Which of the following shows the correct order of decreasing basicity in aqueous medium?

(C H_{3})_{3} N > (C H_{3})_{2} N H > C H_{3} N H_{2} > N H_{3}

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(C H_{3})_{2} N H > (C H_{3})_{3} N > C H_{3} N H_{2} > N H_{3}

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(C H_{3})_{2} N H > C H_{3} N H_{2} > (C H_{3})_{3} N > N H_{3}

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(C H_{3})_{2} N H > C H_{3} N H_{2} > N H_{3} > (C H_{3})_{3} N

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Solution

The correct option is C $(C H_{3})_{2} N H > C H_{3} N H_{2} > (C H_{3})_{3} N > N H_{3}$
$+ I$ effect of $C H_{3}$ group increases electron density on $N$ and makes it a stronger base. So, the order of basic strength would be expected to be $3^{o} > 2^{o} > 1^{o} > N H_{3}$

But in an aqueous medium presence of hydrogen bonding and bulkier groups affects the basicity.

In $(C H_{3})_{3} N$ , alkyl groups hinder the attack of the proton on $N$ . Therefore, it becomes less basic and due to the additive effects of steric hindrance and induction, $(C H_{3})_{2} N H$ is the strongest base in the aqueous medium.

Thus, for an aqueous medium order of basicity is $2^{o} > 1^{o} > 3^{o} > N H_{3}$

Hence the correct option is C.

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Similar questions

C H_{3} N H_{2}, (C H_{3})_{2} N H, (C H_{3})_{3} N, N H_{3}

Arrange the following in the decreasing order of their basic strength in aqueous solution:

$C H_{3} N H_{2}, (C H_{3})_{2} N H, (C H_{3})_{3} N$ and $N H_{3}$

C H_{3} N H_{2}, (C H_{3})_{2} N H, (C H_{3})_{3} N

N H_{3}

C H_{3} N H_{2}, (C H_{3})_{3} N, (C H_{3})_{2} N H

Arrange the following in increasing order of their basic strength:

(i) C₂H₅NH₂, C₆H₅NH₂, NH₃, C₆H₅CH₂NH₂ and (C₂H₅)₂NH

(ii) C₂H₅NH₂, (C₂H₅)₂NH, (C₂H₅)₃N, C₆H₅NH₂

(iii) CH₃NH₂, (CH₃)₂NH, (CH₃)₃N, C₆H₅NH₂, C₆H₅CH₂NH₂.

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