Which of the following compounds are benzylic alcohols? (Two or more options may be correct.) (a) C6H5-CH2-CH2OH (2-phenylethanol). (b) C6H5-CH2OH (benzyl alcohol). (c) C6H5-CH(OH)-CH3 (1-phenylethanol). (d) C6H5-CH2-CH(OH)-CH3 (a compound in which the hydroxyl-bearing carbon is separated from the ring by a CH2 group). (i) compound (a), C6H5-CH2-CH2OH (ii) compound (b), C6H5-CH2OH (iii) compound (c), C6H5-CH(OH)-CH3 (iv) compound (d), C6H5-CH2-CH(OH)-CH3

Which of the following compounds are benzylic alcohols? (Two or more options may be correct.) (a) C6H5-CH2-CH2

NCERT Exemplar · Q60

Q. Which of the following compounds are benzylic alcohols? (Two or more options may be correct.)

(a) C6H5-CH2-CH2OH (2-phenylethanol).

(b) C6H5-CH2OH (benzyl alcohol).

(d) C6H5-CH2-CH(OH)-CH3 (a compound in which the hydroxyl-bearing carbon is separated from the ring by a CH2 group).

(i) compound (a), C6H5-CH2-CH2OH

(ii) compound (b), C6H5-CH2OH

(iii) compound (c), C6H5-CH(OH)-CH3

(iv) compound (d), C6H5-CH2-CH(OH)-CH3

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Concept understanding — Allylic Alcohol Identification

Allylic Alcohol Identification – From Intuition to Precision

Imagine you have a carbon–carbon double bond, like the one in an alkene. Now picture an –OH group (alcohol) attached to a carbon that is one carbon away from that double bond. That specific arrangement — an alcohol sitting on the carbon next to a double bond — is what we call an allylic alcohol.

The name comes from the allyl group:

$CH_2=CH-CH_2-$

If you replace the terminal hydrogen of that group with an –OH, you get allyl alcohol:

$CH_2=CH-CH_2OH$

That is the simplest example. But the concept extends to any alcohol where the carbon bearing the –OH is directly adjacent to a C=C double bond.

The Intuition: Why "Allylic" Matters

The carbon next to a double bond (the allylic carbon) is special. The double bond's π electrons can "talk" to that carbon through resonance. When an –OH is on that allylic carbon, the molecule gains unique chemical behaviour:

The –OH can be oxidised more easily than in a normal alcohol (e.g., to an aldehyde or ketone).
The C–OH bond can break to form a stable allylic carbocation (resonance-stabilised), making these alcohols reactive in substitution reactions.
They give characteristic colour tests (like the Lucas test or chromic acid test) that help identify them in the lab.

So the identification is not just about naming — it's about predicting reactivity.

The Precise Definition

An allylic alcohol is any compound in which an -OH group is attached to a sp³-hybridised carbon that is directly bonded to a carbon-carbon double bond (C=C).

In other words:

$R-CH=CH-CH_2OH$ or $R_2C=CR-CH_2OH$

where the –OH is on a carbon adjacent to the double bond.

How to Identify One – Step by Step

Find the double bond (C=C) in the structure.
Look at the carbons directly attached to either end of that double bond.
Check if any of those adjacent carbons carry an –OH group.
If yes — that is an allylic alcohol.

Tip

A quick mental shortcut: allylic = next to a double bond. If the –OH is on a carbon that is one bond away from a C=C, it's allylic.

Examples and Non-Examples

Structure	Allylic?	Reason
$CH_2=CH-CH_2OH$	✓ Yes	–OH on carbon adjacent to C=C
$CH_3-CH=CH-CH_2OH$	✓ Yes	–OH on carbon next to C=C
$CH_2=CH-CH(OH)-CH_3$	✓ Yes	–OH on the allylic carbon (the one directly attached to the double bond)
$CH_2=CH-CH_2-CH_2OH$	✗ No	–OH is two carbons away from C=C (this is a homoallylic alcohol)
$CH_3-CH_2-CH_2OH$	✗ No	No double bond at all

A Common Mistake to Avoid

Watch out

Do not confuse "allylic" with "vinylic".

A vinylic carbon is one of the two carbons in the double bond (sp²).

An allylic carbon is the sp³ carbon next to the double bond.

An –OH on a vinylic carbon (like $CH_2=CH-OH$ ) is an enol, not an allylic alcohol. Enols are unstable and tautomerise to carbonyl compounds.

--- …

Why this formula?

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:

\ce{CH2=CH-CH2-OH ->[oxidation] CH2=CH-CH=O + H2O}

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:

\ce{R-CH=CH-CH2-OH}

Then oxidation gives:

\ce{R-CH=CH-CHO} \quad (\text{an } \alpha,\beta\text{-unsaturated aldehyde})

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

\ce{R-CH=CH-C(=O)-R'} \quad (\text{an } \alpha,\beta\text{-unsaturated ketone})

Why this formula holds — the mechanism:

Chromate ester formation: The $-OH$ attacks the chromium reagent, forming a chromate ester.
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.
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? …

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A benzylic alcohol is one in which the -OH is bonded to the benzylic carbon, i.e. the sp3 carbon that is itself directly attached to the aromatic ring. Compounds (b) benzyl alcohol and (c) 1-phenylethanol satisfy this; (a) and (d) have the -OH on a carbon that is one atom away from the ring, so they are not benzylic.

Concept

The benzylic carbon is the ring-attached sp3 carbon. If the hydroxyl group is on that carbon, the compound is a benzylic alcohol.

Checking each compound

(a) C6H5-CH2-CH2OH: the ring-attached carbon is -CH2-, but the -OH is on the NEXT carbon -> not benzylic.
(b) C6H5-CH2OH: -OH is on the ring-attached -CH2- carbon -> benzylic. Correct. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
$-OH$
›Reveal solutionSolution
The question asks for the count of -OH groups in bithionol, terpineol, and chloroxylenol. By recalling their structures, we find the counts are 2, 1, and 1 respectively, so the correct option is (A).

Concept & Intuition

This is a test of your memory of common organic compounds used as antiseptics or in fragrances. The key is to know the core skeleton of each molecule and where hydroxyl (-OH) groups attach. Instead of memorizing every atom, focus on the functional groups: bithionol is a bisphenol (two phenol rings, each with an -OH), terpineol is a monoterpenoid alcohol (one -OH on a cyclohexene ring), and chloroxylenol is a chlorinated phenol (one -OH on a benzene ring). Counting -OH groups then becomes straightforward.

Step-by-step reasoning

Bithionol

Bithionol is a disinfectant. Its structure consists of two phenol rings linked by a sulfur atom (or sometimes a methylene bridge, but the common form has a sulfur bridge). Each phenol ring carries one -OH group.

Therefore, bithionol has 2 hydroxyl groups.

Terpineol

Terpineol is a naturally occurring alcohol with a pine-like odor. It is a monoterpene with a cyclohexene ring and a hydroxyl group attached to a carbon that also bears a methyl group (the typical structure is α-terpineol).

There is only one -OH group in terpineol.

Chloroxylenol

Chloroxylenol is the active ingredient in Dettol. It is a phenol derivative with one -OH group, one chlorine atom, and two methyl groups on the benzene ring. …
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TG EAPCET 2023Set ap-2023-05-11-AN1 markMCQ
$Q. Compounds that participate in self aldol condensation are (A) A, B, C & E only (B) A, C, D & E only (C) B, D & E only (D) B & D only$
›Reveal solutionSolution
Self-aldol condensation requires an $\alpha$ -hydrogen. Only acetophenone (B) and the indanone (D) have one; benzaldehyde, formaldehyde and benzoquinone do not. So the answer is "B & D only" — option (D).

The concept first: no $\alpha$ -H, no enolate, no aldol

Every aldol reaction begins the same way. A base removes a proton from the carbon next to the carbonyl — the $\alpha$ -carbon. It can do so because the resulting negative charge is delocalised onto the electronegative oxygen:
$\mathrm{-\!\overset{\alpha}{C}H\!-\!C(=O)\!-} \xrightarrow{\ \mathrm{OH^-}\ } \left[\mathrm{-\!\overset{\alpha}{C}^{-}\!-\!C(=O)\!-} \longleftrightarrow \mathrm{-\!C\!=\!C\!-\!O^-}\right]$
That resonance-stabilised anion is the enolate, and it is the nucleophile that attacks a second molecule of the carbonyl compound. Then dehydration gives the $\alpha,\beta$ -unsaturated product.

Take away the $\alpha$ -hydrogen and the whole sequence is dead at step one. So the entire question reduces to: which of these five molecules has a hydrogen on a carbon directly attached to a carbonyl carbon? Nothing else matters.

Step-by-step: inspect each compound

A — Benzaldehyde, $\mathrm{C_6H_5CHO}$ . The carbonyl carbon's neighbours are (i) the aldehydic H and (ii) an aromatic ring carbon, which bears no removable sp $^3$ hydrogen. There is no $\alpha$ -hydrogen. Benzaldehyde therefore cannot self-condense; with concentrated alkali it undergoes the Cannizzaro disproportionation instead. ✗

B — Acetophenone, $\mathrm{C_6H_5COCH_3}$ . The methyl group sits directly on the carbonyl carbon and carries three $\alpha$ -hydrogens. Base removes one, forming the enolate, and self-aldol condensation proceeds (giving dypnone on dehydration). ✓

C — Formaldehyde, $\mathrm{HCHO}$ . Both substituents on the carbonyl carbon are hydrogens — but those are aldehydic hydrogens, not $\alpha$ -hydrogens. There is no $\alpha$ -carbon at all. So no enolate, no self-aldol. (Like benzaldehyde, HCHO undergoes Cannizzaro — and in the crossed Cannizzaro it is always the one that gets oxidised.) ✗ …
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Q.Which of the following compounds are benzylic alcohols? (Two or more options may be correct.)

Concept understanding — Allylic Alcohol Identification

Allylic Alcohol Identification – From Intuition to Precision

The Intuition: Why "Allylic" Matters

The Precise Definition

How to Identify One – Step by Step

Examples and Non-Examples

A Common Mistake to Avoid

Why this formula?

Allylic Alcohol Identification: Understanding the "Why"

1. What Makes an Allylic Alcohol Special?

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

Why does this test work?

The reasoning behind the selectivity:

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

General formula for product:

Why this formula holds — the mechanism:

4. Another Key Test: Bromine Water Decolorization

Why does this test work? …

Concept

Checking each compound

Method: Benzylic-Carbon Identification Method

Core Concept

Steps

The concept first: no $\alpha$ -H, no enolate, no aldol

Step-by-step: inspect each compound

Q.Which of the following compounds are benzylic alcohols? (Two or more options may be correct.)

Concept

Checking each compound

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