Consider the above reaction, and choose the correct statement :
Show Hint
In acid-catalyzed dehydration of alcohols (E1 mechanism), the reaction proceeds through the most stable carbocation intermediate. The major product is typically the most stable alkene, which is often the most substituted one (Zaitsev's rule) or one where the double bond is in conjugation with a ring or another multiple bond.
The reaction shown is the acid-catalyzed dehydration of an alcohol. The reactant is 1-phenyl-1-propanol.
The mechanism is E1 elimination, which proceeds via a carbocation intermediate.
Step 1: The hydroxyl group (-OH) is protonated by the strong acid (H$_2$SO$_4$) to form a good leaving group, water (-OH$_2^+$).
Step 2: The water molecule leaves, forming a carbocation. The carbocation formed is `C$_6$H$_5$-CH$^+$-CH$_2$CH$_3`$. This is a secondary benzylic carbocation, which is highly stabilized by resonance with the benzene ring.
Step 3: A proton is eliminated from a carbon atom adjacent to the positively charged carbon to form a double bond. The adjacent carbon is C-2 (the -CH$_2$- group).
Step 4: Elimination of a proton from C-2 leads to the formation of 1-phenylprop-1-ene, `C$_6$H$_5$-CH=CH-CH$_3`. This is Compound A.
The double bond formed in Compound A is conjugated with the benzene ring, which provides extra stability to the molecule. According to Zaitsev's rule, the more substituted (and generally more stable) alkene is the major product. Here, the stability is determined by conjugation.
Compound B, `C$\_6$H$\_5$-CH$\_2$-CH=CH$\_2$` (3-phenylprop-1-ene), cannot be formed from the given reactant via a simple elimination reaction, as it would require removal of a proton from the non-adjacent C-3.
Therefore, Compound A is the only expected product, and it will be the major product.
