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The thorough analysis of the reaction mixture tells us that the result is the major production (80%) of an optically active alcohol,  
together with a series of optically inactive alkenes.
It is reasonable to think that the nucleophile (HO-) has attacked the haloalkane by the electron-deficient carbon of 2-bromooctane, that bonded to the bromine.
Consequently, the alcohol should be 2-octanol.
Let's check a chemical catalog, where we can find the following data:

These results take us to the conclusions that follows:

• The halogen was mostly substituted by an OH group.
  
  
• Alcohol configuration is mostly S, contrary to that of the starting haloalkane.
  
  
• Therefore, the substitution reaction mostly happened with INVERSION of configuration.
  
  
• The production of olefins shows that the halogen was in part ELIMINATED.
  
  
• The elimination process must be complex because a number of different olefins is formed.
  
  
• A small amount of alcohol R was formed suggesting that a small amount of RACEMIZATION took place.
  
  
• The reaction rate depends on the two reagents, haloalkane and nucleophile, thus suggesting that the two of them must intervene in the rate-limiting step of the reaction.