Compounds Containing Nitrogen
The mass of benzanilide obtained from the benzoylation reaction of 5.8 g of aniline, if yield of product is $82 \%$, is $\_\_\_\_$ g (nearest integer).
(Given molar mass in $\mathrm{g} \mathrm{mol}^{-1} \mathrm{H}: 1, \mathrm{C}: 12, \mathrm{~N}: 14, \mathrm{O}: 16$ )
Explanation:
Consider the following reaction sequence
The percentage of nitrogen in product ' T ' formed is $\_\_\_\_$ %. (Nearest integer)
(Given molar mass in $\mathrm{g} \mathrm{mol}^{-1} \mathrm{H}: 1, \mathrm{C}: 12, \mathrm{~N}: 14, \mathrm{O}: 16$ )
Explanation:
$\begin{aligned} & \text { Mol. } \mathrm{wt}=6 \times 12+(6 \times 1)+(2 \times 14)+(2 \times 16) \\ & =138 \\ & \% \mathrm{~N}=\frac{28}{138} \times 100=20.29 \%\end{aligned}$
Given below are some nitrogen containing compounds :
Each of them is treated with HCl separately. 1.0 g of the most basic compound will consume _______ mg of HCl.
(Given molar mass in g mol−1 C : 12, H : 1, O : 16, Cl : 35.5)
Explanation:
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
Consider the following sequence of reactions.
Total number of $\mathrm{sp}^3$ hybridised carbon atoms in the major product C formed is _________.
Explanation:
Consider the following sequence of reactions to produce major product (A)
$\begin{aligned} & \text { Molar mass of product }(\mathrm{A}) \text { is } \mathrm{g} \mathrm{~mol}^{-1} \text {. } \\ & \text { (Given molar mass in } \mathrm{g} \mathrm{~mol}^{-1} \text { of } \mathrm{C}: 12, \mathrm{H}: 1, \mathrm{O}: 16, \mathrm{Br}: 80, \mathrm{~N}: 14, \mathrm{P}: 31 \text { ) }\end{aligned}$
Explanation:
Molar mass of product $\left(\mathrm{C}_7 \mathrm{H}_7 \mathrm{Br}\right)(\mathrm{A})$ is $171 \mathrm{~g} \mathrm{~mol}^{-1}$
Consider the following sequence of reactions :
Molar mass of the product formed $(\mathrm{A})$ is __________ $\mathrm{g} \mathrm{~mol}^{-1}$.
Explanation:
Molar mass of $A=154 \frac{\mathrm{gm}}{\mathrm{mole}}$
Number of compounds from the following which cannot undergo Friedel-Crafts reactions is: _________
toluene, nitrobenzene, xylene, cumene, aniline, chlorobenzene, $m$-nitroaniline, $m$-dinitrobenzene
Explanation:
Compounds which can not undergo Friedel Crafts reaction are
If $279 \mathrm{~g}$ of aniline is reacted with one equivalent of benzenediazonium chloride, the maximum amount of aniline yellow formed will be ________ g. (nearest integer)
(consider complete conversion).
Explanation:
Mole of aniline $=\frac{279}{93}=3 \mathrm{~mol}$
Mole of aniline yellow formed $=3 \mathrm{~mol}$
$\begin{gathered} \text { Mass }=3 \times 197 \\ =591 \mathrm{~g} \end{gathered}$
Number of amine compounds from the following giving solids which are soluble in $\mathrm{NaOH}$ upon reaction with Hinsberg's reagent is _________.
Explanation:
$-\mathrm{NH}_2$ 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.
An amine $(\mathrm{X})$ is prepared by ammonolysis of benzyl chloride. On adding p-toluenesulphonyl chloride to it the solution remains clear. Molar mass of the amine $(\mathrm{X})$ formed is _________ $\mathrm{g} \mathrm{mol}^{-1}$.
(Given molar mass in $\mathrm{gmol}^{-1} \mathrm{C}: 12, \mathrm{H}: 1, \mathrm{O}: 16, \mathrm{~N}: 14$)
Explanation:
Molar mass of amine $=287 \mathrm{~g} / \mathrm{mol}$
$9.3 \mathrm{~g}$ of pure aniline upon diazotisation followed by coupling with phenol gives an orange dye. The mass of orange dye produced (assume 100% yield/conversion) is ________ g. (nearest integer)
Explanation:
$\begin{aligned} & \text { Moles of aniline }=0.1 \\ & \text { Moles of dye }=0.1 \\ & \text { Mass of dye }=0.1 \times 198=19.8 \mathrm{~gm} \\ & \approx 20 \\ & \end{aligned}$
$\mathrm{X} \mathrm{~g}$ of ethanamine was subjected to reaction with $\mathrm{NaNO}_2 / \mathrm{HCl}$ followed by hydrolysis to liberate $\mathrm{N}_2$ and $\mathrm{HCl}$. The $\mathrm{HCl}$ generated was completely neutralised by 0.2 moles of $\mathrm{NaOH} . \mathrm{X}$ is _________ g.
Explanation:
$\begin{aligned} & \mathrm{HCl}+\mathrm{NaOH} \rightarrow \mathrm{NaCl}+\mathrm{H}_2 \mathrm{O} \\ & \mathrm{mol} \text { of } \mathrm{HCl}=0.2 \mathrm{~mol} \end{aligned}$
$\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{NH}_2 \xrightarrow[+\mathrm{HCl}]{\mathrm{NaNO}_2} \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{N}_2^{+} \mathrm{Cl}^{-} \xrightarrow{\mathrm{H}_2 \mathrm{O}} \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}+\mathrm{N}_2+\mathrm{HCl}$
$0.2 \mathrm{~mol} \mathrm{~HCl}$ would be generated by $0.2 \mathrm{~mol}$ $\mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2$
$\mathrm{x}=0.2 \times 45=9 \mathrm{~g}$
$9.3 \mathrm{~g}$ of pure aniline is treated with bromine water at room temperature to give a white precipitate of the product '$\mathrm{P}$'. The mass of product '$\mathrm{P}$' obtained is $26.4 \mathrm{~g}$. The percentage yield is ________ %.
Explanation:
moles of aniline taken $=\frac{9.3}{93}=0.1$
$\% \text { yield }=\frac{26.4}{0.1 \times 330} \times 100=80 \%$
Phthalimide is made to undergo following sequence of reactions.
Total number of $\pi$ bonds present in product 'P' is/are ________.
Explanation:
Total number of $\pi$ bonds in P = 8
The number of the correct reaction(s) among the following is _______.
Explanation:
Only reaction in option (C) is correct.
Explanation:
A compound $(x)$ with molar mass $108 \mathrm{~g} \mathrm{~mol}^{-1}$ undergoes acetylation to give product with molar mass $192 \mathrm{~g} \mathrm{~mol}^{-1}$. The number of amino groups in the compound $(x)$ is ___________.
Explanation:
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 $x=\frac{84}{42}=2$
The compound formed by the reaction of ethanal with semicarbazide contains _________ number of nitrogen atoms.
Explanation:
When ethanal reacts with semicarbazide, semi-carbazone is formed via condensation, number of nitrogen atoms present in product is 3.
Number of isomeric aromatic amines with molecular formula $\mathrm{C}_{8} \mathrm{H}_{11} \mathrm{~N}$, which can be synthesized by Gabriel Phthalimide synthesis is ____________.
Explanation:
How many of the transformations given below would result in aromatic amines?
Explanation:
Gabriel phthalimide synthesis cannot be used to prepare Aniline.
The number of sp3 hybridised carbons in an acyclic neutral compound with molecular formula C4H5N is ___________.
Explanation:
$ \begin{aligned} \mathrm{DBE} & =(\mathrm{C}+1)-\left(\frac{\mathrm{H}+\mathrm{X}-\mathrm{N}}{2}\right) \\\\ & =4+1-\left(\frac{5-1}{2}\right)=5-2=3 \end{aligned} $
3 double bond equivalent are present in compound
Only 1 $s p^{3}$ hybridised carbon is there
(Keeping compound as acyclic)
I. Sn $-$ HCl
II. Sn $-$ NH4OH
III. Fe $-$ HCl
IV. Zn $-$ HCl
V. H2 $-$ Pd
VI. H2 $-$ Raney Nickel
Explanation:
(i) Sn + HCl
(ii) Fe + HCl
(iii) Zn + HCl
(iv) H2 $-$ Pd
(v) H2 (Raney Ni)
Explanation:
Explanation:
Three nitrogen atoms are present as per structure below
Explanation:
Gabriel phthalimide synthesis is used to prepare 1o 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.
Explanation:
Explanation:
Chemical formula of Histamine : C5H9N3
$ \therefore $ % by mas of N = ${{3 \times 14} \over {5 \times 12 + 1 \times 9 + 3 \times 14}} \times 100$
= ${{42} \over {111}} \times 100$
= 37.84%
[Use : Molar mass (in $\left.\mathrm{g} \mathrm{mol}^{-1}\right)$ : $\mathrm{H}=1, \mathrm{C}=12, \mathrm{~N}=14, \mathrm{O}=16, \mathrm{Br}=80, \mathrm{Cl}=35.5$
Atoms other than $\mathrm{C}$ and $\mathrm{H}$ are considered as heteroatoms]
Explanation:
Number of Heteroatoms in $R$ is 9 .
[Use : Molar mass (in $\mathrm{g} \mathrm{mol}^{-1}$ ): $\mathrm{H}=1, \mathrm{C}=12, \mathrm{~N}=14, \mathrm{O}=16, \mathrm{Br}=80, \mathrm{Cl}=35.5$
Atoms other than $\mathrm{C}$ and $\mathrm{H}$ are considered as heteroatoms]
Explanation:
Number of Carbon atoms + Number of Heteroatoms $=51$
Explanation:
(Use Molar masses (in g mol$-$1) of H, C, N, O, Cl and Sn as 1, 12, 14, 16, 35 and 119, respectively).
The value of x is _________.
Explanation:
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
(Use Molar masses (in g mol$-$1) of H, C, N, O, Cl and Sn as 1, 12, 14, 16, 35 and 119, respectively).
The value of y is _________.
Explanation:
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
Explanation:
$\beta $-naphthol couples with phenyldiazonium electrophile to produce an intense orange-red dye (Q) as major product.
Given that, volume of aniline (P) = 9.3 mL (density of P = 1.00 g/mL)
So, mass of aniline = 9.3 g
Molecular mass of aniline (C6H7N) = 93 g/mol
Therefore, moles of aniline = 9.3 / 93 = 0.1 mol of P.
Molecular mass of 2 napthol aniline orange dye (Q) = 248 g/mol
$ \Rightarrow $ 0.1 mol of aniline (P) will produce 0.1 mol of compound (Q).
But, according to the question the major product Q from P is 75%.
Therefore, mass of 'Q' produced
= (0.1 $ \times $ 248 $ \times $ 0.75)g = 18.60 g
Explanation:
In the following reaction, X is optically active.
$\underset{\mathbf{X}}{\mathrm{C}_5 \mathrm{H}_{13}} \mathrm{~N} \xrightarrow[\mathrm{~N}_2]{\mathrm{NaNO}_2 / \mathrm{HCl}} \mathbf{Y}($ Tertiary alcohol $)+$ Other products
Find X and Y. Is y optically active? Write all the intermediate steps.
Explanation:
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.
In the following reaction sequence :
Identify A, B, C and D. Also write chemical equations for cons version of A to B and A to C.
Explanation:
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$_2^+$
