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A solution of (-)1-chloro-1-phenylethane in toluene racemises slowly in the presence of a small amount of $$SbCl_5$$, due to the formation of :
We are given that $$(-)-1$$-chloro-1-phenylethane dissolves in toluene and racemises slowly when a small amount of $$SbCl_5$$ is added.
First, let us understand the structure. 1-Chloro-1-phenylethane is:
$$C_6H_5 - CHCl - CH_3$$
The carbon bearing the chlorine atom (C-1) has four different groups: $$-C_6H_5$$, $$-CH_3$$, $$-Cl$$, and $$-H$$. This makes it a chiral centre, and the compound is optically active (it is given as the ($$-$$) enantiomer).
Now, $$SbCl_5$$ is a strong Lewis acid. It can accept the lone pair from the chlorine atom of the organic substrate:
$$C_6H_5CHClCH_3 + SbCl_5 \rightarrow C_6H_5CH^+CH_3 + SbCl_6^-$$
Here, $$SbCl_5$$ abstracts the $$Cl^-$$ from the chiral carbon, generating a carbocation at that carbon.
The resulting carbocation $$C_6H_5CH^+CH_3$$ is a secondary benzylic carbocation. The positive charge is stabilised by resonance with the phenyl ring. Crucially, this carbocation is $$sp^2$$ hybridised and therefore has a planar geometry (trigonal planar around the positively charged carbon).
Because the carbocation is planar, when the $$Cl^-$$ (from $$SbCl_6^-$$) attacks back, it can approach from either face of the plane with equal probability:
(i) Attack from one face regenerates the original $$(-)$$ enantiomer.
(ii) Attack from the opposite face generates the $$(+)$$ enantiomer.
Over time, a 50:50 mixture of both enantiomers (a racemic mixture) is formed, and the optical rotation drops to zero. This is the observed racemisation.
The key intermediate responsible for racemisation is the planar carbocation.
Note: A free radical (Option B) would require homolytic cleavage, but $$SbCl_5$$ is a Lewis acid that promotes heterolytic (ionic) cleavage. A carbanion (Option C) would require removal of $$H^+$$, not $$Cl^-$$. A carbene (Option D) would require loss of $$HCl$$, which is not the mechanism here.
The correct answer is Option A: Carbocation.
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