Question: Would you expect the 2-octanol formed by SN2 hydrolysis of (+) - 2-bromooctane to be optically active? If so, what will be its absolute configuration

Would you expect the 2-octanol formed by SN2 hydrolysis of (+) - 2-bromooctane to be optically active? If so, what will be its absolute configuration and sign of rotation? What about the 2-octanol formed by hydrolysis of racemic 2-bromooctane?

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