The correct order of basicity is
Show Hint
The Correct Option is C
Solution and Explanation
\(\mathrm{CH_3NH_2}\): the lone pair on N is not delocalised and is pushed by the +I effect of methyl, so it is the strongest base.
In anilines the lone pair is delocalised into the ring, reducing basicity. p-Methoxy (+R) increases electron density on N, so p-anisidine is more basic than aniline (\(pK_b\) about 8.7 vs 9.4).
p-Nitro (-R, -I) withdraws electron density strongly, so p-nitroaniline is the weakest base (\(pK_b\) about 13).
Order: p-nitroaniline \(\lt\) aniline \(\lt\) p-methoxyaniline \(\lt\) \(\mathrm{CH_3NH_2}\), which is option (3).
Answer: (3) p-nitroaniline < aniline < p-methoxyaniline < CH3NH2
Top JEE Main Chemistry Questions
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- At equilibrium the concentrations of and in a sealed vessel at . What will be for the reaction
- Find out the major products from the following reactions
Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are
- The correct order of atomic radii is:
Top JEE Main Amines Questions
- Statement-I : Aniline on reaction with concentrated \(H_2SO_4\) at 475 K gives p-amino benzene sulphonic acid. This gives blood red colour with Lassaigne’s test.
Statement-II : Aniline forms a salt with anhydrus \(AlCl_3\) in Friedel Craft’s reaction. - Identify A and B in the following reaction sequence.
- The azo-dye (Y) formed in the following reactions is: Sulphanilic acid + NaNO$_2$ + CH$_3$COOH → X
- Given below are two statements:
Statement I: Aniline reacts with con. \( H_2SO_4 \) 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-Craft's alkylation and acylation reactions, aniline forms salt with the \( AlCl_3 \) catalyst. Due to this, nitrogen of aniline acquires a positive charge and acts as a deactivating group.
In the light of the above statements, choose the correct answer from the options given below: - A compound (\( x \)) with molar mass \( 108 \, \text{g mol}^{-1} \) undergoes acetylation to give a product with molar mass \( 192 \, \text{g mol}^{-1} \). The number of amino groups in the compound (\( x \)) is ______.
Top JEE Main Questions
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,