Question

When iodine react with hot naoh product react with nal so which second compound is formed

Answer 1

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The reaction of iodine and base with methyl ketones is so reliable that the "iodoform test" (the appearance of a yellow precipitate) is used to probe the presence of a methyl ketone. This is also the case when testing for specific secondary alcohols containing at least one methyl group in alpha-position.

Answer 2

Explanation:

Answer:

Explanation:As we learnt that

Haloform reaction -

CH_{3}COR\:\:is\:oxidised\:by\:\:NaOI\:\:to\:form\:iodoform\:and\:\:RCO_{2}Na

- wherein

C_{2}H_{5}COCH_{3}+NaOH+I_{2}\rightarrow C_{2}H_{5}CO_{2}Na+CHI_{3}

 

 

 

Option 1)

This is correct.

Option 2)

This is incorrect.

Option 3)

This is incorrect.

Option 4)

This is incorrect.