
What is X in the following sequence of reactions?
\[X\mathop \to \limits_{ - {H_2}}^{Na} Y\mathop \to \limits_{CaO}^{NaOH} C{H_4}\]
(A) Methanoic acid
(B) Ethanoic acid
(C) Propane
(D) Methane
Answer
233.1k+ views
Hint: It is a significant reaction that when carboxylic acids react with strong alkali metals, a strong basic salt is formed with the release of hydrogen gas. When calcium oxide is added to sodium hydroxide, soda lime is formed which results in decarboxylation of the salt.
Complete step-by-step solution:
We know that acids react with strong alkali metals like sodium and potassium that react with organic acids to produce strong basic salts along with release of hydrogen gas. In the reaction given above, ethanoic acid reacts with sodium metal that results in the formation of sodium acetate or sodium ethanoate and hydrogen gas is released. This reaction is highly exothermic in nature.
\[C{H_3}COOH + Na \to C{H_3}COONa + {H_2}\]
In decarboxylation, the – COOH or – COONa group is removed and replaced by a hydrogen atom and carbonate ion. This is performed by soda lime which is manufactured by adding sodium hydroxide solution to quicklime (CaO). Sometimes, this mixture might include calcium hydroxide as well.
The sodium salt of carboxylic acid when mixed with soda lime at high temperature results in methane gas and sodium carbonate. Thus, we get Y as sodium acetate or sodium ethanoate. This process is called decarboxylation of carboxylic acid and their salts.
\[C{H_3}COONa + NaOH + CaO \to C{H_4} + N{a_2}C{O_3}\]
Methanoic acid is also an organic acid, but on reaction with sodium metal it forms sodium methoxide and hydrogen gas, which can’t produce methane on reaction with soda lime.
Hence, the correct option is (B).
Note: In some reactions, only sodium hydroxide is used in place of soda lime. But solid sodium hydroxide absorbs water from the surrounding and ends up with puddles of extremely corrosive hydroxide solution when exposed to air. Therefore, soda lime is used as it is easy to handle and does not absorb water.
Complete step-by-step solution:
We know that acids react with strong alkali metals like sodium and potassium that react with organic acids to produce strong basic salts along with release of hydrogen gas. In the reaction given above, ethanoic acid reacts with sodium metal that results in the formation of sodium acetate or sodium ethanoate and hydrogen gas is released. This reaction is highly exothermic in nature.
\[C{H_3}COOH + Na \to C{H_3}COONa + {H_2}\]
In decarboxylation, the – COOH or – COONa group is removed and replaced by a hydrogen atom and carbonate ion. This is performed by soda lime which is manufactured by adding sodium hydroxide solution to quicklime (CaO). Sometimes, this mixture might include calcium hydroxide as well.
The sodium salt of carboxylic acid when mixed with soda lime at high temperature results in methane gas and sodium carbonate. Thus, we get Y as sodium acetate or sodium ethanoate. This process is called decarboxylation of carboxylic acid and their salts.
\[C{H_3}COONa + NaOH + CaO \to C{H_4} + N{a_2}C{O_3}\]
Methanoic acid is also an organic acid, but on reaction with sodium metal it forms sodium methoxide and hydrogen gas, which can’t produce methane on reaction with soda lime.
Hence, the correct option is (B).
Note: In some reactions, only sodium hydroxide is used in place of soda lime. But solid sodium hydroxide absorbs water from the surrounding and ends up with puddles of extremely corrosive hydroxide solution when exposed to air. Therefore, soda lime is used as it is easy to handle and does not absorb water.
Recently Updated Pages
JEE Main 2023 April 6 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 6 Shift 2 Question Paper with Answer Key

JEE Main 2023 (January 31 Evening Shift) Question Paper with Solutions [PDF]

JEE Main 2023 January 30 Shift 2 Question Paper with Answer Key

JEE Main 2023 January 25 Shift 1 Question Paper with Answer Key

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Key

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Understanding the Electric Field of a Uniformly Charged Ring

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

NCERT Solutions For Class 12 Chemistry Chapter 1 Solutions (2025-26)

Solutions Class 12 Chemistry Chapter 1 CBSE Notes - 2025-26

NCERT Solutions For Class 12 Chemistry Chapter 4 The d and f Block Elements (2025-26)

Biomolecules Class 12 Chemistry Chapter 10 CBSE Notes - 2025-26

NCERT Solutions For Class 12 Chemistry Chapter 10 Biomolecules (2025-26)

