
Chlorine reacts with ethanol to give
A) Ethyl chloride
B) Chloroform
C) Acetaldehyde
D) Chloral
Answer
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Hint: Here chlorine is an oxidising agent. In the whole process, two reactions take place, one is oxidation reaction and second is halogenation reaction. The same reaction takes place in the the haloform test of ethanol, however, in the given reaction base is missing.
Complete step-by-step answer:Chlorine reacts with ethanol to form Chloral and hydrogen chloride. Let’s see how it is happening.
In the first step,
In this step, Ethanol is converted to ethanal aldehyde by oxidation of chlorine.
\[C{l_2}{\text{ }} + {\text{ }}C{H_3}C{H_2}OH{\text{ }} \to C{H_3}CHO{\text{ }} + {\text{ }}2HCl\]
In the Second Step,
\[3C{l_2}{\text{ }} + {\text{ }}C{H_3}CHO{\text{ }} \to CC{l_3}CHO{\text{ }} + {\text{ }}3HCl\]
Overall Reaction is --
\[4C{l_2}{\text{ }} + {\text{ }}C{H_3}C{H_2}OH{\text{ }} \to CC{l_3}CHO{\text{ }} + {\text{ }}5HCl\]
The final product is Chloral.
Additional information The formula of chloral is \[C{l_3}CCHO\]. The C=O bond in choral is polar bond so that it is easy for OH to attack the bond. Choral when reacted with alcohol or water gives chloral hydrates which are unstable. The chloral is formed as intermediate in the haloform test using chlorine. It forms when methyl carbonyl compounds react with chlorine and strong bases such as sodium hydroxide, calcium hydroxide and potassium hydroxide. Due to its instability, this intermediate compound breaks to give a positive haloform test by forming chloroform. The similar type of reaction can happen with compounds with molecular formula \[C{l_3}CCRO\], where R is an alkyl group.
Option ‘D’ is correct
Note: If in the reaction with chlorine and ethanol a base such as sodium hydroxide or calcium hydroxide was also present then the reaction will give chloroform as the final product instead of chloral.
Complete step-by-step answer:Chlorine reacts with ethanol to form Chloral and hydrogen chloride. Let’s see how it is happening.
In the first step,
In this step, Ethanol is converted to ethanal aldehyde by oxidation of chlorine.
\[C{l_2}{\text{ }} + {\text{ }}C{H_3}C{H_2}OH{\text{ }} \to C{H_3}CHO{\text{ }} + {\text{ }}2HCl\]
In the Second Step,
\[3C{l_2}{\text{ }} + {\text{ }}C{H_3}CHO{\text{ }} \to CC{l_3}CHO{\text{ }} + {\text{ }}3HCl\]
Overall Reaction is --
\[4C{l_2}{\text{ }} + {\text{ }}C{H_3}C{H_2}OH{\text{ }} \to CC{l_3}CHO{\text{ }} + {\text{ }}5HCl\]
The final product is Chloral.
Additional information The formula of chloral is \[C{l_3}CCHO\]. The C=O bond in choral is polar bond so that it is easy for OH to attack the bond. Choral when reacted with alcohol or water gives chloral hydrates which are unstable. The chloral is formed as intermediate in the haloform test using chlorine. It forms when methyl carbonyl compounds react with chlorine and strong bases such as sodium hydroxide, calcium hydroxide and potassium hydroxide. Due to its instability, this intermediate compound breaks to give a positive haloform test by forming chloroform. The similar type of reaction can happen with compounds with molecular formula \[C{l_3}CCRO\], where R is an alkyl group.
Option ‘D’ is correct
Note: If in the reaction with chlorine and ethanol a base such as sodium hydroxide or calcium hydroxide was also present then the reaction will give chloroform as the final product instead of chloral.
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