
The product formed by the reaction of ethyl alcohol with sodium is/ are.
A. Ethene and hydrogen.
B. Sodium ethoxide and Hydrogen
C. Sodium ethoxide and oxygen.
D. Sodium hydroxide and sodium ethoxide.
Answer
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Hint: We know that ethyl alcohol is an organic compound and the chemical formula of ethyl alcohol is ${C_2}{H_5}OH$ and sodium is an alkali metal with atomic number $11$ and the symbol of sodium is$Na$. Sodium is a highly reactive metal.
Complete step by step answer: First, we see the Reaction between Sodium Metal and Ethanol.
When we add a piece of sodium into ethanol, sodium reacts gradually to give hydrogen gas and leaves a colorless solution of sodium ethoxide.
The chemical reaction between sodium metal and ethyl alcohol is,
${C_2}{H_5}OH\left( l \right) + Na\left( s \right)\xrightarrow{{}}{C_2}{H_5}ONa\left( {aq} \right) + 1/2{H_2}\left( {aq} \right)$
So, the correct answer is “Option B”.
Note: Now we can see the uses of above reaction as,
To dispose of small amounts of sodium: If we spilled the sodium metal on the bench or have a small amount left over from a reaction it cannot simply disposed in the sink because sodium has the great tendency to react explosively with the water, on the other hand it reacts gently with ethanol so ethanol is hence used to dissolve small quantities of waste sodium.
To test for the -OH group in alcohols: We can use the sodium for the detection of presence of alcohol in which the sodium reacts aggressively with acids and produce a salt and hydrogen, be sure that the liquid testing was neutral and be clear that there was no trace of water because sodium reacts with the -OH group in water still better than with the one in an alcohol. With those conditions, if a tiny piece of sodium is added to a neutral liquid free of water and bubbles of hydrogen produced, then the liquid is an alcohol.
Complete step by step answer: First, we see the Reaction between Sodium Metal and Ethanol.
When we add a piece of sodium into ethanol, sodium reacts gradually to give hydrogen gas and leaves a colorless solution of sodium ethoxide.
The chemical reaction between sodium metal and ethyl alcohol is,
${C_2}{H_5}OH\left( l \right) + Na\left( s \right)\xrightarrow{{}}{C_2}{H_5}ONa\left( {aq} \right) + 1/2{H_2}\left( {aq} \right)$
So, the correct answer is “Option B”.
Note: Now we can see the uses of above reaction as,
To dispose of small amounts of sodium: If we spilled the sodium metal on the bench or have a small amount left over from a reaction it cannot simply disposed in the sink because sodium has the great tendency to react explosively with the water, on the other hand it reacts gently with ethanol so ethanol is hence used to dissolve small quantities of waste sodium.
To test for the -OH group in alcohols: We can use the sodium for the detection of presence of alcohol in which the sodium reacts aggressively with acids and produce a salt and hydrogen, be sure that the liquid testing was neutral and be clear that there was no trace of water because sodium reacts with the -OH group in water still better than with the one in an alcohol. With those conditions, if a tiny piece of sodium is added to a neutral liquid free of water and bubbles of hydrogen produced, then the liquid is an alcohol.
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