Scheme past year semester 2- 1 Edition

Chapter 16 Hydroxyl Compounds

Question Section Answer

Jan 991.

i)

ii)

iii)

Explain briefly the following observations:

Phenol dissolves in aqueous sodium hydroxide but cyclohexane does not.

Phenol dissolve in aqueous NaOH because it is a weak acid. It react with strong base NaOH to form soluble salt.

Cyclohexane is a weak base does not react with NaOH

Ethanol reacts with iodine in aqueous solution of sodium hydroxide forming yellow precipitate but 1-propanol does not

Ethanol gives +ve test with Iodoform reagent, because it has methyl alcohol structure, when it oxidised with Iodoform reagent a methyl carbonyl structure compound is formed.

1-propanol gives –ve test with Iodoform reagent, no methyl alcohol structure.

Write a chemical equation for the reaction between ethanol and iodine in aqueous solution of sodium hydroxide.

June 992.

a) Give suitable chemical tests to distinguish the following pairs of compounds by specifying the necessary reagents, appropriate observations and chemical equations involved.

i. C6H5CH2OH and C6H5OH ii. CH3CH(OH)CH2CH2CH3 and (CH3)2C(OH)CH2CH2CH3

i) Benzyl alcohol react with Lucas’s reagent, solution turns cloudy immediately, 1

Scheme past year semester 2- 1 Edition

phenol does not react with Lucas’s reagent.

ii) 2-pentanol reacts with acidified KMnO4, purple colour disappear, brown precipitate is formed. 2-methyl-2-pentanol does not react, purple colour remain.

b) Give chemical equations involved in the following conversions:i. 1,2-propanediol from 1-propanolii. cyclohexylmethanol from chlorocyclohexane

iii. 1-methylcyclohexanol from cyclohexanone

i)

ii)

iii)

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Scheme past year semester 2- 1 Edition

June 003.

4.

Based on the molecular formula of C6H14O, draw the structural formulae for two tertiary alcohols. Then draw and label the structural formula of an alcohol which contains primary and secondary carbons atoms. State whether this alcohol is optically active

CH3CCH2CH2CH3

CH3

OH OH

CH3

CH3CH2CCH2CH3

(I) (II)

C6H14O

CH3CHCH2CHCH2OH

CH3

OH20

10

C

CH2OH

HOH

CH2CH3OH

H

CH2OH

CH3CH2C

Optically active because : 1) molecule contained a chiral center/chiral carbon 2) Molecule not super imposable with its mirror image

The following equation shows the dehydration of an alcohol forming an alkene.

CH3 CH3 CH3

OH CH3 Major product

Show the mechanism for the formation of the major product. Draw the structure of another possible product and state the rule applied in determining the major product.

Mechanism:

3

Conc. H2SO4

Scheme past year semester 2- 1 Edition

2001/02Session

5.

Other product:

Rule applied: Saytzeff’s rule – determines the main product.

A compound AA, C3H8O reacts with phosporus pentachloride, PCl5 to form compound BB. Compound AA forms a cloudy solution in 5 minutes after the addition of Lucas reagent. Compound BB reacts with magnesium in ether solvent to produce CC. The reaction between CC and carbon dioxide, followed by hydrolysis yields DD. By giving appropriate reasons, draw the structural formulae of AA to DD and the structural formula for an isomer of AA.

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Scheme past year semester 2- 1 Edition

2002/03 Session

6.

i)ii)iii)iv)

i)ii)

iii)

Reasons:* AA is a secondary alcohol as it react with Lucas’s reagent moderately (cloudy after 5 mins).

* BB is a secondary haloalkane, OH group of alcohol AA is substitute with Cl when the alcohol react with PCl5.

* CC is a Gridnard reagent as haloalkane reacts with magnesium.

* DD is a carboxylic acid as Gridnard reagent react with CO2 followed by hydrolysis.

Isomer AA is CH3CH2CH2OH

The structure of alcohol A is shown as follows:

Name compound A according to the IUPAC nomenclature.Is A optically active? Explain.Classify alcohol A.Give the major product of the dehydration of A.

4,6-dimethyl-2-octanol

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Scheme past year semester 2- 1 Edition

7.

iv)

(i)

Yes, A is optically active because at least there are 3 chiral carbon, at carbon number 3, 4 and 6.A is secondary alcohol.Product of the dehydration of A:

Major product:

Given below are the observations from experiments carried out on an alcohol, C4H10O which has several isomers.

(i) Isomer L does not show any changes upon mixing with Lucas reagent but it forms butanoic acid upon oxidation with acidified potassium dichromate (IV).(ii) Isomer M is optically active and gives positive test with I2/NaOH.(iii)Isomer N reacts instantly with Lucas reagent but it does not change the purple

solution of acidified potassium manganate (VII).

Explain the observations above and deduce the isomers L, M and N.

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Scheme past year semester 2- 1 Edition

(ii)

(iii)

2007/08 Session

8

The structure of 4-hydroxyazobenzene is as shown

Suggest how the above compound can be synthesized using benzene and phenol

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