Correct order of basic strength of given amines is:

M e_{2} N M 2^{o} > M e N H_{2} 1^{o} > M e_{3} N > N H_{3} 3^{o}

(Protic solvent)

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E t_{2} N H 2^{o} > E t_{3} H 3^{o} > E t N H_{2} 1^{o} > N H_{3}

(Protic solvent)

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M e_{3} N > M e_{2} N H > M e - N H_{2} > N H_{3}

(Gas phase)

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All are correct

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Solution

The correct option is D All are correct
Refer to Image
Basic strength (in gas phase)
Because Me is an electron donating group it increases the $e^{-}$ density on N and hence increase its tendency to donate lone pair.
Basicity of amine is directly related to the stability of amine after gaining $H^{+}$ .
In the aqueous phase, the substituted ammonium cations gets stabilized not only by e^- releasing effect of the alkyl group (+I) but also by selvation with water molecules. The greater the size of ions, lesser will be the selvation and he less stabilised is the ion. Secondly, when the alkyl group is small like $- C H_{3}$ group, there is no steric hindrance to $- H$ bounding.
$∴ \begin{matrix} M e_{2} N H 2^{o} \end{matrix} > \begin{matrix} M e_{2} N H_{2} 1^{o} \end{matrix} > \begin{matrix} M e_{3} N 3^{o} \end{matrix} > N H_{3}$
$∴ \begin{matrix} E t_{2} N H 2^{o} \end{matrix} > \begin{matrix} E t_{3} N 3^{o} \end{matrix} > \begin{matrix} E t N H_{2} 1^{o} \end{matrix} > N H_{3}$

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Similar questions

1^{o}, 2^{o}, 3^{o}

N H_{3}

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3^{o}

(1^{o})

(2^{o})

(3^{o})

(1^{o}, 2^{o}

3^{o})

1^{o} \times 2^{o} + 3^{o} \times 4^{o} + 5^{o} \times 6^{o}

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