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Question: A solution of \[\left( + \right)-1-chloro-1-phenylethane\] in toluene racemises slowly in the presen...

A solution of (+)−1−chloro−1−phenylethane\left( + \right)-1-chloro-1-phenylethane in toluene racemises slowly in the presence of the small amount of SbCl5SbC{{l}_{5}} due to the formation of:
Al.Carbanion.
B.Carbene.
C.Free-Radical.
D.Carbocation.

Explanation

Solution

We know that for this question, we have to write the complete reaction in which firstly the antimony pentachloride will remove the chlorine atom from the given compound. Due to which the formation of the intermediate will take place. Haloalkanes are hydrocarbons consisting of aliphatic alkanes with one or more hydrogen atom/s replaced by halogens.

Complete answer:
Haloarenes are also hydrocarbons which consist of aromatic ring/s with one or more hydrogen atoms/s replaced by halogens. In haloarenes, the halogen atom gets attached to sp3s{{p}^{3}}hybridized carbon atoms of the alkyl group. Haloarenes are also known as aryl halides. They show the following physical and chemical properties: Most important member of haloarenes is aryl chlorides. Aryl chlorides are a wide class of haloarenes. They give many derivatives as well. Chlorobenzene is colourless haloarene. It is liquid at room temperature. It gives a sweet almond-like odor. It is insoluble in water and has higher density than water.
Cl−CH(Ph)−CH3→TolueneSbCl5[Ph−C+ H−CH3]SbCl6−Cl-CH(Ph)-C{{H}_{3}}\xrightarrow[Toluene]{SbC{{l}_{5}}}[Ph-\overset{+}{\mathop{C}}\,H-C{{H}_{3}}]SbCl_{6}^{-}
We have to determine the correct intermediate which will be formed when (+)−1−chloro−1−phenylethane\left( + \right)-1-chloro-1-phenylethane will react with antimony pentachloride. As we know that the toluene is the six membered ring structures on which one methyl group is also attached. Now, when antimony pentafluoride reacts with the (+)−1−chloro−1−phenylethane\left( + \right)-1-chloro-1-phenylethane which is present in the solution of toluene, it removes the chlorine molecule which is attached with the carbon atom. A solution of (+)−1−chloro−1−phenylethane\left( + \right)-1-chloro-1-phenylethane in toluene racemises slowly in the presence of a small amount of SbCl5SbC{{l}_{5}} due to the formation of carbocation.
[Ph−C+ H−CH3]SbCl6−→Ph−CH(Cl)−CH3+SbCl5[Ph-\overset{+}{\mathop{C}}\,H-C{{H}_{3}}]SbCl_{6}^{-}\to Ph-CH(Cl)-C{{H}_{3}}+SbC{{l}_{5}}
Carbocation has planar geometry. Hence, the nucleophile can attack from either side. Here, the carbocation formed is known to have planar geometry. Also, the central carbon atom is known as chiral carbon because it is attached to the four different groups that are chlorine, phenyl, hydrogen and methyl groups. Now, when an electrophile attaches to the carbocation then it will form two types of the compound and form a racemic mixture.
Therefore, the correct answer is option D.

Note:
Remember that the haloarene undergoes sulfonation. As free radical is the unpaired pair of electrons whereas carbanion is the carbon species which consist of a negative charge or electrons. For example, on heating chlorobenzene reacts with conc. Sulfuric acid gives 2−2- chlorobenzenesulfonic acid and 4−4- chlorobenzenesulfonic acid. In carbene, the carbon molecule consists of two valence electrons and two unshared electrons.