Which of the following is an incorrect match | vedclass

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(A) Let's analyze each option:$1$. Benzene and Benzaldehyde: Fehling solution is used to distinguish aliphatic aldehydes from ketones. Benzaldehyde is

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Compounds to DifferentiateReagent$1$. Benzene and BenzaldehydeFehling solution

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(A) Let's analyze each option:$1$. Benzene and Benzaldehyde: Fehling solution is used to distinguish aliphatic aldehydes from ketones. Benzaldehyde is an aromatic aldehyde and does not give a positive Fehling test. However,benzene is an inert hydrocarbon. This pair is generally not differentiated by Fehling solution as neither reacts positively in a way that distinguishes them effectively,making this an incorrect match.$2$. $CH_3OH$ and $C_2H_5OH$: $C_2H_5OH$ gives a positive iodoform test ($CHI_3$ yellow precipitate) with $NaOH/I_2$,while $CH_3OH$ does not. This is a correct match.$3$. $CH_3-C\equiv CH$ and $HC\equiv CH$: Both are terminal alkynes and will form precipitates with Tollens' reagent or ammoniacal cuprous chloride. They cannot be distinguished by this reagent. Wait,actually,both react. Thus,this is also problematic,but let's re-evaluate the most 'incorrect' one. Actually,$CH_3-C\equiv CH$ and $HC\equiv CH$ both react with $(Cu_2Cl_2 + NH_4OH)$ to form red precipitates. So this is also an incorrect match.However,in standard chemistry problems of this type,option $1$ is the most fundamentally incorrect as Fehling solution is specifically for aliphatic aldehydes.Actually,looking at the options provided,$1$ is the most clearly incorrect match.

Which of the following is an incorrect match for differentiating the given compounds?

Similar Questions

Acetaldehyde on treatment with dil. $NaOH$ followed by heating gives:

Which of the following products is formed when benzaldehyde is treated with $CH_3MgBr$ and the addition product so obtained is subjected to acid hydrolysis?

Which reagent is suitable for the following transformation?

Consider the following chemical reaction.
$CH \equiv CH \xrightarrow[{(2) CO, HCl, AlCl_3}]{{(1) Red hot Fe tube, 873 K}} \text{Product}$
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The same product $(A)$ is formed in both of the following reactions:
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Compounds to Differentiate	Reagent
$1$. Benzene and Benzaldehyde	Fehling solution

Compounds to Differentiate	Reagent
$2$. $CH_3OH$ and $C_2H_5OH$	$NaOH / I_2$

Compounds to Differentiate	Reagent
$3$. $CH_3-C\equiv CH$ and $HC\equiv CH$	$(Cu_2Cl_2 + NH_4OH)$

Compounds to Differentiate	Reagent
$4$. $(CH_3)_3COH$ and $(CH_3)_2CHCH_2OH$	$[HCl_{(conc.)} + ZnCl_{2(anhyd.)}]$