Category: Class 12 Chemistry

NCERT Solutions For Class 12 Chemistry Chapter 13 Amines

Topics and Subtopics in NCERT Solutions for Class 12 Chemistry Chapter 13 Amines:

Section Name	Topic Name
13	Amines
13.1	Structure of Amines
13.2	Classification
13.3	Nomenclature
13.4	Preparation of Amines
13.5	Physical Properties
13.6	Chemical Reactions
13.7	Method of Preparation of Diazonium Salts
13.8	Physical Properties
13.9	Chemical Reactions
13.10	Importance of Diazonium Salts in Synthesis of Aromatic Compounds

NCERT INTEXT QUESTIONS

13.1. Classify the following amines as primary, secondary and tertiary:

Ans: (i) 1° (ii) -3° (iii) 1° (iv) 2°

13.2. Write the structures of different isomeric amines corresponding to the molecular formula, C₄H₁₁N.
(i) Write the IUPAC names of all the isomers
(ii) What type of isomerism is exhibited by different types of amines?
Ans: Eight isomeric amines are possible

Isomerism exhibited by different amines

• Chain isomers: (i) and (ii) ; (iii) and (iv) ; (i) and (iv)
• Position isomers: (ii) and (iii) ; (ii) and (iv)
• Metamers: (v) and (vi) ; (vii) and (viii)
• Functional isomers: All the three types of amines are the functional isomers of each other.

13.3. How will you convert:
(i) Benzene into aniline
(ii) Benzene into N,N-dimethylaniline
(iii) Cl-(CH₂)₄-Cl into Hexane -1,6- diamine
Ans:

13.4. Arrange the following in increasing order of their basic strength :
(i) C₂H₅NH₂, C₆H₅NH₂, NH₃, C₆H₅CH₂NH₂, (C₂H₅)₂NH
(ii) C₂H₅NH₂, (C₂H₅)₂NH, (C₂H₅)₃N, C₆H₅NH₂
(iii) CH₃NH₂, (CH₃)₂NH, (CH₃)₃N, C₆H_sNH₂, C₆H₅CH₂NH₂
Ans:
In general, the basic character of ammonia (NH₃) and the amines is linked with the availability of the lone electrons pair on the nitrogen atom. In other words, these are all Lewis bases.
Amines act as Lewis bases due to the presence of lone electron paîr on the nitrogen atom. Since the nitrogen atom is sp³ hybridised, its electron attracting tendency is considerably reduced. It can readily lose its electron pair and acts as a base. For example, amines form hydroxides with water.

Here K_t is called dissociation constant for the base. Greater the K_b value stronger will be the base. The basic strength
of amines can also be expressed as pK_b value which is related to K_b as :

The K_b values are :

13.5. Complete the following acid-base reactions and name the products:
(i) CH₃CH₂CH₂NH₂+HCl ——–>
(ii) (C₂H₅)₃ N+HCl ——–>
Ans:

13.6. Write reactions of the final alkylation product of aniline with excess of methyl iodide in the presence of sodium carbonate solution.
Ans:

13.7. Write chemical reaction of aniline with benzoyl chloride and write the name of the product obtained.
Ans:

13.8. Write structures of different isomers corresponding to the molecular formula, C₃H₉N. Write IUPAC names of the isomers which will liberated N₂ gas on treatment with nitrons acid.
Ans: In ‘all, four structural isomers are possible. These are:

13.9. Convert:
(i) 3-Methylanilineinto3-nitrotoluene
(ii) Aniline into 1,3,5- Tribromo benzene
Ans:

NCERT EXRECISES

13.1. Write IUPAC names of the following compounds and classify them into primary, secondary, and tertiary amines. 
(i) (CH₃)₂ CHNH₂ (ii) CH₃(CH₂)₂NH₂ (iii) CH₃NHCH(CH₃)₂
(iv) (CH₃)₃ CNH₂ (v) C₆H₅NHCH₃(vi) (CH₃CH₂)₂NCH₃
(vii)m-BrC₆H₄NH₂
Ans: (i) Propan-2-amine(1°)
(ii) Propan-1-amine (1°),
(iii) N-Methylpropan-2-amine (2°).
(iv) 2-Methylpropan-2-amine(l°)
(v) N-MethylbenzenamineorN-methylaniline(2°)
(vi) N-Ethyl-N-methylethanamine (3°)
(vii) 3-Bromobenzenamine or 3-bromoaniline (1°)

13.2. Give one chemical test to distinguish between the following pairs of compounds:
(i)Methylamine and dimethylamine
(ii) Secondary and tertiary amines
(iii) Ethylamine and aniline
(iv) Aniline and benzylamine
(v) Aniline and N-Methylaniline.
Ans:

13.3. Account for the following
(i) pKb of aniline is more than that of methylamine
(ii) Ethylamine is soluble in water whereas aniline is not.
(iii) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide.
(iv) Although amino group is o and p – directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m-nitroaniline.
(v) Aniline does not undergo Friedel-Crafts reaction.
(vi) Diazonium salts of aromatic amines are more stable than those of aliphatic amines.
(vii) Gabriel phthalimide synthesis is preferred for synthesising primary amines.
Ans: (i) In aniline, the lone pair of electrons on the N-atom is delocalised over the benzene ring.
As a result, electron density on the nitrogen . atom decreases. Whereas in CH₃NH₂,+ I-effect of -CH₃ group increases the electron density on the N-atom. Therefore, aniline is a weaker base than methylamine and hence its pKb value is higher than that of methylamine.
(ii) Ethylamine dissolves in water due to intermolecular H-bonding. However, in case of aniline, due to the large hydrophobic part, i.e., hydrocarbon part, the extent of H-bonding is very less therefore aniline is insoluble in water.

(iii) Methylamine being more basic than water, accepts a proton from water liberating OH^– ions,

(iv) Nitration is usually carried out with a mixture of cone HNO₃ + cone H₂SO₄. In presence of these acids, most of aniline gets protonated to form ahilinium ion. Therefore, in presence of acids, the reaction mixture consist of aniline and anilinium ion. Now, -NH2 group in aniline is activating and o, p-directing while the -+NH₃ group in anilinium ion is deactivating and rw-directing: Nitration of aniline (due to steric hindrance at o-position) mainly gives p-nitroaniline, the nitration of anilinium ion gives m-nitroaniline. In actual practice, approx a 1:1 mixture of p-nitroaniline and m-nitroaniline is obtained. Thus, nitration of aniline gives a substantial amount of m-nitroaniline due to protonation of the amino group.

13.4. Arrange the following:
(i) In decreasing order of pKb values:
C₂H₅NH₂,C₆H₅NHCH₃,(C₂H₅)₂NH and C₆H₅NH₂
(ii) In increasing order of basic strength:
C₆H₅NH₂, C₆H₅N(CH₃)₂, (C₂H₅)₂ NH and CH₃NH₂.
(iii) In increasing order of basic strength:
(а)Aniline,p-nitroaniline andp-toluidine
(b)C₆H₅NH₂, C₆H₅NHCH₃, C₆H₅CH₂NH₂
(iv) In decreasing order of basic strength in gas phase:
C₂H₅NH₂, (C₂H₅)₂NH, (C₂H₅)₃N and NH₃
(v) In increasing order of boiling point:
C₂H₅OH, (CH₃)₂NH, C₂H₅NH₂
(vi) In increasing order of solubility in water:
C₆H₅NH₂,(C₂H₅)₂NH,C₂H₅NH₂
Ans: (i) Due to delocalisation of lone pair of electrons of the N-atom over the benzene ring,C₆H₅NH₂ and C₆H₅NHCH₃ are far less basic than C₂H₅NH₂ and (C₂H,)₂NH. Due to +I-effect of the -CH₃ group, C₆H₅NHCH₃ is little more basic that C₆H₅NH₂. Among C₂H₅NH₂ and (C₂H₅)₂NH, (C₂H₅)₂NH is more basic than C₂H₅NH₂ due to greater+I-effect of two -C₂H₅ groups. Therefore correct order of decreasing pKb values is:

(ii) Among CH₃NH₂ and (C₂H₅)₂NH, primarily due to the greater +I-effect of the two -C₂H₅ groups over one -CH₃ group, (C₂H₅)₂NH is more basic than CH₃NH₂.In both C₆H₅NH2 and C₆H₅N(CH₃)₂ lone pair of electrons present on N-atom is delocalized over the benzene ring but C₆H₅N(CH₃)₂ is more basic due to +1 effect of two-CH₃ groups.

(iii) (a) The presence of electron donating -CH₃ group increases while the presence of electron withdrawing -NO₂ group decreases the basic strength of amines.

(b) In C₆H₅NH₂ and C₆H₅NHCH₃, N is directly attached to the benzene ring. As a result, the lone pair of electrons on the N-atom is delocalised over the benzene ring. Therefore, both C₆H₅NH₂ and C₆H₅NHCH₃ are weaker base in comparison to C₆H₅CH₂NH₂. Among C₆H₅NH₂ and C₆H₅NHCH₃, due to +1 effect of-CH₃ group C₆H₅NHCH₃ is more basic.

(iv) In gas phase or in non-aqueous solvents such as chlorobenzene etc, the solvation effects i. e., the stabilization of the conjugate acid due to H-bonding are absent. Therefore, basic strength depends only upon the +I-effect of the alkyl groups. The +I-effect increases with increase in number of alkyl groups.Thus correct order of decreasing basic strength in gas phase is,

(v) Since the electronegativity of O is higher than thalof N, therefore, alcohols form stronger H-bonds than amines. Also, the extent of H-bonding depends upon flie number of H-atoms on the N-atom, thus the extent of H-bonding is greater in primary amine than secondary amine.

(vi) Solubility decreases with increase in molecular mass of amines due to increase in the size of the hydrophobic hydrocarbon part and with decrease is the number of H-atoms on the N-atom which undergo H-bonding.

13.5. How will you convert:
(i) Ethanoic acid into methanamine
(ii) Hexanenitrile into 1-aminopentane
(iii) Methanol to ethanoic acid.
(iv) Ethanamine into methanamine
(v) Ethanoic acid into propanoic acid
(vi) Methanamine into ethanamine
(vii) Nitromethane into dimethylamine
(viii) Propanoic acid into ethanoic acid?
Ans:

13.6. Describe the method for the identification of primary, secondary and tertiary amines. Also write chemical equations of the reactions involved.
Answer:
The distinction in the three types of amines can be done by the following methods :
(i) Hinsberg’s Test:
This is a very useful test for the distinction of primary, secondary and tertiary amines. An amine is shaken with
Hinsberg’s reagent (benzene suiphonyl chloride) in the presence of excess of aqueous KOH solution. The reactions taking
place are given on the next page.

1.  A primary amine forms N – alkyl benzene suiphonamide which dissolves in aqueous KOH solution to form potassium salt and upon acidification with dilute HCI regenerates the insoluble suiphonamide.
   
2.  A secondary amine forms N, N – dialkylbenzene suiphonamide which remains insoluble in aqueous KOH and even after acidification with dilute HCl
   
3. A tertiary amine does not react with benzene suiphonyl chloride and remains insoluble in aqueous KOH.
   However, on acidification with dilute HCI it gives a clear solution due to the formation of the ammonium salt.
   

(ii) Reaction with nitrous acid:
All the three types of amines, aliphatic as well as aromatic, react with nitrous acid under different conditions to form variety of products. Since nitrous acid is highly unstable, it is prepared in situ by the action of dilute hydrochloric acid on sodium nitrite.

(a) Primary aliphatic amines react with nitrous acid at low temperature (cold conditions) to form primary alcohol and
nitrogen gas accompanied by brisk effervescence. Nitrous acid is unstable in nature and is prepared in situ by reacting sodium
nitrite with dilute hydrochloric acid. For example,

The reaction is used as a tesijôr primary aliphadc amines as no other amine evolves nitrogen with nurous acid.

(b) Primary aromatic amines such as aniline react with nitrous acid under ice cold conditions (273 – 278 K) to form benzen diazonium salt. The reaction is known as diazotisation reaction.

in case, the temperature is allowed to rise above 278 K, benzene diazortium chloride is decomposed by water to form phenol.

Aliphatic primary amines also react with nitrous acid to form alkyl diazonium salts in a similar manner. But these are
quite unstable and decompose to form a mixture of alcohols, alkenes and alkyl halides along with the evolution of N₂ gas.

(c) Secondary amines (both aliphatic and aromatic) react with nitrous acid to form nitrosoamines which separate as
Yellow oily liquids.

(d) Tertiary aliphatic amines dissolve in a cold solution of nitrous acid to form salts which decompose on warming to
give nitrosoamine and alcohol. For example,

(e) Tertiary aromatic amines react with nitrous acid to give a coloured nitrosoderivative. This reaction is called
nitrosation and as a result, a hydrogen atom in the para position gets replaced by a nitroso (-NO) group. For example,

13.7. Write short notes on the following:
(i) Carbylamine reaction
(ii) Diazotisation
(iii) ‘Hofmann’s bromamide reaction
(iv) Coupling reaction
(v) Ammonolysis
(vi) Acetylation
(vii) Gabriel phthalimide synthesis
Ans: (i) Carbylamine reaction: Both aliphatic and aromatic primary amines when warmed with chloroform and an alcoholic solution of KOH, produces isocyanides or carbylamines which have very unpleasant odours. This reaction is called carbylamine reaction.

(ii) Diazotisation: The process of conversion of a primary aromatic amino compound into a diazonium salt, is known as diazotisation. This process is carried out by adding an aqueous solution of sodium nitrite to a solution of primary aromatic amine (e.g., aniline) in excess of HCl at a temperature below 5°C.

(iii) Hoffmann’s bromamide reaction: When an amide is treated with bromine in alkali solution, it is converted to a primary amine that has one carbon atom less than the starting amide. This reaction is known as Hoffinann’s bromamide degradation reaction.

(iv) Coupling reaction: In this reaction, arene diazonium salt reacts with aromatic amino compound (in acidic medium) or a phenol (in alkaline medium) to form brightly coloured azo compounds. The reaction generally takes place at para position to the hydroxy or amino group. If para position is blocked, it occurs at ortho position and if both ortho and para positions are occupied, than no coupling takes place.

(v) Ammonolysis: It is a process of replacement of either halogen atom in alkyl halides (or aryl halides) or hydroxyl group in alcohols (or phenols) by amino group. The reagent used for ammonolysis is alcoholic ammonia. Generally, a mixture of primary, secondary and tertiary amine is formed.

(vi) Acetylation: The process of introducing an acetyl (CH₃CO-) group into molecule using acetyl chloride or acetic anhydride is called acetylation.

(vii) Gabriel phthalimide synthesis: It is a method of preparation of pure aliphatic and aralkyl primary amines. Phthalimide on treatment with ethanolic KOH gives potassium phathalimide which on heating with a suitable alkyl Or aralkyl halides gives N-substituted phthalimides, which on hydrolysis with dil HCI or with alkali give primary amines.

13.8. Accomplish the following conversions:
(i) Nitrobenzene to benzoic acid
(ii) Benzene to m-bromophenol
(iii) Benzoic acid to aniline
(iv) Aniline to 2,4,6-tribromofluorobenzene
(v) Benzyl chloride to 2-phenylethanamine
(vi) Chlorobenzene to p-Chloroaniline
(vii) Aniline to p-bromoaniIine
(viii)Benzamide to toluene
(ix) Aniline to benzyl alcohol.
Ans:

13.9. Give the structures of A,B and C in the following reaction:


Ans:

13.10. An aromatic compound ‘A’ on treatment with aqueous ammonia and heating forms compound ‘B’ which on heating with Br₂ and KOH forms a compound ‘C’ of molecular formula C₆H₇N. Write the structures and IUPAC names of compounds A, B and C.
Ans:
From the available information, we find that ‘B’ upon heating with Br₂ and KOH forms a compound ‘C’. The compound ‘B’ is
expected to be an acid amide. Since ‘B’ has been formed upon heating compound ‘A’ with aqueous ammonia, the compound ‘A’ is an aromatic acid.
It is benzoic acid. The reactions involved are given as follows:

13.11. Complete the following reactions:

Ans:

13.12. Why cannot aromatic primary amines be prepared by Gabriel phthalimide synthesis?
Ans: The success of Gabriel phthalimide reaction depends upon the nucleophilic attack by the phthalimide anion on the organic halogen compound.
Since aryl halides do not undergo nucleophilic substitution reactions easily, therefore, arylamines, i.e., aromatic, primary amines cannot be prepared by Gabriel phthalimide reaction.

13.13. Write the reactions of (i) aromatic and (ii) aliphatic primary amines with nitrous acid.
Ans: Both aromatic and aliphatic primary amines react with HNO₂ at 273-278 K to form aromatic and aliphatic diazonium salts respectively. But aliphatic diazonium salts are unstable even at this low temperature and thus decompose readily to form a mixture of compounds. Aromatic and aliphatic primary amines react with

13.14. Give plausible explanation for each of the following:
(i) Why are amines less acidic than alcohols of comparable molecular masses?
(ii) Why do primary amines have higher boiling point than tertiary amines?
(iii) Why are aliphatic amines stronger bases than aromatic amines?
Ans: (i) Loss of proton from an amine gives an amide ion while loss of a proton from alcohol give an alkoxide ion.
R—NH₂—>R—NH^– +H⁺
R—O —H—>R— O^– +H⁺ .
Since O is more electronegative than N, so it wijl attract positive species more strongly in comparison to N. Thus, RO~ is more stable than RNH®. Thus, alcohols are more acidic than amines. Conversely, amines are less acidic than alcohols.
(ii) Due to the presence of two H-atoms on N-atom of primary amines, they undergo extensive intermolecular H-bonding while tertiary amines due to the absence of H-atom on the N-atom do not undergo H-bonding. As a result, primary amines have higher boiling points than tertiary amines of comparable molecular mass.
(iii) Aromatic amines are far less basic than ammonia and aliphatic amines because of following reasons:
(a) Due to resonance in aniline and other aromatic amines, the lone pair of electrons on the nitrogen atom gets delocalised over the benzene ring and thus it is less easily available for protonation. Therefore, aromatic amines are weaker bases than ammonia and aliphatic amines.
(b) Aromatic amines arS more stable than corresponding protonated ion; Hence, they hag very less tendency to combine with a proton to form corresponding protonated ion, and thus they are less basic.

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NCERT Solutions For Class 12 Chemistry Chapter 12 Aldehydes Ketones and Carboxylic Acids

Topics and Subtopics in NCERT Solutions for Class 12 Chemistry Chapter 12 Aldehydes Ketones and Carboxylic Acids:

Section Name	Topic Name
12	Aldehydes, Ketones and Carboxylic Acids
12.1	Nomenclature and Structure of Carbonyl Group
12.2	Preparation of Aldehydes and Ketones
12.3	Physical Properties
12.4	Chemical Reactions
12.5	Uses of Aldehydes and Ketones
12.6	Nomenclature and Structure of Carboxyl Group
12.7	Methods of Preparation of Carboxylic Acids
12.8	Physical Properties
12.9	Chemical Reactions
12.10	Uses of Carboxylic Acids

NCERT INTEXT QUESTION

12.1. Write the structures of the following compounds:
(i) α-Methoxypropionaldehyde
(ii) 3-Hydroxybutanal
(iii) 2-Hydroxycyclopentane carbaldehyde
(iv) 4-OxopentanaI
(v) Di-sec.butylketone
(vi) 4-fluoroaeetophenone
Ans:

12.2. Write the structures of the products of the following reactions:

Ans:

12.3. Arrange the following compounds in increasing order of their boiling points:
CH₃CHO, CH₃CH₂OH, CH₃OCH₃, CH₃CH₂CH₃
Ans: The order is : CH₃CH₂CH₃ < CH₃OCH₃ < CH₃CHO <CH₃CH₂OH
All these compounds have comparable molecular masses CH₃CH₂OH undergoes extensive intermolecular Il-bonding and thus its b.pt. is the highest. CH₃CHO is more pdlar than CH₃OCH₃ so that dipole-dipoie interactions in CH₃CHO are greater than in CH₃OCH₃. Thus, b.pt. of CH₃CHO > CH₃OCH₃. CH₃CH₂CH₃ has only weak van der waals forces between its molecules and hence has the lowest b.pt.

12.4. Arrange the following carbonyl compounds in increasing order of their reactivity in nucleophilic addition reactions :
(a) Ethanal, propanal, propanone, butanone
(b) Benzaldehyde, p-tolualdehyde, p-nitrobenzaldehyde, acetophenone
Ans: (a) The increasing order of reactivity of the carbonyl compounds towards nucleophilic addition reactions is :
butanone < propanone < propanal < ethanal
The reactivity is based upon two factors. These are: steric factors and electronic factors. 

(b) The increasing order of reactivity is :
acetophenone < p-tolualdehyde < benzaldehyde < p-nitrobenzaldehyde

Explanation: Acetophenone being a ketone is the least reactive towards nucleophilic addition. All others are aldehydes. Among them, p-tolualdehyde is less reactive than benzaldehyde because CH₃ group present at the para position w.r.t.  -CHO group will increase the electron density on the carbonyl carbon atom due to hyper conjugation effect. As a result, the nucleophile attack occurs to lesser extent as compared to benzaldehyde.

In p-nitrobenzaldehyde, the nitro group has an opposing effect. It is electron withdrawing in nature due to -I effect as well as -R effect. The electron density on the carbonyl carbon atom decreases and this favours the nucleophile attack.

12.5. Predict the products of the following reactions:


Ans:

12.6. Give the 1UPAC names of the following compounds:
(i) PhCH₂CH₂COOH
(ii) (CH₃)₂ C=CHCOOH

Ans: (i) 3 – Phenylpropanoic acid
(ii) 3 – Methylbut-2-enoic acid
(iii) 2-Methylcyclohexanecarboxylic acid
(iv) 2,4,6 – Trinitrobenzoic acid

12.7. Show how each of the following compounds can be converted into benzoic acid.
(i) Ethylbenzene
(ii) Acetophenone
(iii) Bromobenzene
(iv) Phenylethene (styrene)
Ans:

12.8. Which acid from each of the following pairs would you expect to be a stronger acid?
(i) CH₃COOH or CH₂FCOOH
(ii) CH₂FCOOH or CH₂ClCOOH
(iii) CH₂FCH₂CH₂COOH or CH₃CHFCH₂COOH

Ans:
Explanation: CH₃ group with +I effect increases the electron density on the oxygen atom in O – H bond in the carboxyl group and cleavage of bond becomes diffcult. It therefore, decreases the acidic strength. The F atom has very strong -I effect, i.e., electron withdrawing influence. It decreases the electron density on the oxygen atom and cleavage of bond becomes easy. The acidic character therefore, increases. It is further related to the

1. No. of F atoms present in the molecule.
2.  Relative position of the F atom in the carbon atom chain.

In the light of the above discussion.
(i) CH₂FCOOH is a stronger acid.
(ii) CH₂FCOOH is a stronger acid.
(iii) CH₃CHFCH₂COOH is a stronger acid.

NCERT EXERCISES

12.1. What is meant by the following terms? Give an example of the reaction in each case.
(i) Cyanohydrin
(ii) Acetal
(iii) Semicarbazone
(iv) Aldol
(v) Hemiacetal
(vi) Oxime
(vii) Ketal
(viii) Imine
(ix) 2,4-DNP derivative
(x) Schiff’s base.
Ans: (i) Cyanohydrin: gem-Hydroxynitriles, i.e., compounds possessing hydroxyl and cyano groups on the same carbon atom are called cyanohydrins. These are produced by addition of HCN to aldehydes or ketones in a weakly basic medium.

(ii) gem – Dialkoxy compounds in which the two alkoxy groups are present on the terminal carbon atom are called acetals. These are produced by the action of an aldehyde with two equivalents of a monohydric alcohol in presence of dry HCl gas.

These are easily hydrolysed by dilute mineral acids to regenerate the original aldehydes. Therefore, these are used for the protection of aldehyde group in organic synthesis.

(iii) Semicarbazones are derivatives of aldehydes and ketones and are produced by action of semicarbazide on them in acidic medium.

(iv) Aldols are P-hydroxy aldehydes or ketones and are produced by the condensation of two molecules of the same or one molecule each of two different aldehydes or ketones in presence of a dilute aqueous base. For example,

(v) gem – Alkoxyalcohols are called hemiacetals. These are produced by addition of one molecule of a monohydric alcohol to an aldehyde in presence of dry HCl gas.

(vi) Oximes are produced when aldehydes or ketones react with hydroxyl amine in weakly acidic medium.

(vii) Ketals are produced when a ketone is heated with dihydric alcohols like ethylene glycol in presence of dry HCl gas or /3-toluene sulphonic acid (PTS).

These are easily hydrolysed by dilute mineral acids to regenerate the original ketones. Therefore, ketals are used for protecting keto groups in organic synthesis.

(viii) Compounds containing -C = N – group are called imines. These are produced when aldehydes and ketones react with ammonia derivatives.

(ix)2, 4-Dinitrophenyl hydrazone (i.e., 2,4-DNP derivatives) are produced when aldehydes or ketones react with 2,4-dinitrophenyl hydrazine in weakly acidic medium.

(x) Aldehydes and ketones react with primary aliphatic or aromatic amines to form azomethines or SchifFs bases.

12.2. Name the following compounds according to IUPAC system of nomenclature:
(i) CH₃CH (CH₃)—CH₂ CH₂—CHO
(ii) CH₃CH₂COCH(C₂H₅)CH₂CH₂Cl
(iii) CH₃CH=CHCHO
(iv) CH₃COCH₂COCH₃
(v) CH₃CH(CH₃)CH₂C(CH3)2COCH₃
(vi) (CH₃)₃CCH₂COOH.
(vii) OHCC₆H₄CHO-p
Ans: (i) 4-Methyl pentanal
(ii) 6-Chloro-4-ethylhexan-3-one
(iii) But-2-en-l-al
(iv) Pentane-2,4-dione
(v) 3,3,5-Trimethyl-hexan-2-one
(vi) 3,3-Dimethyl butanoic acid
(vii) Benzene-1,4-dicarbaldehyde

12.3. Draw the structures of the following compounds.
(i) 3-Methylbutanal
(ii) p-Methylbenzaldehyde
(iii) 4-Chloropentan-2-one
(iv) p, p’-Dihydroxybenzophenone
(v) p-Nitropropiophenone
(vi) 4-Methylpent-3-en-2-one.
(vii) 3-Bromo-4-phenylpentanoic acid
(viii) Hex-2-en-4-ynoic acid
Ans:

12.4. Write the IUPAC names of the following ketones and aldehydes. Wherever possible, give also common names.
(i) CH₃CO(CH₂)₄CH₃
(ii) CH₃CH₂CH BrCH₂CH(CH₃)CHO
(iii) CH₃(CH₂)₅CHO
(iv) Ph—CH=CH—CHO

Ans:

12.5. Draw structures of the following derivatives:
(i) The 2,4-dinitrophenylhydrazone of benzaldehyde
(ii) Cydopropanone oxime
(iii) Acetaldehydedimethylacetal
(iv) The semicarbazone of cyclobutanone
(v) The ethylene ketal of hexan-3-one
(vi) The methyl hemiacetal of formaldehyde
Ans:

12.6. Predict the product when cyclohexanecarbaldehyde reacts with following reagents :
(i) C₆H₅MgBr followed by H₃0⁺
(ii) Tollen’s reagent
(iii) Semicarbazide in the weakly acidic medium
(iv) Excess of ethanol in the presence of acid
(v) Zinc amalgam and Cyclohexanecarbaldehyde Semicarbazide
Ans:

12.7. Which of the following compounds would undergo aldol condensation, which the Cannizzaro reaction and which neither? Write the structures of the expected products of aldol condensation and Cannizzaro reaction.
(i) Methanal
(ii) 2-Methylpentanal
(iii) Benzaldehyde.
(iv) Benzophenone
(v) Cyclohexanone
(vi) 1-Phenylpropanone
(vii) Phenylacetaldehyde
(viii) Butan-l-ol 1
(ix) 2,2-Dimethylbutanal
Ans: 2-Methylpertfanal, cyclohexanone, 1-phenylpropanone and phenylacetaldehyde contain one or more a-hydrogen and hence undergo aldol condensation. The reactions and the structures of the expected products are given below:

12.8. How will you convert ethanal into the following compounds?
(i) Butane-1,3-diol
(ii) But-2-enal
(iii) But-2-enoic acid
Ans:

12.9. Write structural formulas and names of four possible aldol condensation products from propanal and butanal. In each case, indicate which aldehyde acts as nucleophile and which as electrophile.
Ans:

12.10. An organic compound with the molecular formula C₉H₁₀O forms 2,4-DNP derivative, reduces Tollen’s reagent, and undergoes Cannizzaro reaction. On vigorous oxidation, it gives 1,2-benzenedicarboxylic acid. Identify the compound. 
Ans: Since the given compound with molecular formula C₉H₁₀O forms a 2,4-DNP derivative and reduces Tollen’s reagent, it must be an aldehyde. Since it undergoes Cannizzaro reaction, therefore, CHO group is directly attached to die benzene ring.
Since on vigorous oxidation, it gives 1, 2-benzene dicarboxylic acid, therefore, it must be an ortho- substituted benzaldehyde. The only o-substituted aromatic aldehyde having molecular formula C₉H₁₀O is o-ethyl benzaldehyde. Ail the reactions can now be explained on the basis of this structure.

12.11. An organic compound (A) (molecular formula C₈H₁₆O₂) was hydrolysed with dilute sulphuric acid to give a carboxylic acid (B} and an alcohol (C). Oxidation of (C) with chromic acid produced (B). (Q on dehydration gives but-l-ene. Write equations for the reactions involved.
Ans: Since an ester A with molecular formula C₈H₁₆O₂ upon hydrolysis gives carboxylic acid B and the alcohol C and oxidation of C with chromic acid produces the acid B, therefore, both the carboxylic acid B and alcohol C must contain the same number of carbon atoms.
Further, since ester A contains eight carbon atoms, therefore, both the carboxylic acid B and the alcohol C must contain four carbon atoms each.
Since the alcohol C on dehydration gives but-l-ene, therefore, C must be a straight chain alcohol, i.e., butan-l-ol.
If C is butan-l-ol, then the acid B must be butanoic acid and the ester A must be butyl butanoate.The chemical equations are as follows:

12.12. Arrange the following in increasing order of the property indicated :
(i) Acetaldehyde, Acetone, Di tert. butyl ketone, Methyl tert. butyl ketone (reactivity towards HCN). (C.B.S.E. Sample Paper 2011, 2015, C.B.S.E. Delhi 2012)
(ii) CH₃CH₂CH(Br)COOH, CH₃CH(Br)CH₂COOH, (CH₃)₂CHCOOH, CH₃CH₂CH₂COOH (acid strength) (C.B.S.E. Delhi2008)
(iii) Benzoic acid, 4-Nitrobenzoic acid, 3, 5-Dinitrobenzoic acid, 4-Methoxybenzoic acid (acid strength) (C.B.S.E. Sample Paper 2011, 2015; C.B.S.E. Delhi 2012, C.B.S.E. Outside Delhi 2015, Rajasthan Board 2015)
Ans: (i) Cyanohydrin derivatives are formed as a result of the reaction in which the nucleophile (CN^– ion) attacks the carbon atom of the carbonyl group. The order of reactivity

• decreases with increase in +I effect of the alkyl group.
• decreases with increase in steric hindrance due to the size as well as number of the alkyl groups. In the light of the above information, the decreasing order of reactivity is :
  

(ii) We know that alkyl group with +I effect decreases the acidic strength. The +I effect of isopropyl group is more than that of n-propyl group. Similarly, bromine (Br) with -I-effect increases the acidic strength. Closer its position in the carbon atom chain w.r.t., carboxyl (COOH) group, more will be its -I-effect and stronger will be the acid. In the light of this, the increasing order of acidic strength is :
(CH₃)₂CHCOOH< CH₃CH₂CH₂COOH < CH₃CH(Br)CH₂COOH < CH₃CH₂CH(Br) COOH
(iii) We have learnt that the electron donating group (OCH₃) decreases the acidic strength of the benzoic acid. At the same time, the electron withdrawing group (N0₂) increases the same. Keeping this in mind, the increasing order of acidic strength is:

12.13. Give simple chemical tests to distinguish between the following pairs of compounds.
(i) PropanalandPropanone
(ii) Acetophenone and Benzophenone
(iii) Phenol and Benzoic acid
(iv) Benzoic acid and Ethyl benzoate
(v) Pentan-2-one and Pentan-3-one
(vi) Benzaldehyde and Acetophenone.
(vii) EthanalandPropanal
Ans:

12.14. Row will you prepare the following compounds from benzene? You may use any inorganic reagent and any organic reagent having not more than one carbon atom.
(i) Methyl benzoate
(ii) m-nitrobenzoic acid
(iii) p-nitrobenzoic acid
(iv) Phenylaceticacid
(v) p-nitrobenzaldehyde
Ans:

12.15. How will you bring about the following conversions in not more than two steps?
(i) PropanonetoPropene
(ii) Benzoic acid to Benzaldehyde
(iii) Ethanol to 3-Hydroxybutanal
(iv) Benzene to m-Nitroacetophenone
(v) Benzaldehyde to Benzophenone –
(vi) Bromobenzeneto 1-PhenylethanoL
(vii) Benzaldehyde to 3-Phenylpropan-1-ol.
(viii) Benzaldehyde to α Hydroxyphenylacetk acid
(ix) Benzoic acid to m-Nitrobenzy 1 alcohol
Ans:

12.16. Describe the following:
(i) Acetylation
(ii) Cannizzaro reaction
(iii) Cross aldol condensation
(iv) Decarboxylation
Ans: (i) Acetylation refers to the process of introducing an acetyl group into a compound namely, the substitution of an acetyl group for an active hydrogen atom. Acetylation is usually carried out in presence of a base such as pyridine, dimethylanitine, etc.

(ii) Cannizzaro reaction : Aldehydes which do not contain an a-hydrogen atom, when treated with concentrated alkali solution undergo disproportionation, i.e., self oxidation reduction. As a result, one molecule of the aldehyde is reduced to the corresponding alcohol at the cost of the other which is oxidised to the corresponding carboxylic acid. This reaction is called Cannizzaro reaction.

(iii) Cross aldol condensation: Aldol condensation between two different aldehydes is called cross aldol condensation.If both aldehydes contain a-hydrogens, It gives a mixture of four products.

(iv) Decarboxylation: The process of removal of a molecule of CO₂ from a carboxylic acid is called decarboxylation. Sodium salts of carboxylic acids when heated with soda-lime undergoes decarboxylation to yield alkanes.

12.17. Complete each synthesis by giving missing starting material, reagent or products.

Ans:

12.18. Give plausible explanation for each of the following:
(i) Cyclohexanone forms cyanohydrin in good yield but 2,2, fctrimethylcyclohexanone does not
(ii) There are two – NH₂ groups in semicarbazide. However, only one is involved in the formation of semicarbazones.
(iii)During the preparation of esters from a carboxylic acid and an alcohol in the presence of an acid catalyst, the water or the ester should be removed as soon as it is formed.
Ans:

The yield of second reaction is very low because of the presence of three methyl groups at ex-positions with respect to the C = O, the nucleophilic attack by the CN^– ion does not occur due to steric hinderance. Since there is no such steric hindrance in cyclohexanone, therefore, nucleophilic attack by the CN^– ion occurs readily and hence cyclohexanone cyanohydrin is obtained in good yield.

Although semicarbazide has two – NH₂ groups but one of them (i.e., which is directly attached to C = O) is involved in resonance as shown above. As a result, electron density on N of this -NH₂ group decreases and hence it does not act as a nucleophile. In contrast, the other -NH₂ group (i.e.. attached to NH) is not involved in resonance and hence lone pair of electrons present on N atom of this -NH₂ group is available for nucleophilic attack on the C = O group of aldehydes and ketones.’
(iii) The formation of esters from a carboxylic acid and an alcohol in presence of an acid catalyst is a reversible reaction.

Thus to shift the equilibrium in the forward direction, the water or the ester formed should be removed as fast as it is formed.

12.19. An organic compound contains 69-77% carbon, 11-63 % hydrogen and rest oxygen. The molecular mass of the compound is 86. It does not reduce Tottens’ reagent but forms an addition compound with sodium hydrogensulphite and give positive iodoform test. On vigorous oxidation, it gives ethanoic and propanoic acid. Write the possible structure of the compound.
Ans:

Since the compound form sodium hydrogen sulphite addition product, therefore, it must be either an – aldehyde or methyl/ cyclic ketone. Since the compound does not reduce Tollens’ reagent therefore, it cannot be an aldehyde. Since the compound gives positive iodoform test, therefore, the given compound is a methyl ketone. Since the given compound on vigorous oxidation gives a mixture ofethanoic acid and propanoic acid, therefore, the methyl ketone is pentan-2-one, i.e.,

12.20. Although phenoxide ion has more number of resonating structures than carboxylate ion, carboxylic acid is a stronger acid than on phenol. Why?
Ans: Consider the resonating structures of carboxylate ion and phenoxide ion.

In case of phenoxide ion, structures (V – VII) carry a negative charge on the less electronegative carbon atom.Therefore, their contribution towards the resonance stabilization of phenoxide ion is very small.
In structures I and II, (carboxylate ion), the negative charge is delocalized over two oxygen atoms while in structures III and IV, the negative charge on the oxygen atom remains localized only the electrons of the benzene ring are delocalized. Since delocalization of benzene electrons contributes little towards the stability of phenoxide ion therefore, carboxylate ion is much more resonance stabilized than phenoxide ion. Thus, the release of a proton from carboxylic acids is much easier than from phenols. In other words, carboxylic acids are stronger acids than phenols.

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NCERT Solutions For Class 12 Chemistry Chapter 11 Alcohols Phenols and Ether

Topics and Subtopics in NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols Phenols and Ether:

Section Name	Topic Name
11	Alcohols, Phenols and Ethers
11.1	Classification
11.2	Nomenclature
11.3	Structures of Functional Groups
11.4	Alcohols and Phenols
11.5	Some Commercially Important Alcohols
11.6	Ethers

NCERT Solutions CBSE Sample Papers ChemistryClass 12 Chemistry

NCERT TEXTBOOK QUESTIONS SOLVED

11.1. Classify the following as primary, secondary and tertiary alcohols.

Ans: Primary alcohols: (i), (ii), (iii)
Secondary alcohols: (iv), (v)
Tertiary alcohols: (vi)

11.2. Identify aliylic alcohols in the above examples.
Ans: (ii) and (iv) i.e. H₂C=CH – CH₂OH and

11.3. Name the following compounds according to IUPAC system.

Ans:

11.4. Show how are the following alcohols prepared by the reaction of a suitable  Grignard reagent on methanal ?

Ans:

11.5. Write structures of the products of the following reactions:

Ans:

11.6. Give structures of the products you would expect when each of the following alcohol reacts with (a)HCl-ZnCl₂ (b)HBrand (c) SOCl₂
(i)Butan-1-ol
(ii)2-Methylbutan-2-ol
Ans:

11.7. Predict the major product of acid catalysed dehydration of
(i) 1-nicthylcyclohcxanoland
(ii) butan-1-ol
Ans:

11.8. Ortho and para nitrophenols are more acidic than phenol. Draw the resonance structures of the corresponding phenoxide ions.
Ans:
The resonance structures of o-and p- nitrophenoxide ions and phenoxide ion are given below:

11.9; Write the equations involved in the following reactions:
(i) Reimer-Tiemann reaction
(ii) Kolbe’s reaction
Ans: (i) Reimer-Tiemann reaction

11.10. Write the reactions of Williamson synthesis of 2-ethoxy-3-methylpentane starting from ethanol and 3-methylpentan-2-ol.
Ans: In Williamsons’s synthesis, the alkyl halide should be primary. Thus, the alkyl halide should be derived from ethanol and the alkoxide ion from 3-methylpentan-2-ol. The synthesis is as follows

11.11. Which of the following is an appropriate set of reactants for the preparation of l-methoxy-4- nitrobenzene and why?

Ans:

11.12. Predict the products of the following reactions:

Ans:

NCERT EXERCISES

11.1. Write IUPAC names of the following compounds:

Ans: (i) 2,2,4-Trimethylpentan-3-ol
(ii) 5-Ethylheptane-2,4-dioI
(iii) Butane-2,3-diol
(iv) Propane-1,2,3-triol
(v) 2-Methylphenol
(vi) 4-Methylphenol
(vii) 2,5-DimethylphenoI
(viii) 2,6-Dimethylphenol
(ix) 1-Methoxy-2-methylpropane
(x) Ethoxybenzene
(xi) 1-Phenoxyheptane
(xii) 2-Ethoxybutane

11.2. Write structures of the compounds whose IUPAC names are as follows:
(i)2-Methylbutan-2-ol
(ii)l-Phcnylpropan-2-ol
(iii)3,5-DimethyIhexane-l,3,5-triol
(iv)2,3-Dicthylphenol
(v)1-Ethoxypropane
(vi)2-Ethoxy-3-methylpentane
(vii) Cyclohexylmethanol
(viii) 3-Cyclohexylpcntan-3-ol
(ix)Cyclopcnt-3-en-l-ol
(x)4-ChIoro-3-ethylbutan-l-ol
Ans:

11. 3. (a) Draw the structural formulas and write IUPAC names of all the isomeric alkanols with the molecular formula C_sH₁₂O
(b) Classify the isomers of alcohols given in part (a) as primary, secondary and tertiary alcohols.
Ans:
(a) The molecular formula C₅H₁₂0 represents eight isomeric alkanols. These are :

(b) Praimary: (i), (ii), (iii), (iv) ; Secondary :(v), (vi), (viii) ; Tertiary : (vii)

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 the reaction for the preparation of phenol from cumene.
Ans: This process has a great industrial importance because it gives the preparation of two very useful compounds i.e. phenol and acetone. The raw materials are benzene and propene and it initially proceeds by Friedel Crafts alkylation of benzene.

Oxygen is bubbled through the above solution to form cumene hydroperoxide which is decomposed with aqueous acid
solution to form phenol and acetone as follows:

11.10. Write chemical reaction for the preparation of phenol from chlorobenzene.
Ans:

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.
Ans:

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?
Ans:

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 orthonitrophenol more acidic than orthomethoxyphenol?
Ans: Nitro (NO₂) group is an electron withdrawing group while methoxy (OCH₃) group is electron releasing in nature. The release of H⁺ ion is therefore, easier from o-nitrophenol while it is quite difficult from o-methoxyphenol. Apart form that, o-nitrophenoxide ion is stabilised due to resonance o-nitrophenol is steam volatile while p-nitrophenol is not. This is on account of intramolecular hydrogen bonding in the molecules of o-nitrophenol. As a result, its boiling point is less than that of p-nitrophenol in which the molecules are linked by intermolecular hydrogen bonding.
It is interesting to note that in the substituted phenols, the nature and position of the substituent influences the boiling point of phenol.
For example: .o-nitrophenol is steam volatile while p-nitrophenol is not. This is supported by the fact that the boiling point temperature of o-nitrophenol (100°C) is less than that of p-nitrophenol, (279°C). In o-nitrophenol, there is intramolecular hydrogen bonding in OH and NO₂ groups placed in a adjacent positions. However, these are linked by intermolecular hydrogen bonding in the p-isomers. It is quite obvious that extra energy is needed to the cleave the hydrogen bonds in the p-isomer. Consequently, its boiling point is more.

o-nnrophenol with lower boiling point is steam volatile while p-nitrophenol is not. This helps in the separation of the two isomers present in the liquid mixture.  On passing steam, o-nitropbenol volatilises and its vapours rise alongwith steam and after condensation, collect in the receiver p-nitrophenol is left behind in the distillation flask. e-nkrophenol p-nnrophenol.
On the contrary, o-methoxyphenoxide is destabilised since the electron density on the negatively charged oxygen tends to increase due to the electron releasing tendency of the methoxy(OCH₃) group.

In the light of the above discussion, we may conclude that o-nitrophenol is a stronger acid (pKa = 7-23) than o-methoxyphenl  (pK_a = 9.98)

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 phenol
(iv) Treating phenol with chloroform in presence of aqueous NaOH.
Ans:

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, ethyl methyl ether, methyl phenyl ether, 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) Propane → Propan-2-ol
(ii) Benzyl chloride → Benzyl alcohol
(iii) Ethyl mag. chloride → Propan-1-ol
(iv) Methyl mag. bromide → 2-Methylpropan-2-ol.
Ans:

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.
(iii) Brominationofphenolto2,4,6-tribromophenol
(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⁺Cr₂O₇^2-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-methoxy ethane
(iii) 4-Nitroanisole
(iv) 1-Methoxypropane
(v) 1 -Ethoxy-4 -4 – dimethyl cyclohexane
(vi)Ethoxybenzene

11,24. Write the names of the reagents and equations for the preparation of the following ethers by Williamson’s synthesis :
(i) 1-Propoxypropane
(ii) 2-Methoxy-2-methylpropane
(iii) Ethoxybenzene
(iv) Methoxyethane.
Ans:

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 propane-1-ol? Write mechanism of the reaction. (C.B.S.E. Sarnie Paper 2015)
Ans: Two methods can be used for the synthesis of 1-propoxypropane from propan-1-ol
By Williamson’s synthesis
The halogen derivative such as bromoderivative and sodium salt of the alcohol take part in the 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
Ans:

11.29. Explain the fact that in alkyl aryl ethers, alkoxy group :
(i) activates the benzene ring towards electrophilic substitution.
(ii) directs the incoming substituents towards ortho and para positions in the ring.
Ans:
(i) The alkoxy group (RO -) with lone electron pairs on the oxygen atom activates the ortho and para positions in the ring by + M (or + R) effect as shown below :

As the ortho and para positions in the ring become points of high electron density, the electrophiles prefer to attack these positions.

(ii) The alkoxy group directs the incoming group which is an electrophile towards the ortho and para positions in the ring. As a result, a mixture of isomeric products is formed.

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.
Ans:

11.32. Show how will you synthesise the following from appropriate alkenes.

Ans: All the alcohols are formed by the hydration of alkenes in the acidic medium. The addition follows Markownikov’s rule. 1-Methylcyclohexene can be used in the reaction.

(ii)  4-Methylpent-3-ene upon hydration in the acidic medium will give the desired alcohol.

(iii) Pent-2-ene gives the desired alcohol upon hydration in the presence of acid.

(iv) The cyclic alkene used in this reaction is 2-cyclohexylbut-2-ene.

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.)
Ans:

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.

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