In Williamson synthesis of mixed ether having a primary and a tertiary alkyl group if tertiary halide is used, then :

We first recall the essence of the Williamson ether synthesis. It is an $$S_N2$$ nucleophilic substitution reaction in which an alkoxide ion $$RO^-$$ attacks an alkyl halide $$R'X$$ to give an ether:

$$RO^- + R'X \;\rightarrow\; ROR' + X^-$$

The fundamental requirement for a smooth $$S_N2$$ reaction is that the carbon atom bearing the leaving group $$X$$ should be unhindered; in other words, a primary alkyl halide is ideal because the nucleophile can approach the carbon from the back side and displace the halide ion in a single concerted step.

Now we analyse the situation given in the question. We wish to prepare a mixed ether that contains one primary alkyl group and one tertiary alkyl group. The chemist has two choices for the halide partner:

1. Take the primary alkyl group as the halide and the tertiary alkyl group as the alkoxide, or

2. Take the tertiary alkyl group as the halide and the primary alkyl group as the alkoxide.

The question specifically says that the tertiary halide is used. So we are in case 2. Let us write the reaction that we would be attempting:

$$R_3C{-}X \;+\; R' O^- \;\longrightarrow\; R_3C{-}O{-}R' \;+\; X^-$$

Here $$R_3C{-}X$$ is a tertiary alkyl halide, and $$R' O^-$$ is a primary alkoxide ion. Because the carbon $$R_3C{-}$$ is tertiary, it is heavily crowded. An $$S_N2$$ attack requires the alkoxide to approach from the back side of the $$C{-}X$$ bond, but steric bulk blocks this route. Therefore the rate of $$S_N2$$ substitution becomes extremely slow.

At the same time, the alkoxide is a strong base. Whenever a strong base meets a tertiary alkyl halide, an elimination pathway is highly favourable. The base abstracts a β-hydrogen, the electrons form a double bond, and the halide ion departs. This is the $$E2$$ mechanism. We can symbolise it as follows (for simplicity a single β-hydrogen is shown):

$$\begin{array}{c} \text{Base }(R'O^-) + R_3C{-}CH_2{-}X \;\;\xrightarrow{E2}\;\; R_3C=CH_2 + R'OH + X^-\\[6pt] \text{(alkene)}\hspace{45pt}\text{(recovered alcohol)} \end{array}$$

Because the $$E2$$ pathway has a much lower steric requirement (the base merely needs to abstract a hydrogen and does not have to approach the crowded carbon atom directly), elimination overwhelmingly dominates over substitution when a tertiary halide is present.

Thus, instead of the desired mixed ether, the chief product that comes out of the reaction mixture is an alkene derived from the tertiary alkyl halide. The ether yield is negligible.

Comparing this conclusion with the options given:

A. “Rate of reaction will be slow due to slow cleavage of carbon-halogen bond.”  →  While the $$S_N2$$ path is slow, this statement ignores the dominant elimination and is therefore not the best choice.

B. “Alkene will be the main product.”  →  This matches our analysis exactly.

C. “Simple ether will form instead of mixed ether.”  →  Not supported; elimination predominates.

D. “Expected mixed ether will be formed.”  →  Contrary to the mechanistic reasoning.

Therefore the correct statement is option B.

Hence, the correct answer is Option B.