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(D) The starting material is a haloalkane containing both a $Br$ and a $Cl$ atom. $NaSH$ is a strong nucleophile. The $Br$ atom is a better leaving gr

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(D) The starting material is a haloalkane containing both a $Br$ and a $Cl$ atom. $NaSH$ is a strong nucleophile. The $Br$ atom is a better leaving group than the $Cl$ atom,so $NaSH$ will selectively attack the carbon attached to $Br$ via an $S_N2$ mechanism to form a thiol $(-SH)$. After the formation of the thiol,the molecule contains both a thiol group and a chloro group. The thiol group $(-SH)$ is a strong nucleophile. In the presence of $NaOH$ (a base),the thiol is deprotonated to form a thiolate ion $(-S^-)$. This internal thiolate ion then performs an intramolecular $S_N2$ reaction by attacking the carbon attached to the $Cl$ atom,displacing the $Cl^-$ ion and forming a cyclic sulfide (a thiacycloalkane). The resulting product is a five-membered ring containing a sulfur atom,with a methyl group and an ethyl group attached to the carbons adjacent to the sulfur atom.

The major product obtained in the above reaction is

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