Explain the use of the following in this laboratory. a. Boiling stones b. Charco

Question

Explain the use of the following in this laboratory. a. Boiling stones b. Charcoal c. Allyl alcohol d. Concentrated HCl Benzaldehyde (C_6H_5CHO) is oxidized to benzoic acid by potassium permanganate in an acid medium. The rate of the reaction is found to be directly proportional to the concentrations of H^+, benzaldehyde, and KMnO_4. When the oxidation is carried out with C_6H_5CHO, the rate is much slower than with C_6H_5CHO. The oxidation of C_6H_5CHO in H_2O^18 and MnO_4^- gives the product C_6H_5COOH, whereas the same reaction in the presence of Mn^18O_4^- gave benzoic acid containing^18 O. What important mechanistic conclusions can you draw from the above experimental observations and results?

Explanation / Answer

3)

a. boiling chip

A boiling chip is a boiling stone, or anti-bumping granule is a tiny, unevenly shaped piece of substance added to liquids to make them boil more calmly. Boiling chipsare frequently employed in distillation and heating.

b. Charcoal

Charcoal is used as decolourising agent to remove colouring matter from substance containing coloured impurities, charcoal id good adsorbent.

c. Allyl alcohol

Allyl alcohol is an organic compound with the structural formula CH2=CHCH2OH. it is a water-soluble, colourless liquid, but it is more toxic than typical small alcohols. Allyl alcohol is used as a raw material for the production of glycerol, but is also used as a precursor in the synthesis of compounds such as flame-resistant materials, drying oils, and plasticizers

d. Concentrated HCl

Concentrated HCl is common laboratory reagent used in laboratory, it is used in acid-base titrations, preparation of dilute acids, buffers, detection of carbonates and organic synthesis

4.

The formation of C6H5COOH from C6H5CHO involves

breacking of C-H bond in benzaldehyde which can be rate determining step, Which can be concluded by oxidation of C6H5CDO.

since rate of reaction depends on concentration of H+, C6H5CHO and KMnO4 it is trimolecular reaction

The Oxidation of C6H5COOH in H2O18 and MnO4- gives the product C6H5COOH, whereas the same reaction in the presence of Mn18O4- give benzoic acid containing 18O. This shows that Oxygen in introduced into benzoic acid from KMnO4

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