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 option(s) with correct sequence of reagents for the conversion of $\mathbf{P}$ to $\mathbf{Q}$ is(are)
i) Lindlar's catalyst, $\mathrm{H}_2$;
ii) $\mathrm{SnCl}_2 / \mathrm{HCl}$;
iii) $\mathrm{NaBH}_4$;
iv) $\mathrm{H}_3 \mathrm{O}^{+}$
i) Lindlar's catalyst, $\mathrm{H}_2$;
ii) $\mathrm{H}_3 \mathrm{O}^{+}$;
iii) $\mathrm{SnCl}_2 / \mathrm{HCl}$;
iv) $\mathrm{NaBH}_4$
i) $\mathrm{NaBH}_4$;
ii) $\mathrm{SnCl}_2 / \mathrm{HCl}$;
iii) $\mathrm{H}_3 \mathrm{O}^{+}$;
iv) Lindlar's catalyst, $\mathrm{H}_2$
i) Lindlar's catalyst, $\mathrm{H}_2$;
ii) $\mathrm{NaBH}_4$;
iii) $\mathrm{SnCl}_2 / \mathrm{HCl}$;
iv) $\mathrm{H}_3 \mathrm{O}^{+}$
Considering the following reaction sequence,
the correct option(s) is(are)
2. $\mathrm{KMnO}_{4}-\mathrm{KOH}$, heat
In the following reactions, the major product W is
The major product of the reaction is
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)
List-I contains various reaction sequences and List-II contains different phenolic compounds. Match each entry in List-I with the appropriate entry in List-II and choose the correct option.
| List-I | List-II |
|---|---|
|
|
|
|
|
|
|
|
|
Match the compounds in LIST-I with the observations in LIST-II, and choose the correct option.
| List-I | List-II |
|---|---|
| (I) Aniline |
(P) Sodium fusion extract of the compound on boiling with $\mathrm{FeSO}_{4}$, followed by acidification with conc. $\mathrm{H}_{2} \mathrm{SO}_{4}$, gives Prussian blue color. |
| (II) $o$-Cresol | (Q) Sodium fusion extract of the compound on treatment with sodium nitroprusside gives blood red color. |
| (III) Cysteine |
(R) Addition of the compound to a saturated solution of $\mathrm{NaHCO}_{3}$ results in effervescence. |
| (IV) Caprolactam |
(S) The compound reacts with bromine water to give a white precipitate. |
| (T) Treating the compound with neutral $\mathrm{FeCl}_{3}$ solution produces violet color. |
The compound R is
The compound T is
Match the four starting materials (P, Q, R, S) given in List I with the corresponding reaction schemes (I, II, III, IV) provided in List II and select the correct answer using the code given below the lists.
The major product of the following reaction is
Amongst the compounds given, the one that would from a brilliant coloured dye on treatment with NaNO2 in dil. HCl followed by addition to an alkaline solution of $\beta$-naphthlol is
In the reaction,
The structure of the product T is :
| Column I | Column II |
|---|---|
(A) 
|
(P) Racemic mixture |
(B) 
|
(Q) Addition reaction |
(C) 
|
(R) Substitution reaction |
(D) 
|
(S) Coupling reaction |
Match each of the compounds in Column I with its characteristic reaction(s) in Column II.
| Column I | Column II | ||
|---|---|---|---|
| (A) | $C{H_3}C{H_2}C{H_2}CN$ | (P) | Reduction with $Pd - C/{H_2}$ |
| (B) | $C{H_3}C{H_2}OCOC{H_3}$ | (Q) | Reduction with $SnC{l_2}/HCl$ |
| (C) | $C{H_3} - CH = CH - C{H_2}OH$ | (R) | Development of foul smell on treatment with chloroform and alcoholic KOH |
| (D) | $C{H_3}C{H_2}C{H_2}C{H_2}N{H_2}$ | (S) | Reduction with diisobutylaluminium hydride (DIBAL-H) |
| (T) | Alkaline hydrolysis |
Statement 1 : Aniline on reaction with NaNO$_2$/HCl at 0$^\circ$C followed by coupling with $\beta$-naphthol gives a dark blue coloured precipitate.
Statement 2 : The colour of the compound formed in the reaction of aniline with NaNO$_2$/HCl at 0$^\circ$C followed by coupling with $\beta$-naphthol is due to the extended conjugation.
Match the compounds in Column I with their characteristic test(s)/reaction(s) given in Column II. Indicate your answer by darkening the appropriate bubbles of the 4 $\times$ 4 matrix given in the ORS.
| Column I | Column II | ||
|---|---|---|---|
| (A) |  |
(P) | sodium fusion extract of the compound gives Prussian blue colour with FeSO$_4$. |
| (B) |  |
(Q) | gives positive FeCl$_3$ test. |
| (C) |  |
(R) | gives white precipitate with AgNO$_3$. |
| (D) |  |
(S) | reacts with aldehydes to form the corresponding hydrazone derivative. |
$\mathrm{CH}_3 \mathrm{NH}_2+\mathrm{CHCl}_3+\mathrm{KOH} \rightarrow$ Nitrogen containing compound $+\mathrm{KCl}+\mathrm{H}_2 \mathrm{O}$.
Nitrogen containing compound is :
$\mathrm{CH}_3-\mathrm{C} \equiv \mathrm{N}$
$\mathrm{CH}_3-\mathrm{NH}-\mathrm{CH}_3$
$\mathrm{CH}_3-\overline{\mathrm{N}} \equiv \stackrel{+}{\mathrm{C}}$
$\mathrm{CH}_3-\stackrel{+}{\mathrm{N}} \equiv \stackrel{-}{\mathrm{C}}$
How can the conversion of (i) to (ii) be brought about?
KBr
$\mathrm{KBr}+\mathrm{CH}_3 \mathrm{ONa}$
$\mathrm{KBr}+\mathrm{KOH}$
$\mathrm{Br}_2+\mathrm{KOH}$
Which is the rate determining step in Hofmann bromamide degradation?
Formation of (i)
Formation of (ii)
Formation of (iii)
Formation of (iv)
The compound (iii) will loose $\mathrm{Br}^{-}$ion and lone pair of N atom will transfer to adjacent C atom forming a double bond. The compound (iv) is formed from compound (iii) through rearrangement which is slow and rate determining step of the reaction. The reacts with methanol.What are the constituent amines formed when the mixture of (i) and (ii) undergoes Hofmann bromamide degradation?
[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$_2^+$
