In imine formation, why are the reaction conducted at $p H = 5 - 6$ $pH=5-6$ ?

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In imine formation, why are the reaction conducted at $p H = 5 - 6$ $pH=5-6$ ?

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Answer 1 · 2017-01-31T08:06:00

Because such a $p H$ $pH$ is optimal for bond-making, and for the elimination of water.

Explanation:

The general equation for the formation of an imine is:

$R_{1} C (= O) R_{2} + H_{2} N R_{3} \to R_{1} C (- O - H) (N H R_{3}) R_{2}$ $R_1C(=O)R_2 + H_2NR_3rarrR_1C(-O-H)(NHR_3)R_2$

With arrow pushing, the lone pair on the nitrogen is conceived to make a $C - N$ $C-N$ bond, and one of the originally nitrogen bound hydrogens transfer to the oxygen centre. If this oxygen centre becomes protonated to give:

$R_{1} C (-^{+} O H_{2}) (N H R_{3}) R_{2}$ $R_1C(-""^+OH_2)(NHR_3)R_2$

We now have a potential leaving group, that can be lost to give a $C = N$ $C=N$ bond:

$R_{1} C {= N (R_{3})} R_{2} + H_{2} O$ $R_1C{=N(R_3)}R_2 + H_2O$

The given $p H$ $pH$ optimizes bond formation, the formation of the $N - C$ $N-C$ bond, and is acidic enough to allow the elimination of water.

Most of this will be discussed in greater detail, and with much greater sofistication in your organic text. I urge you to consult it.

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In imine formation, why are the reaction conducted at $p H = 5 - 6$ $pH=5-6$ ?

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In imine formation, why are the reaction conducted at pH=5-6pH=5−6pH=5-6?

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In imine formation, why are the reaction conducted at $p H = 5 - 6$ $pH=5-6$ ?