Q.Complete the following reaction: 4-hydroxybenzyl alcohol (a benzene ring with '$CH_2OH$' at one position and '$HO$—' at the position directly opposite/para to it) $+ HCl \rightarrow ?$

Allylic Alcohol Identification: Understanding the "Why"

Allylic alcohols are a specific class of organic compounds where a hydroxyl group ($-OH$) is attached to a carbon atom that is adjacent to a carbon-carbon double bond ($C=C$). The key to identifying them lies in understanding their unique reactivity — and that reactivity stems from the allylic position.

1. What Makes an Allylic Alcohol Special?

Consider the general structure:

R-CH=CH-CH2-OH

Here, the carbon bearing the $-OH$ is allylic (the carbon next to a double bond). This arrangement creates two important effects:

• Resonance stabilization of any intermediate carbocation formed at the allylic carbon.
• Increased acidity of the allylic C–H bonds (not the O–H bond).

Key idea: The double bond "communicates" with the allylic carbon through resonance, making reactions at that position faster and more selective.

2. The Core Identification Test: Oxidation with PCC or Jones Reagent

Why does this test work?

Reaction:

Allylic alcohols are oxidized to α,β-unsaturated aldehydes or ketones (enals/enones) under mild conditions.

• PCC (Pyridinium Chlorochromate) in $CH_2Cl_2$:

  $\text{Allylic alcohol} \xrightarrow{PCC} \alpha,\beta\text{-unsaturated carbonyl}$

• Jones reagent ($CrO_3/H_2SO_4$) :

  Same product, but harsher — may over-oxidize sensitive substrates.

Why this is diagnostic:

• Simple alcohols (non-allylic) give saturated aldehydes/ketones.
• Allylic alcohols give conjugated carbonyls, which have a distinct UV-Vis absorption (longer wavelength) and can be detected by NMR or chemical tests.

The reasoning behind the selectivity:

The allylic C–H bond is weaker than a typical sp³ C–H bond because the resulting radical or cation is resonance-stabilized:

The transition state for oxidation is lower in energy for allylic alcohols due to delocalization of electron density into the $\pi$ system.

3. The "Why" Behind the Key Formula: Oxidation Product

General formula for product:

If the allylic alcohol is:

Then oxidation gives:

If the alcohol is secondary (e.g., $\ce{R-CH=CH-CH(OH)-R'}$), the product is:

Why this formula holds — the mechanism:

1. Chromate ester formation: The $-OH$ attacks the chromium reagent, forming a chromate ester.
2. Elimination: A base (e.g., pyridine in PCC) abstracts the allylic C–H (not the O–H), breaking the C–H bond and forming a $C=O$ double bond.
3. Resonance drives the reaction: The developing positive charge on the allylic carbon is stabilized by the adjacent double bond, making this elimination much faster than for a simple alcohol.

Key takeaway: The formula is not arbitrary — it follows directly from the fact that the allylic C–H is the one removed, and the double bond remains intact, shifting to conjugation with the new carbonyl.

4. Another Key Test: Bromine Water Decolorization

Why does this test work? …