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NEET CHEMISTRYHydrocarbonsMedium

Question

Predict the correct intermediate and product in the following reaction: H3C−C≡CH→H2O,H2SO4,HgSO4Intermediate→ProductH_3C-C \equiv CH \xrightarrow{H_2O, H_2SO_4, HgSO_4} \text{Intermediate} \to \text{Product}

A

(Structure of Intermediate and Product for Option A)

B

(Structure of Intermediate and Product for Option B)

C

(Structure of Intermediate and Product for Option C)

D

Intermediate: H3C−C(OH)=CH2H_3C-C(OH)=CH_2 (Enol), Product: H3C−CO−CH3H_3C-CO-CH_3 (Propanone)

Step-by-Step Solution

  1. Reaction Type: The reaction is the hydration of an alkyne in the presence of dilute sulphuric acid (H2SO4H_2SO_4) and mercuric sulphate (HgSO4HgSO_4) at 333 K .
  2. Mechanism (Markovnikov's Addition): Water adds to the triple bond according to Markovnikov's rule. The negative part of the addendum (OH−OH^-) attaches to the carbon atom with fewer hydrogen atoms (the central carbon in propyne).
  3. Intermediate Formation: The addition results in the formation of an unstable enol intermediate: CH3−C≡CH+H−OH→CH3−C(OH)=CH2CH_3-C \equiv CH + H-OH \to CH_3-C(OH)=CH_2 This intermediate is prop-1-en-2-ol.
  4. Tautomerisation: The unstable enol undergoes rapid tautomerisation (rearrangement of bonds and protons) to form the more stable keto form (carbonyl compound). CH3−C(OH)=CH2⇌CH3−C(=O)−CH3CH_3-C(OH)=CH_2 \rightleftharpoons CH_3-C(=O)-CH_3
  5. Final Product: The final product is Propanone (Acetone). Note: Only ethyne yields an aldehyde (ethanal); all other alkynes yield ketones in this reaction .

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Hydrocarbons. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYHydrocarbonspredictcorrectintermediateproductfollowing

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Phenylacetylene undergoes a reaction with dilute $\text{H}_2\text{SO}_4$ in the presence of $\text{HgSO}_4$ to yield:

A.Option 1 (Structure missing)
B.Option 2 (Structure missing)
C.Option 3 (Structure missing)
D.Option 4 (Structure missing)
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A.(Option 1 Structure Missing)
B.(Option 2 Structure Missing)
C.4
D.(Option 4 Structure Missing)
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A.(i) > (ii) > (iii) > (iv)
B.(ii) > (iv) > (i) > (iii)
C.(iv) > (iii) > (ii) > (i)
D.(iv) > (ii) > (i) > (iii)
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X and Y in the above-mentioned reaction are respectively:

A.X = 2–Butyne; Y = 3–Hexyne
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D.X = 1-Butyne; Y = 3–Hexyne
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Match the bond line structures of hydrocarbons given in List-I with the corresponding boiling points given in List-II. List-II (Boiling point in K): (i) 300.9 (ii) 282.5 (iii) 309.1 (iv) 341.9 Choose the correct answer from the options given below:

A.(a)-(i), (b)-(iv), (c)-(iii), (d)-(ii)
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A.$60^{\circ}$
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A.0
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The correct statement regarding the comparison of staggered and eclipsed conformations of ethane is:

A.The eclipsed conformation of ethane is more stable than staggered conformation because eclipsed conformation has no torsional strain.
B.The eclipsed conformation of ethane is more stable than staggered conformation even though the eclipsed conformation has a torsional strain.
C.The staggered conformation of ethane is more stable than eclipsed conformation because staggered conformation has no torsional strain.
D.The staggered conformation of ethane is less stable than eclipsed conformation because staggered conformation has a torsional strain.
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