Hydrocarbons

Given reaction sequence:

L → (i) PhMgBr → (ii) H₃O⁺ → M → CrO₃/H⁺ → N → Ph₃P=CH₂ → Ph₂C=CH₂

Step-by-step interpretation:

Step 1: L + PhMgBr → M (Grignard reaction)
This is a nucleophilic addition of Grignard reagent (PhMgBr) to a carbonyl compound. For the Grignard reagent to work, L must be a compound with a carbonyl group (like an aldehyde or ketone). If L = Benzaldehyde (Ph-CHO), the reaction proceeds as:
Ph-CHO + PhMgBr → Ph-CH(OH)-Ph (a secondary alcohol, M)

Step 2: M → N using CrO₃/H⁺
This step is oxidation. The alcohol (Ph-CH(OH)-Ph) is oxidized to a ketone:
Ph-CH(OH)-Ph → Ph-CO-Ph (benzophenone, N)

Step 3: N → Final product using Ph₃P=CH₂ (Wittig reaction)
The ketone (benzophenone) reacts with Ph₃P=CH₂ to form an alkene:
Ph-CO-Ph + Ph₃P=CH₂ → Ph₂C=CH₂

Therefore, the correct identity of L is: Benzaldehyde

Correct option: (A): Benzaldehyde

There are two common qualitative methods used to test for unsaturation in compounds: the reaction with bromine in carbon tetrachloride and the reaction with potassium permanganate. In both cases, a positive result is indicated by the fading of the reagent's color.

In the bromine in carbon tetrachloride test, alkenes (with C=C) and alkynes (with C≡C) react with bromine to form dibromo- or tetrabromo-compounds. This uses up molecular bromine, which causes the dark red-brown color to fade. Thus, rapid disappearance of the bromine color confirms unsaturation.

The Baeyer test uses potassium permanganate ( ), an oxidizing agent. It oxidizes:

• Alkenes to glycols (alkanediols)
• Alkynes to carboxylic acids

In both cases, the purple color of fades, indicating a positive test.

Only ethene (an alkene) shows positive results in both tests. Ethane (a saturated hydrocarbon) does not react.

So, the correct answer is option(s):
(A): (positive bromine test)
(D):