NCERT Grade 12-Haloalkanes and Haloarenes-Answers

NCERT Solutions for Class 12 Chemistry

Find 100% accurate solutions for NCERT Class XII Chemistry. All questions have been explained and solved step-by-step, to help you understand thoroughly. Free Download option available!

Sign Up to Download as PDF

1.    (i) 2, 2, 4-Trimethylpentan-3-ol

(ii) 5-Ethylheptane-2, 4-diol

(iii) Butane-2, 3-diol

(iv) Propane-1, 2, 3-triol

(v) 2-Methylphenol

(vi) 4-Methylphenol

(vii) 2, 5-Dimethylphenol

(viii) 2, 6-Dimethylphenol

(ix) 1-Methoxy-2-methylpropane

(x) Ethoxybenzene

(xi) 1-Phenoxyheptane

(xii) 2-Ethoxybutane.

2.    

3.    (i) The structures of all isomeric alcohols of molecular formula, are C5H12O shown below:

(a) CH3 – CH2 – CH2 – CH2 – CH2 – OH

Pentan-1-ol (1°)

(ii) Primary alcohol: Pentan-1-ol; 2-Methylbutan-1-ol;

3-Methylbutan-1-ol; 2, 2 – Dimethylpropan-1-ol

Secondary alcohol: Pentan-2-ol; 3-Methylbutan-2-ol;

Pentan-3-ol

Tertiary alcohol: 2-methylbutan-2-ol.

4.     Propanol undergoes intermolecular H-bonding due to which it’s boiling point is higher than butane which is non-polar and held by weak van der Wall’s forces of attraction, and it is so because of the presence of -OH group.

Therefore, extra energy is required to break hydrogen bonds. For this reason, propanol has a higher boiling point than hydrocarbon butane.

5.    Alcohols form H-bonds with water due to the presence of -OH group which prevails formation of H-bonding with water. However, hydrocarbons cannot form H-bonds with water as they are non- polar.

As a result, alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses.

6.    The addition of borane B4H6 followed by oxidation to form propan-1-ol, and is known as the hydroboration-oxidation reaction. For example, propan-1-ol is produced by the hydroboration-oxidation reaction of propene. In this reaction, propene reacts with diborane (BH3)2 to form trialkyl borane as an addition product. This addition product is oxidized to alcohol by hydrogen peroxide in the presence of aqueous sodium hydroxide.

7.    

8.   Intramolecular H-bonding is present in o-nitrophenol. O-Nitrophenol is steam volatile because of weak intra molecular H-Bonding. In p-nitrophenol, the molecules are strongly associated due to the presence of intermolecular bonding. Hence, o-nitrophenol is steam volatile whereas p-nitrophenol is not.

9.    To prepare phenol, cumene is first oxidized in the presence of air to form cumene hydro-peroxide.

Then, cumene hydroperoxide is treated with dilute acid to prepare phenol and acetone as byproducts.

10.  Chlorobenzene is fused with NaOH (at 623 K and 320 atm pressure) to produce sodium phenoxide, which gives phenol on acidification by nucleophilic sustitution reaction.

11.   The mechanism of hydration of ethene to form carbocation by electrophilic attack of hydronium ion involves three steps.

Step 1:

Protonation of ethene to form carbocation by electrophilic attack of H3O+:

12.  

13.  (i) By acid-catalyzed hydration of ethylbenzene (styrene), 1-phenylethanol can be synthesized.

(ii) When chloromethylcyclohexane is treated with sodium hydroxide, cyclohexymethanol is obtained.

(iii) Whn 1-chloropentane is treated with NaOH, pentan-1-ol is produced.

14.  The acidic nature of phenol can be represented by the following two reactions:

(i) Phenol reacts with sodium to give sodium phenoxide and release H2 gas as by- product.

2 phenol+2Na——–> 2 sodium phenoxide + H2

(ii) Phenol reacts with sodium hydroxide to give sodium phenoxide and water as by-product.

The acidity of phenol is more than that of ethanol. This is because after losing a proton, the phenoxide ion undergoes resonance and gets stabilized whereas ethoxide ion does not undergo resonance. Resonance leads to distribution of charge due to which the compound gets stable unlike concentration of charge at single point like in case of ethoxide ion which promotes unstability.

15.  

The nitro-group is an electron-withdrawing group. The presence of this group in the ortho position decreases the electron density in the O-H bond and increase positive charge on Oxygen. As a result, it is easier to lose a proton. Also, the o-nitrophenoxide ion formed after the loss of protons is stabilized by resonance. Hence, ortho nitrophenol is a stronger acid.

On the other hand, methoxy group is an electron-releasing group and it decreases positive charge on oxygen . Thus, it increases the electron density in the O-H bond and hence, the proton cannot be given out easily. It makes it less acidic.

For this reason, ortho-nitrophenol is more acidic than ortho-methoxyphenol.

16.  The -OH group is an electron-donating group. Thus, it increases the electron density in the benzene ring as shown in the given resonance structure of phenol.

As a result, the benzene ring is activated towards electrophilic substitution.

17.  Propan-1-ol is treated with alcoholic KMnO4 followed by hydrolysis to form propanoic acid as follows

(iii) By the process of nitration, as follows

18.  (i) Kolbe’s reaction: When phenol is treated with sodium hydroxide, sodium phenoxide is produced. This sodium phenoxide when treated with carbon dioxide, at 400K under 3-7 atm.pressure followed by acidification, undergoes electrophilic substitution to give ortho-hydroxybenzoic acid as the main product also known as salicylic acid. This reaction is known as Kolbe’s reaction.

(ii) Reimer-Tiemann reaction: When phenol is treated with chloroform (CHCl3) in the presence of sodium hydroxide or potassium hydroxide, a -CHO group is introduced at the ortho position of the benzene ring and 3KCl and 2 molecules of water are formed as by product.

This reaction is known as the Reimer-Tiemann reaction.

The intermediate is hydrolyzed in the presence of alkalis to produce salicyldehyde.

(iii) Williamson ether synthesis: Williamson ether synthesis is a laboratory method to prepare symmetrical and unsymmetrical ethers by allowing alkyl halides to react with sodium alkoxides.

This reaction involves SN2 attack of the alkoxide ion on the alkyl halide. Better results are obtained in case of primary alkyl halides.

If the alkyl halide is secondary or tertiary, then elimination competes over substitution.

(iv) Unsymmetrical ether: An unsymmetrical ether is an ether where two groups on the two sides of an oxygen atom differ (i.e., have an unequal number of carbon atoms). For example: ethyl methyl ether (CH3O-CH2CH3).

19.  The mechanism of acid dehydration of ethanol to yield ethene involves the following three steps:

Step 1:

Protonation of ethanol to form ethyl oxonium ion:

Step 2:

Formation of carbocation (rate determining step):

Step 3:

Elimination of proton to form ethene:

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

20.  (i) If propene is allowed to react with water in the presence of an acid as a catalyst, then propan-2-ol is obtained.

(ii) If benzyl chloride is treated with NaOH (followed by acidification) then benzyl alcohol is produced.

(iii) When ethyl magnesium chloride is treated with methanal, an adduct is the produced which gives propan-1-ol and MgClOH on hydrolysis.

(iv) When methyl magnesium bromide is treated with propane, an adduct is the product which gives 2-methylpropan-2-ol on hydrolysis.

21.  (i) Acidified potassium permanganate

(ii) Pyridinium chlorochromate (PCC)

(iii) Bromine water

(iv) Acidified potassium permanganate

(v) 85% phosphoric acid or conc. Sulphuric acid at 443K.

(vi) NaBH4 or LiAlH4 or H2/Ni.

22.  Ethanol undergoes intermolecular H-bonding due to the presence of -OH group, resulting in the association of molecules. Extra energy is required to break these hydrogen bonds. On the other hand, methoxymethane does not undergo intermolecular H-bonding. Hence, the boiling point of ethanol is higher than that of methoxymethane.

23.  (i) 1-Ethoxy-2-methylpropane

(ii) 2-Chloro-1-methoxyethane

(iii) 4-Nitro-1-methoxybenzene

(iv) 1-Methoxypropane

(v) 1-Ethoxy-4,4-dimethylcyclohexane

(vi) Ethoxybenzene.

24.  

25.  The reaction of Williamson synthesis involves SN2 attack of an alkoxide ion on a primary alkyl halide because tertiary halide undergoes elimination reaction instead of substitution reaction.

That’s why if secondary or tertiary alkyl halides are taken in place of primary alkyl halides, then elimination would compete over substitution. As a result, alkenes would be produced. This is because alkoxides are nucleophiles as well as strong bases. Hence, they react with alkyl halides, which results in an elimination reaction.

26.  1-propoxypropane can be synthesized from propan-1-ol by dehydration.

Propan-1-ol undergoes dehydration in the presence of protic acids (such as H2SO4, H3PO4) to give 1-propoxypropane.

The mechanism of this reaction invloves the following three steps:

27.  The formation of ethers by dehydration of secondary or tertiary alcohol is a bimolecular reaction (SN2) involving the attack of an alcohol molecule on a protonated alcohol molecule very quickly. In the method, the alkyl group should be unhindered. In case of secondary or tertiary alcohols, the alkyl group is hindered. As a result, elimination dominates substitution. Hence, in place of ethers, alkenes are formed.

28.  

29.  (i) In aryl alkyl ethers, due to the +R effect of the alkoxy group, the electron density in the benzene ring increases as shown in the following resonance structure.

Thus, benzene is activated towards electrophilic substitution by the alkoxy group and electrophile will attack at O- and p- positions.

(ii) It can also be observed from the resonance structures that the electron density increases more at the ortho and para positions than at the meta position. As a result, the incoming substituents are directed to the ortho and para positions in the benzene ring.

30.  The mechanism of the reaction of HI with methoxymethane involves the following steps:

Step 3: When HI is in excess and the reaction is carried out at a high temperature, the methanol formed in the second step reacts with another HI molecule and gets converted to methyl iodide

31.  

32.  The given alcohols can be synthesized by applying Markovnikov’s rule of acid-catalyzed hydration of appropriate alkenes.

Acid-catalyzed hydration of pent-2-ene also produces pentan-2-ol but along with pentan-3-ol.

Thus, the first reaction is preferred over the second one to get pentan-2-ol.

33.  The mechanism of the given reaction involves the following steps:

Step 1: Protonation

Step 2: Formation of 2° carbocation by the elimination of a water molecule

Step 3: Re-arrangement by the hydride-ion shift

Step 4: Nucleophilic attack

34.  Primary alcohol → (i), (ii) (iii)

Secondary alcohol → (iv), (v)

Tertiary alcohol → (vi)

35.  The alcohols given in (ii) and (vi) are allylic alcohols.

36.  (i) 3-Chloromethyl-2-isopropylpentan-1-ol

(ii) 2, 5-Dimethylhexane-1, 3-diol

(iii) 3-Bromocyclohexanol

(iv) Hex-1-en-3-ol

(v) 2-Bromo-3-methylbut-2-en-1-ol.

37.  

38.  

39.  

Primary alcohol do not react appreciably with Lucas’ reagent (HCl-ZnCl2) at room temperature.

Teritary alcohols react immediately with Lucas’ reagent.

40.  

41.  

Resonance structure of the phenoxide ion

Resonance structures of p-nitrophenoxide ion

Resonance structures of o-nitrophenoxide ion

It can be observed that the presence of nitro groups increases the stability of phenoxide ion.

42.  (i) Reimer–Tiemann reaction: When phenol reacts with CHCl3 and KOH at 60`C, salicylaldehyde is formed.

(ii) Kolbe’s reaction: when sodium phenoxide reacts with CO2 at high pressure, on heating sodium salicylate is formed which on acidification gives salicylic acid.

43.  In Williamson synthesis, an alkyl halide reacts with an alkoxide ion. Also, it is an SN2 reaction. In the reaction, alkyl halides should be primary having the least steric hindrance. Hence, an alkyl halide is obtained from ethanol and alkoxide ion from 3-methylpentan-2-ol.

44.  Set (ii) is an appropriate set of reactants for the preparation of 1-methoxy-4nitrobenzene.

In set (i), sodium methoxide (CH3ONa) is a strong nucleophile as well as a strong base. Hence, an elimination reaction predominates over a substitution reaction.

45.  

MySchoolPage connects you with exceptional, certified chemistry tutors who help you stay focused, understand concepts better and score well in exams!

Have a Question?




Mathematics - Videos


Physics - Videos


Biology - Videos


Chemistry - Videos