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In the following reaction the products formed are not enantiomers but are diastereomoers.So there is no racemisation occured.So is it always correct to say that SN1 result in racemisation?

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This doesn't qualify as racemisation, it is epimerization. Racemization would require both stereocentres to invert to form the enantiomer and - as you rightly suppose - the stereocentre bearing the methyl group is untouched by this reaction.

It is also unlikely to lead to a 1:1 mixture of products since the two diastereomers (and the transition states leading to them) are different in energy, so under either thermodynamic or kinetic control one would expect a non-1:1 mixture. This contrasts with the case of a molecule undergoing true racemization in which the two enantiomeric products will be equal in energy so will be formed in equal quantity.

Your title also poses an interesting and more general question - do SN1 reactions always lead to racemisation even in molecules containing only one stereocentre? I was winding up to write an answer to this question, before finding the outstanding answer previously given by NotEvans here - well worth a read. (hint: the answer is - not necessarily).

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It results in racemisation if only one chiral carbon atom is present in the compound. If there are many, it merely leads to formation of compounds with both R and S configurations of the chiral carbon on which substitution takes place.

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