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I could find two products: The first one where is $\ce{Cl}$ is replaced by $\ce{CN}$. AND the second one where $\ce{CH3CH2O}$ replaces $\ce{Cl}$. Did I make a mistake or miss anyone?

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  • $\begingroup$ What about elimination? $\endgroup$ – K_P Nov 1 '14 at 15:17
  • $\begingroup$ Ok, so there would be another two that are alkenes. $\endgroup$ – yolo123 Nov 1 '14 at 15:25
  • $\begingroup$ OUPS, three others. $\endgroup$ – yolo123 Nov 1 '14 at 15:27
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$\ce{CN-}$ is a significantly stronger nucleophile than $\ce{EtOH}$ and will therefore primarily displace the chloride. The facts that substitution takes place at a tertiary benzylic carbon, and that a polar protic solvent (ethanol) is used indicate that the reaction follows the $\ce{S_{N}1}$ mechanism (sources: 1, 2, 3). Due to the achiral intermediate (carbocation), the nitrile will be obtained as a racemic mixture, i.e. 2 products.

Elimination is also a possible reaction pathway which needs to be considered, since cyanide is a good nucleophile but a weak base, which favors an E2 reaction unless an excess of $\ce{CN-}$ is present. In this case, the product with the more substituted double bond should be favored.

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nucleophilicity $CN->I->EtO-$ which implies cyanide attacks, no heating means no elimination generally

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