Which of the following compound on treatment with $L i A l H_{4}$ will give a product that will give positive iodoform test?

C H_{3} C H_{2} C H O

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C H_{3} C H_{2} C O_{2} C H_{3}

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C H_{3} C H_{2} O C H_{2} C H_{3}

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C H_{3} C O C H_{3}

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Solution

The correct option is D $C H_{3} C O C H_{3}$
$C H_{3} C H_{2} C H O L i A l H_{4} - --- \to C H_{3} C H_{2} C H_{2} O H$

$C H_{3} C H_{2} - O - C H_{2} C H_{2} C H_{3} L i A l H_{4} - --- \to No reduction possible$

Iodoform test is used to check the presence of carbonyl compounds with the structure $R - C O - C H_{3}$ or alcohols with the structure $R - C H (O H) - C H_{3}$ in a given unknown substance.

When Iodine and sodium hydroxide are added to a compound that contains either a methyl ketone or a secondary alcohol with a methyl group in the alpha position, a pale yellow precipitate of iodoform or triiodomethane is formed. It can be used to identify aldehydes or ketones. If an aldehyde gives a positive iodoform test, then it must be acetaldehyde since it is the only aldehyde with a $C H_{3} C = O$ group.

Hence the last compound gives positive iodoform test.
Therefore, option (d) is correct.

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