Solved: CHEMISTRY Question for NEET

NEET CHEMISTRYMedium

Compound A ( $C_8H_{10}O$ ) is found to react with NaOI (produced by reacting Y with NaOH) and yield a yellow precipitate with a characteristic smell. A and Y are respectively?

2-Phenylethanol and Iodine

Acetophenone and Iodine

1-Phenylethanol and Chlorine

1-Phenylethanol and Iodine

Step-by-Step Solution

Identify the Reaction: The formation of a yellow precipitate with a characteristic smell upon reaction with sodium hypoiodite (NaOI) is the specific confirmatory test for Iodoform ( $CHI_3$ ).
Identify Reagent Y: NaOI is prepared in situ by the reaction of Iodine ( $I_2$ ) with Sodium Hydroxide (NaOH). Therefore, Y is Iodine. ( $I_2 + 2NaOH \rightarrow NaOI + NaI + H_2O$ )
Identify Compound A: The iodoform test is positive for compounds containing the methyl ketone group ( $CH_3-C=O$ ) or the methyl carbinol group ( $CH_3-CH(OH)-$ ) linked to a carbon or hydrogen. The molecular formula is given as $C_8H_{10}O$ . 1-Phenylethanol ( $C_6H_5-CH(OH)-CH_3$ ) fits this formula ( $C_8H_{10}O$ ) and contains the required secondary alcohol group ( $CH_3-CH(OH)-$ ) attached to a phenyl ring. Upon oxidation by NaOI, it converts to acetophenone and then undergoes the haloform reaction to yield iodoform ( $CHI_3$ ) and sodium benzoate. Note: 2-Phenylethanol ( $C_6H_5-CH_2-CH_2-OH$ ) does not have the required grouping. Acetophenone ( $C_6H_5COCH_3$ ) reacts but its formula is $C_8H_8O$ , not $C_8H_{10}O$ .

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