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Chapter 10: Problem 20

The hydrocarbon obtained by treating sodium ethanoate with soda lime is

Short Answer

Expert verified
Answer: Methane (CH4)

Step by step solution

01

Write the given compounds and their formulas

Sodium ethanoate has the formula CH3COONa, and soda lime is a mixture of sodium hydroxide (NaOH) and calcium oxide (CaO).
02

Determine the reaction between sodium ethanoate and soda lime

Sodium ethanoate reacts with soda lime (NaOH) in a decarboxylation process. The carboxyl group (-COO-) is removed from the sodium ethanoate molecule, leaving behind a R- (alkyl group) connected to a hydrogen atom. The general formula for decarboxylation is: R-COO^- + NaOH -> R-H + CO2 + Na^+ In our case, sodium ethanoate (CH3COONa) reacts with soda lime (NaOH), as follows: CH3COO^- + NaOH -> CH3-H + CO2 + Na^+ which can be simplified as: CH3COONa + NaOH -> CH4 + CO2 + Na^+
03

Identify the resulting hydrocarbon

The hydrocarbon obtained from the decarboxylation of sodium ethanoate with soda lime is CH4, known as methane.

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Most popular questions from this chapter

Which property of carbon gives rise to the structural isomerism of organic compounds?

Observe the molecular formulae and classify the following hydrocarbons into alkanes, alkenes and alkynes. (i) \(\mathrm{C}_{2} \mathrm{H}_{4}\) (ii) \(\mathrm{C}_{2} \mathrm{H}_{6}\) (iii) \(\mathrm{C}_{4} \mathrm{H}_{10}\) (iv) \(\mathrm{CH}_{4}\) (v) \(\mathrm{C}_{3} \mathrm{H}_{6}\) (vi) \(\mathrm{C}_{3} \mathrm{H}_{8}\)

(i) Describe indetail one test to distinguish saturated hydrocarbon from unsaturated hydrocarbon.(ii) Describe in detail one test to distinguish ethene 7 and acetylene.

Which characteristics of carbon are responsible for the existence of a large number of organic compounds?

Organic compounds having the same molecular formula, but different structural formulae are called
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