11.1 Write IUPAC names of the following compounds:
Ans. (i) 2,2,4-Trimethylpentan-3-ol
11.2 Write structures of the compounds whose IUPAC names are as follows:
11.3(i) Draw the structures of all isomeric alcohols of molecular formula C5HI20 and give their IUPAC names.
(ii) Classify the isomers of alcohols in question 11.3 (i)as primary, secondary and tertiary alcohols.
Ans.Eight isomers are possible. These are:
11.4 Explain why propanol has higher boiling point than that of the hydrocarbon, butane?
Ans. The molecules of butane are held together by weak van der Waal’s forces of attraction while those of propanol are held together by stronger intermolecular hydrogen bonding.
11.5 Alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses. Explain this fact.
Ans. Alcohols can form hydrogen bonds with water and by breaking the hydrogen bonds already existing between water molecules. Therefore, they are soluble in water.
On die other hand, hydrocarbons cannot from hydrogen bonds with water and hence are insoluble in water.
11.6 What is meant by hydroboration-oxidation reaction? Illustrate it with an example.
Ans. The addition of diborane to alkenes to form trialkyl boranes followed by their oxidation with alkaline hydrogen peroxide to form alcohols is called hydroboration-oxidation. For example,
11.7 Give the structures and IUPAC names of monohydric phenols of molecular formula, C7H8O.
Ans.The three isomers are:
11.8 While separating a mixture of ortho and para nitrophenols by steam distillation, name the isomer which will be steam volatile. Give reason.
Ans. 0-N itrophenol is steam volatile due to chelation (intramolecular H – bonding) and hence can be separated by steam distillation from/Miitrophenol which is hot steam volatile because of intermolecular H-bonding.
11.9 Give the equations of reactions for the preparation of phenol from cumene.
11.10 Write chemical reaction for the preparation of phenol from chlorobenzene.
11.11 Write the mechanism of hydration of ethene to yield ethanol.
Ans.Direct addition of H20 to ethene in presence of an acid does not occur. Indirectly, ethene is first passed through concentrated H2S04, when ethyl hydrogen sulphate is formed.
11.12 You are given benzene, cone. H2S04and NaOH. Write the equations for the preparation of phenol using these reagents.
11.13 Show how will you synthesise
(i)1 -phenylethanol from a suitable alkene.
(ii)cyclohexylmethanol using an alkyl halide by an SN2 reaction.
(iii) Pentan-l-ol using a suitable alkyl halide?
11.14 Give two reactions that show the acidic nature of phenol. Compare its acidity with that of ethanol.
Ans. The reactions showing acidic nature of phenol are:
(a) Reaction with sodium: Phenol reacts with active metals like sodium to liberate H, gas.
(b) Reaction with NaOH: Phenol dissolves in NaOH to form sodium phenoxide and water.
Phenol is more acidic than ethanol. This is due to the reason that phenoxide ion left after the loss of a proton from phenol is stabilized by resonance, while ethoxide ion left after less of a proton from ethanol, is not.
11.15 Explain why is ortho-nitrophenol more acidic than ortho-methoxyphenol?
Ans. Due to strong -R and – I-effect of the -N02 group, electron density of the O – H bond decreases and hence the loss of a proton becomes easy.
11.16 Explain how does the – OH group attached to a carbon of benzene ring activate it towards electrophilic substitution?
Ans. Phenol may be regarded as a resonance hybrid of structures I-V, shown below.
As a result of +R effect of the -OH group, the electron density in the benzene ring increases thereby facilitating the attack of an electrophile. In other words, presence of -OH group, activates the benzene ring towards electrophilic substitution reactions. Further, since the electron density is relatively higher at the two o-and one p-position, therefore electrophilic substitution occurs mainly at o-and p-positions.
11.17 Give equations of the following reactions:
(i)Oxidation of propan-l-ol with alkaline KMnO4 solution.
(ii)Bromine in CS2 with phenol.
(iii)Dilute HNO3 acid with phehoL
(iv)Treating phenol with chloroform in presence of aqueous NaOH.
11.18 Explain the following with an example
(i) Kolbe’s reaction (ii) Reimer – Tiemann reaction –
(iii) Williamson ether synthesis (iv) Unsymmetrical ether
Ans. (i) Kolbe’s reaction: Sodium phenoxide when heated with C02 at 400K under a pressure of 4-7 atmospheres followed by acidification gives 2-hydroxybenzoic acid (salicylic acid) as the major product along with a small amount of 4-hydroxybenzoic acid.This reaction is called Kolbe’s reaction.
(ii)Reimer-Tiemann reaction: Treatment of phenol with CHC13 in presence of aqueous sodium or potassium hydroxide at 340 K followed by hydrolysis of the resulting product gives 2-hydroxybenzaldehyde (salicyialdehyde) as the major product. This reaction is called Reimer-Tiemann reaction.
(iii) Williamson’s ether synthesis: It involves the treatment of an alkyl halide with a suitable sodium alkoxide to obtain ethers. The sodium alkoxide needed for the purpose is prepared by the action of sodium on a suitable alcohol. In this reaction alkyl halide should primary. Secondary and tertiary halides will predominantly give an alkene.
(iv)Unsymmetrical ether: If the alkyl or aryl groups attached to the oxygen atom are different, ethers are called unsymmetrical ethers. For example, ethylmethylether, methylphenylether, 4-chlorophenyl- 4-nitrophenyl ether, etc.
11.19 Write the mechanism of acid dehydration of ethanol to yield ethene.
Ans.The mechanism of dehydration of alcohols to form alkenes occur by the following three steps:
11.20 How are the following conversions carried out?
(i) Propene —> Propan -2-ol
(ii) Benzyl chloride —> Benzyl alcohol
(iii) Ethyl magnesium chloride —> Propan-l-ol
(iv) Methyl magnesium bromide —> 2-Methylpropan-2-ol
11.21 Name the reagents used in the following reactions:
(i)Oxidation of a primary alcohol to carboxylic acid.
(ii)Oxidation of a primary alcohol to aldehyde.
(iv)Benzyl alcohol to benzoic acid.
(v)Dehydration of propan-2-oI to propene.
(vi)Butan-2-one to butan-2-oL .
Ans. (i) Acidified potassium dichromate or neutral/ acidic/ alkaline potassium permanganate.
(ii)Pyridinium chlorochromate (PCC), (C5H5NH)+ ClCrO3– in CH2Cl2
or Pyridinium dichromate (PDC),[(C5H5NH)2]2+Cr2O72-in CH2Cl2
(iii)Aqueous bromine, i.e., Br2/H2O.
(iv)Acidified or alkaline potassium permanganate.
(v)85% H2S04 at 440 K.
(vi)Ni/H2 or NaBH4 or LiAlH4.
11.22 Give reason for the higher boiling point of ethanol in comparison to methoxymethane.
Ans. Ethanol undergoes intermolecular H-bonding due to the presence of a hydrogen atom attached to the electronegative oxygen atom. As a result, ethanol exists as associated molecules.
Consequently, a large amount of energy is required to break these hydrogen bonds. Therefore, the boiling point of ethanol is higher than that of methoxymethane which does not form H-bonds.
11.23 Give IUPAC names of the following ethers.
Ans. (i)1-Ethoxy-2-methylpropane (ii)2-Chlorlo-l-methoxyethane
(v)1 -Ethoxy-4 -4 – dimethylcyclohexane (vi)Ethoxybenzene
11.24 Write the names of reagents and equations for the preparation of the following ethers by Williamson’s synthesis:
11.25 Illustrate with examples the limitations of Willamson synthesis for the preparation of certain types of ethers.
Ans. Williamson’s synthesis is a versatile method for the synthesis of both symmetrical and unsymmetrical ethers. However, for the synthesis of unsymmetrical ethers, a proper choice of reactants is necessary. Since Williamson’s synthesis occurs by SN2 mechanism and primary alkyl halides are most reactive in Sn2 reaction, therefore, best yields of unsymmetrical ethers are obtained when the alkyl halides are primary and the alkoxide may be primary, secondary or tertiary. For example, tert-butylethyl ether is prepared by treating ethyl bromide with sodium tert-butoxide.
11.26 How is 1-propoxypropane synthesised from propan-l-ol? Write the mechanism of this reaction.
Ans.(a) Williamson’s synthesis
11.27 Preparation of ethers by acid dehydration of secondary or tertiary alcohols is not a suitable method.Give reason.
Ans. Acid catalysed dehydration of primary alcohols to ethers occurs by SN2 reaction involving nucleophilic attack by the alcohol molecule on the protonated alcohol molecule.
Under these conditions, 2° and 3° alcohols, however, give alkenes rather than ethers. The reason being that due to steric hindrance, nucleophilic attack by the alcohol molecule on the protonated alcohol molecule does not occur. Instead protonated 2° and 3° alcohols lose a molecule of water to form stable 2° and 3° carbocation. These carbocations prefer to lose a proton to form alkenes rather than undergoing nucleophilic attack by alcohol molecules to form ethers.
11.28 Write the equation of the reaction of hydrogen iodide with (i)1-propoxypropane (ii)methoxybenzene, and (iii)benzyl ethyl ether
11.29 Explain the fact that in aryl alkyl ethers (i) the alkoxy group activates the benzene ring towards electrophilic substitution and (ii) it directs the incoming substituents to ortho and para positions in benzene ring.
Ans. In aryl alkyl ethers, the +R-effect of the alkoxy (OR) group increases the electron density in the benzene ring, thereby activating the benzene ring towards electrophilic substitution reaction.
Since the electron density increases more at the two ortho and one para position as compared to meta position therefore, electrophilic substitution reactions mainly occur at o-and -positions.
11.30 Write the mechanism of the reaction of HI with methoxymethane.
Ans. When equimolar amounts of HI and methoxy methane are reacted, a mixture of methyl alcohol and methyl iodide is formed by the following mechanism:
11.31 Write equations of the following reactions:
(i) Friedel-Crafts reaction -alkylation of anisole
(ii)Nitration of anisole.
(iii)Bromination of anisole in ethanoic acid medium
(iv) Friedel-Craft’s acetylation of anisole.
11.32 Show how would you synthesise the following alcohols from appropriate alkanes?
11.33 When 3-methylbutant 2-ol is treated with HBr, the following reaction takes place:
Give a mechanism for this reaction.
(Hint: The secondary carbocation formed in step II rearranges to a more stable tertiary carbocation by a hydride ion shift from 3rd carbon atom.)
Protonation of the given alcohol followed by loss of water gives a 2° carbocation(I), which being unstable rearranges by 1,2-hydride shift to form the more stable 3° carbocation (II). Nucleophilic attack by Br ion on this carbocation (II) gives the final product.