The reaction that you will conduct in this experiment is known as a "haloform" reaction. When a methyl ketone is used as a subst
rate, what is it converted to in this reaction?
1 answer:
It is converted into a CARBOXYLIC ACID
Haloform reaction mechanism.
First step
The base (hydroxide ion) takes out the alpha hydrogen producing enolate. Then, the reaction between the enolate and the halogen occurs, leading to the formation of the halogenated ketone along with the halogens corresponding anion.
Second step
Step 1 is repeated twice to yield a tri-halogenated ketone. The net reaction till the formation of the tri-halogenated ketone can be written as
The hydroxide ion acts as a nucleophile and attacks the electrophilic carbon which is doubly bonded to oxygen. This carbon-oxygen double bond becomes a single bond making the oxygen atom anionic. This makes the reformation of the carbon oxygen double bond favourable and the carbon attached to three halogens is displaced, leaving us with the carboxylic acid. An acid base reaction ensues, the carboxylic acid donates a proton to the tri-halomethyl anion giving the required haloform product.
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<h2>6 m/s</h2>Explanation:
The speed of the man can be found by using the formula
d is the distance
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From the question we have
We have the final answer as
<h3>6 m/s</h3>Hope this helps you