Compounds Containing Nitrogen
A student performed analysis of aliphatic organic compound ‘X’ which on analysis gave C = 61.01%, H = 15.25%, N = 23.74%.
This compound, on treatment with HNO2/H2O produced another compound ‘Y’ which did not contain any nitrogen atom. However, the compound ‘Y’ upon controlled oxidation produced another compound ‘Z’ that responded to iodoform test.
The structure of ‘X’ is :
CH3–CH2CH(NH2)–CH3
CH3CH2CH2NH2
Ph–CH(CH3)–NH2
Total number of alkali insoluble solid sulphonamides obtained by reaction of given amines with Hinsberg's reagent is ________.
Aniline, N-Methylaniline, Methanamine, N,N-Dimethylmethanamine, N-Methyl methanamine, Phenylmethanamine, N-propylaniline, N-phenylaniline, N,N-Dimethylaniline, Allyl amine, Isopropyl amine
8
5
2
4
Given below are two statements :
Statement I : The dipole moment of R-CN is greater than R-NC and R-NC can
Statement II : R-CN hydrolyses under acidic medium to produce a compound which on treatment with $\mathrm{SOCl}_2$, followed by the addition of $\mathrm{NH}_3$ gives another compound $(\mathrm{x})$. This compound $(\mathrm{x})$ on treatment with $\mathrm{NaOCl} / \mathrm{NaOH}$ gives a product, that on treatment with $\mathrm{CHCl}_3 / \mathrm{KOH} / \Delta$ produces R-NC
In the light of the above statements, choose the correct answer from the options given below
Both Statement I and Statement II are false
Statement I is false but Statement II is true
Statement I is true but Statement II is false
Both Statement I and Statement II are true
A student has planned to prepare acetanilide from aniline using acetic anhydride.
The student has started from 9.3 g of aniline. However, the student has managed to obtain 11 g of dry acetanilide.
The % yield of this reaction is :-
59.5%
$72.5 %$
$97.5 %$
81.5%
$ \text { The correct stability order of the following diazonium salts is } $
$C>D>B>A$
$A>C>D>B$
$\mathrm{A}>\mathrm{B}>\mathrm{C}>\mathrm{D}$
$\mathrm{C}>\mathrm{A}>\mathrm{D}>\mathrm{B}$
$ \text { Given below are two statements : } $
In the light of the above statements, choose the correct answer from the options given below
Both Statement I and Statement II are true
Statement I is false but Statement II is true
Statement I is true but Statement II is false
Both Statement I and Statement II are false
A student has been given a compound " $x$ " of molecular formula- $\mathrm{C}_6 \mathrm{H}_7 \mathrm{~N}$. ' $x$ ' is sparingly soluble in water. However, on addition of dilute mineral acid, ' $x$ ' becomes soluble in water. ' $x$ ' when treated with $\mathrm{CHCl}_3$ and $\mathrm{KOH}(\mathrm{alc})$, ' $y$ ' is produced. ' $y$ ' has a specific unpleasant smell. On treatment with benzenesulphonyl chloride, ' $x$ ' gives a compound ' $z$ ' which is soluble in alkali. The number of different " H " atoms present in ' z ' is:-
8
4
5
7
Consider the following sequence of reactions.
Assuming that the reaction proceeds to completion, then 137 mg of 4-nitrotoluene will produce
$\_\_\_\_$ mg of $B$.
(Given molar mass in $\mathrm{g} \mathrm{mol}^{-1} \mathrm{H}: 1, \mathrm{C}: 12, \mathrm{~N}: 14, \mathrm{O}: 16, \mathrm{Br}: 80$ )
208
228
146
301
'A' is a neutral organic compound (M. F : $\mathrm{C}_8 \mathrm{H}_9 \mathrm{ON}$ ). On treatment with aqueous $\mathrm{Br}_2 / \mathrm{HO}^{(-)}$, ' A ' forms a compound ' B ' which is soluble in dilute acid. ' B ' on treatment with aqueous $\mathrm{NaNO}_2 / \mathrm{HCl}\left(0-5^{\circ} \mathrm{C}\right)$ produces a compound ' C ' which on treatment with $\mathrm{CuCN} / \mathrm{NaCN}$ produces ' D '. Hydrolysis of ' D ' produces ' E ' which is also obtainable from the hydrolysis of ' A '. ' E ' on treatment with acidified $\mathrm{KMnO}_4$ produces ' F '. ' F ' contains two different types of hydrogen atoms. The structure of ' A ' is
Consider the above sequence of reactions. The number of bromine atom(s) in the final product (P) will be :
6
5
1
3
An organic compound $(\mathrm{P})$ on treatment with aqueous ammonia under hot condition forms compound $(\mathrm{Q})$ which on heating with $\mathrm{Br}_2$ and KOH forms compound $(\mathrm{R})$ having molecular formula $\mathrm{C}_6 \mathrm{H}_7 \mathrm{~N}$. Names of P, Q and R respectively are.
Toluic acid, methylbenzamide, 2-methylaniline
Benzoic acid, benzamide, aniline
Benzoic acid, 4-methylbenzamide, 4-methylaniline
Phenylethanoic acid, phenylethanamide, benzamine
A hydrocarbon ' P ' $\left(\mathrm{C}_4 \mathrm{H}_8\right)$ on reaction with HCl gives an optically active compound ' Q ' $\left(\mathrm{C}_4 \mathrm{H}_9 \mathrm{Cl}\right)$ which on reaction with one mole of ammonia gives compound ' $\mathrm{R}^{\prime}\left(\mathrm{C}_4 \mathrm{H}_{11} \mathrm{~N}\right)$. ' $\mathrm{R}^{\prime}$ on diazotization followed by hydrolysis gives ' S '. Identify $\mathrm{P}, \mathrm{Q}, \mathrm{R}$ and S .
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:
$\begin{aligned} & \text { Mol. } \mathrm{wt}=6 \times 12+(6 \times 1)+(2 \times 14)+(2 \times 16) \\ & =138 \\ & \% \mathrm{~N}=\frac{28}{138} \times 100=20.29 \%\end{aligned}$
In the following reaction sequence, P, Q, S and T are the major products.
The correct statement(s) about P, Q, S and T is(are) :
Q on treatment with ethanol generates an aromatic aldehyde.
S gives positive phthalein dye test.
P is a dinitro compound.
T is a coloured compound.
$\mathrm{A} \xrightarrow[\text { (i) } \mathrm{H}_3 \mathrm{O}^{+}]{\text {(i) } \mathrm{NaH}} \mathrm{B} \xrightarrow[\text { (ii) } \mathrm{H}_2 \mathrm{SO}_4, \Delta]{\text { (i) } \mathrm{EOH}} \mathrm{C}$
' A ' shows positive Lassaign's test for N and its molar mass is 121 .
' B ' gives effervescence with aq $\mathrm{NaHCO}_3$.
'C' gives fruity smell.
Identify $\mathrm{A}, \mathrm{B}$ and C from the following.
Match List - I with List - II.
| List - I (Conversion) | List - II (Reagents, Conditions used) |
|---|---|
 |
(I) Warm H2O |
 |
(II) (a) NaOH, 368K; (b) H3O+ |
 |
(III) (a) NaOH, 443K; (b) H3O+ |
 |
(IV) (a) NaOH, 623K, 300 atm; (b) H3O+ |
Choose the correct answer from the options given below :
(A)-(II), (B)-(III), (C)-(I), (D)-(IV)
(A)-(III), (B)-(IV), (C)-(II), (D)-(I)
(A)-(IV), (B)-(III), (C)-(I), (D)-(II)
(A)-(IV), (B)-(III), (C)-(II), (D)-(I)
Which of the following amine (s) show (s) positive carbylamine test?
B. $\left(\mathrm{CH}_3\right)_2 \mathrm{NH}$
C. $\mathrm{CH}_3 \mathrm{NH}_2$
D. $\left(\mathrm{CH}_3\right)_3 \mathrm{~N}$
Choose the correct answer from the options given below:
The major product (A) formed in the following reaction sequence is
The sequence from the following that would result in giving predominantly $3,4,5-$ Tribromoaniline is :
$ \text {Identify }[\mathrm{A}],[\mathrm{B}] \text { and }[\mathrm{C}] \text {, respectively in the following reaction sequence : } $
$ \text { In the following reactions, which one is NOT correct? } $
When a concentrated solution of sulphanilic acid and 1-naphthylamine is treated with nitrous acid $(273 \mathrm{~K})$ and acidified with acetic acid, the mass $(\mathrm{g})$ of 0.1 mole of product formed is :
(Given molar mass in $\mathrm{g} \mathrm{mol}^{-1} \mathrm{H}: 1, \mathrm{C}: 12, \mathrm{~N}: 14, \mathrm{O}: 16, \mathrm{~S}: 32$ )
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 :
Which one of the following reaction sequences will give an azo dye?
In the following substitution reaction :
Product 'P' formed is :
The steam volatile compounds among the following are :
$\text { Choose the correct answer from the options given below : }$
(B) and (D) Only
(A), (B) and (C) Only
(A) and (B) Only
(A) and (C) Only
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 :
(B) and (C) only
(D) and (E) only
(A) and (B) only
(B) and (D) only
For reaction
The correct order of set of reagents for the above conversion is :
Given below are two statements I and II.
Statement I : Dumas method is used for estimation of "Nitrogen" in an organic compound.
Statement II : Dumas method involves the formation of ammonium sulphate by heating the organic compound with conc $\mathrm{H}_2 \mathrm{SO}_4$.
In the light of the above statements, choose the correct answer from the options given below
Which among the following react with Hinsberg's reagent?
A. 
B. 
$\text { C. } \mathrm{CH}_3-\mathrm{NH}_2$
D. $\mathrm{N}\left(\mathrm{CH}_3\right)_3$
E. 
Choose the correct answer from the options given below :
Match the Compounds (List - I) with the appropriate Catalyst/ Reagents (List - II) for their reduction into corresponding amines.
| List - I (Compounds) |
List - II (Catalyst/Reagents) |
||
|---|---|---|---|
| (A) |  |
(I) | $ \mathrm{NaOH} \text { (aqueous) } $ |
| (B) |  |
(II) | $ \mathrm{H}_2 / \mathrm{Ni} $ |
| (C) | $ \mathrm{R}-\mathrm{C} \equiv \mathrm{~N} $ |
(III) | $ \mathrm{LiAlH}_4, \mathrm{H}_2 \mathrm{O} $ |
| (D) |  |
(IV) | $ \mathrm{Sn}, \mathrm{HCl} $ |
Choose the correct answer from the options given below :
The products formed in the following reaction sequence are :
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 $A=154 \frac{\mathrm{gm}}{\mathrm{mole}}$
For the reaction sequence given below, the correct statement(s) is(are):
Both X and Y are oxygen containing compounds.
Y on heating with CHCl$_3$/KOH forms isocyanide.
Z reacts with Hinsberg’s reagent.
Z is an aromatic primary amine.
The major products obtained from the reactions in List-II are the reactants for the named reactions mentioned in List-I. Match each entry in List-I with the appropriate entry in List-II and choose the correct option.
| List–I | List–II |
|---|---|
| (P) Stephen reaction | (1) $ \text { Toluene } \xrightarrow{\begin{array}{l} \text { (i) } \mathrm{CrO}_2 \mathrm{Cl}_2 / \mathrm{CS}_2 \\ \text { (ii) } \mathrm{H}_3 \mathrm{O}^{+} \end{array}} $ |
| (Q) Sandmeyer reaction | (2) $ \text { Benzoic acid } \xrightarrow{\substack{\text { (i) } \mathrm{PCl}_5 \\ \text { (ii) } \mathrm{NH}_3 \\ \text { (iii) } \mathrm{P}_4 \mathrm{O}_{10}, \Delta}} $ |
| (R) Hoffmann bromamide degradation reaction | (3) $ \text { Nitrobenzene } \xrightarrow{\begin{array}{l} \text { (i) } \mathrm{Fe}, \mathrm{HCl} \\ \text { (ii) } \mathrm{HCl}, \mathrm{NaNO}_2 \\ (273-278 \mathrm{~K}), \mathrm{H}_2 \mathrm{O} \end{array}} $ |
| (S) Cannizzaro reaction | (4) $ \text { Toluene } \xrightarrow{\begin{array}{ll} \text { (i) } \mathrm{Cl}_2 / \mathrm{h\nu}, \mathrm{H}_2 \mathrm{O} \\ \text { (ii) Tollen's reagent } \\ \text { (iii) } \mathrm{SO}_2 \mathrm{Cl}_2 \\ \text { (iv) } \mathrm{NH}_3 \end{array}} $ |
| (5) $ \text { Aniline } \xrightarrow{\begin{array}{l} \text { (i) }\left(\mathrm{CH}_3 \mathrm{CO}\right)_2 \mathrm{O}, \text { Pyridine } \\ \text { (ii) } \mathrm{HNO}_3, \mathrm{H}_2 \mathrm{SO}_4, 288 \mathrm{~K} \\ \text { (iii) aq. } \mathrm{NaOH} \end{array}} $ |
P → 2; Q → 4; R → 1; S → 3
P → 2; Q → 3; R → 4; S → 1
P → 5; Q → 3; R → 4; S → 2
P → 5; Q → 4; R → 2; S → 1
This reaction is known as Etard reaction $-\mathrm{CH}_3$ group in toluene is oxidized by oxidizing agent chromylchloside $\left(\mathrm{CrO}_2 \mathrm{Cl}_2\right)$ in carbon disulfide $\left(C S_2\right)$ as solvent, to form a chromiumcomplex. This complex on arid hydrolysis gives the major product benzaldehyde.$
\text { (2) }
$$
\text { Benzoic acid } \xrightarrow[\substack{\text { 2) } \mathrm{NH}_3 \\ \text { 3) } \mathrm{P}_4 \mathrm{O}_{10}, \Delta}]{\text { 1) } \mathrm{Pcl}_5} \text { Benzaldehyde }
$Benzoic acid reacts with $\mathrm{PCl}_5$ to form benzoyl chloride. This step converts the - COOH group to. -COCl gorup.
Then, benzoyl chloride reacts with ammonia. This reaction converts the avid chloride to an amide, -COCl to $-\mathrm{CONH}_2$. It is the nucleophilic acyl substitution reaction where amonia acts as the nucleophile. So, this step gives benzamide.
In the next step, the amide is treated with $P_4 O_{10}$ and heat $P_4 O_{10}$ is a dehydrating agent, which removes water from the amide molecule, resulting in the formation of a nitrile. So, $-\mathrm{CONH}_2$ convert to -CN group.
The overall reaction converts benzoic acid to benzonitrile$
\text { (3) }
$
$
\text { Nitrobenzene } \xrightarrow[\substack{2) \mathrm{Hcl}, \mathrm{NaNO}_2 \\(273-278 \mathrm{~K}), \mathrm{H}_2 \mathrm{O}}]{\text { 1) } \mathrm{Fe}, \mathrm{Hcl}} \text { Phenol }
$$
\text { In the first step, nitrobenzene convert to aniline }
$
In the next step, aniline reacts with HCl and $\mathrm{NaNO}_2$ gives benzenediazonium chloride. The reaction condition is low temperature $273-278 \mathrm{~K}$ $\left(\mathrm{O}^{\circ} \mathrm{C}-5^{\circ} \mathrm{C}\right)$. It is called diazotization. $\mathrm{H}_2 \mathrm{O}$ works as proton donoe
Diazonium salt formation occurs at low temperature only.So, the temperature condition $273-278 k$ is the low temperature and it is the required temperature for the, diazotization reaction.For diazotization reaction, aniline (aromatic amine) is the reactant. $\mathrm{HCl}+\mathrm{NaNO}_2$ is the reagent. Or $\mathrm{HNO}_2$ can also be used for the diazotization reaction The reaction involves the formation of nitrosonium ion ( $\mathrm{NO}^{+}$) which attacks the aromatic ring, leading to the formation of diazonium salt.
The final product is diazonium salt$
\text { (4) }
$
First, toluene reacts with $\mathrm{Cl}_2$ in the presence of light. Side chain chlorination occurs in toluene, results in benzol chloride. It is then hydrolysis$
\text { (reaction with water) gives benzaldehyde. }
$
In the next step benzaldehyde reacts with Tollen's reagent and forms benzoic acid - CHO convert to -COOH 
In the third step, benzoic acid reacts with $\mathrm{SO}_2 \mathrm{Cl}_2$ and gives benzoyl chloride.
In the next step, benzoyl chloride reaction with ammonia gives benzamide.
The fined major product of the overall reaction is benzamide.$
\text { (5) }
$$
\text { Aniline } \xrightarrow[\text { 3)aq, } \mathrm{NaOH}]{\substack{\text { 1) }\left(\mathrm{CH}_3 \mathrm{CO}\right)_2 \mathrm{O}, \text { Pyridine } \\ \text { 2) } \mathrm{HNO}_3, \mathrm{H}_2 \mathrm{SO}_4, 288 \mathrm{~K}}}
$Aniline reacts with acetic anhydride $\left(\mathrm{CH}_3 \mathrm{CO}\right)_2 \mathrm{O}$ in the presence of pyridine gives actylation of amingroup of aniline, forming a cetanilide.Pyridine acts as a base to neutralize the acid produced during the reaction
Then, acetanilide undergoes nitration by reacting with a mixture of $\mathrm{HNO}_3$ and $\mathrm{H}_2 \mathrm{SO}_4$ at 288 K gives p-nitroacetanilide. $-\mathrm{NO}_2$ group is substituted at the para position. $\mathrm{H}_2 \mathrm{SO}_4$ acts as catalyst.
In the third step, para-nitroacetanilide reaction with aqueous NaOH give para-nitroaniline. The amide bond is hydrolyzed, removing the acetyl group and regenerating the amino group
$
\text { Final product is p-nitroaniline }
$$
\text { Analyzing the list-I reactions, }
$$
\text { (P) Stephen, reaction }
$In this reaction nitriles are converted into aldehydes.
The reactant of stephen reaction is nitriles. In List - II, reaction (2) product is benzonitrile So, P matches with reaction (2)$
P-2
$(Q) Standmeyer reaction
In this reaction an aromatic amine converts to an aryl halide, cyanide, or hydroxilated compound by using diazonium salts and copper (1) salts as catalysts.
Aromatic amine to diazonium salt reaction is known as Dicizotization.
Diazonium salt reaction with Cu(1) salt is the Sandmeyer reaction.
The reactant of Sandmeyer reaction is diazonium salt.In List- II, reaction (3) gives diazonium salt product.$
\text { So, Q matches with reaction (3) }
$$
Q-3
$(R) Hoffman Bromamide degradation reaction. This reaction converts amides to primary amines
$
\mathrm{R}-\mathrm{CONH} \xrightarrow{\mathrm{Br}_2, \mathrm{NaOH}} \mathrm{R}-\mathrm{NH}_2
$The reactant fol this reaction is an amide.
List -II, reaction (4) product is an amide ( benzamide).
So, $R$ matches reaction (4)$
R-4
$$
\text { (5) Cannizzaro reaction }
$It is the reaction of base -catalyzed disproportionation of aldehydes that have no $\alpha$-hydrogens.
$
2 \,\,\mathrm{R}-\mathrm{CHO}\frac{\mathrm{conc} \cdot \mathrm{NaOH}}{\Delta} \mathrm{R}-\mathrm{CH}_2 \mathrm{OH}+\mathrm{R}-\mathrm{COONa}
$$
(R=H \text {, Ar or Alkyl group without } \alpha-H)
$Reaction (1) product benzaldehyde is an aldehyde that lacks $\alpha$-hydrogen. So, $S$ matches with reaction (1).$
S-1
$
Answer: Correct matching is
$
P \rightarrow 2, Q \rightarrow 3, R \rightarrow 4, S \rightarrow 1
$
Option (B)
What is the product ' $Z$ ' in the given sequence of reactions?
What is the major product ' $Z$ ' in the given reaction sequence?
The reagent which is used to distinguish primary, secondary and tertiary amines from the mixture is
Fehling's reagent
Tollen's reagent
Lucas reagent
Hinsberg's reagent
$ \text { The product ' } C \text { ' in the given reaction sequence is } $
The amine/salt of amine which gives positive test with a mixture of chloroform and alcoholic KOH solution is
The major products $P$ and $Q$ from the following reactions are
$ P=\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2 ; Q=\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{NH}_2 $
$ P=\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{NH}_2 ; Q=\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2 $
$ P=\mathrm{C}_6 \mathrm{H}_5 \mathrm{CN} ; Q=\mathrm{C}_6 \mathrm{H}_5 \mathrm{Br} $
Which of the following sets of reagents convert aniline to chlorobenzene?
$\mathrm{NaNO}_2 / \mathrm{HCl}, 273-278 \mathrm{~K} ; \mathrm{Cu}_2 \mathrm{Cl}_2 / \mathrm{HCl}$
$\mathrm{NaNO}_2 / \mathrm{HCl}, 293-298 \mathrm{~K} ; \mathrm{Cu}_2 \mathrm{Cl}_2 / \mathrm{HCl}$
$ \mathrm{C}_6 \mathrm{H}_5 \mathrm{~N}_2^{+} \mathrm{Cl}^{-} \longrightarrow \mathrm{C}_6 \mathrm{H}_5 \mathrm{Cl}+\mathrm{N}_2 $The \% of carbon in ' $Z$ ' is (At.wt. C=12u,H=1u,N=14u $\mathrm{O}=16 \mathrm{u})$
71.3
51.3
61.3
48.3
Molar mass of $Z$ i.e.,What are $B$ and $C$ respectively in the following set of reactions?
What are $X$ and $Y$ respectively in the following set of reactions?
Benzyl amine can be prepared from which of the following reactions?
$\mathrm{C}_6 \mathrm{H}_5 \mathrm{Cl} \xrightarrow{\mathrm{CH}_3 \mathrm{NH}_2}$
$\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{Cl} \xrightarrow[\text { (si) } \mathrm{H}_2 / \mathrm{Ni}]{\text { (N) } \mathrm{AgCN}}$
$\mathrm{C}_6 \mathrm{H}_5 \mathrm{CONH}_2 \xrightarrow{\mathrm{NaOH} / \mathrm{Br}_2}$
$\mathrm{C}_6 \mathrm{H}_5 \mathrm{CONH}_2 \xrightarrow[\text { (si) } \mathrm{H}_2 \mathrm{O}]{\text { (I) } \mathrm{LAH}_4}$
