Identify the incorrect match with respect to | vedclass

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Question

(A) In option $A$,both $CH_3OH$ and $CH_3CH_2OH$ do not give the iodoform test. The iodoform test is given by compounds containing the $CH_3CO-$ group

Vedclass · Accepted Answer

$CH_3OH, CH_3CH_2OH \longrightarrow (I_2 + NaOH 	ext{ solution})$

Vedclass · Answer

(A) In option $A$,both $CH_3OH$ and $CH_3CH_2OH$ do not give the iodoform test. The iodoform test is given by compounds containing the $CH_3CO-$ group or $CH_3CH(OH)-$ group. Since neither compound contains these groups,they cannot be distinguished by $(I_2 + NaOH)$.In option $B$,Lucas reagent $(	ext{Anhydrous } ZnCl_2 + 	ext{Conc. } HCl)$ distinguishes primary,secondary,and tertiary alcohols. $CH_3CH_2OH$ is a primary alcohol (reacts slowly upon heating),while $CH_3-C(CH_3)_2-OH$ is a tertiary alcohol (reacts immediately). This is a correct match.In option $C$,terminal alkynes like $CH_3-C \equiv CH$ react with $Na$ to release $H_2$ gas due to the acidic hydrogen,whereas internal alkynes like $CH_3-C \equiv C-CH_3$ do not. This is a correct match.In option $D$,$2,4-DNP$ is used to identify carbonyl compounds. However,both aldehydes and ketones react with $2,4-DNP$ to form a yellow/orange precipitate,so it cannot distinguish between them. Wait,the question asks for the *incorrect* match. Actually,$2,4-DNP$ is a general test for carbonyls,not a distinguishing test. However,in the context of distinguishing,$A$ is definitely incorrect as neither reacts,and $D$ is also incorrect as both react. Re-evaluating: $A$ is the most fundamentally incorrect as the reagent is completely ineffective for both.

Identify the incorrect match with respect to compounds to be distinguished and the reagent used.

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