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October 13, 2022 admin Off Uncategorized

INTRODUCTION

Hydroxy compound can be classified in the following three categories.

Aliphatic hydroxy compound

Monohydric i.e.
Dihydric
Polyhydric

Contain three and more than three hydroxyl group.

Monohydric alcohols are of three types.

Illustration 1. Classify the following into primary, secondary and tertiary alcohols:

(a)

(b)

(c)

Solution: (a) Tertiary

(b) Secondary

PREPARATION OF ALCOHOLS

From Alkanes

Alkanes having tertiary carbon on oxidation with cold alkaline KMnO4 give tertiary alcohol.

From Alkenes

Alkenes can be converted into alcohol by the following reactions:

From alkyl halides

Alkyl halides give alcohol with KOH/NaOH or with moist Ag2O.

Reduction of aldehydes and ketones

(a) Reduction by reducing agents

(i) Aldehyde gives primary alcohol

(ii) Ketone gives secondary alcohol

Reducing agents

(i) LiAlH4 (ii) NaBH4

(iii) Na/C2H5OH (iv) Metal (Zn, Fe or Sn)/Acid (HCl, dil H2SO4 or CH3COOH)

(v) (a) Aluminium isopropoxide/isopropylalcohol (b) H2O

(vi) H2/Ni

NaBH4 and aluminium isopropoxide reduces only carbonyl group and has no effect on any other group.

Reduction with aluminium isopropoxide is known as Meerwein – Ponndorf Verley (MPV) reduction.
LiAlH4 has no effect on double and triple bonds but if compound is β – aryl, α, β – unsaturated carbonyl compound then double bond also undergoes reduction.

(b) Reduction by Grignard reagents

Addition followed by hydrolysis

Methanol can not be prepared by this method.

Reduction of carboxylic acid, Acid chlorides and esters:

(a) Reduction by LiAlH4

G = OH (acid)

G = Cl (acid chloride)

G = OR′ (ester)

(b) Reduction by BH3

Carboxylic acids and esters are reduced in to primary alcohol by BH3.

*Illustration 2.*
*Solution:*
	If compound has ⎯COOH as well as ester group then reactivity of ester is more than acid towards LiAlH4.

	Acid is more reactive than ester towards BH3.

Exercise 1. Find A and B.

(i)

(ii)

From aliphatic primary amines

It react with nitrous acid to give alcohol.

Nature of alcohol depends on the nature of carbon having ⎯NH2 group.
Reaction proceeds through carbocation hence rearranged alcohol is obtained.

From Oxiranes

Oxiranes react with Grignard reagent to give mono hydric alcohol. Nature of G.R is basic hence it attack on less hindered carbon of oxirane ring.

Illustration 3. (a) Find A, B, C, D, E.

(b)

*Solution:* (a)
(b)

Fermentation of carbohydrates

*Illustration 4.*	(i)
	(ii)

*Solution:*	(i)
	(ii)

Exercise2.

(i) Identify A, B & C

*(ii)*

	*Account the reason for the above reactions.*

PHYSICAL PROPERTIES

Solubility

Alcohols are soluble in water due to formation of H – bonding between water & them. As the molecular mass increases, the alkyl group become larger which resists the formation of
H – bonds with water molecules and hence the solubility goes an decreasing.

Boiling Point

Intermolecular H – bonding is present between alcohol molecules. This makes high boiling point.

Amongst the isomeric alcohols, the order of boiling point is 1° > 2° > 3° alcohol.

CHEMICAL PROPERTIES

Chemical properties of alcohols can be discussed under following categories:

(A) Reaction involving breaking of carbon – oxygen bond.

(B) Reaction involving breaking of oxygen – hydrogen bond.

(D) Dehydrogenation of alcohols.

(E) Some miscellaneous reactions of monohydric alcohol.

(A) Reaction involving breaking of carbon – oxygen bond

Order of reactivity of alcohol. 3° > 2° > 1°

(i) SN reaction

(ii) Dehydration of alcohol

Dehydration of alcohol to give alkene.

(a) Dehydrating agents are

Conc H2SO4/Δ, KHSO4/Δ, H3PO4/Δ, Anhyd Al2O3/Δ, Anhyd PCl5/Δ, Anhyd ZnCl2/Δ, BF3/Δ, P2O5/Δ.

(b) Reactivity of alcohols. (Ease of dehydration)

3° > 2° > 1°

* Alcohols on acetylation gives acetyl derivative which on pyrolytic elimination always gives Hofmann product.

Mechanism in presence of acidic medium

E1 mechanism: follow saytzeff’s rule.

Exercise 3.

Write mechanism

(B) Reactions due to breaking of oxygen hydrogen bond. (Reactions due to acidic character of alcohols)

(a) Alcohols are acidic in nature because hydrogen is present on electro negative oxygen atom.

(b) Alcohol is weaker acid

acidity ∝ stability of acid anions.

Acidity of 1° > 2° > 3°

Alcohols give following reactions due to breaking of oxygen – hydrogen bond.

(i) Reaction with metal

M = 1st group metal.

M = Al, Mg, Zn

Aluminium alkoxide

(ii) Esterification (With carboxylic acid)

It is reversible acid catalysed reaction. It follow SN1 mechanism.

Increasing the size of alkyl group on alcohol part decreases the nucleophilic character because steric hindrance increases.

Order of reactivity of alcohols CH3OH > 1° alc > 2° alc > 3° alc

(iii) Ester formation with proton acid having –OH group: to give inorganic ester.

(a)
(b)

(iv) Alkylation of Alcohol

Methylation is mainly used for determination of hydroxyl groups in an unknown compound.

Exercise 4.

Arrange the following in increasing order of acidic strength.

(a)

(b)

Arrange the following in increasing order of esterification:

MeCOOH EtCOOH (Et)2CHCOOH

(i) (ii) (iii)

Oxidation of alcohol is dehydrogenation reaction which is 1, 2 – elimination reaction.

So oxidation of alcohol ∝ numbers of α – hydrogen atom.

(a) With mild oxidising agents:

(i) X2

(ii) Fenton reagent [FeSO4/H2O2]

(iii) Jones reagent / CH3COCH3 [CrO3/dil. BaSO4]

(iv) K2Cr2O7/H+ cold

Note:

PCC (Pyridinium chloro chromate) is a selective reagent which converts 1° alc to aldehyde.

(b) With strong oxidising agent

Oxidising agents are

(i) (ii)

(iii) (iv)

(D) Dehydrogenation with Cu/573K or Ag/573K

(a) 1° alcohol ⎯⎯→ aldehyde

(b) 2° alcohol ⎯⎯→ ketone

Reduction

(E) Miscellaneous reactions of mono hydric alcohol

(i) Methylation with CH2N2 in presence of BF3.

(ii) Haloform reaction

Ethyl alcohol and 2° methyl alcohol gives haloform reaction.

Exercise 5.

(i) Out of these compound which gives iodoform test.

(a) CH3 ⎯ CH2 ⎯ CHOHCH3 (b) PhCH2CHOHCH3

(c) PhCHOHCH3 (d) CH3CH2OH

(e) CH3COCH2 ⎯ COOC2H5

(ii)

Distinguishing 1°, 2°, 3° alcohol

Test	1° alc	2° alc	3° alc
(I) Lucas test [ZnCl2 + HCl]	No reaction at room temperature	White turbidity after 5 – 10 min.	White turbidity instantaneously
(II) Victor Meyer test (P/I2, AgNO2, HNO2, NaOH)	Red colour	Blue colour	Colourless

PERIODATE OXIDATION

Compounds that have hydroxyl group on adjacent atoms undergo oxidation cleavage when they are treated with aq. Periodic acid (HIO4). The reaction breaks carbon carbon bonds and produced carbonyl compounds (aldehyde, ketones or acids)

It takes place through a cyclic intermediate.

Other examples

This oxidation is useful in determination of structure.

Illustration 5. Write the products of the reaction of t-butyl alcohol with PBr3, conc. H2SO4, CH3COCl, Na, CH3MgBr, Na2Cr2O7/H2SO4.

Solution: .

Illustration 6. Write products

(a)

(b)

Solution: (a) 2HCHO + CO2

(b) No reaction (as it is not a vicinal diol)

Exercise 6.

(a) Arrange the following alcohols in order of ease of dehydration.

(b) Find product

*(a)*		*(b)*
*(c)*

PINACOL – PINACOLONE REARRANGEMENT

Action of H2SO4 on 1, 2 diols.

Ditertiory – 1, 2 diols convert in to ketones on treatment wit H2SO4.

Mechanism

Migratory preference of the group

Migration depends on the stability of Transition state.

In general migration of C6H5 > alkyl

Illustration 7. List three methods with chemical equations for the preparation of alcohols from alkenes.

Solution: Hydration, hydroboration and oxymercuration – demercuration of alkenes.

Hydration:

Hydroboration

Oxymercuration – demercuration

Illustration 8. Which of the following alcohols would react fastest with Lucas reagent?

Solution: , it being a tertiary alcohol.

Exercise 7.

Find out

(i)

(ii)

Exercise 8.

Give a possible structure for the substance C5H10O2 behaving in the following manner.

ETHER

* R ⎯ O ⎯ R′ Alkoxy alkane (Di alkyl ether)

* R = R′ Symmetrical ether.

R ≠ R′ Unsymmetrical or mixed ether.

‘O’ is to be counted with least number of C atom.

Example:

CH3 ⎯ O ⎯ C2H5 Methoxy ethane

CH3 ⎯ O ⎯ C6H5 Methoxy benzene

There are various types of cyclic ethers also.

PREPARATION OF ETHERS

(i) From 1° alcohol

(a) With H2SO4

Order of dehydration 1° > 2° > 3° alcohol

(b) With diazomethane

(c) Alcohol having at least one hydrogen at fourth carbon gives five membered cyclic ether with Pb(OAC)4. The reaction is free radical reaction which is initiated by heat or light.

Williamson’s synthesis

reaction of a sodium alkoxide with alkyl halide, alkyl sulphonate or alkyl sulphate is known as Williamson synthesis of ethers.

In this reaction alkoxide may be alkoxide of primary, secondary as well as tertiary alcohol.
Alkyl halide must be primary.
In case of tertiary alkyl halide, elimination occurs giving alkenes
With a secondary alkyl halide, both elimination and substitution products are obtained.

Illustration 9. Write the product

(i)
(ii)

Solution:

(i)

(ii)

Exercise 9.

Find product

(i)

(ii)

(3) From Alkane

(a)

(b)

(4) From Grignard reagent

Higher ethers can be prepared by treating α – halo ethers with suitable reagents.

(5) From Alkyl halide

PROPERTIES OF ETHERS

Dipole nature of ether

Ethers have a tetrahedral geometry i.e. oxygen is sp3 hybridized. The C ⎯ O ⎯ C bond angle in ether is 110°. Because of the greater electronegativity of oxygen than carbon, the C ⎯ O bonds are slightly polar and are inclined to each other at an angle of 110°C, resulting in a net dipole moment.

The bond angle is slightly greater than the tetrahedral angle due to repulsive interaction between the two bulky groups.

Chemical Reaction

Dialkyl ethers reacts with very few reagents other than acids. The only active site for other reagents are the C ⎯ H bonds of the alkyls. Ethers has ability to solvate cations (electrophile) by donating an electron pair from their oxygen atom. These properties make ether as solvents for many reactions.

On standing in contact with air, most aliphatic ethers are converted slowly into unstable peroxides.

Ether gives following reactions:

Nucleophilic substitution reactions

Note:

Type of ethers also make a difference in the mechanism followed during the cleavage of C—O by HI/HBr.

Combinations	Mechanism follows
1°R + 2°R	Less sterically hindered ⇒ SN2
2°R + 3°R	More sterically hindered ⇒ SN1
1°R + 3°R	Nature of mechanism decoded by nature of solvent.

Methyl cation is stabler than phenyl cations

(B) Dehydration with H2SO4/Δ and Anhy Al2O3/Δ

(i) When both alkyl groups has β – hydrogen.

(ii) When only alkyl group has β – hydrogen.

Hot conc. H2SO4 react with secondary and tertiary ethers to give a mixture of alcohols and alkenes.

(CH3)3CO—CH3 (CH3)2—C=CH2 + CH3OH

(1) Halogenation:

Monohalogenation takes place at α carbon (with small amount)

(a)
(b)

(2) Reaction with CO: give ester

Illustration 10. Explain why sometimes explosion occurs while distilling ethers.

Solution: It is due to formation of peroxide

Illustration 11. The basicity of the ethers towards BF3 has the following order, explain.

Solution: There are steric effects in the Lewis acid-Lewis base complex formation between BF3 and the respective ethers.

Illustration 12. What are crown ethers? How can the following reaction be made to proceed?

Solution: Crown ethers are large ring polyethers and are basically cyclic oligomers of oxirane which may have annulated rings. They are designated according to ring size and the number of complexing oxygen atoms, thus 18-crown – 6 denotes an 18-membered ring with 6-oxygens. The molecule is shaped like a “doughnut”, and has a hole in the middle.

These are phase transfer catalysts. This is a unique example of “host-guest relationship”. The crown ether is the host, the cation is the guest. The cavity is well suited to fit a K+ or Rb+ which is held as a complex. Interaction between host and guest in all these complexes are mainly through electrostatic forces and hydrogen bonds.

The reaction can be made to process by using catalytic amount of crown ether, 18-crown-6.

Illustration 13. Explain why

is much more soluble than furan in water.

Solution: THF is more soluble than furan. In THF, in contrast to furan the electron pairs are available for H-bonding with water which makes it more soluble in water.

Exercise 10.

What chemical methods can be used to distinguish between the following pairs of compounds?

(a) Ethoxy ethanol and methyl isopropyl ether.

(b) Butyl iodide and butyl ethyl ether.

(c) Ethyl propyl ether and ethyl allyl ether.

Exercise 11.

Ether A cleaves much faster than B with conc. HI. Explain.

PHENOL

These are organic compounds a hydroxyl group attached directly to a benzene ring.

Preparation

Industrial Method

(i) From chloro benzene (Dow’s process)

Chlorobenzene is heated with NaOH at 673 K and under pressure of 300 atm to produced sodium phenoxide which on acidification yields phenol.

(ii) Cumene Process

Cumene obtained from propene & benzene cumene on air oxidation followed by acidification with H2SO4 gives phenol & acetone.

(iii) From benzene sulphonic acid

It is fused with NaOH gives sodium salt of phenol.

(iv) From benzene diazonium chloride

This gives Ar SN1 reaction with H2O to form phenol.

Illustration 14. Starting from 1-methyl cyclohexene, prepare the following:

(a)

(b)

*Solution:*	(a)
	(b)

Exercise 12.

Starting from 1-methyl cyclohexene, prepare the following:

(a)

(b)

PHYSICAL PROPERTIES

Phenol is needle shaped solid, soon liquefies due to high hygroscopic nature. It is less soluble in water, but readily soluble in organic solvents.

Phenol has high boiling point due to presence of hydrogen bonding.

Acidity of phenol

Phenol is weak acid. It reacts with aqueous NaOH to form sodium phenoxide, but does not react with sodium bicarbonate.

The acidity of phenol is due to the stability of the phenoxide ion, which is resonance stabilized as shown below:

In substituted phenols, the presence of electron withdrawing groups at ortho and para positions such as nitro group, stabilizes the phenoxide ion resulting in an increase in acid strength. It is due to this reason that ortho and para nitro phenols are more acidic than phenol.

On the other hand, electron releasing groups such as alkyl group, do not favour the formation of phenoxide ion – resulting in decrease in acid strength.

For example: (cresol are less acidic then phenol)

Exercise 13.

Arrange each group of compounds in order of decreasing acidity:

CHEMICAL REACTIONS

(A) Reaction due to breaking of O – H bond

Phenol is more reactive than alcohol for this reaction because phenoxide ion is more stable than the alkoxide ion.

Reactions of phenol due to breaking of ⎯O ⎯ H bond are given below:

Acylation

Fries rearrangement

Phenolic esters are converted in to o ⎯ and p ⎯ hydroxy ketones in the presence of anhydrous AlCl3. Generally low temperature favours the formation of p – isomer and higher temperature favour the o – isomer.

(B) Reactions due to breaking of carbon- Oxygen bond

Nucleophilic substitution reaction

Phenols are less reactive than aliphatic compound because:

(i) ⎯OH group is present on sp2 hybridised carbon. This makes C ⎯ O bond stronger.

(ii) ‘O’ is more electronegative than halogens. This also makes C ⎯ O bond stronger than
C ⎯ X.

(iii) There is some double bond character between carbon and oxygen due to the resonance. This also makes C ⎯ O bond stronger.

However it give SN under drastic condition.

MERCURATION

Mercuric acetate cation. [HgOAC]+ is a weak electrophile which substitutes in ortho and para position of phenol. Usually donating product is O–acetoxy mercuriphenol. The mercuric compound can be converted to iodophenol.

Miscellaneous reaction

(i) Reaction with Zn dust

(ii) Oxidation

(iii) Condensation with phthalic anhydride

Exercise 14.

Discuss the product formed in the bromination of p-phenol sulfonic acid.

Mechanism of some important reactions

Reimer Tieman reaction

The electrophile is the dichloro carbene, :CCl2, formation of carbene is an example of
α – elimination.

(i)
(ii)
(iii)

Kolbe’s reaction

Mechanism

Illustration 15. How will you convert?

(i) phenol to aspirin (ii) phenol to salol.

(iii) phenol to oil of winter green. (iv) phenol to benzoic acid.

*Solution:*	(i)
	(ii)
	(iii)
	(iv)

Illustration 16. What product would you expect in the following reaction? Explain.

Solution:

It is an ‘abnormal’ product formed in the Reimer-Tiemann reaction when the dienone cannot tautomerize to regenerate a phenolic system.

Illustration 17. How would you distinguish between the following pairs?

(a) Phenol and cyclohexanol

(b) Ethyl alcohol and methyl alcohol

Solution: (a) Phenol gives coloration with FeCl3 solution

(b) Ethyl alcohol responds to the iodoform test

Exercise 15.

How would you distinguish between the following pairs?

(a) 2-Pentanol and 3-pentanol

(b) 1-Propanol and phenol

Exercise 16.

Offer explanation for the following observations:

(a) Why is phenol unstable in the keto-form?

(b) The following dehydration is extremely facile:

(c) Why does thionyl chloride provide alkyl chlorides of high purity?

(d) 2-Methyl -2- pentanol dehydrates faster than 2 – methyl – 1 – pentanol.

(e) Phenol is acidic but ethyl alcohol is neutral.

(f) Ethanol responds to Iodoform test but tert- butanol does not.

(g) A tertiary alcohol reacts faster than a primary alcohol in the Lucas test.

Exercise 17.

How will you effect the following conversion?

*(a)*
*(b)*	C2H5OH ⎯→ CH3CH2CH2OH

Exercise 18.

How will you effect the following conversion?

*(a)*
*(b)*
*(c)*
*(d)*

Exercise 19.

Write mechanism

Some Commercially Important Alcohols And Phenols

(i) Methanol: Methanol is also called wood spirit since originally it was obtained by destructive distillation of wood. Now a days it is prepared by catalytic hydrogenation of water gas.

Uses: It is largely used as:

(a) a solvent for paints, varnishes and celluloids.

(b) for manufacturing of formaldehyde.

(c) for denaturing ethyl alcohol, i.e. to make it unfit for drinking purpose. Denatured alcohols is called methylated spirit.

(d) in manufacture of perfumes and drugs.

Ethanol: Ethanol is mainly prepared by hydration of ethene formation of carbohydrates gives only 95% alcohol the rest being water. This is called rectified spirit.

Uses: It is largely used as an

(a) antiseptic.

(b) solvent for paints, lacquers, varnishes, dyes, cosmetics, perfumes, tinctures, cough syrups etc.

(d) As an important beverages.

(e) As power alcohol a mixture of 20% absolute alcohol and 80% petrol (gasoline) with benzene or tetralin as a co-solvent.

(f) As an antifreeze in automobile radiators.

Absolute alcohol: Absolute alcohol is 100% ethanol prepared from rectified spirit 95.5% alcohol as follows:

In laboratory absolute alcohol is prepared by keeping the rectified spirit in contact with calculated amount of quick lime for few hours and then refluxing and distilling it.

Phenol or Carbolic Acid

Uses

(i) As an antiseptic and disinfectant in soaps and lotions.

(ii) In manufacture of drugs like, aspirin, salol, salicylic acid, phenacetin.

(iii) In the manufacture of bakelite.

(iv) In the manufacture of picric acid, phenolphthalein, azo dyes.

(v) As a preservative for ink.

Ethylene Glycol: Ethane 1, 2 diol

Preparation: Lab preparation by hydroxylation.

(i)

Manufacture

Physical properties:

It is highly viscous because of the presence of two OH bond it undergoes extensive intermolecular H-bonding. Same reason owes to high solubility in water and high boiling point.

Chemical properties

Reaction with sodium

(i)
(ii)
(iii)
(iv)
(v)
(vi)	Oxidation: Ethyelene glycol upon oxidation gives different products with different oxidising agents. For example.
	(a)
	(b)	With periodic acid HIO4 or lead tetra acetate.
		Also called malapride reaction.

Dehydration

(a)
(b)
(c)
(d)

Uses

(a) As a solvent.

(b) Antifreeze in the radiators of cars and aeroplanes.

Exercise 20.

How would you convert cyclohexane to 1, 6 – hexanediol?

Glycerol (Propane 1, 2, 3 triol)

One of the most important trihydric alcohol.

Preparation:

(i) By Saponification of oils and fats.

(ii) From Propylene

(iii) Synthesis from its elements

Physical Properties

Highly viscous due to three ⎯OH group due to which it undergoes extensive intermolecular
H-bonding.

Chemical Properties

(i) It undergoes reaction of both secondary and primary alcoholic group.

(ii)

To replace the third hydroxyl group in either of two dichlorohydrins, PCl5 or PCl3 is fused.

(iii)

Reaction with concentrated nitric acid:

A mixture of glyceryl trinitrate and glyceryl dinitrate absorbed on Kieselguhr is called dynamite.

Reaction with KHSO4 – Dehydration.

Oxidation.

(i) With dil. HNO3, a mixture of glyceric acid and tartronic acid is produced.

(ii) With conc. HNO3, mainly glyceric acid is obtained.

(iii) With bismuth nitrate, only mesoxalic acid is formed.

(iv) Mild oxidising agents like bromine water, sodium hypobromite (Br2/NaOH) and Fenton’s reagent (H2O2 + FeSO4) give a mixture of glyceraldehyde and dihydroxyacetone. The mixture is called glycerose.

(v) With periodic (HIO4) acid.

(vi) With acidified potassium permanganate.

Reaction with phosphorous halides.
Reaction with monocarboxylic acids. Glycerol reacts with monocarboxylic acids to form mono-, di- and tri- ester depending upon the amount of the acid used and the temperature of the reaction. An excess of the acid and high temperature favour the formation of tri-esters. For example, with acetic acid, glycerol monoacetate, diacetate and triacetate may be formed.
Acetylation. When treated with acetyl chloride, glycerol forms glycerol triacetate.

Reaction with oxalic acid

(i)
(ii)

Uses: Glycerol is used:

In the preparation of nitroglycerine used in making dynamite. Nitroglycerine is also used for treatment of angina pectoris.
As an antifreeze in automobile radiators.
In medicines like cough syrups lotions etc.
In the production of glyptal or alkyl resin (a cross – linked polyester obtained by the condensation polymerization of glycerol and phthalic acid) which is used in the manufacture of paints and lacquers.
In making non-drying printing inks, stamp colours, shoes polishes etc.
In the manufacture of high class toilet soaps and cosmetics since it does not allow them to dry due to its hydroscopic nature.
As a preservative for fruits and other eatables.
As a sweetening agent in beverages and confectionary.

Exercise 21.

How does glycerol react with a. HI, b. (COOH)2 and c. conc HNO3?

ANSWER TO EXERCISES

Exercise 1:

(i)

(ii) A = CH3CH2CH2OH

B = CH3OH

Exercise 2:

(i)

(ii) The R– (carbanion) pulls acidic hydrogen from OH group making it an acid-base reaction instead of usual nucleophilic addition reaction. On further hydrolysis it produces reactant. But in the second case you have two moles of RMgX that completes the reaction because there are two R– for two different sites.

Exercise 3:

Exercise 4:

(a) (iii) > (ii) > (i) > (iv)

(b) (i) > (ii) > (iii)

Exercise 5:

(i) Only (c) and (d) gives iodoform test.

(ii)

Exercise 6:

(a)

(b) (i) No reaction

(ii) 6HCOOH

(iii) RCHO + R′CHO + NH3

Exercise 7:

(i)		(ii)
	Here phenyl migration takes place.	Here methyl migration takes place.

Exercise 8:

Considering the reactions of C5H10O2, it is likely that C6H5NHNH3, I2, NaOH will react only with a keto group while CH3COCl and oxidation are reactions of an alcohol. Therefore two isomeric structures are possible for the original compound.

Exercise 9:

	α, w – halo alcohols undergoes intramolecular Williamson ether.
	Synthesis in the presence of NaOH by reaction.

Exercise 10:

(a) Ethoxyethanol give the Iodoform test.

(b) Butyl iodide with AgNO3 yields AgI precipitate.

Exercise 11:

The cleavage reaction of an ether by HI is initiated by protonation of the ether oxygen.

In the second step the attack by Br− has to take place by SN1 or SN2 process. Since the system B is rigid either attack is very slow. The ether A does not pose such a problem

Exercise 12:

(a)
(b)

Exercise 13:

Exercise 14:

A phenol is highly reactive towards electrophilic substitution. Treatment of phenol with aqueous solution of bromine results in the replacement of every hydrogen ortho and para to the ⎯OH group, and it even cause the displacement of the sulfonic group to yield tribromophenol.

Exercise 15:

(a) 2-Pentanol responds to the Iodoform test

(b) Phenol gives coloration with FeCl3 solution

Exercise 16:

(a) Phenol loses the aromatic stabilization in the keto form.

(b) The alkene formed is more stable due to resonance.

(d) 2-Methyl-2-pentanol yield a stable alkene

(e) The phenoxide ion formed from phenol is stabilized by resonance.

(f)

(g) A tertiary alcohol form a stable tertiary carbonation.

Exercise 17:

	(a)
	(b)

Exercise 18:

	(a)
	(b)
	(c)
	(d)

Exercise 19:

Exercise 20:

Exercise 21:

See the text

MISCELLANEOUS EXERCISES

Exercise 1: Write a note on: (a) Oxo process (b) absolute alcohol

Exercise 2: What happens when?

(a) ethanol vapours are passed over alumina at 600 K,

(b) excess of ethanol is heated with conc. H2SO4 at 413 K,

(c) phenol is treated with acetyl chloride?

Exercise 3: How is glycerol prepared from fats and oils? Why it is highly viscous? Give its uses?

Exercise 4: Discuss the following reactions:

(a) Reimer and Tiemann reaction

(b) Kolbe’s reaction

Exercise 5: Complete the following by putting structures of A, B, C, D.

Complete the following.

Exercise 6: Compete the following by drawing correct structures.

Exercise 7: How the sugar is converted to ethanol?

Exercise 8: Write the structure of organic compounds A to F in the following sequence of reactions:

Exercise 9: Predict the products of the following reactions:

(a)

(b)

ANSWER TO MISCELLANEOUS EXERCISES

Exercise 1: (a) Oxo process: It is used for the preparation of alcohols. When alkenes are treated with carbon monoxide and hydrogen in presence of octa-carbonyl dicobalt at high temperature and pressure, aldehyde are formed, which on reduction with H2/Ni give alcohols.

(b) Absolute alcohol: 100% ethanol is known as absolute alcohol. 95.6% alcohol and 4.4% water form azeotropic mixture and boil at same temperature. To prepare absolute alcohol, rectified spirit (95% alcohol) and benzene mixture is fractionally distilled. The first fraction obtained at 331.8 K contains the azeotropic mixture of total water (7.5%) some alcohol (18.5%) and maximum benzene (74%). The second fraction, which is an azeoptroopic mixture of remaining benzene (nearly 68%) and some alcohol (32%) is obtained at 341.2 K. The third and last fraction consists of absolute alcohol which is obtained at 351K.

Exercise 2: (a) Dehydration with take place and ethylene is formed.

(b) Diethylether is formed.

Exercise 3: Glycerol is prepared by the Saponification of oils and fats i.e., hydrolysis of oils and fats with NaOH. Fats and oils are the esters of glycerol and higher fatty acids i.e. glycerides of fatty acids.

Glycerol is very viscous with extensive intermolecular hydrogen bonding. Glycerol molecule contains three −OH groups which form extensive hydrogen bonding. Hence the intermolecular forces are very high and molecules are highly associated. Due to this reason it is very viscous and posses high boiling point (563K).

Uses:

(i) Used in the preparation of nitroglycerine which is main constituent of dynamite,

(ii) Used as antifreezing agent in automobile radiators.

(iii) Used as a sweetening agent in confectionery and beverages.

Exercise 4: (a) Reimer and Tiemann reaction: When phenol is refluxed with chloroform and sodium hydroxide at 340 K followed by hydrolysis, gives o-hydroxy benzaldehyde (main product) and p-hydroxy benzaldehyde (minor product). This reaction is known as Reimer and Tiemann reaction.

If CCl4 is used in place of CHCl3 the main product will be salicylic acid.

(b) Kolbe’s Reaction: When sodium phenoxide is treated with carbon dioxide under pressure (4 to 7) and at 400 K, sodium salicylate is formed which on acidic hydrolysis gives salicylic acid. This reaction is known as Kolbe’s reaction.

Exercise 5:

Exercise 6:

Exercise 7: Molasses, which is the main source of sugar is treated with yeast. The enzymes present in yeast ferment the sugar and ethanol is obtained. Sugar is converted to glucose and fructose by the invertase enzyme. Glucose is further converted to ethanol by the zymase enzyme.

Exercise 8:

*Exercise 9:*	(a)
	(b)

SOLVED PROBLEMS

Subjective:

Board Type Questions

Prob 1. What happens when?

(i) Vapours of ethanol are passed over heated alumina at 523 K.

(ii) Ethyl bromide is heated with dry silver oxide.

(iii) Methyl magnesium bromide is treated with methoxy chloro methane.

(iv) Ethyl alcohol is heated with diazomethane in presence of HBF4.

(v) 2 methyl propene is heated with methanol in the presence of dil. H2SO4.

*Sol.*	(i)	Diethyl ether is formed.

	(ii)
	(iii)	Ethyl methyl ether is formed.

	(iv)	Ethyl methyl ether is formed

	(v)	Tertiary – butyl methyl ether

Prob 2. Why phenol has smaller dipole moment than methanol?

Sol. In case of phenol, the electron withdrawing inductive effect of oxygen is opposed by electron releasing resonance effect. Hence, phenol has smaller dipole moment. In case of methanol only electron withdrawing inductive effect is operative. Hence, it has higher dipole moment.

Prob 3. Why di tert – butyl ether can not be obtained by Williamson’s method?

Sol. In order to prepare the above ether the reagents to be used are tert – butyl bromide and tert – butoxide. Since the tertiary bromide prefers to under go elimination, therefore, major product of the reaction shall be iso butylene and not di tert – butyl ether.

Prob 4. Identify the product in the following reaction

*Sol.*

Prob 5. Write the various product when ethanol reacts with sulphuric acid in suitable conditions.

Sol.

IIT Level Questions

Prob 6. Give products expected when each of the following diols are reacted with periodic acid HIO4 .

(a)

(b)

*Sol.*	(a)
	(b)

Prob 7. Effect the following conversion

(a)
(b)
*Sol.*	(a)
	(b)

Prob 8. Postulate a mechanism for the following reactions:

Sol.

Prob 9. Conversion

(a)
(b)
*Sol.*	(a)
	(b)

Prob 10. Identify the products in the following reaction sequence.

Sol.

Prob 11. Distinguish between each pair of compounds by a simple test.

(a) (CH3)2CHOH and (CH3)2CHSH (b) CH3CH2CH2OH and (CH3)2CHOH

Sol. (a) The thiol gives a precipitate with heavy metal cations like

(b) (CH3)2CHOH (2° alc) gives a yellow precipitate of CHI3 with I2/OH− (iodoform test)

Prob 12. Write mechanism of the following reaction

Sol.

Prob 13. Compound (A) C6H14O is insoluble in water and gives negative Lucas test. On treatment with conc. H2SO4 followed by hydrolysis it gives only one compound (B) (C3H8O). Compound (B) is soluble in water and gives red colour with ceric ammonium nitrate. (B) gives yellow precipitate (C) and compound (D) on treatment with I2/Na2CO3 followed by acidification. Identify the compound A, B, C and D. Give the reasons.

*Sol.*
	C = CHI3 D = CH3COOH

Prob 14. Explain the low a boiling point and decreased water solubility by o – nitro phenol and o – hydroxy benzaldehyde as compared with this m and p – isomers.

Sol. Intramolecular H-bonding (chelation) in the o-isomers inhibits intermolecular attraction, lowering the boiling point and reduces H-bonding with H2O, decreasing water solubility. Intramolecular chelation can not occurs in m – and p – isomers.

Objective:

Prob 1. When ether is exposed to air for some time an explosive substance is produced

(A) peroxide (B) TNT

(C) oxide (D) super oxide

Sol. (A) Peroxide

Prob 2. What is the major product obtained when phenol is treated with chloroform and aqueous alkali?

	(A)		(B)
	(C)		(D)

Sol. (A) Salicyldehyde

*Prob 3.*
	(A)		(B)
	(C)	Both	(D)	none

Sol. Stability of carbocation is 3° > 2° > 1°

∴ (A)

Prob 4. Organic acid without a carboxylic acid group is ……………………..

(A) ascorbic acid (B) vinegar

(C) oxalic acid (D) picric acid

Sol. Picric acid is a phenol i.e. 2, 4, 6 tri nitro phenol

∴ (D)

Prob 5. The product of the following reaction

(A) 1 – pentanol (B) 2 – pentanol

(C) pentane (D) 1, 2 – pentane diol

Sol. Anti Markowvnikov product.

∴ (A)

Prob 6. Product in the reaction

(A) propane (B) ethyl iodide

(C) butanol (D) methoxy ethane

Sol.

∴ (D)

Prob 7. Phenol can be distinguished from alcohol with

(A) Tollen’s reagent (B) Schiff’s base

(C) Neutral FeCl3 (D) HCl

Sol. Phenol gives violet colour with neutral FeCl3.

∴ (C)

Prob 8. Tert – butyl methyl ether on heating with HI of one molar concentration gives

(A) CH3OH + (CH3)3C−I (B) CH3I + (CH3)3COH

(C) CH3I + (CH3)3CI (D) none of these

Sol.

∴ (A)

Prob 9. Which of the following is the strongest acid?

	(A)		(B)
	(C)		(D)

Sol. Due to greater electron withdrawing effect of NO2 group than Cl atom nitrophenols are stronger acids than p–chloro phenol among nitro phenols P nitro phenol is the strongest acid.

∴ (C)

Prob 10. Phenol is heated with phthalic anhydride in presence of conc. H2SO4. The product is

(A) Phenolphthalein (B) bakelite

(C) salicylic acid (D) fluorescein

Sol Phenolphthalein

∴ (A)

Prob 11. An organic compound X of molecular formula C4H10O undergoes oxidation to give a compound Y of molecular formula C4H8O2. X could be

(A) CH3CH2CH2CH2OH (B) CH3CH2CH(OH)CH3

(C) (CH3)2CHCH2OH (D) (CH3)3COH

Sol. (A) & (C)

Prob 12. Which of the following methods is/are not the industrial methods to prepare methanol?

(A) catalytic reduction of carbon monoxide in presence of ZnO-Cr2O3.

(B) reacting methane with steam at 900°C with a Ni catalyst.

(C) reducing formaldehyde with lithium aluminium hydride.

(D) reacting formaldehyde with aqueous sodium hydroxide solution.

Sol. (B), (C) & (D)

Prob 13. Which of the following reagent (s) or test (s), can be used to distinguish 2 methyl propanol and 2-methyl propanol – 2?

(A) I2/NaOH (B) HCl/anhyd. ZnCl2

(C) Victor-Meyer test (D) oxidation with Cu at 573 K

Sol. (B), (C) & (D)

Prob 14. Phenol has higher pka value than

(A) acetic acid (B) p-methoxy phenol

(C) p-nitrophenol (D) ethanol

Sol. (A) & (C)

Prob 15. Which of the following reactions can be used to prepare methyl phenyl ether?

(A) reacting sodium phenoxide with methyl chloride

(B) reacting phenol with diazomethane (CH2N2)

(C) reacting sodium methoxide with chlorobenzene

(D) reacting C6H5OMgCl with chloromethane

Sol. (A) & (B)

Fill in the Blanks

Prob 16. Williamson’s synthesis involves the reaction of an………………..with an……………….

Sol. Alkyl halide, alkoxide

Prob 17. An ether is more volatile than alcohol having the same formula due to absence of………………..

Sol. Intermolecular hydrogen bonding

Prob 18. Anisole reacts with HI to form……………………

Sol. Methyl iodide and phenol

Prob 19. Power alcohol is a mixture of absolute alcohol and…………….in the ratio…………….

Sol. Petrol, 20 : 80

Prob 20. Glycerol on reaction with potassium hydrogen sulphite and heat yields a product which…………to form acraldehyde.

Sol. Tautomerise

Prob 21. Phenol on exposure to air produces a………………..coloured product known as……………………..

Sol. Red, phenoquione

True and False

Prob 22. t-butyl alcohol reacts less rapidly with metallic sodium than the primary alcohol.

Sol. True

Prob 23. o-nitro phenol is more soluble in water than p-nitrophenol.

Sol. False

Prob 24. 2-methyl – 2- propanol gives cloudiness with HCl and ZnCl2 immediately at room temperature.

Sol. True

Prob 25. The hydroboration oxidation process gives product corresponding to Markonikov’s addition of water to the carbon – carbon double bond.

Sol. False

Prob 26. p-amino phenol has higher pka value than phenol.

Sol. True

ASSIGNMENT PROBLEMS

Subjective:

Level – O

Write all the canonical forms of phenol.

Account for the following:

(i) When HI react with methyl phenyl ether, phenol and methyl iodide are formed and not methyl alcohol and iodobenzene?

(ii) An ether would posses a dipole moment even if the alkyl groups present in it are identical.

O – nitro phenol is less soluble in water than p – nitro phenol. Why?

3, 3 – dimethyl butane 2 – ol loses a molecule of water in the presence of concentrated sulphuric acid to give tetra methyl ethylene as a major product.

Suggest a suitable mechanism.

Complete the following sequences of reactions by supplying X, Y and Z:

(a)

(b)

How will you convert phenol to salol?

Name the reagents used in the following reactions:

(i) Benzyl alcohol to benzoic acid. (ii) Phenol to 2, 4, 6 tribromo phenol.

(iii) Ethanol to ethanal (iv) Ethene to ethane – 1, 2 diol.

Arrange the following set of compound in order of their increasing boiling points.

(a) Pentan – 1 – ol, butan – 1- ol, butan – 2- ol, ethanol, propan – 1 – ol, methanol.

(b) Pentan – 1 – ol, n – butane, pentanal, ethoxyethane.

Arrange the following compounds in increasing order of their acid strength:

Propan – 1 – ol, 2, 4, 6 – trinitrophenol, 3 – nitrophenol, 3, 5 – dinitrophenol, phenol,
4 – methyl phenol.

(a) Explain why a non symmetrical ether is usually prepared by heating a mixture of ROH and R′OH in acid.

(b) Would you get any di-tert-butyl ether from this reaction? Explain.

How the following conversion carries out?

(i) Propane ⎯⎯⎯→ Propan – 2 – ol

(ii) benzyl chloride ⎯⎯⎯→ benzyl alcohol

(iii) Ethyl magnesium chloride ⎯⎯⎯→ propanol -1

(iv) Methanol to Ethanol

Identify the major products in the following reactions.

(i)

(ii)

Give one test to distinguish

(a) Ethanol and methanol (b) Phenol & Benzyl alcohol

Compound (A) C7H8O is insoluble in water and dilute NaHCO3 but dissolve in dilute NaOH solution. On treatment with Br2 water it readily gives a precipitate of C7H5OBr3. Write down the structure of the compound.

Give the mechanism of Reimer Tiemann reaction on phenol.

Level – I

Write the product of the following reactions

Write the product A & B

Write the product A & B

Convert nitro benzene to m – nitro phenol.

An unknown compound A (C4H10O2) reacts with sodium metal to liberate hydrogen gas.
A is inert towards periodic acid, it react with CrO3 to form B (C4H6O3). Identify
A and B.

Convert

Write mechanism of above conversion.

Identify compounds A – D in the following reaction.

Give mechanism of the following reaction

Identify the product in the given reaction

10.

Identify X and Y.

Identify A & B

Pyrole

Give a simple test that distinguish between the compounds.

(a) Allyl and n – propyl alcohol.

(b) Benzyl methyl ether and benzyl alcohol.

How will you synthesize?

An organic compound (A) gives positive Libermann reaction and on treatment with CCl4/KOH followed by hydrogenation gives (B). B on reduction gives (C) C7H8O2. (B) react with (CH3CO)2O/CH3COOH give a pain reliever (D). Identify A to D.
Provide suitable mechanism

Level – II

An ester A (C4H8O2) on treatment with excess of methyl magnesium bromide followed by acidification gives an alcohol (B) as the sole organic product. Alcohol (B) on oxidation with NaOCl followed by acidification gives acetic acid.

Deduce the structure of A and B and show the reactions involved.

Propose a mechanism for the following reaction.

	(a)
	(b)

3.	Give products

A on ozonolysis gives nonane – 2, 8 dione. What is A and how is it formed?

An organic compounds (P) having molecular formula C5H10O treated with dilute H2SO4 gives two compounds Q & R both gives positive iodoform test. The reaction of C5H10O with dilute H2SO4 gives reaction 1015 times faster than ethylene. Identify organic compound of Q & R. Give the reason for the extra stability of P.

Complete the following reaction.
(a) Diphenyl ether is not very easily prepared using Williamsons sysnthsis. Provide an useful synthetic route for the same.

(b) When phenol is treated with Br2 presence of H2O. 2, 4, 6 tri bromo phenol is obtained whereas on treatment with Br2 in presence of CCl4. p – bromo phenol is obtained. Explain.

Write mechanistic step of the following reaction?
Compound (A) gives positive Lucas test in 5 minutes when 6.0 gm of (A) heated with Na metal, 1120 ml of H2 is evolved at STP. Assuming (A) to contain one of oxygen per molecule, write structural formula of (A). Compound (A) when heated with PBr3 gives (B) which when treated with benzene in presence of anhydrous AlCl3 gives (C). What are (A), (B) and (C)?

10.	Carry out the following conversion.

Treatment of compound (A) C8H10O with chromic acid & pyridine gives (B) C8H8O. Treatment of (B) with two equivalent Br2 yields (C) C8H6OBr2 which on treatment with caustic soda followed by acidifications gives a compound (D) C8H8O3. (D) liberates CO2 on treatment with NaHCO3 and is resolvable.

Compound X, Y and Z are isomeric alcohol with formula C5H12O. X and Y react wit chromic acid solution, Y forming an acid A. The three isomers react with HBr with decreasing relative rates of Z > X >> Y, all giving the same C5H11Br (B) in varying yields. X alone can be oxidised by I2/OH− to C. Write the structure of X, Y, Z with proper explanations.
Predict the product (s) and write the mechanism of each of the following reactions

(a)

(b)

Find product

(a)
(b)
(c)
(d)

15.

		(Given negative test with Fehling solution but responds to iodoform)
			(both give positive Tollen’s test but do not give iodoform test)

	Identify A to G and Z.

Objective:

Level – I

Which one of the following has the maximum acidic strength?

(A) Phenol (B) o-nitro phenol

The boiling point to isomeric alcohols follows the order

(A) primary > secondary > tertiary (B) tertiary > secondary > primary

A mixture of benzoic acid and phenol may be separate by treatment with

(A) NaHCO3 (B) NaOH

In the lucas test of alcohols, appearance of cloudiness is due to the formation of

(A) aldehyde (B) ketone

The dehydration of 1 – butanol gives

(A) 1 – butene as the main product (B) 2 – butene as the main product

Ethyl alcohol is obtained when ethyl chloride is boiled with

(A) alc. KOH (B) aq. KOH

The number of methoxy groups in a compound can be determined by treating with

(A) Na2CO3 (B) NaOH

Diethyl ether absorbs oxygen to form

(A) red coloured sweet smelling compound (B) CH3COOH

Which of the following compounds is oxidised to prepare methyl – ethyl ketone?

(A) 2 – propanol (B) 1 – butanol

Order of reactivity of HX towards ROH is

(A) HI > HBr > HCl (B) HBr > HI > HCl

Glycerol has

(A) one 1° and one 2° alcoholic groups (B) one 1° and two 2° alcoholic groups

Ethyl iodide reacts with moist silver oxide to produce

(A) ethane (B) propane

Reaction of tertiary butyl alcohol with hot Cu at 350° C produces

(A) butanol (B) butanal

1° alcohol can be converted to aldehyde by using the reagent

(A) pyridinium chloro chromate (B) potassium di chromate

Reaction of ethanol with H2SO4 and suitable conditions can lead to the formation of

(A) C2H5HSO4 (B) ethene

Level – II

2 – phenyl propene on acidic hydration gives

(A) 2 – phenyl – 2 propanol (B) 2 phenyl – 1 – propanol

The order of reactivity of phenyl magnesium bromide with the following compound is

(A) II > III > I (B) I > III > II

Which one is the stronger base?

(A) CH3CH2O− (B) CF3CH2O−

The acidic character of 1°, 2°, 3° alcohols H2O and RC ≡ CH is in the order

(A) H2O > 1° > 2° > 3° > RC ≡ CH (B) RC ≡ CH > 3° > 2° > 1° > H2O

Choose the correct statement (s) for the reaction

(A) (b) is formed more rapidly at higher temperature

(B) (b) is more volatile than (a)

(D) (a) is formed higher yields at lower temperature

Predict the major product

(A) HO ⎯ CH2 ⎯ CH2 ⎯ CH2 ⎯ CH2 ⎯ I

(B) HO ⎯ CH2 ⎯ CH2 ⎯ CH2 ⎯ CH2 ⎯ OH

(D) no reaction

Dipole moment of CH3CH2CH3, CH3CH2OH and CH3CH2F is in order

(I) (II) (III)

(A) I < II < III (B) I > II > III

3 – methyl – 3- hexanol can be prepared by

(A) CH3MgI and 3 – hexanone, followed by hydrolysis

(B) C2H5MgI and 2 – pentanone, followed by hydrolysis

(D) any of the method above

Alcohol B reacts fastest with Lucas reagent. Hence A and B are

	(A)		(B)
	(C)		(D)

Vinyl carbinol is

(A) HO ⎯ CH2 ⎯ CH = CH2 (B) CH3C(OH) = CH2

The reaction of elemental sulphur with Grignard reagent followed by acidification leads to the formation of

(A) mercaptan (B) sulphoxide

Conversion of chloro benzene into phenol by Dow’s process is an example of

(A) free radical substation (B) nucleophilic substitution

For the preparation of tert-butyl methyl ether by Williamson’s method the correct choice of reagents is

(A) methoxide and tert – butyl bromide (B) methanol and 2 – bromobutane

Allyl alcohol is obtained when glycerol reacts with the following at 260°C

(A) formic acid (B) oxalic acid

The correct decreasing order of acidic strength is

(A) C6H5OH > C6H5CH2OH > C6H5COOH > C6H5SO3H

(B) C6H5CH2OH > C6H5OH > C6H5SO3H > C6H5OH

(D) C6H5SO3H > C6H5COOH > C6H5OH > C6H5CH2OH

ANSWERS TO ASSIGNMENT PROBLEMS

Subjective:

Level – O

Refer to text.

(i) In methyl phenyl ether.

The Ist step is protonation of ether to form phenyl methyl oxonium ion.

Phenyl carbocation is highly unstable and bond between C6H5 ⎯ O is stronger than
O⎯ CH3 so I− attacks on CH3 to form CH3I & phenol.

(ii) Oxygen of ether is sp3 hybridised, therefore ethers have tetrahedral geometry. Due to more electronegativity of oxygen than carbon, C ⎯ O bonds of ether are polar in nature.

The two dipoles do not cancel each other hence the molecule would posses dipole moment.

In O – nitro phenol there is intramolecular hydrogen bonding. This inhibits its hydrogen bonding with water and reduces its solubility in water.

(a) X = CH3OH, Y = CH3Br, Z = CH3CH2Cl

(b)

X = CH3CH = CH2, Y = CH3COCH3, Z =

(i) Alkaline KMnO4

(ii) Br2 water

(iii) P.C.C

(iv) 1% alkaline KMnO4

(a) Methanol < ethanol < propan – 1- ol < butan – 2- ol < butan – 1 – ol < pentan – 1- ol.

(b) n – butane < ethoxy ethane < pentanal < pentan – 1- ol.

Propan–1–ol < 4 methoxyphenol < phenol < 3 nitro phenol < 3, 5– dinitrophenol < 2, 4, 6 – trinitrophenol.

(a) A mixture of three ethers, R ⎯ O ⎯ R, R ⎯ O ⎯ R′ and R′ ⎯ O ⎯ R′ is obtained.

(b) No t–butanol does not solvate the 3° carbocation readily because of steric hindrance.

12.

(i)

(ii)

(a) Iodoform test given by ethanol.

(b) Neutral FeCl3 test given by phenol.

14.

15.

Level – I

2.
3.
4.
5.

6.
7.

10.

11.

(a) Add Br2/CCl4 – Allyl decolourises orange colour

(b) Add small piece of Na.

H2 releases from alcohol.

13.

14.

15.

Level – II

2.	(a)
	(b)

A ⎯⎯→ Nonane 2, 8 dione (No loss of C)

6.


7. (a)
(b)		H2O ionises phenol into the most reactive phenoxide anion and hence tri substitution occurs. Moreover H2O can stabilizing the arenium ion intermediate formed as well as the electrophile Br+. Hence the rate of reaction is fast and hence tri – substitution results. Hence the rate of the reaction is slow and hence now substitution results.

(A) is secondary alcohol.

CnH2n+1OH = Molecular weight

Molecular weight of (A) =

CnH2n+1OH = 60 (n = 3)

(A) C3H7OH

10.

11.

12.

13.

14.	(a)	(b)
	(c)	(d)	E = 2HCHO + F = CO2

15.

Objective:

Level – I

D 2. A 3. A
D 5. B 6. B
C 8. C 9. C
A 11. C 12. C
D 14. A 15. D

Level – II

A 2. C 3. A
A 5. C 6. C
A 8. D 9. A
A 11. A 12. B
D 14. B 15. D

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