Consider the above reaction, and choose the correct statement:

Reaction Analysis: Acid-Catalysed Dehydration

The reaction involves a secondary alcohol, $$\mathrm{1\text{-}phenylpropan\text{-}1\text{-}ol}$$, treated with concentrated sulfuric acid and heat.

$$\mathrm{1\text{-}Phenylpropan\text{-}1\text{-}ol \xrightarrow[\ \Delta\ ]{Conc.\ H_2SO_4} Alkene\ Products}$$

These conditions favour dehydration through an $$\mathrm{E1}$$ elimination mechanism.

Step 1: Protonation and Leaving Group Formation

The hydroxyl group gets protonated by the acid, converting it into a better leaving group.

$$\mathrm{Ph-CH(OH)-CH_2-CH_3 + H^+ \longrightarrow Ph-CH(OH_2^+)-CH_2-CH_3}$$

Step 2: Carbocation Formation

The protonated alcohol loses water to form a carbocation intermediate.

$$\mathrm{Ph-CH(OH_2^+)-CH_2-CH_3 \longrightarrow Ph-CH^+-CH_2-CH_3 + H_2O}$$

The carbocation formed is a benzylic carbocation.

This intermediate is highly stable because the positive charge is delocalised through resonance with the aromatic ring.

Step 3: Elimination (Zaitsev’s Rule)

A weak base such as $$\mathrm{H_2O}$$ or $$\mathrm{HSO_4^-}$$ removes a beta-hydrogen from the adjacent carbon.

Formation of the double bond gives an alkene conjugated with the benzene ring.

$$\mathrm{Ph-CH^+-CH_2-CH_3 \longrightarrow Ph-CH=CH-CH_3 + H^+}$$

Formation of Products $$\mathrm{A}$$ and $$\mathrm{B}$$

The newly formed double bond gives rise to geometrical isomers.

Product $$\mathrm{A}$$ (Trans-Isomer):

In this isomer, the phenyl group and methyl group are present on opposite sides of the double bond.

$$\mathrm{trans\text{-}Ph-CH=CH-CH_3}$$

Product $$\mathrm{B}$$ (Cis-Isomer):

In this isomer, the phenyl group and methyl group are present on the same side of the double bond.

$$\mathrm{cis\text{-}Ph-CH=CH-CH_3}$$

Conclusion:

The trans-isomer is more stable because bulky groups remain farther apart, reducing steric hindrance.

The cis-isomer experiences greater steric repulsion because the phenyl and methyl groups are on the same side.

Therefore:

$$\mathrm{Product\ A\ (trans\text{-}isomer)\ is\ the\ major\ product}$$

while

$$\mathrm{Product\ B\ (cis\text{-}isomer)\ is\ the\ minor\ product}$$

Correct Statement:

$$\mathrm{A\ is\ the\ major\ product\ (because\ the\ trans\text{-}alkene\ is\ more\ stable\ due\ to\ less\ steric\ hindrance)}$$