NCERT Solutions For Class 12 Chemistry Chapter 11 Alcohols Phenols and Ether: Download Free PDF

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• A11.1 It is primary alcohol because carbon which carries the –OH group is only attached to one alkyl group.
• It is primary alcohol because carbon which carries the –OH group is only attached to one alkene group.
• It is primary alcohol because the carbon which carries the –OH group is only attached to one propyl group.
• It is secondary alcohol because the carbon which carries the –OH group is joined directly to methyl and benzene.
• It is secondary alcohol because the carbon which carries the –OH group is joined directly to two different alkyl groups.

A 11.2 

Allylic alcohol is an organic compound which has the structural formula CH₂ = CHCH₂OH. In other words, in these alcohols, the the-OH group is attached to sp² hybridized carbon next to the carbon-carbon double bond, that is to an allylic carbon. Therefore, in the above examples, the following are the allylic alcohols.

(ii) H₂C = CH – CH₂OH and

A 11.3

• 3-chloroethyl-2-isopropylpentan-1-ol
• 2,5-Dimethylhexane-1,3-diol
• 3-Bromocyclohexanol
• Hex-1-en-3-ol
• 2-Bromo-3-methylbut-2-en-1-ol

A 11.4

In the Grignard reagent reaction, the first step of the reaction is the nucleophilic addition of Grignard reagent to the carbonyl group to form an adduct, Hydrolysis of adduct results in the formation of alcohol.

Here, is the general reaction with Grignard reagent below:-

From here, it is clear that HCHO gives CH₂OH groups, so R of Grignard reagent is the remaining part of given alcohols. Thus, select the suitable Grignard reagent by substituting the value of R. Now we can see the reaction given below:-

Methanal reacts with iso-propyl magnesium bromide, in presence of dry ether gives an additional compound. And this additional compound on reaction with H₂O /H⁺ gives iso-butyl alcohol (i.e., 2-methylpropane-1-ol) as one of the final product.

Methanal reacts with cyclohexyl magnesium bromide, in presence of dry ether, which gives an intermediate product. This intermediate product when reacts with given reagent, as shown above, gives cyclohexyl methanol as a product.

1. A 11.5 In this reaction, when propene reacts with the given reagent then the double bond of propene breaks down with charges on them. So, H⁺ gets placed on the carbon which already has two hydrogen atom and OH^- gets substituted on center carbon because it has the more positive charge which attracts OH^-. Thus we get propene-2-or as a
2. In this reaction, when Methyl ( 2-oxocyclohexyl) ethanoate reacts with the given reagent then the double bond between the oxygen atom and cyclohexyl gets breaks down, such that O has a negative charge and that particular carbon will have a positive charge on it. So, to neutralize it, H⁺ gets substituted to that carbon and with O^- to form the structure of alcohol. Thus we get Methyl (2-hydroxycyclohexyl) ethanoate as a
3. In this reaction, when 2-Methylbutanal reacts with the given reagent NaBH₄, then the double bond between carbon and oxygen gets to break down in the above aldehyde compound, where carbon has positive charge and oxygen will have a negative charge on So, H⁺ gets placed on the carbon to complete its octet and H⁺ gets substituted on O^- to form OH, i.e, an alcohol. Thus we get 2-Methylbutan-1-ol as a product.

A 11.6

(a)(i) Primary alcohols do no react appreciably with Lucas’ reagent (HCl –ZnCl₂) at room temperature.

(ii) Tertiary alcohol reacts immediately with Lucas ‘reagent.

A 11.7 

A 11.8

Resonating structures of o-nitrophenoxide ions that are formed by the loss of a proton from o-nitrophenol are as follows:

Resonating structures of p-nitrophenoxide ions that are formed by the loss of a proton from p- nitrophenol are as follows:

Resonating structures of phenoxide ions that are formed by the loss of a proton from phenol are as follows:

It is clearly evident from the above structures that due to —R-effect of— NO₂NO₂ group, o-and p-nitrophenoxide ions are more stable than phenoxide ions. Consequently, o- and p- nitrophenols are more acidic than phenols.

A 11.9

1. Phenol on mixing with chloroform and NaOH at 340K followed by Acidic hydrolysis, salicyl aldehyde is formed. When carbon tetrachloride (CCl₄) is used at the place of chloroform salicylic acid is This type of reaction is known as Reimer - Tiemann reaction.
2. The sodium phenoxide reacts with CO₂ under pressure 4-7 atm at a 400K temperature to form sodium salicylate, which on acidification yields salicylic acid. This type of reaction is known as Kolbe’s

A 11.10

During Williamson synthesis of ethers, an alkyl halide reacts with an alkoxide(ion with –ve charge on the oxygen of alcohol and + ve charge on alkali metal like Na) ion. it is an S_N2 reaction. In the reaction, alkyl halides should be least hindered. Hence, an alkyl halide is obtained from ethanol and alkoxide ion from 3-methylpentan-2-ol. The reactions are shown below:

A 11.11 Set (ii) is appropriate Because CH₃Br is only a nucleophile whereas CH₃ONa is nucleophile as well as strong base, so the elimination reaction can occur,

A 11.12

1. When N –propyl methyl ether reacts with HBr, it forms propanol and bromomethane, n- propyl methyl ether will cleave at O . and H⁺ will attack at O, and Br^- will attack CH ⁺

2. When Ethoxybenzene Reacts with HBr, it forms Phenol and bromoethane, Ethoxybenzene cleaves at H⁺ will attack at O, and Br^- will attack C H ⁺

3. When nitrating mixture reacts with ethoxy benzene introduction of nitro group is occurred at para position as it will give the stable product without
4. As HI is a strong nucleophile it will protonate the oxygen , to form a good leaving And I^- will attack at C(CH₃)₃⁺ to give tert- butyl iodide and ethanol.

A 11.1

1. 2,2,4-Trimethylpentan-3-ol

The naming of the compound usually starts with numbering the carbons in the chain. The lower set of locants are chosen for this purpose, while in this case numbering under this condition is done from the left side. Once the carbons are mentioned, the position of -OH group is numbered and -ol is added as the suffix.

2.      5-Ethylheptane-2,4-diol

Here the longest chain is the straight chain. For such situations, numbering should be such that the functional groups should be denoted by the smallest number. Ethyl group is named at the beginning as it’s a side chain.

3.      Butane-2,3-diol

In this system, glycols are called as Diols and their class name is Alkane diols. The two-hydroxyl group position is indicated by Arabic numerals. Firstly, the carbon chain is named, the -OH positions are numbered and suffixed with -diol (depends on the number of -OH groups.

4.      Propane-1,2,3-triol

Initially carbon chain is named, -OH groups are numbered (can be done from any side as it’s a symmetrical compound) and suffixed with triol.

5.      2-Methylphenol

In the IUPAC system, the position of the substituent w.r.t –OH group is indicated by an Arabic numeral, with the carbon carrying the OH group being numbered 1. So, methyl is numbered at the minimum position and phenol is added. The common name is o-cresol (ortho-cresol).

6.      4-Methylphenol

Methyl group is numbered 4 w.r.t to -OH position. The compound is symmetrical. The compound is known by its common name p-cresol (para cresol).

7.      2,5-Dimethylphenol

The position of the substituent w.r.t –OH group is indicated by an Arabic numeral, with the carbon carrying the OH group being numbered 1. The lowest number chain is further chosen. The positions of the methyl group is 2 and 5.

8.      1-Methoxy-2-methylpropane 

Longest carbon chain attached to oxygen is chosen. It is called the principal chain and the other carbon chain is named with -oxy(Alkoxy) suffix at the end. Now the position of alkoxy group is 1

w.r.t the principal chain(propane) and the methyl group

Is 2. Further alphabetically methoxy comes before methyl, therefore named as such.

9.      Ethoxybenzene

The principal chain is benzene, so the short chain is named as ethoxy and suffixed with benzene

10.  1-Phenoxyheptane

Principal chain is heptane group having 7 carbon groups. Phenol group is called as substituent alkoxy group and is named.

11.  2-Ethoxybutane

The longest chain here is butane and the ethyl group is attached to the 2^nd carbon.

A 11.2 Butane is the longest chain.

Propane is principle chain and substituents are numbered accordingly.

Hexane is the longest chain. There are three -OH substituents and 2 methyl groups.

-OH of phenol is numbered a 1.

Propane is the principle chain and the alkoxy group is ethyl group.

Longest chain is pentane and substituents are numbered accordingly.

In such cases, cyclo groups are named first, followed by conventional naming methods.

The cyclo group is named first, followed by conventional naming methods.

Cyclo group is named first. Pentane having a double bond is named then with the suffix -OL added to the end.

The longest chain is butane when looked for minimum numbering case.

A 11.3 The structures of all isomeric alcohols of C₅H₁₂O are given below:

Naming is done by the conventional method. The -OH group is attached on the first carbon.

(b) 3-Methylbutan-1-ol

Butane is the longest chain and methyl is the substituent group.

(c) 3-Methylbutan-1-ol

Longest chain is butane and conventional naming method is used.

(d) 2,2-Dimethylpropan-1-ol

Isomer is made by transforming the principal carbon into tertiary type. The longest chain is butane and named accordingly.

(e) Pentan-2-ol

Longest chain is pentane and the numbering is chosen from the minimum position.

(f) 3-Methylbutan-2-ol

Butane is the longest chain, further numbering is done by choosing a minimum position for the

-OH group followed by other substituents.

(g) Pentan-3-ol

Conventional method of naming is used.

(h) 2-Methylbutan-2-ol

Butane is the longest chain and substituents are named accordingly by looking into minimum position.

A 11.4

• Here, propanol undergoes intermolecular H-bonding because of the presence of -OH group while butane has no such property

(intermolecular Hydrogen bonding in propanol)

Therefore, extra energy will be required to break those hydrogen bonds which in turn causes higher boiling point for propanol when compared to butane.

A 11.5 Due to the presence of –OH group, alcohols form hydrogen-bonds with water but hydrocarbons cannot form hydrogen-bonds with water.

Due to inter moleculer hydrogen bonding between Alcohol and water molecular they remain tightly bounded to water molecules and have higher solubility. Whereas in case of hydrocarbon there is no chance of hydrogen bonding.

A 11.6

The hydroboration-oxidation reaction is a two-step reaction that converts an alkene into a neutral alcohol by the net addition of water across the double bond. The hydrogen and hydroxyl group are added in a syn addition leading to the cis configuration. Hydroboration- oxidation is an anti-Markovnikov reaction, with the hydroxyl group attaching to the less substituted carbon. In first step Addition of Hydroborate group is done and in next step, it is oxidized by hydrogen peroxide.

For example: - When propene undergoes hydroboration-oxidation reaction, then it produces propan-1-ol as product. In this reaction diborane i.e., (BH₃)₂ reacts with propene, which in result generates trialkyl borane as an addition product. Then trialkyl borane is oxidised, by using hydrogen peroxide in the presence of aqueous sodium hydroxide to form alcohol, as final product.

A 11.7 The different forms of cresol is formed with given molecular formula:

• 2-methylphenol
• 3-methylphenol
• 4-methylphenol

A 11.8 In ortho nitrophenol there is intra-molecular H bonding, whereas in para-nitrophenol there is inter-molecular H bonding, as shown below:

And because of that para-nitrophenol get tightly bounded with water and ortho nitrophenol is steam volatile and it will leave the solution.

A 11.9 The conversion of Phenol from Cumene requires the air oxidation of

The air oxidation of cumene (isopropyl benzene) leads to the production of both phenol and acetone (costlier than phenol).

The air oxidation of cumene gives cumene hydro peroxide as an intermediate which on further hydrolysis (H₃O⁺) gives phenol and acetone.

A 11.10 There are many ways to this conversion. Two of them are given below:-

(a) In the above conversion, the chlorobenzene is treated with a base such as NaOH, KOH etc. (strong base). The base abstracts the hydrogen from the C-2 position (it can also abstract the hydrogen from the C-6 position, as both are equally acidic) leaving the negative charge at that position.

In the next step Cl^- leaves, leaving behind the positive charge at that carbon. Both the negative charge and positive charge forms a bond resulting Benzyne as the intermediate.

After the formation of the Benzyne intermediate OH^- of the base attacks at the C-1 position and further, the H⁺ attacks to stabilize the negative charge thus resulting in the phenol.

The reaction can start from benzene also. Chlorination of benzene gives Chlorobenzene. When it is further treated with NaOH and H₂O at 350°C it results in the formation of sodium phenoxide which on further treatment H⁺ gives Phenol as the final product.

A 11.11 Step 1:- Protonation of ethene to form carbocation by electrophilic attack of H₃O⁺.

Step 2:-Nucleophilic attack of water on carbocation.

Step 3:- Deprotonation to form ethanol.

A 11.12 The reaction given below is:

Benzene reacts with concern. H₂SO₄ and undergoes the following mechanism:-

Step 1: The equilibrium produces SO₃ in concentrated H₂SO₄, as shown below:

Step 2: SO₃ is the electrophile which reacts with benzene to form arenium ion, as shown below:

Step 3: A proton is removed from the arenium ion to form benzenesulfonate ion.

Step 4: The benzenesulfonate ion accepts a proton to become benzene-sulphonic acid, as shown below:

Step 5: The benzene sulphonic acid then reacts with NaOH to give phenol as the final product, as shown below:

A 11.13

1. 1-phenylethanol from a suitable  - The addition of water takes place according to Markovnikov rule. The alkene taken is styrene. And According to the rule, the positive charge i.e. H⁺ goes to the carbon of the double bond which has more number of hydrogens and the negative part i.e. OH- goes to the carbon that has less number of hydrogens. Therefore resulting the final product as 1-phenyl ethanol.

2.In the above conversion, NaOH gets dissociated into Na⁺ and OH^-and Na⁺ then combines with Cl of chloromethylcyclohexane forming NaCl and thus the final product Cyclohexylmethanol is obtained.

3.In this conversion also, Na forms the compound with Cl i.e. NaCl and to compensate the negative charge formed by Cl, OH attacks at that position resulting in pentan-1-ol.

A 11.14 The acidic nature of phenol can be represented by the following two reactions:-

(a) Phenols react with sodium to give sodium phenoxide, liberating H₂.

(b) Phenols react with sodium hydroxide to give sodium phenoxide and water as a by-product.

The acidity of phenol is more than that of ethanol. This is because phenol after losing a proton becomes phenoxide ion which undergoes resonance and is stabilized whereas ethoxide ion does not.

The resonating structures of phenoxide ion are shown as below:

The lone pair of electrons on oxygen delocalizes into the benzene (mesomeric effect) which reduces the electron density in the O-H bond. The O-H bonds are weaker and therefore breaks easily whereas in ethanol the electron releasing inductive effect of the alkyl group increases the electron density on the O-H bond. This strengthens the bond so the bond does not break easily. Therefore making ethanol less acidic than phenol.

A 11.15 ortho-nitro phenol is more acidic than ortho-methoxy phenol.

Explanation: Due to strong –R and –I effect of NO₂ group, electron density in the O-H bond decreases and hence the loss of a proton becomes easy.

Now after the loss of a proton, the o-nitrophenoxide ion left behind is stabilized by resonance and thus making o-nitro phenol a stronger acid.

In contrast, due to the +R effect of methoxy group increases the electron density in the O-H bond. Thereby making the loss of proton difficult.

Now, the o-methoxyphenoxide ion left after the loss of a proton is destabilized by resonance. The two negative charges repel each other, thereby destabilizing the o-methoxyphenoxide ion.

Therefore, o-nitrophenol is more acidic than o-methoxyphenyl.

A 11.16 The –OH group is an electron donating group. Thus, it increases the electron density in the benzene ring as shown by its resonating structure of phenol.

As a result benzene ring is activated towards electrophilic substitution.

A 11.17 Oxidation of propane-1-ol with alkaline KMnO₄ solution gives propanoic acid as the product. As the oxidation of primary alcohol gives carboxylic acid as the major product in the presence of a strong oxidizing reagent. And here KMnO₄ is a very strong oxidizing agent.

A mixture of o-bromo phenol and p-bromo phenol is formed.

The formation of 2 products depends totally on the reaction conditions.

Dilute HNO₃ with phenol.

Only dilute acid will be required for the nitration of phenol, nitric acid contains a small amount of nitrous acid which because of the activation of the ring will be more than enough to nitrate the phenol. Two products are formed 2-nitrophenol (ortho) and 4-nitrophenol (para).

Treating Phenol with chloroform in presence of aqueous NaOH results in the formation of salicylaldehyde(Reimer-Tiemann reaction)

A 11.18 Kolbe’s Reaction: it is a carboxylation chemical reaction that proceeds by heating sodium phenoxide(the sodium salt of phenol)with carbon dioxide under pressure(100 atm,125°C), then treating the product with a sulphuric acid. The final product is salicylic acid (the precursor to aspirin).

The reaction is given as:

The mechanism is given below:

Reimer-Tiemann reaction: The Reimer Tiemann reaction is a chemical reaction used for the ortho-formylation of phenols, with the simplest example being the conversion

of phenol to salicylaldehyde.

When phenol is treated at 340K with chloroform and alkali, it forms salicylaldehyde.

Williamson ether synthesis: It is an organic reaction forming ether from an organohalide and a deprotonated alcohol (alkoxide). Typically it involves the reaction of an alkoxide ion with primary alkyl halide via S_N2 reaction.

During this reaction, the main bonds broken is the C-Br bond and the new bonds formed are “C- O” bond.

Unsymmetrical ether: It is an ether in the molecule of which the two ligands on the ether group are different.

Eg.

A 11.19 The mechanism of acid dehydration of ethanol to yield ethane involves three steps:

Step 1:- Protonation of ethanol to form ethyl oxonium ion.

Step 2:- Formation of carbocation (rate determining step).

Step 3:-Elimination of a proton to form ethane

 The acid consumed in step 1 is released in step 3. After the formation of ethane, it is removed to shift the equilibrium in a forward direction.

A 11.20

1. The conversion of Propene to propane-2-ol takes place according to markovnikoff rule. The positive part of H2O that is H+ goes to the carbon which has more hydrogen and the negative part that is OH-goes to carbon that has less number of carbons
2. NaOH act as a base in the conversion of benzyl chloride to benzyl On hydrolysis removal of NaCl takes place and OH is inserted in place of Cl.
3. Ethyl magnesium chloride (Grignard reagent) attacks on the carbon of the (The partial positive and negative charge is because of the electronegativity difference). After the formation of the addition product, hydrolysis takes place which further results in the formation of propane-1-ol and Mg(OH)Cl as the by product.
4. Attack of methyl magnesium bromide (Grignard reagent) on carbonyl carbon results in the formation of adduct, which has partial charges due to electronegativity differences. The adduct on hydrolysis yields 2-methylpropan2-ol.

A 11.21

1. Acidified KMnO₄ (potassium permanganate)

Potassium permanganate is a strong oxidant and is able to react with many functional groups. Here KMnO₄ will readily react with primary carbon (where hydrogen is attached) and transforms that to acid.

2. PCC (Pyridinium Chlorochromate)

This is actually a milder version of chromic acid. This works as a sort of elimination reaction. The formation of aldehyde occurs because of the action chromium (a good leaving group) which will be replaced when the C-H bond is broken.

3. Bromine water

Bromine water is actually an aqueous form of bromine. Here in aqueous solution, phenol ionizes to form phenoxide ion due to the presence of negative charge, the oxygen of phenoxide ion donates electron on benzene ring to a large extent, as a result, the ring gets highly activated and hence tri-substituion occurs.

4. Acidified KMnO₄ (potassium permanganate)

Potassium permanganate is a strong oxidant and oxidizes benzyl alcohol to benzoic acid. The compound is used due to its high oxidation power and the reaction proceeds due to the presence of hydrogen attached to the carbon group (Benzylic position)

5. Sulphuric acid or concentrated Phosphoric acid

Alcohols are amphoteric. So, the lone pair of oxygen atoms makes the -OH group weakly basic. Thus, in the presence of a strong acid, R—OH acts as a base and protonates into the very acidic alkyloxonium ion ⁺OH2.This basic characteristic of alcohol necessarily helps in conversion to propene when dehydrated with a strong acid.

6. L_iAlH₄ (lithium aluminum hydride) or NaBH₄

Both compounds are best-reducing agents containing 4 Hydrogen atoms each. In reaction with a ketone, the double bond is reduced and hydrogen atoms are substituted to form alcohol.

A 11.22

• Due to the presence of -OH group, ethanol undergoes intermolecular hydrogen bonding which results in the association of molecules.
• Therefore, extra energy is required to break those hydrogen bonds. Whereas methoxymethane does not undergo those hydrogen bonding which implies ethanol has a higher boiling point than that of methoxymethane.

A 11.23

• Ethoxy-2-methylpropane
• Chloro-1-methoxyethane 4-Nitroanisole

1-Methoxypropane

• Ethoxy-4,4-dimethylcyclohexane Ethoxy benzene

A 11.24 During the reaction, an alkoxide ion is formed which is then added to an alkyl halide to form the ether via S_N2 mechanism.

In the primary step, the nucleophile is formed (O^- ) which will the approach to the alkyl halide and after the transition stage, the substitution takes place.

During the reaction, an alkoxide ion is formed which is then added to an alkyl halide to form the ether via S_N2 mechanism.

During the reaction, an alkoxide ion is formed which is then added to an alkyl halide to form the ether via S_N2 mechanism.

During the reaction, an alkoxide ion is formed which is then added to an alkyl halide to form the ether via S_N2 mechanism.

A 11.25 Williamson synthesis is basically a S_N2 reaction of a primary alkyl halide with an alkoxide ion. The basic mechanism for this reaction is

Now consider this reaction,

This reaction proceeds as conventional Williamson synthesis. But if secondary or tertiary alkyl halides are taken in place of primary alkyl halides, then elimination would compete over substitution reaction, which will result in the formation of alkenes. The reason is alkoxides are better nucleophiles as well as strong bases. Therefore, they react with alkyl halides resulting in an elimination reaction.

A 11.26 According to the question we have to perform the following conversion: -

The mechanism of the above reaction is as follows : - The mechanism is given below: -

In the first step, the alcohol gets protonated by the acid present to give a protonated alcohol.

In the second step, the nucleophilic attack of another alcohol molecule on the protonated alcohol gives us 1-propoxypropane as the desired product.

A 11.27 The preparation of ether by acid dehydration of primary alcohol involves the nucleophilic addition of alcohol molecule to the protonated alcohol molecule as shown below: -

However, under these conditions secondary and tertiary alcohols forms alkenes rather than ethers. The reason for this being that due to stearic hindrance, nucleophilic attack by the alcohol molecule on the protonated alcohol molecule does not take place. Instead protonated 2⁰ and 3⁰ alcohols lose a molecule of water to form stable carbocations. The stable carbocations so formed prefers to lose a proton to form alkenes instead of forming ethers by undergoing nucleophilic attack by another alcohol molecule. This could be seen clearly from the following reactions : -

A 11.28 1. 1-propoxypropane reacts with hydrogen iodide to give propan-1-ol and 1-iodopropane as the products.

• 2. Methoxybenzene reacts with hydrogen iodide to give phenol and iodomethane
• Benzyl ethyl ether reacts with hydrogen iodide to give benzyl iodide and ethanol

A 11.29 (i) In aryl alkyl ethers the +R effect of the alkoxy group leads to an increase in the electron density of the benzene ring as they push electrons into the ring making the benzene ring activated towards electrophilic substitution reactions. This could be understood more clearly from the following resonating structures : -

(ii) It could be clearly seen from the above resonating structures that the electron density increases more at the ortho and para positions as compared to the meta positions. Hence, we can conclude that the alkoxy group directs the incoming substituents to ortho and para positions in the benzene ring.

For example –

A 11.30 The reaction of HI with methoxymethane yields two different sets of products depending upon the initial amount of HI taken.

• When equal moles of HI and methoxymethane are taken, a mixture of methyl alcohol and methyl iodide is

The mechanism is given below:

In the first step, methoxymethane reacts with hydrogen iodide to extract a proton to give the dimethyloxonium ion.

In the second step of the reaction, the Dimethyloxonium ion reacts with the iodide ion present to yield methyl iodide and methyl alcohol as the product via SN² pathway.

• If an excess of HI is used the methyl alcohol formed in Step II is also converted into methyl iodide by the following mechanism : -

In the first step, methoxymethane reacts with hydrogen iodide to extract a proton to give the dimethyloxonium ion.

In the second step of the reaction the Dimethyloxonium ion reacts with the iodide ion present to yield methyl iodide and methyl alcohol as the product via SN2 pathway.

In the third step of the reaction Methyl alcohol formed above reacts with hydrogen iodide to extract a proton to give the protonated methyl alcohol which finally reacts in the fourth step with the iodide ion via SN² pathway to give methyl iodide and water as the product.

A 11.31

The driving force of all the reactions given to the question is that the alkoxy group is an ortho and para directing group because it exerts its +R effect in the benzene ring. Para position being comparatively more stable than the ortho position is usually preferred because ortho position leads to stearic hindrance, hence the major product is mostly the para- substituted compound.

As seen from the resonating structures above the structure in which the negative charge is in the para position will form a more stable product when attacked by an electrophile. Hence in the following reactions, we will be considering that resonating structure as the starting material.

The mechanism is given below:

• The driving force of all the reactions given to the question is that the alkoxy group is an ortho and para directing group because it exerts its +R effect in the benzene ring. Para position being comparatively more stable than the ortho position is usually preferred because ortho position leads to stearic hindrance, hence the major product is mostly the para-substituted

• As seen from the resonating structures above the structure in which the negative charge is in the para position will form a more stable product when attacked by an electrophile. Hence in the following reactions, we will be considering that resonating structure as the starting material.

  Nitration of anisole gives p-nitroanisole as the major product.

• The mechanism is given below:

• • The driving force of all the reactions given to the question is that the alkoxy group is an ortho and para directing group because it exerts its +R effect in the benzene ring. Para position being comparatively more stable than the ortho position is usually preferred because ortho position leads to stearic hindrance, hence the major product is mostly the para-substituted
  • • The driving force of all the reactions given to the question is that the alkoxy group is an ortho

      
      As seen from the resonating structures above the structure in which the negative charge is in the para position will form a more stable product when attacked by an electrophile. Hence in the following reactions, we will be considering that resonating structure as the starting material. and para directing group because it exerts its +R effect in the benzene ring. Para position being comparatively more stable than the ortho position is usually preferred because ortho position leads to stearic hindrance, hence the major product is mostly the para-substituted compound.

    The Friedel-Craft’s acetylation of anisole gives 4- methoxyacetophenone as the major product.

     

7

A 11.32

1. We know that the addition and elimination reactions are opposite of each other.Hence, for solving the above questions our approach should be to first dehydrate a suitable alcohol to give either a single alkene or a mixture of an alkene, if we obtain a mixture of alkene then we would have to detect which of the alkene will give us the desired alcohol. Wherever required the acid- catalyzed addition of water to alkenes will follow Markovnikov’s rule.

2. We know that the addition and elimination reactions are opposite of each other.Hence, for solving the above questions our approach should be to first dehydrate a suitable alcohol to give either a single alkene or a mixture of an alkene, if we obtain a mixture of alkene then we would have to detect which of the alkene will give us the desired alcohol. Wherever required the acid- catalyzed addition of water to alkenes will follow Markovnikov’s rule

 

3. We know that the addition and elimination reactions are opposite of each other.Hence, for solving the above questions our approach should be to first dehydrate a suitable alcohol to give either a single alkene or a mixture of an alkene, if we obtain a mixture of alkene then we would have to detect which of the alkene will give us the desired alcohol. Wherever required the acid- catalyzed addition of water to alkenes will follow Markovnikov’s rule.

But the addition of H₂O to pent-2-ene gives us a mixture of two alcohols out of which one is desired and the other one undesired.

Hence, the alkene to be chosen to arrive at the desired alcohol as the product should be pent- 1-ene as it gives only a single product which is the desired alcohol.

4. We know that the addition and elimination reactions are opposite of each other.Hence, for solving the above questions our approach should be to first dehydrate a suitable alcohol to give either a single alkene or a mixture of an alkene, if we obtain a mixture of alkene then we would have to detect which of the alkene will give us the desired alcohol. Wherever required the acid- catalyzed addition of water to alkenes will follow Markovnikov’s rule.

As addition is the reverse of elimination we first dehydrate our required product to obtain –

Now, if we add H₂O to either of the three alkenes in presence of an acid we get the desired alcohol.

 

A 11.33  1. The first step in the mechanism of the given reaction is protonation of the alcohol followed by loss of water to give a 2⁰ carbocation.

2. The next step is a rearrangement of the 2⁰ carbocations formed in the above step is less stable it rearranges by a 1,2-hydride shift to form more stable 3° carbocations.

3. The last step of the reaction is the nucleophilic attack of Br^- ion on the 3° carbocations giving the final product.