friedel Crafts reaction are

Mole of aniline $=\frac{279}{93}=3 \mathrm{~mol}$

Mole of aniline yellow formed $=3 \mathrm{~mol}$

$\begin{gathered} \text { Mass }=3 \times 197 \\ =591 \mathrm{~g} \end{gathered}$

$-\mathrm{NH}_2$ group containing compound can give solid with Hinsberg's reagent, which is soluble in $\mathrm{NaOH}$ solution

can give solid with Heinsberg reagent, which is soluble in NaOH solution.

Molar mass of amine $=287 \mathrm{~g} / \mathrm{mol}$

$\begin{aligned} & \text { Moles of aniline }=0.1 \\ & \text { Moles of dye }=0.1 \\ & \text { Mass of dye }=0.1 \times 198=19.8 \mathrm{~gm} \\ & \approx 20 \\ & \end{aligned}$

$\begin{aligned} & \mathrm{HCl}+\mathrm{NaOH} \rightarrow \mathrm{NaCl}+\mathrm{H}_2 \mathrm{O} \\ & \mathrm{mol} \text { of } \mathrm{HCl}=0.2 \mathrm{~mol} \end{aligned}$

$\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{NH}_2 \xrightarrow[+\mathrm{HCl}]{\mathrm{NaNO}_2} \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{N}_2^{+} \mathrm{Cl}^{-} \xrightarrow{\mathrm{H}_2 \mathrm{O}} \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}+\mathrm{N}_2+\mathrm{HCl}$

$0.2 \mathrm{~mol} \mathrm{~HCl}$ would be generated by $0.2 \mathrm{~mol}$ $\mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2$

$\mathrm{x}=0.2 \times 45=9 \mathrm{~g}$

moles of aniline taken $=\frac{9.3}{93}=0.1$

$\% \text { yield }=\frac{26.4}{0.1 \times 330} \times 100=80 \%$

Total number of $\pi$ bonds in P = 8

Only reaction in option (C) is correct.

Gain in molecular weight after acylation with one $-\mathrm{NH}_2$ group is 42.

Total increase in molecular weight $=84$

$\therefore$ Number of amino group in $x=\frac{84}{42}=2$

Three nitrogen atoms are present as per structure below

Gabriel phthalimide synthesis is used to prepare 1^o aliphatic or alicyclic amine. Hence, amine which can be synthesised by Gabriel phthalimide synthesis method contains $\alpha $-carbon.

Aniline (C₆H₅NH₂) does not contain $\alpha $-C cannot be prepared by Gabriel reaction.

Remaining amines all contain $\alpha $-C in its respective position, hence they can easily give Gabriel phthalimide reaction.

Mass of organic salt produced (aniline) = 1.29 g

Molar mass of organic salt (aniline)

= 12 × 6 + 1 × 8 + 14 × 1 + 35 × 1

= 72 + 8 + 14 + 35

= 129 g/mol

$Moles\ of\ organic\ salt=\frac{Mass\ of\ organic\ salt}{Molar\ mass} $

$ =\frac{1.29}{129} =0.01\ mol $

From reaction 1 moles of salt is produced from 3 mole of Sn. So, 0.01 mole of organic salt is produced by 0.03 mole Sn. Atomic mass of Sn = 119 g mol⁻¹

Mass of Sn = x = mole of Sn × Molar mass

x = 0.03 × 119 $ \Rightarrow $ x = 3.57 g

The value of x is 3.57

1 mole of organic salt is produced by 1 mole of nitrobenzene 0.01 mole of organic salt is produced by 0.01 mole nitrobenzene.

Molar mass of nitrobenzene

= 12 × 6 + 1 × 5 + 14 × 1 + 16 × 2

= 72 + 5 + 14 + 32= 123 g mol⁻¹

Mass of nitrobenzene required,

y = moles of nitrobenzene × molar mass

= 0.01 × 123 = 1.23 g

Given,

${C_5}{H_{13}}N\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{{N_2}}^{NaN{O_2}/HCl}} Y$ (Tertiary alcohol) + Other products

When primary amine reacts with NaNO$_2$ + HCl, the amine undergoes diazotisation which then hydrolysed to an alcohol.

$R - N{H_2}\buildrel {NaN{O_2} + HCl} \over \longrightarrow R - OH + {N_2}(g) \uparrow $

If nitrogen gas is evolved during the reaction, the amine must be a primary amine. The alcohol Y is a tertiary alcohol. Thus, we can say that X is a primary amine.

There are 2 possibilities for deducing the structure of X.

(1) The possible structure of C$_5$H$_{13}$N is shown below, in which 4 C substituents are arranged on a single C atom. But, the compound is optically inactive.

Hence, this is not the structure of compound X.

(2) In the given structure, the C atoms are arranged in chain form with 2 methyl substituents. The C atom directly attached to amine group is chiral C. Therefore, the compound is optically active.

Hence, this is the structure of compound X. The formation of Y from X with its intermediate steps is shown below.

The compound X when reacts with NaNO$_2$ + HCl, it undergoes diazotisation which on removal of N$_2$ gas, form a secondary carbocation. The secondary carbocation undergoes hydride shift, producing a more stable tertiary carbocation. The tertiary carbocation on hydrolysis produces a tertiary alcohol which is optically inactive. The chemical reaction for formation of Y from X is

Final Answer :

Compound Y is optically inactive.

Hints :

The evolution of nitrogen gas during the reaction confirms that X is a primary amine.

The given reaction of compound A are as follows :

Compound A on reaction with NaBr + MnO$_2$ gives brown fumes with pungent smell with the formation of compound B. Also, compound A reacts with conc. HNO$_3$ to produce an intermediate C which on reaction with toluene producing compound D.

The reaction of compound A with NaBr + MnO$_2$ is an example of redox reaction. The compound A is sulphuric acid, H$_2$SO$_4$. The products formed during the reaction are Na$_2$SO$_4$, MnSO$_4$, Br$_2$ and H$_2$O. The bromine gas is brown in colour and have pungent smell. The chemical reaction for the same is shown below.

The compound A that is sulphuric acid on reaction with conc. HNO$_3$ yields nitronium ion as shown below. The compound C formed is an intermediate and is nitronium ion.

$\mathrm{\mathop {{H_2}S{O_4}}\limits_{(A)} + HN{O_3} \to HSO_4^ - + \mathop {NO_2^ \oplus }\limits_{(C)} + {H_2}O}$

The nitronium ion on reaction with toluene, causes its nitration, producing compound D. The nitronium ion is an electrophile and causes substitution of 3 H atoms at two ortho and one para position of benzene ring by nitro group. The compound D is formed which is trinitrotoluene or TNT. The TNT is explosive in nature.

The complete reaction sequence is as shown below.

Final Answer :

A: H$_2$SO$_4$

B : Br$_2$

C: NO$_2^+$