Given below are two statements:
Statement I: Aniline reacts with con. H_2SO_4$H_2SO_4$ followed by heating at 453-473 K gives p-aminobenzene sulphonic acid, which gives blood red colour in the 'Lassaigne's test'.
Statement II: In Friedel-Crafts alkylation and acylation reactions, aniline forms salt with the AlCl_3$AlCl_3$ catalyst. Due to this, nitrogen of aniline acquires a positive charge and acts as deactivating group.
In the light of the above statements, choose the correct answer from the options given below:
A.text(1) Statement I is false but statement II is true$\text{(1) Statement I is false but statement II is true}$
B.text(2) Both statement I and statement II are false$\text{(2) Both statement I and statement II are false}$
C.text(3) Statement I is true but statement II is false$\text{(3) Statement I is true but statement II is false}$
D.text(4) Both statement I and statement II are true$\text{(4) Both statement I and statement II are true}$
Solution & Explanation
### Core Logic
Statement I: Aniline reacting with concentrated H_2SO_4$H_2SO_4$ gives anilinium hydrogensulphate, which on heating at 453-473 K produces sulphanilic acid (p-aminobenzene sulphonic acid). Because sulphanilic acid contains both Nitrogen and Sulphur, it gives a blood-red colouration in Lassaigne's test due to the formation of thiocyanate ion SCN^-$SCN^-$ which reacts with Fe^3+$Fe^{3+}$ to form [Fe(SCN)]^2+$[Fe(SCN)]^{2+}$. Thus, Statement I is true.
Statement II: In Friedel-Crafts reactions, the Lewis acid catalyst AlCl_3$AlCl_3$ reacts with the lone pair on the nitrogen atom of aniline to form a salt. This generates a positive charge on the nitrogen, transforming the -NH_2$-NH_2$ group from a strong activating group into a strong deactivating group, thus preventing the Friedel-Crafts reaction from occurring. Thus, Statement II is true.
Chemical Reactions of Amines diagram for Q70 - JEE Main 2024 Evening
### Step 1: Final Conclusion
Both Statement I and Statement II are true. Option (4) is correct.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: Amines
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
Keywords:#Chemical Reactions of Amines#JEE Main 2024 Evening Q70#Amines JEE Main 2024#Chemical Reactions of Amines JEE Main 2024
More Amines Previous-Year Questions — Page 5
Q69jee_main_2024_31_jan_eveningReactions of Diazonium Salts
The azo-dye (Y) formed in the following reactions is
textSulphanilic acid + NaNO_2 + CH_3COOH rightarrow X$\text{Sulphanilic acid } + NaNO_2 + CH_3COOH \rightarrow X$The image shows the reaction of intermediate X with N,N-dimethylaniline to form an azo dye (Y).
### Core Logic
1) Sulphanilic acid reacts with NaNO_2$NaNO_2$ and CH_3COOH$CH_3COOH$ to form a diazonium salt (X).
2) The diazonium salt (X) then reacts with N,N-dimethylaniline (given in the coupling step image). The coupling takes place at the para position of the highly activated N,N-dimethylaniline ring.
3) This coupling yields Methyl Orange, an azo dye. Its structure is p$p$-dimethylaminoazobenzenesulphonic acid.
The image shows the reaction of intermediate X with N,N-dimethylaniline to form an azo dye (Y).
### Step 1: Final Identification
The final product (Y) matches option (4) structurally, containing the sulphonic acid group on one ring, the azo linkage, and the N,N-dimethylamine group on the para position of the other ring.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Amines
Q83jee_main_2024_31_jan_eveningAcylation of Amines
A compound (x) with molar mass 108mathrm~g\,mol^-1$108\mathrm{~g\,mol^{-1}}$ undergoes acetylation to give product with molar mass 192mathrm~g\,mol^-1$192\mathrm{~g\,mol^{-1}}$. The number of amino groups in the compound (x) is ________.
Numerical Answer.Answer: 2 to 2
Solution
### Related Formula
R-NH_2 + CH_3COCl rightarrow R-NH-COCH_3 + HCl$R-NH_2 + CH_3COCl \rightarrow R-NH-COCH_3 + HCl$
### Core Logic
During the acetylation of an amino group, one hydrogen atom (mass = 1text g/mol$1\text{ g/mol}$) is replaced by an acetyl group (-COCH_3$-COCH_3$, mass = 43text g/mol$43\text{ g/mol}$).
Gain in molecular weight for every one -NH_2$-NH_2$ group acetylated = 43 - 1 = 42text g/mol$43 - 1 = 42\text{ g/mol}$.
### Step 1: Calculating Number of Groups
Total increase in molecular weight = Final mass - Initial mass = 192 - 108 = 84text g/mol$192 - 108 = 84\text{ g/mol}$.
textNumber of amino groups = fractextTotal mass increasetextMass increase per group = frac8442 = 2$\text{Number of amino groups} = \frac{\text{Total mass increase}}{\text{Mass increase per group}} = \frac{84}{42} = 2$
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Amines
More Amines Questions — jee_main_2024_31_jan_evening
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