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(C) The reaction involves a nucleophilic substitution with $NaN_3$ (azide ion,$N_3^-$) acting as the nucleophile.$Br^-$ is a better leaving group than

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(C) The reaction involves a nucleophilic substitution with $NaN_3$ (azide ion,$N_3^-$) acting as the nucleophile.$Br^-$ is a better leaving group than $Cl^-$,so the substitution occurs at the carbon atom attached to the bromine atom.Since $NaN_3$ is a strong nucleophile and the reaction proceeds via an $S_N2$ mechanism,the nucleophile attacks from the side opposite to the leaving group,resulting in an inversion of configuration at the chiral center.Therefore,the $-N_3$ group replaces the $-Br$ group with inversion of configuration.

What is the principal product of the following reaction?
$CH_3-CH(Br)-CH_2-CH_2-CH(Cl)-CH_3 + NaN_3 \rightarrow \text{Product}$

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What is the principal product of the following reaction?$CH_3-CH(Br)-CH_2-CH_2-CH(Cl)-CH_3 + NaN_3 \rightarrow \text{Product}$