When ethers are stored in the presence of atmospheric oxygen, they slowly oxidise to form hydroperoxides and dialkylperoxides. These are explosive in nature. Such a spontaneous oxidation by atmospheric oxygen is called autooxidation.

Inter molecular dehydration of alcohol: We have already learnt that when ethanol is treated with con.HSO₂ 4 at 443K, elimination takes place to form ethene. If the same reaction is carried out at 413K, substitution competes over elimination to form ethers.

Saytzeff's rule: During intramolecular dehydration, if there is a possibility to form a carbon-carbon double bond at different locations, the preferred location is the one that gives the more (highly) substituted alkene l.e., the stable alkene.
For example, the dehydration of 3,3 - dimethyl - 2- butanol gives a mixture of alkenes. The secondary carbocation formed in this reaction undergoes rearrangement to form a more stable tertiary carbocation.

An organic compound A (C₂H₆O₂) liberates hydrogen with metallic sodium.

(ii) Compound A when heated with anhydrous zinc chloride ultimately gives B (C₂H₄O)

(iii) Compound A when heated with conc. phosphoric acid gives C (C₄H₁₀O₃).

(iv) Compound A on oxidation with acidified K₂Cr₂O₇ gives compound D (CH₂O₂).

Compound	Compound Name	Formula
A	Ethylene glycol	\(\overset { { CH }_{ 2 }OH }{ \underset { { CH }_{ 2 }OH }{ | } } \)
B	Acetaldehyde	CH3-CHO
C	Diethylene glycol	\(\overset { HCOOH }{ \underset { HCOOH }{ + } } \)
D	Formic acid

(i) An aromatic hydrocarbon A reacts with propene in the presence of anhydrous AICl₃ to give a compound B.

(ii) Compound B undergoes oxidation in the presence of air and hydrogen peroxide to give C.

(iii) Compound C decomposes in HCI acid solution to give compound D and acetone. Identify A, B, C and D. Explain the reactions.

Compound	Compound Name	Formula
A	Benzene	C6H6
B	Cumene
C	Cumene hydroperoxide
D	Phenol	C6H5OH

(i) An organic compound A (C₂H₆O) liberates hydrogen with sodium metal.
\(\underset{(A)}{2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}}+2 \mathrm{Na} \longrightarrow 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{ONa}+\mathrm{H}_{2}\) 
(ii) A when heated with alumina at 620 K gives III an alkene B which when passed through Bayer'sreagent gives C(C₂H₆O₂)
\(\underset{(A)}{{ C }_{ 2 }{ H }_{ 5 }OH}\overset { { Al }_{ 2 }{ O }_{ 3 } }{ \underset { 620K }{ \longrightarrow } } \underset{(B)}{ { CH }_{ 2 }={ CH }_{ 2 } }\)
\(\underset{(B)}{{ CH }_{ 2 }=CH_{ 2 }+{ H }_{ 2 }O}+\left( O \right) \overset { ColdColddill,\ alkaline \ KMnO_{ 4 } }{ \underset { (Baeyer's \ Reagent) }{ \longrightarrow } } \overset { { CH }_{ 2 }OH }{ \underset { \underset { (C) }{ { CH }_{ 2 }OH } }{ | } } \) 
(iii) C reacts with PI₃ and gives back B. Identify A, B and C. Write the reactions

Compound	Compound Name	Formula
A	Ethanol	C2H5OH
B	Ethylene	CH2 = CH2
C	Ethylene glycol

(i) When alkaline solution of (A) is refluxed with chloroform.isomers (B) and (C) are obtained

(ii) (B) on oxidation gives acid (D),

(iii) The acid (D) is also obtained by treating sodium salt of (A) with CO₂ under pressure followed by hydrolysis.

Compound	Compound Name	Formula
A	Phenol	C6H5OH
B	Ortho hydroxy benzaldehyde
C	Para hydroxy benzaldehyde
D	Ortha hydroxybenzoic acid

(i) Compound (A) molecular formula C₃H₈O is isopropyl alcoho
\({ CH }_{ 3 }-\underset { \overset { | }{ OH } }{ C } H-{ CH }_{ 3 }\) 
(ii) Isopropyl alcohol reacts with P/I₂ to give Isopropyl iodide (B).

(iii) (B) reacts with AgNO₂ to form (C). (C) in turn reacts with nitrous acid to form pseudonitrol (blue colour).

(i) Compound (A) (C₃H₈O) answers Lucas test within 5-10 min is a 2° alcohol.

(ii) Compound (A) on oxidation gives (B) C₃H₆O.

(iii) Compound (B) on further oxidation gives (C) C₂H₄O₂.
\({ H }_{ 3 }C-\overset { \underset { | }{ { CH }_{ 3 } } }{ C } =O\overset { (O) }{ \longrightarrow } \underset { (C) }{ { CH }_{ 3 }COOH } \)
(iv) Compound (B) undergoes iodoform reactions
\({ CH }_{ 3 }-\overset { \underset { | }{ { CH }_{ 3 } } }{ C } H-OH\overset { { I }_{ 2 }/KOH }{ \longrightarrow } { CH }_{ 3 }{ COCH }_{ 3 }\overset { { I }_{ 2 } }{ \longrightarrow } { Cl }_{ 3 }{ COCH }_{ 3 }\overset { KOH }{ \longrightarrow } { CHI }_{ 3 }+{ CH }_{ 3 }COOK\)

(i) An organic compound (A) is identified as toluene from its molecular formula.
(ii) (A) on reaction with chlorine gives benzyl chloride which on further reaction with NaOH produces (B).
\({ C }_{ 6 }{ H }_{ 5 }-{ { { CH }_{ 3 } } }\overset { { Cl }_{ 2 } }{ \underset { -HCl }{ \longrightarrow } } { C }_{ 6 }{ H }_{ 5 }{ CH }_{ 2 }Cl\overset { NaOH }{ \underset { -Nacl }{ \longrightarrow } } \underset { (B) }{ { C }_{ 6 }H_{ 5 }{ CH }_{ 2 }OH } \) 
(iii) B) on oxidation with acidified K₂Cr₂O₇ gives (C).
\({ C }_{ 6 }{ H }_{ 5 }{ CH }_{ 2 }\overset { { K }_{ 2 }{ Cr }_{ 2 }{ O }_{ 7 }/{ H }^{ + } }{ \underset { (O) }{ \longrightarrow } } \underset { (C) }{ { C }_{ 6 }{ H }_{ 5 }CHO+{ { H }_{ 2 }O } } \) 
(iv) Benzaldehyde on treatment with 50% caustic soda gives (D) and (B).
\(\underset { (C) }{ { C }_{ 6 }{ H }_{ 5 }CHO } +{ C }_{ 6 }{ H }_{ 5 }CHO\overset { 50 \% }{ \underset { NaOH }{ \longrightarrow } } \underset { (B) }{ { C }_{ 6 }{ H }_{ 5 }{ CH }_{ 2 }OH } +\underset { (D) }{ { C }_{ 6 }{ H }_{ 5 }COOH } \) 

(i) Isopropyl alcohol (A).
(ii) (B) on oxidation gives (B)

(B) On further oxidation gives (C) which gives brisk efferescencewith Na₂CO₃.
\({ CH }_{ 3 }-\underset { \overset { || }{ \underset { (B) }{ O } } }{ C } -{ CH }_{ 3 }+{ I }_{ 2 }+KOH\longrightarrow \underset { (C) }{ { CH }I_{ 3 }+{ CH }_{ 3 }COOH } \) 

(i) Compound (A) is carbonyl compound, it is acetone
(ii) (A) on reduction with LiAlH₄ gives (B) it gives blue colour in Victor Meyer'stest.
\({ CH }_{ 3 }-\underset { \overset { || }{ \underset { (A) }{ O } } }{ C } -{ CH }_{ 3 }\overset { { LiAIH }_{ 4 } }{ \underset { \left[ H \right] }{ \longrightarrow } } { CH }_{ 3 }-{ CH }_{ 3 }-\underset { \overset { | }{ \underset { (B) }{ OH } } }{ CH } -{ CH }_{ 3 }\) 
(iii) (B) reacts with SOCl₂to give (C).

(iv) (C) on treatment with alcoholic KOH, forms (D) by elimination reaction.
\({ CH }_{ 3 }-\underset { \overset { | }{ \underset { (C) }{ Cl } } }{ C } H-{ CH }_{ 3 }\overset { alc.KOH }{ \longrightarrow } \underset { (D) }{ { CH }_{ 3 }CH={ CH }_{ 2 }+HCl } \) 

(i) Compound A is phenol. (C₆H₅OH)
Phenols on reaction with metallic Na compound (B) ie., Sodium Phenoxide
\(\underset { (A) }{ { 3C }_{ 6 }{ H }_{ 5 }OH+2Na } \longrightarrow \underset { (B) }{ { 2C }_{ 6 }{ H }_{ 5 }ONa+{ H }_{ 2 } } \) 
(ii) (B) is heated with CO₂ at 400K under pressure, to give (C). This is decomposed by dil.HCl, and (D) is formed

Compound	Compound Name	Formula
A	Phenol	C6H5OH
B	Sodium Phenoxide	C6H5ONa
C	Sodium Salicylate
D	Salicylicacid

(i) \( \underset { (A) }{2 \mathrm{C}_{2} \mathrm{H}_{5}} \mathrm{OH}+2 \mathrm{Na} \longrightarrow 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{ONa}+\mathrm{H}_{2} \uparrow \) 
(ii) Ethyl alcohol on mild oxidation gives acetaldehyde (B) It answers idoform test.
\({ CH }_{ 3 }-\underset{(A)}{{ CH }_{ 2 }OH}\overset { Mild\left( O \right) }{ \longrightarrow } \underset { (B) }{ CH_{ 3 }CHO } +{ H }_{ 2 }O\) 
\( \underset {(B)}{\mathrm{CH}_{3} \mathrm{CHO}+}3 \mathrm{I}_{2}+\mathrm{NaOH} \overset { B }{ \underset { Iodoform }{ \longrightarrow } } \mathrm{CHI}_{3}+\mathrm{HCOONa}+3 \mathrm{HI}\)
(iii) Acetaldehyde when treated with Cone H₂SO₄ undergoes polymerisation and the product obtained is paraldehyde (C).

Compound	Compound Name	Formula
A	Ethyl alcohol	CH3CH2OH
B	Acetaldehyde	CH3CHO
C	Paraldehyde

(i) (A) that gives turbidity within 5-10 min on reaction with anhydrous ZnCl₂/HCl means it must be secondary alcohol.
\({ CH }_{ 3 }\underset { \overset { | }{ OH } }{ CH } -{ CH }_{ 3 }\overset { anhydrous\ ZnCl_{ 2 }/HCl }{ \longrightarrow } { CH }_{ 3 }-\underset { \overset { | }{ CH } }{ CH } -{ CH }_{ 3 }\) 
(ii) Iso propyl alcohol on treatment with sodium hypochlorite, undergoes oxidation to give acetone compound (B).
\({ CH }_{ 3 }-\underset { \overset { | }{ \underset { (A) }{ OH } } }{ C } H-{ CH }_{ 3 }\overset { (O) }{ \longrightarrow } { CH }_{ 3 }-\underset { \overset { | }{ \underset { (B) }{ O } } }{ C } -{ CH }_{ 3 }\)