The product of the following reaction is:

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(B) The reaction involves the treatment of $2-(1-bromo-1-methylethyl)phenol$ with $NaOH$. $NaOH$ acts as a base and deprotonates the phenolic $-OH$ gr

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(B) The reaction involves the treatment of $2-(1-bromo-1-methylethyl)phenol$ with $NaOH$. $NaOH$ acts as a base and deprotonates the phenolic $-OH$ group to form a phenoxide ion. The phenoxide oxygen then acts as a nucleophile and attacks the carbon atom attached to the bromine atom via an intramolecular $S_N2$ or $S_N2$-like mechanism,displacing the bromide ion. This results in the formation of a cyclic ether (a chroman derivative).

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