The major product [B] in the following reaction is:

To determine the major product of the given reaction sequence, we first analyze the cleavage of the ether with $$HI$$ and then examine the dehydration of the alcohol formed in the first step.

The starting compound is the asymmetric ether 1-ethoxy-2-methylbutane, having the structure

$$CH_3-CH_2-CH(CH_3)-CH_2-O-CH_2-CH_3.$$

When ethers are treated with excess $$HI$$ under heating, the oxygen atom is protonated and the $$C-O$$ bond undergoes cleavage. The iodide ion preferentially attacks the less sterically hindered alkyl group via an $$S_N2$$ mechanism. Consequently, the ethyl group forms ethyl iodide, while the larger alkyl fragment is converted into the corresponding alcohol.

Thus, the intermediate formed is 2-methylbutan-1-ol, having the structure

$$CH_3-CH_2-CH(CH_3)-CH_2OH.$$

On treatment with concentrated $$H_2SO_4$$ and heat, this alcohol undergoes acid-catalyzed dehydration. Initially, the hydroxyl group is protonated and leaves as water, generating a primary carbocation. This unstable carbocation undergoes a hydride shift to produce a more stable tertiary carbocation.

The rearranged tertiary carbocation then eliminates a proton according to Zaitsev's rule, leading to the formation of the most substituted and thermodynamically most stable alkene.

Hence, the major product obtained is 2-methylbut-2-ene, whose structure is

$$CH_3-CH=C(CH_3)-CH_3.$$

Therefore, the correct answer is

$$CH_3-CH=C(CH_3)-CH_3.$$