Chemistry NCERT Exemplar Solutions Class 12th Chapter Eleven: Overview, Questions, Preparation

1. Monochlorination of toluene in sunlight followed by hydrolysis with aq. NaOH yields.

I. o-Cresol

II. m-Cresol

III. 2, 4-Dihydroxytoluene

IV. Benzyl alcohol

Ans: IV Benzyl alcohol

The chlorine atom is substituted by the hydrogen atom of  methyl hydrocarbon present in toluene to form benzyl chloride on monochlorination of toluene in  presence of sunlight by free radical pathway

C₆H₅CH₃    +   Cl₂   C₆H₅CH₂Cl

Toluene                        Benzyl Chloride

Benzyl chloride (monochlorinated toluene product) on hydrolysis with aq. NaOH yields benzyl alcohol by substituting the chloride ion by OH^- ion by nucleophilic substitution reaction.

C6H5CH2Cl C6H5CH2OH

Benzyl                    Benzyl

Chloride                Alcohol

2. How many alcohols with molecular formula C₄H₁₀O are chiral in nature? 

I. 1

II. 2 

III. 3

IV. 4 

Ans: 1

The alcohol with molecular formula C₄H₁₀O is butanol has 4 isomers, these are:

• CH₃-CH₂CH₂-CH₂OH   (Butane-1-ol)
• CH₃-C^*H-CH₂-CH₃          (Butane-2-ol)

              |

            OH

• CH₃-CH-CH₂-OH         (2-methylpropane-1-ol)

               |

            CH₃

            CH3

              | 

• CH₃- C-CH3          (2-methylpropan-2-ol)

   |

           OH

Butane-2-ol has chiral carbon which has all different substituted groups attached, so butane-2-ol alcohol with molecular formula C₄H₁₀O is chiral in nature.

3. What is the correct order of reactivity of alcohols in the following reaction? 

R - OH   +    HCl R - Cl   +   

I. 1° > 2° > 3°

II. 1° < 2° > 3° 

III. 3° > 2° > 1°

IV. 3° > 1° > 2°

Ans:  III 3° > 2° > 1° 

Alcohols are classified as primary, secondary and tertiary by lucas reagent which is a mixture of concentrated HCl and dry anhydrous ZnCl₂. The carbocation is formed as the intermediate during the lucas test of alcohol.More the stability of intermediate carbocation, more will be the reactivity of alcohol. 

Tertiary carbocation is most stable then the secondary alcohol is stable  due to hyperconjugation and the primary alcohol is the least stable.

Stability of intermediate carbocation:

3° > 2° > 1° 

Reactivity of alcohol:

3° > 2° > 1° 

4. CH₃CH₂OH can be converted into CH₃CHO by ________________ .

I. catalytic hydrogenation

II. treatment with LiA1H₄

III. treatment with pyridinium chlorochromate

IV. treatment with KMnO₄

Ans: III. treatment with pyridinium chlorochromate 

The pyridinium chlorochromate is a complex of chromium trioxide with pyridine. This reagent is preferred for mild oxidation to form aldehydes and hence the oxidation to carboxylic acid is prevented.

Primary alcohols are oxidized to form aldehyde, whereas secondary and tertiary alcohols are oxidized to form ketones. The conversion of a hydroxyl group to aldehyde can be done by the removal of hydrogen atoms (Oxidation), looking at the following options; pyridinium chlorochromate and potassium dichromate are both oxidizing agents. However, when pyridinium chlorochromate is used, it is a weaker oxidizing agent compared to potassium permanganate; hence, it reduces ethanol to acetaldehyde (ethanal).

CH₃CH₂OH  CH₃CHO

5. The process of converting alkyl halides into alcohols

I. addition reaction

II. substitution reaction 

III. dehydrohalogenation reaction

IV. rearrangement reaction 

Ans: II. substitution reaction 

Alkyl halides undergo substitution reactions to form corresponding alcohol.The halide ion, X^- is substituted by OH^- ion  to form alcohol by nucleophilic substitution reaction.

R-X → R-OH

6. Which of the following compounds is aromatic alcohol? 

I. A, B, C, D

II. A, D 

III. B, C

IV. A 

Ans: III. B, C 

The aromatic alcohols or aryl-alcohols are a class of chemical compounds containing a hydroxyl group (—OH) bonded directly to an aromatic hydrocarbon group.

In the given question, the aromatic alcohols are those compounds in which the hydroxyl group is not directly attached to the benzene ring but is linked to a carbon atom situated in a side-chain. In short, we will search for that figure in which the hydroxyl group is linked to sp³ hybridised carbon.

7. Give IUPAC name of the compound given below. 

        CH₃—CH—CH₂--CH₂--CH—CH₃

                    |                                |

                  Cl                             OH

I. 2-Chloro-5-hydroxyhexane

II. 2-Hydroxy-5-chlorohexane 

III. 5-Chlorohexan-2-ol

IV. 2-Chlorohexan-5-ol 

Ans:  III. 5-Chlorohexan-2-ol 

The IUPAC naming of the given compound are as follows:

Word root: It depends upon the number of carbon atoms in the longest continuous carbon chain selected, called the parent chain. Depending upon the number of carbons in the chain the compound is assigned a word.  The given compound contain 6 carbon attached in the continuous longest chain and word root for 6 carbon atoms is ‘hex’.

The suffix- A suffix is added after the word root to indicate the nature of the carbon-carbon bond. The given carbon chain contains a single bond then a suffix will be added as “ane”.

Prefix- The groups which are not regarded as functional but present in the carbon chain as substituents are written before the word root as a prefix. Such groups are fluorine, chlorine, nitro, etc.

According to the IUPAC rule, the prefix (Chlorine) will be written first followed by the word root and primary suffix and secondary suffix. As the compound contains chlorine which will be regarded as substituents and will be written as a prefix. The compound contains six carbon atoms due to which it will be considered as hexane and the word root will be “hex”. Since it is an alkane compound, the primary suffix will be “ane” and the secondary suffix will be “ol”. Therefore, the IUPAC name of the given compound is 5-Chlorohexan-2-ol.

8. IUPAC name of m-cresol is

I. 3-methylphenol

II. 3-chlorophenol 

III. 3-methoxyphenol

IV. benzene- 1.3-diol 

Ans: I.  3-methylphenol 

The m-cresol is an organic compound which contains a benzene ring, a -OH group and a methyl group. It has a methyl group substituted to meta position in the phenol ring. According to the IUPAC system, the -OH group is given more priority than the methyl group.

The cresols are also called hydroxytoluene or the methylphenols. These contain a benzene ring with one methyl and one phenol group substituted. There are three forms of cresols which are o-cresol, m-cresol and p-cresol. m- cresol is a colourless liquid and is viscous in nature.

In case of m-cresol, the methyl group and the hydroxyl group at 1st and 3rd position. The structure of m-cresol can be written as-

Now, if we see the IUPAC nomenclature then the -OH group should be given 1st place because it has more priority and the methyl group should be at 3rd position i.e. at meta to OH group.

 

Thus, the IUPAC name is- 3-Methylphenol.

9. IUPAC name of the compound CH₃—CH—OCH₃ is  _______________ .

                                                                       |

                                                                      CH3

I. 1-methoxy-1-methylethane

II. 2-methoxy-2-methylethane

III. 2-methoxypropane

IV. isopropylmethyl ether 

Ans: II. 2-methoxy-2-methylethane

IUPAC provides consistency to the names of organic compounds. It enables every compound to possess a unique name, which otherwise is not plausible with the common names. The chemical structure of the given compound is shown below:

The given compound is an alkane so suffix in our case will be –ane. An ether functional group is also present so prefix will be Alkoxy-.

To determine the name of a compound, identify the longest and continuous chain of carbon containing the functional group and count the number of carbon atoms in the chain as shown in the table below.

 

In the given compound, the number of carbon atoms in the longest chain is 2. That means the prefix of alkane in our case is eth-.

 

Numbering of the carbons in the longest chain of carbon (Remember that If the organic molecule is not an alkane (and has a functional group) you have to initiate numbering such that the functional group is placed on the carbon having the lowest possible number).

 

In the given compound, we can number as shown below:

In the given compound, either the group and methyl group present on the second carbon will be treated like a branched group. It will be written as 2-methyl methoxy before the main chain name.

 

Finally, combine the elements of the compound name in the specified order as: branched groups, prefix, suffix according to the functional group and its location along the longest carbon chain.

 

Therefore, the given compound will have IUPAC name as: 2-methyl methoxy ethane.

 

Hence, the correct answer is Option (B), the IUPAC name of the given compound is 2-Methyl methoxy ethane.  

10. Which of the following species can act as the strongest base?

^{I. Θ}OH

^{II. Θ}OR 

III. ^ΘOC₆H₅ 

IV.

Ans: II. ^ΘOR 

The species which is the weakest acid has the strongest conjugate base.  The ROH is the weakest acid among the given compounds  due to the inductive effect of the alkyl group. Therefore,  the strongest conjugate base is ^ΘOR.

11. Which of the following compounds will react with sodium hydroxide solution in water?

I. C₆H₅OH

II. C₆H₅CH₂OH 

III. (CH₃)₃COH

IV. C₂H₅OH 

Ans: I. C₆H₅OH

C₆H₅OH, phenol  is a strong acid (due to resonance) which will react with  sodium hydroxide solution in water to form sodium phenoxide and water. Phenol is more acidic than sp³ hybridized attached alcohols. 

12. Phenol is less acidic than 

I. ethanol

II. o-nitrophenol 

III. o-

IV. o-methoxyphenol 

Ans: IV. o-methoxy phenol

The stability of the conjugate base  determines the acidic character of the compound.  Higher the stability of the conjugate base, the higher is the acidic character.

Phenol, O-nitrophenol, O-methyl phenol, and O-methoxy phenol all are aromatic compounds. In aromatic compounds, the negative charge of the conjugate base charge is delocalised. The delocalisation of negative charge gives additional stability to the compound. 

Ethanol is not aromatic and it cannot delocalise the negative charge formed after the  H⁺ ion leaves the c. So, it is the least acidic among the given options. Thus, we can eliminate it.

There are two types of groups, the electron-withdrawing, and the electron releasing group. The methyl group and the methoxy group are electron releasing groups. They release the electrons and hence decreases the stability of the negatively charged conjugate base..

On the other hand, a nitro group is an electron-withdrawing group. They withdraw the electrons from the aromatic ring. Withdrawing the negative charge from the ring will increase the stability of the compound in this case.

Phenoxide is stabilised by resonance effect only and it does not have any electron releasing group. So, it is more acidic than o-methyl phenol and o-methoxy phenol. But the absence of an electron-withdrawing group makes it less acidic than o-nitrophenol.

13. Which of the following is most acidic?

I. Benzyl alcohol

II. Cyclohexanol 

III. Phenol

IV. m-Chlorophenol 

Ans: IV. m-Chlorophenol 

Phenols are more acidic than sp³ and sp² hybridized alcohol.  The m-Chlorophenol is more acidic than phenol due to the -I effect (dominated by the + R effect) of chlorine atom increasing the acidity of phenol.

14. Mark the correct order of decreasing acid strength of the following compounds. 

I. e > d > b > a > c

II. b > d > a > c > e 

III. d > e > c > b > a

IV. e > d > c > b > a 

Ans: II. b > d > a > c > e 

The acidic strength decreases with increasing stability of the conjugate base of the given alcohol. -NO₂ group at the para position in compound (b) is the most acidic due to the -M effect of the -NO₂ group.

Compound (d) has -I effect on the conjugate base, whereas the -OCH₃ group has + M effect on the conjugate base and hence decreases the acidity of compounds (c) and (e).

15. Mark the correct increasing order of reactivity of the following compounds with HBr/HC1. 

I. a < b < c

II. b < a < c 

III. b < c < a

IV. c < b < a 

Ans:  III. b < c < a 

The reaction will proceed by SN¹ mechanism to form the carbocation intermediate, more the stability of carbocation more will be the reactivity of the given compound with HBr/HC1.

-NO₂ group is an electron withdrawing group and due to the -M effect, increases the stability of carbocation. Cl group at para position shows + M effect due to it, carbocation is least stable and hence, the compound (c)  is the least reactive.

16. Arrange the following compounds in increasing order of boiling point. Propan- 1-ol. butan-l-ol. butan-2-ol. pentan- 1-ol

I. Propan- 1-ol. butan-2-ol. butan- 1-ol. pentan- 1-ol

II. Propan- 1 -ol. butan- 1 -ol. butan-2-ol. pentan- 1 -ol 

III. Pentan-l-ol. butan-2-ol. butan- 1 -ol. propan- 1-ol

IV. Pentan-l-ol. butan- 1-ol. butan-2-ol. propan- 1-ol 

Ans: I. Propan- 1-ol. butan-2-ol. butan- 1-ol. pentan- 1-ol 

The compounds Propan-l-ol, butan-2-ol, butan-1-ol ,pentan-1-ol belong to the alcohol family. The structure of the given  alcohols are as shown below:

With increase in the molecular weight, the boiling points of alcohols increase. Hence, butan-1-ol has higher boiling point than propan-1-ol. Similarly, pentan-1-ol has higher boiling point than butan-1-ol.

Hence, the increasing order of the boiling points of alcohols is propan-1-ol < butan-1-ol < pentan-1-ol.

With increase in the branching of alkyl groups, the boiling point of alcohols decreases. Thus, butan-2-ol has lower boiling point than butan-1-ol. The increasing order of the boiling points of given alcohols is

I. H₂/Pd

II. LiA1H₄

III. NaBH₄

Ans:

I. H₂/Pd 

II. LiA1H₄

III. NaBH₄

Aldehydes and ketones are converted to alcohols by catalytic hydrogenation.The reagents used to convert the aldehydes to alcohol are by H₂/Pd , lithium aluminium hydride (LiA1H₄), and sodium borohydride (NaBH₄). 

18. Which of the following reactions will yield phenol?

Phenol is prepared by:

• Dow’s process: The chlorobenzene on heating with NaOH, form sodium phenoxide which on hydrolysis with water of acid from phenol.
• Diazotization of aniline: Aniline on reacting with nitrous acid at 0-5^o C followed by hydrolysis with water form phenol.
• Benzene undergoes sulphonation on reacting with oleum to form benzene sulphonic acid which on heating with sodium hydroxide forms sodium phenoxide and then phenol on hydrolysis.

19. Which of the following reagents can be used to oxidise primary alcohols to aldehydes?

I. CrO₃ in anhydrous medium.

II. KMnO₄ in an acidic medium. 

III. Pyridinium chlorochromate.

IV. Heat in the presence of Cu at 573K.

Ans:

I. CrO₃ in anhydrous medium. 

II. Pyridinium chlorochromate. 

III.  Heat in the presence of Cu at 573K. 

KMnO₄ is a strong oxidising agent and oxidises the primary alcohol to ketone. Hence cannot be used as a reagent to oxidise primary alcohols to aldehydes.

20. Phenol can be distinguished from ethanol by the reactions with 

I. Br₂/water

II. Na 

III. Neutral FeCl₃

IV. All the above 

Ans: 

I. Br₂/water 

III. Neutral FeCl₃ 

Phenol does not give the bromine water (Br₂/water ) test whereas ethanol gives the bromine water.

Phenol on reaction with neutral FeCl₃ forms a colored complex with Fe₃⁺ ion whereas ethanol does not form the colored complex with neutral FeCl₃.

21. Which of the following arc benzylic alcohols?

I. C₆H₅—CH₂—CH₂—OH

II. C₆H₅—CH₂—OH

III. C₆H₅—CH—OH

              |

             CH₃

IV. C₆H₅—CH—CH—OH

                        |

                       CH₃

Ans: 

I. C₆H₅—CH₂—OH

III. C₆H₅—CH—OH

              |

             CH₃

Benzylic alcohol are those alcohols which have phenyl group attached to the alpha  carbon  of alcohol.

  R—CαH₂—OH                   :     Benzylic Alcohol

The compound II and III have phenyl group attached at the Cα of the alcohol and thence they are benzylic alcohols

Ans: In case of anisole, methylphenyl oxonium ion,

is formed by protonation of ether. The bond between O—CH ₃ is weaker than the bond between O—C ₆H ₅ because the carbon of phenyl group is sp2 hybridised and there is partial double bond character. Therefore, the attack by I- ion breaks O—CH ₃ bond to form CH ₃ I. Phenols do not react further to give halides because the sp2 hybridised carbon of phenol cannot undergo nucleophilic substitution reaction needed for conversion to the halide.

2. (a) Name the starting material used in the industrial preparation of phenol.

Ans: Cumene is the beginning element for the industrial production of phenol.

(b) Write complete reaction for the bromination of phenol  in aqueous and non-aqueous medium.

Ans:

When bromine water is used to treat phenol. As a whitish precipitate, 2,4,6-tribromophenol is formed

                                                            2.4.6 Tribromophenol

(c) Explain why Lewis acid is not required in bromination of phenol?

Ans: Benzene is normally halogenated in the presence of a Lewis acid, such as FeBr3, which polarises the halogen molecule. Even in the absence of Lewis acid, the polarisation of the bromine molecule occurs in the case of phenol. It's because the benzene ring's —OH group has a strong activating impact.

4. Explain a process in which a biocatalyst is used in industrial preparation of a compound known to you.

Ans: Commercially, ethanol (C₂H₅OH) is made by fermenting carbohydrates, which is the earliest process. In the presence of an enzyme called invertase, sugar in molasses, sugarcane, or fruits like grapes is transformed to glucose and fructose (both of which have the formula C₆H₁₂O₆).

In the presence of another enzyme, zymase, present in yeast, glucose and fructose are fermented.

Grapes are used to make wine because they contain sugars and yeast. The amount of sugar in grapes increases as they ripen, and yeast forms on the outer peel. When grapes are crushed, sugar and enzymes collide, causing fermentation to begin. Fermentation takes place in the absence of oxygen in anaerobic circumstances. During fermentation, carbon dioxide is emitted.