Organic Compounds Containing Oxygen Formula Sheet — JEE Main Chemistry

Every key Organic Compounds Containing Oxygen formula, definition and theorem for JEE Main Chemistry in one place — with common examiner traps and worked examples. Free to read; blurt from memory, then check your gaps.

Syllabus — topics coveredNTA · 16 sub-topics

General methods of preparation, properties, reactions and uses
Alcohols: Identification of primary, secondary and tertiary alcohols
Mechanism of dehydration of alcohols
Phenols: Acidic nature, electrophilic substitution reactions
Halogenation, nitration and sulphonation of phenols
Reimer-Tiemann reaction
Ethers: Structure
Aldehydes and Ketones: Nature of carbonyl group
Nucleophilic addition to >C=O group
Relative reactivities of aldehydes and ketones
Nucleophilic addition reactions (HCN, NH3 and derivatives), Grignard reagent
Oxidation, reduction (Wolff-Kishner and Clemmensen)
Acidity of alpha-hydrogen, Aldol condensation
Cannizzaro reaction, Haloform reaction
Chemical tests to distinguish aldehydes and ketones
Carboxylic Acids: Acidic strength and factors affecting it

Alcohols & Phenols — Classification, Structure & Acidity

Alcohols, phenols & ethers: Compounds with a

C-O single bond

Alcohols

have

- OH

on an

s p^{3}

carbon;

phenols

have

- OH

on an

s p^{2}

(aromatic) carbon;

ethers

have an oxygen bridging two carbons (

R - O - R^{'}

In phenol, the O lone pair conjugates into the ring, giving the C–O partial double-bond character.

Classifying alcohols

▸By number of $- OH$ : $mono / di / tri-hydric$ .
▸By the carbon: $1^{\circ}$ , $2^{\circ}$ , $3^{\circ}$ ( $s p^{3}$ , alkyl).
▸ $Allylic / benzylic$ : $- OH$ on $s p^{3}$ C next to $C = C$ / ring.
▸ $Vinylic / aryl$ ( $s p^{2}$ ): on $C = C$ / aromatic ring (= phenol).

★ Remember · Phenol — the C–O is partly double

The oxygen lone pair conjugates into the ring, so phenol's

C - O

shortens to

136 pm

(

s p^{2}

C) and gains partial double-bond character. This makes the ring electron-rich (activates EAS,

o / p

-directing) and the

O - H

more easily ionised.

🎯 Exam · Acidity ladder: acid > phenol > water > alcohol

Order follows conjugate-base stability: carboxylate (two equivalent resonance forms)

>

phenoxide (charge spread onto ring)

>

O H^{-}

>

alkoxide (charge localised). So

- COOH >

phenol

>

water

>

alcohol.

Compound	$p K_{a}$
$p$ -Nitrophenol	7.1
Phenol	10.0
$p$ -Cresol	10.2
Ethanol	15.9

⚠️ Watch out · Substituents flip acidity the opposite way for the two

phenol

: EWG (

- N O_{2}

, esp.

o / p

)

\Rightarrow

stronger acid; EDG (

- C H_{3}

, cresols)

\Rightarrow

weaker. Among

alcohols

, acidity is

1^{\circ} > 2^{\circ} > 3^{\circ}

because the

+ I

effect of alkyl groups destabilises the alkoxide.

🚫 Examiner Trap · Examiner traps

(1) Acidity order: carboxylic acid

>

phenol

>

water

>

alcohol — phenol IS acidic (turns blue litmus red), but a WEAKER acid than carbonic so it does NOT liberate C

O_{2}

from NaHC

O_{3}

(key phenol-vs-acid test). (2) EWG (

- N O_{2}

) on phenol raises acidity (

o, p ≫ m

); EDG (cresols) lowers it. (3) Alcohol acidity

1^{\circ} > 2^{\circ} > 3^{\circ}

(

+

I of alkyl). (4) Phenol's C

-

O is partly double (resonance)

\Rightarrow

shorter, and the ring is activated/o,p-directing.

Reactions of Alcohols, Phenols & Ethers

🎯 Exam · Oxidation tells the three alcohols apart

1^{\circ} PCC

aldehyde

KMn O_{4}

carboxylic acid;

2^{\circ} \to

ketone;

3^{\circ}

resists (no

α

-H on the C–OH carbon). Hot Cu (573 K) dehydrogenates

1^{\circ} / 2^{\circ}

to carbonyls but

dehydrates

3^{\circ}

to an alkene.

★ Remember · Lucas test & dehydration both follow 3° > 2° > 1°

Lucas

(conc. HCl/

ZnC l_{2}

3^{\circ}

turbid at once,

2^{\circ}

in ~5 min,

1^{\circ}

no turbidity cold.

Dehydration

ease

3^{\circ} > 2^{\circ} > 1^{\circ}

(carbocation route, Saytzeff major). Both reflect carbocation stability.

Phenol-only reactions

▸ $Kolbe$ : phenoxide $+ C O_{2} \to$ salicylic acid (o-hydroxybenzoic acid).
▸ $Reimer-Tiemann$ : $CHC l_{3}$ /NaOH $\to$ salicylaldehyde (o-CHO).
▸ $EAS is easy$ : $B r_{2}$ water $\to$ 2,4,6-tribromophenol (white ppt), no catalyst needed.
▸Dilute $HN O_{3} \to o / p$ -nitrophenol; conc. $\to$ picric acid; Zn dust $\to$ benzene.

★ Remember · Ethers: Williamson synthesis & HX cleavage

Williamson

R - X + R^{'} O^{-} N a^{+} S_{N} 2 R - O - R^{'}

— use a

primary

halide (

3^{\circ}

only eliminates).

Cleavage

by hot HX (HI > HBr > HCl): dialkyl ether

\to

R–I + R

^{'}

–OH; aryl alkyl ether

\to

phenol + R-X

(the strong aryl–O bond survives).

⚠️ Watch out · Esterification needs the acid, not the alcohol, activated

R - OH + R^{'} COOH ⇌ H^{+} R^{'} COOR + H_{2} O

is reversible — drive it by removing water. Acetylation of salicylic acid with

(C H_{3} CO)_{2} O

gives

aspirin

🚫 Examiner Trap · Examiner traps

(1) Lucas test (anhydrous ZnC

l_{2}

/HCl):

3^{\circ}

turbid at once,

2^{\circ}

\sim

5 min,

1^{\circ}

no reaction cold — distinguishes the three alcohols. (2) Dehydration ease

3^{\circ} > 2^{\circ} > 1^{\circ}

(carbocation), product follows Saytzeff. (3) Williamson works best with

primary

-

X (a bulky

3^{\circ}

halide ELIMINATES instead). (4) In ether cleavage by HI, the smaller/less-hindered alkyl leaves as R

-

I; phenyl ether gives phenol

+

-

I (aryl

-

O survives).

Aldehydes & Ketones — The Carbonyl & Its Preparation

The carbonyl group >C=O: A trigonal-planar

s p^{2}

carbon (

\sim 12 0^{\circ}

) doubly bonded to oxygen. Because O is more electronegative, the bond is polarised

C δ + = O δ -

— the carbon is electrophilic (open to nucleophiles), the oxygen nucleophilic.

The carbonyl group: a planar sp^2 carbon polarised d+C=d-O. — The carbonyl group: a planar $s p^{2}$ carbon polarised $C δ + = O δ -$ .

Aldehydes — named routes

▸ $Rosenmund$ : $RCOCl H_{2}, Pd - BaS O_{4} RCHO$ (poisoned catalyst stops at aldehyde).
▸ $Stephen$ : $RCN SnC l_{2} /HCl \to RCHO$ ; or DIBAL-H on a nitrile/ester.
▸ $Etard$ : toluene $+ Cr O_{2} C l_{2} \to$ benzaldehyde.
▸ $Gatterman-Koch$ : $C_{6} H_{6} + CO / HCl$ (AlC $l_{3}$ /CuCl) $\to$ benzaldehyde.

Ketones & shared routes

▸Acyl chloride $+ R_{2} Cd \to$ ketone.
▸Nitrile $+ R^{'} MgX H_{3} O^{+}$ ketone.
▸ $Friedel-Crafts$ acylation: arene $+ RCOCl / AlC l_{3} \to$ aryl ketone.
▸ $Both$ : oxidation/dehydrogenation of $1^{\circ} / 2^{\circ}$ alcohols; ozonolysis; alkyne hydration ( $H g^{2 +} / H_{2} S O_{4}$ : ethyne $\to$ ethanal, others $\to$ ketone).

★ Remember · Aldehyde vs ketone — the key difference

An aldehyde has at least one

H

on the carbonyl carbon (

RCHO

); a ketone has two carbon groups (

RCO R^{'}

). That single H is why aldehydes are oxidised far more easily than ketones.

🎯 Exam · Reactivity to nucleophiles: aldehyde > ketone

Two alkyl groups in a ketone (i) crowd the approach of a nucleophile (

steric

) and (ii) push electron density onto the carbonyl C by

+ I

, lowering its

δ +

(

electronic

). Aromatic carbonyls are even less reactive (ring resonance donates).

🚫 Examiner Trap · Examiner traps

(1)

Rosenmund

(

H_{2}

/Pd-BaS

O_{4}

, poisoned) stops at the aldehyde — without the poison it over-reduces. (2) Aldehyde always has

\geq 1

H on the carbonyl C; ketone has two C-substituents

\Rightarrow

different oxidation behaviour. (3) Etard (Cr

O_{2} C l_{2}

) and Gatterman

-

Koch (CO/HCl) make aromatic ALDEHYDES specifically. (4) Reactivity to nucleophiles: aldehyde

>

ketone (less steric crowding, less

+

I donation).

Nucleophilic Addition to the Carbonyl

Nucleophilic addition to >C=O: Nu^- attacks the planar carbonyl to give an sp^3 alkoxide, then H^+ adds. — Nucleophilic addition to $> C =$ O: Nu $^{-}$ attacks the planar carbonyl to give an $s p^{3}$ alkoxide, then H $^{+}$ adds.

Common additions

▸ $+ HCN$ (base-catalysed) $\to$ $cyanohydrin$ ( $- OH$ and $- CN$ on one C).
▸ $+ NaHS O_{3} \to$ crystalline $bisulphite adduct$ — used to separate/purify aldehydes & methyl ketones (regenerate with dilute acid/alkali).
▸ $+ 2 R^{'} OH$ /dry HCl $\to$ $acetal$ (via hemiacetal); ketone $+$ glycol $\to$ cyclic ketal — a protecting group.
▸ $+ RMgX H_{3} O^{+}$ alcohol.

Reagent H₂N–Z	Product
Ammonia	Imine
$1^{\circ}$ amine	Schiff base (subst. imine)
Hydroxylamine	Oxime
Hydrazine	Hydrazone
Phenylhydrazine	Phenylhydrazone
2,4-DNP (Brady's)	2,4-DNP-hydrazone
Semicarbazide	Semicarbazone

🎯 Exam · 2,4-DNP is the carbonyl fingerprint

Aldehydes and ketones give orange-red

2,4-dinitrophenylhydrazone

solids with Brady's reagent — a classic test for the

> C = O

group (but it does

not

distinguish an aldehyde from a ketone).

⚠️ Watch out · These reactions are acid-catalysed but not too acidic

H_{2} N - Z

additions need mild acid: too much acid protonates the nucleophile (killing it), too little slows the dehydration to

> C = N - Z

. The equilibrium is pulled forward by losing water.

🚫 Examiner Trap · Examiner traps

(1) Nucleophile attacks

⊥

to the planar carbonyl; C rehybridises

s p^{2} \to s p^{3}

— net addition of Nu

^{-}

and H

^{+}

across C

=

O. (2) NaHS

O_{3}

adduct works for aldehydes and METHYL ketones only (steric) — used to purify them. (3) 2,4-DNP (Brady's) gives orange-red solids

=

test for

> C =

O (NOT acids/esters). (4) Reactivity aldehyde

>

ketone here too.

α-Hydrogen Reactions, Reduction & Distinguishing Tests

★ Remember · Aldol needs an α-H; Cannizzaro needs none

Aldol

(dil. NaOH,

α

-H required):

2 C H_{3} CHO \to

3-hydroxybutanal

Δ, - H_{2} O

but-2-enal.

Cannizzaro

(no

α

-H, conc. alkali):

2 HCHO \to C H_{3} OH + HCO O^{-}

(one reduced, one oxidised).

Reduction of the carbonyl

▸ $NaB H_{4}$ / $LiAl H_{4}$ (or $H_{2}$ /cat.) $\to$ $1^{\circ}$ / $2^{\circ}$ alcohol.
▸ $Clemmensen$ (Zn-Hg / conc. HCl): $> C = O \to C H_{2}$ — acidic conditions.
▸ $Wolff-Kishner$ ( $N H_{2} N H_{2}$ , then KOH/ $Δ$ ): $> C = O \to C H_{2}$ — basic conditions.
▸Pick by what the rest of the molecule tolerates (acid vs base).

Distinguishing aldehydes from ketones

▸ $Tollens$ : $[Ag (N H_{3})_{2}]^{+} \to$ bright $silver mirror$ (aldehydes only).
▸ $Fehling$ : $C u^{2 +} \to$ red-brown $C u_{2} O ↓$ (aliphatic aldehydes; aromatic aldehydes do $not$ respond).
▸Ketones resist both mild oxidants; strong oxidation cleaves C–C $\to$ smaller acids.
▸ $Iodoform$ ( $I_{2}$ /NaOH): $C H_{3} CO -$ or $C H_{3} CH (OH) -$ $\to$ yellow $CH I_{3} ↓$ .

🎯 Exam · α-H acidity is the engine of condensations

The

α

-H is acidic because the carbonyl withdraws electrons (

- I

) and the resulting carbanion is resonance-stabilised as an

enolate

. No

α

-H

\Rightarrow

no aldol, but Cannizzaro becomes possible.

⚠️ Watch out · Cross-aldol gives a messy mixture

Two different carbonyls that

both

have

α

-H give up to

four

products (two self + two cross). It is only synthetically useful when one partner has

no

α

-H (e.g. benzaldehyde) so it can only be the electrophile.

🚫 Examiner Trap · Examiner traps

(1) Aldol needs an

α

-H; Cannizzaro happens only with NO

α

-H (HCHO, benzaldehyde) — disproportionation. (2)

Tollens/Fehling oxidise aldehydes only

(ketones resist); aromatic aldehydes FAIL Fehling. (3) Iodoform (

I_{2}

/NaOH) is positive for C

H_{3} CO -

or C

H_{3}

CH(OH)

-

groups (and ethanal/ethanol) — flags a methyl-carbonyl, NOT all carbonyls. (4) Use Clemmensen (acidic) vs Wolff

-

Kishner (basic) by the substrate's acid/base sensitivity.

Carboxylic Acids — Preparation & Acidity

Carboxylic acid –COOH: A carbonyl fused to a hydroxyl (

carb

onyl + hydr

oxyl

). The

- OH

lone pair donates into the

C = O

by resonance, so the carboxyl carbon is

less

electrophilic than a plain carbonyl — it does acyl

substitution

, not simple addition.

Preparation of –COOH

▸Oxidation of $1^{\circ}$ alcohols / aldehydes ( $KMn O_{4}$ , $K_{2} C r_{2} O_{7}$ ).
▸Alkylbenzenes $KMn O_{4} /KOH$ ArCOOH (whole side chain $\to$ COOH).
▸Hydrolysis of nitriles / amides ( $H^{+}$ or $O H^{-}$ ).
▸ $Grignard + C O_{2}$ (dry ice) $H_{3} O^{+}$ RCOOH (adds one carbon).
▸Hydrolysis of acyl halides, anhydrides or esters.

★ Remember · Why –COOH beats phenol

Both lose

H^{+}

, but the

carboxylate

has

two equivalent

resonance forms with the charge on two electronegative O atoms, whereas phenoxide spreads charge onto less-electronegative ring carbons. So

- COOH >

phenol

>

water

>

alcohol.

🎯 Exam · Substituents: EWG strengthen, EDG weaken

C F_{3} COOH > CC l_{3} COOH > CHC l_{2} COOH > ClC H_{2} COOH > HCOOH > C H_{3} COOH

. EWG (

- I

) stabilise the carboxylate; more/closer halogens

\Rightarrow

stronger acid. (Acetic

p K_{a}

4.76, benzoic 4.19.)

⚠️ Watch out · Aromatic ring effect can look backwards

Direct attachment of phenyl/vinyl

increases

acidity (the

s p^{2}

carbon is more electronegative). On a benzoic acid,

EWG

raise acidity (4-nitrobenzoic

p K_{a}

3.41) while

EDG

lower it (4-methoxy 4.46 vs benzoic 4.19).

🚫 Examiner Trap · Examiner traps

(1) Acidity: acid

>

phenol

>

water

>

alcohol — carboxylate has TWO equivalent resonance forms (charge on 2 O). (2) EWG (

-

I: F, Cl, N

O_{2}

) RAISE acid strength, and nearer the COOH

=

stronger (CC

l_{3}

COOH

≫

H_{3}

COOH). (3) Acids exist as H-bonded

dimers

\Rightarrow

b.p. higher than alcohols of similar mass. (4) HCOOH is the strongest simple acid (no

+

I alkyl) and also reduces (has a

-

CHO part).

Carboxylic Acids — Reactions & The Carbonyl Toolkit

🎯 Exam · NaHCO₃ test separates acid from phenol

RCOOH + NaHC O_{3} \to RCOONa + H_{2} O + C O_{2} ↑

. The

brisk effervescence

confirms

- COOH

; phenol is too weak to react with bicarbonate, so this cleanly distinguishes the two.

C–OH cleavage (acyl substitution)

▸ $Esterification$ : $+ R^{'} OH / H^{+} ⇌$ ester $+ H_{2} O$ .
▸ $+ PC l_{5} / PC l_{3} / SOCl_{2} \to$ acyl chloride ( $SOC l_{2}$ best — gaseous by-products).
▸ $Anhydride$ : $2 RCOOH P_{2} O_{5}, Δ (RCO)_{2} O$ .
▸ $+ N H_{3} Δ$ amide (via ammonium salt, $- H_{2} O$ ).

–COOH group & the α-carbon

▸ $Reduction$ : $LiAl H_{4}$ or $B_{2} H_{6} \to 1^{\circ}$ alcohol — $NaB H_{4}$ does $not$ touch $- COOH$ .
▸ $Decarboxylation$ : $RCOONa NaOH/CaO, Δ R - H$ (soda-lime).
▸ $HVZ$ : $RC H_{2} COOH X_{2} /red P α$ -halo acid (needs an $α$ -H).
▸Kolbe electrolysis of the salt $\to$ alkane (doubles the carbon count).

★ Remember · Diborane reduces –COOH but not esters

B_{2} H_{6}

(diborane) reduces

- COOH

readily to

1^{\circ}

alcohol but leaves esters, nitro and halo groups alone — a useful selectivity.

LiAl H_{4}

is stronger and less selective.

⚠️ Watch out · No simple addition at –COOH

Unlike aldehydes/ketones, the carboxyl carbon's

δ +

is muted by the

- OH

lone pair (resonance). So nucleophiles trigger

acyl substitution

(replace

- OH

), never a stable tetrahedral addition product.

🚫 Examiner Trap · Examiner traps

(1)

-

COOH does NOT give simple nucleophilic ADDITION (the

-

OH lone pair feeds the carbonyl by resonance) — it does acyl SUBSTITUTION. (2) LiAl

H_{4}

B_{2}

H_{6}

reduce

-

COOH to

1^{\circ}

alcohol, but

\textbf{NaBH}$_4$ does NOT. (3) HVZ ($X_2$/red P) needs an $\alpha$-H $\Rightarrow$ $\alpha$-halo acid. (4) NaHC$O_3$ effervescence (C$O_2

↑

) d i s t in gu i s h es

-$COOH from phenol — fast, diagnostic.

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This Organic Compounds Containing Oxygen formula sheet covers all the high-yield Chemistry formulas, definitions and theorems you need for JEE Main, across General methods of preparation, properties, reactions and uses, Alcohols: Identification of primary, secondary and tertiary alcohols, Mechanism of dehydration of alcohols, Phenols: Acidic nature, electrophilic substitution reactions, Halogenation, nitration and sulphonation of phenols — each shown with the key result and, where useful, a worked example.

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