Answer
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Hint: Start by writing the structure of acetone and chloroform. Then mark the functional groups in both the compounds. Compare and see which all functional groups need to be removed and added to get acetone. Recollect the reagents used in chlorination since chloroform has chlorine atoms and one of the steps will involve chlorination. Write the reaction and then explain it.
Complete step by step answer:
- Acetone is the simplest ketone. Its IUPAC name is 2-propanone.
- Chloroform is a halogen derivative of methane having three chlorine atoms. It is also known as trichloromethane, $CHC{{l}_{3}}$.
- Acetone is a ketone and chloroform is an alkyl halide. So, from acetone we need to remove one acetyl group and two hydrogen atoms and add three chlorine atoms to get chloroform.
- Removing acetyl groups from acetone is difficult so first we need to chlorinate acetone to form 1, 1, 1-trichloro-2-propanone. Now, there are three chlorine atoms on one carbon atom making it electron deficient because chlorine is electronegative.
- So, now acetyl linkage can be easily broken in the presence of a base like calcium hydroxide. When calcium hydroxide is added to the intermediate 1, 1, 1-trichloro-2-propanone, calcium acetate is formed along with chloroform.
- The reaction is given as,
- Therefore, chloroform is formed from acetone by the first chlorination of acetone followed by action of calcium hydroxide.
Note: Remember chlorine is electronegative and therefore it will pull the shared electrons in covalent bond towards itself thus, creating polarity of bond. Because of this polarity, the bond between substituted carbon atoms and acetyl groups weakens. Calcium hydroxide being a base abstracts acetyl cation and trichlorocarban ion accepts a proton to form chloroform.
Complete step by step answer:
- Acetone is the simplest ketone. Its IUPAC name is 2-propanone.
- Chloroform is a halogen derivative of methane having three chlorine atoms. It is also known as trichloromethane, $CHC{{l}_{3}}$.
- Acetone is a ketone and chloroform is an alkyl halide. So, from acetone we need to remove one acetyl group and two hydrogen atoms and add three chlorine atoms to get chloroform.
- Removing acetyl groups from acetone is difficult so first we need to chlorinate acetone to form 1, 1, 1-trichloro-2-propanone. Now, there are three chlorine atoms on one carbon atom making it electron deficient because chlorine is electronegative.
- So, now acetyl linkage can be easily broken in the presence of a base like calcium hydroxide. When calcium hydroxide is added to the intermediate 1, 1, 1-trichloro-2-propanone, calcium acetate is formed along with chloroform.
- The reaction is given as,
- Therefore, chloroform is formed from acetone by the first chlorination of acetone followed by action of calcium hydroxide.
Note: Remember chlorine is electronegative and therefore it will pull the shared electrons in covalent bond towards itself thus, creating polarity of bond. Because of this polarity, the bond between substituted carbon atoms and acetyl groups weakens. Calcium hydroxide being a base abstracts acetyl cation and trichlorocarban ion accepts a proton to form chloroform.
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