The correct order of decreasing basic strengt | vedclass

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Question

(C) The basic strength of amines depends on the availability of the lone pair of electrons on the nitrogen atom.$1$. Aliphatic amines are more basic t

Vedclass · Accepted Answer

$N$-ethylethanamine > ethanamine > $N$-methylaniline > benzenamine

Vedclass · Answer

(C) The basic strength of amines depends on the availability of the lone pair of electrons on the nitrogen atom.$1$. Aliphatic amines are more basic than aromatic amines because the lone pair on the nitrogen atom in aromatic amines is delocalized into the benzene ring.$2$. Among aliphatic amines,secondary amines ($N$-ethylethanamine) are generally more basic than primary amines (ethanamine) due to the electron-releasing inductive effect of the alkyl groups.$3$. Among aromatic amines,$N$-methylaniline is more basic than benzenamine due to the $+I$ effect of the methyl group,which increases the electron density on the nitrogen atom.Therefore,the correct order of decreasing basic strength is: $(C_2H_5)_2NH > C_2H_5NH_2 > C_6H_5NHCH_3 > C_6H_5NH_2$,which corresponds to $N$-ethylethanamine > ethanamine > $N$-methylaniline > benzenamine.

The correct order of decreasing basic strength of the given amines is:

Similar Questions

Which is the weakest Bronsted base?

The relation between $(A)$ and $(B)$ is:
$CH_3-CONH_2 \xrightarrow{KOBr} (A)$
$CH_3-CONH_2 \xrightarrow{LiAlH_4} (B)$

$C_6H_5NH_2 \xrightarrow{Br_2/CCl_4} ?$
What is the product?

Assertion : Acetamide reacts with $Br_2$ in presence of methanolic $CH_3ONa$ to form methyl $N$-methylcarbamate.
Reason : Methyl isocyanate is formed as an intermediate which reacts with methanol to form methyl $N$-methylcarbamate.

The major product of the reaction is:

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