1. Identify the substrate

The given alcohol is

This is a secondary alcohol. In presence of concentrated $H_2SO_4$, dehydration occurs, generally by an $E1$ mechanism.

2. Write the carbon skeleton around the OH-bearing carbon

Let us label the chain:

The carbon bearing $OH$ is adjacent to:

• on the left: $CH_2Ph$
• on the right: $CH(CH_3)_2$

After protonation of $OH$ and loss of water, a carbocation forms at the $OH$-bearing carbon:

3. Consider possible rearrangement

This carbocation is secondary. A more stable carbocation can form by hydride shift from the benzylic carbon ($CH_2Ph$) to the cationic center.

That gives a benzylic carbocation:

This benzylic carbocation is highly stabilized by resonance with the benzene ring, so this rearrangement is favored.

4. Elimination from the rearranged carbocation

Now elimination can occur from the adjacent carbon ($CH_2$), giving the alkene:

This alkene is conjugated with the benzene ring through the benzylic double bond, making it especially stable.

5. Check other possible alkene(s)

Without rearrangement, dehydration could also give alkenes such as:

• $C_6H_5CH_2CH=C(CH_3)_2$
• $C_6H_5CH=CHCH(CH_3)_2$

Among these, the alkene

is more stable because the double bond is directly conjugated with the phenyl ring. Hence it is the main product.

6. Conclusion

The major product is:

So the correct option is A.

7. Comparison with stored answer

Stored correct answer: A

Our derived answer: A

Hence, they agree.