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

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In case of anisole, methylphenyl oxonium ion, is formed by protonation of ether. The bond between O—CH₃ is weaker than the bond between O—C₆H₅ because the carbon of phenyl group is sp2 hybridised and there is partial double bond character. Therefore, the attack by I- ion breaks O—CH₃ bond to form CH₃I. 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.

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(a) Cumene is the beginning element for the industrial production of phenol.

(b)

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

(c) 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.

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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.

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The  structure of glycerol is:

  CH₂—CH—CH₂

   |          |         |

  OH     OH  OH

The IUPAC of glycerol is Propane-1,2,3-triol.

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The IUPAC name of the given compound is:

(A) 3-Ethyl-5-methylhexane-2,4-diol

(B) 1-Methoxy-3-nitrocyclohexane

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3-Methylpent-2-ene-1,2-diol

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The solubility of alcohol in water depends upon the following factors:

(a) Hydrogen bonding: Alcohols form hydrogen bonding with water due to the presence of -OH group in alcohol and hence alcohols are soluble in water. 

(b) Size  of alkyl or aryl group: The alkyl and aryl group are hydrophobic in nature and larger the size of  alkyl or aryl group of alcohol lesser will be the solubility of alcohol in water.

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Alcohols such as ethanol  are made unfit for consumption by adding some additives like copper sulphate and pyridine which gives a foul smell, and  bad taste  to alcohol, it is known as denatured alcohol. 

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The reagent pyridinium chlorochromate (PCC) is a complex formed by CrO₃, pyridine and HCl converts the secondary alcohol to ketone without oxidising the double bond.

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2-chloroethanol is more acidic than ethanol  because the conjugate base  of  2-chloroethanol   is more stable than that of ethanol due to the -I effect of chlorine atom.

Stability:   Cl—CH₂—CH₂—OΘ           >                     CH₃—CH₂—OΘ

Acidity:     Cl—CH₂—CH₂—OH   >                     CH₃—CH₂—OH

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The reagent used for the conversion of ethanol to ethanal is pyridinium chlorochromate (PCC) which is a complex formed by CrO₃, pyridine and HCl which converts the primary alcohols to aldehydes.

CH₃—CH₂—OH CH₃—CHO

Ethanol                                   Ethanal

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The strong oxidising agents such as acidified KMnO₄ or  K₂Cr₂O₇ reagents are used for the conversion of ethanol to ethanoic acid. These reagents are so strong that they convert the alcohols to carboxylic acids.

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Out of o-nitrophenol and p-nitrophenol, o-nitrophenol is more volatile due to presence of intramolecular hydrogen bonding in o-nitrophenol whereas p-nitrophenol has intermolecular hydrogen bonding.

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Out of o-nitrophenol and o-cresol, o-nitrophenol  is more acidic due to the -I effect and -R effect of -NO₂ group as the -NO₂ group is an electron withdrawing group whereas the -CH₃ group is electron releasing group and has +I effect on the conjugate base which increases the electron density and decreases the polarity of O-H bond and hence less acidic.

Fig: The -I effect of NO₂ group and + I effect of CH₃ group on the polarity of O-H bond

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When phenol is treated with bromine water, it forms the white precipitate of 2, 4, 6-tribromophenol. In the bromination reaction of phenol, water is ionized. And also bromine gets ionized to produce bromonium ions to a larger extent. Phenol gets ionized to produce an ortho-para directing phenoxide ion. Bromine water is mostly used as a test for C=C double bond. During this reaction white precipitates form at the end and bromine water is decolourised.

Hence, the formation of bromonium ions and strong ortho-para directing species indicates that the product formation is 2, 4, 6-tribromophenol.

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The increasing order of acidity:

o-cresol  <  Phenol   <  o-nitrophenol

The electron-withdrawing group on the substituted phenol increases the acidity due to increasing the polarity of the O-H bond and thus, the acidity increases  and vice versa. -NO₂ is an electron-withdrawing group whereas the -CH₃ group is an electron-donating group.

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The decreasing order of reactivity of alcohol with sodium metal:

Tertiary Alcohols > Secondary Alcohols   > Primary Alcohols 

When alcohol reacts with active metals  e.g. Na. K etc., the O-H bond of alcohols breaks to form the corresponding alkoxide. The alkyl  group is electron donating and has +I effect due to which the O-H bond becomes strong and hence, the reactivity of alcohol decreases.

Fig: +I effect of alkyl group of alcohol

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When benzene diazonium chloride is heated with water, it forms phenol along with nitrogen and hydrochloric acid as by-products as shown below:

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The decreasing order of acidity:

H₂O > ROH >  HC ≡ CH 

The more the stability of the conjugate base of the given compounds, the more acidity will be.

^? OH > ^? O-R > ^? C ≡ CH 

The negative charge on the electronegative oxygen atom is more stable than carbon atom. The + I effect of the alkyl group does not stabilize the negative charge on the electronegative oxygen atom, hence less stable than ^? OH. 

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The enzyme invertase, which is present in yeast,   is used to convert sucrose to glucose and fructose. The glucose and fructose are further converted to ethanol by zymase enzyme (also present in yeast).

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The propan-2-one can be converted into tert-butyl alcohol by using CH₃MgBr grignard reagent to form the additional product tert-butyl alcohol.

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

(a) CH₃-CH₂CH₂-CH₂OH   (Butane-1-ol)

(b) CH₃-C^*H-CH₂-CH₃  (Butane-2-ol)

            |

          OH

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

            |

          CH₃

          CH₃

             | 

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

             |

           OH

Only butane-2-ol has chiral carbon due to it being the alcohol with molecular formula C₄H₁₀O exhibits optical activity.

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The electron pairs of oxygen atom of hydroxyl (-OH) group  in phenols are in conjugation with the pi electrons of phenyl ring and hence, the bond C-OH bond has double bond character due to which OH group in phenols more strongly held as compared to OH group in alcohols.

Resonance in phenol

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The C-OH bond in phenols has double bond character due to resonance  of electron pairs of oxygen atom with the pi electrons of phenyl ring, which makes the C-OH bond strong and hence, the  nucleophilic substitution of a nucleophile with the -OH group of alcohol is not possible as the C-OH does not break easily.

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Step 1: Protonation of alkene due to the presence of double bond which  attacks the H₃O⁺ ion and forms carbocation. 

Step 2: Water molecule act as nucleophile and attack the carbocation. 

Step 3: Deprotonation occur to get the alcohol and hydronium ion forms 

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In  O? C? O, the dipole moment is zero, hence non-polar while in R—O—R the dipole moment is non-zero due to which it is polar.

The CO₂ is a linear molecule and the dipole moment is equal and in the opposite direction due to which the dipole moment is zero. While in R—O—R, has bent structure and lone pair of electrons due to which the net dipole moment is nonzero and hence, it is a polar molecule.

The CO₂ is a linear molecule and the dipole moment is equal and in the opposite direction due to which the dipole moment is zero. While in R—O—R, has bent structure and lone pair of electrons due to which the net dipole moment is nonzero and hence, it is a polar molecule.

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The reaction of alcohol with the conc. HCl and ZnCl₂ (Lucas reagent) leads to the formation of a carbocation through SN¹  mechanism. The more stable the carbocation, the faster the reaction will be.

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

Stability of carbocation:

Tertiary carbocation > Secondary carbocation  > Primary carbocation 

According to stability of carbocation, the reactivity order of the alcohol is:

Tertiary alcohol > Secondary alcohol  > Primary alcohol 

Tertiary alcohol reacts immediately with Lucas reagent and shows turbidity, secondary alcohol does not show turbidity with Lucas reagent at room temperature but on heating turbidity appears and primary alcohol do not react with lucas reagent even on heating.

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Step 1: Addition of NaOH to obtain sodium phenoxide

Step 2: Sodium Phenoxide undergoes Kolbe’s  reaction at high temperature and pressure in presence of carbon dioxide (CO₂) gas.  The product is  further  hydrolysed  to obtain salicylic acid. 

Step 3: Acetylation of salicylic acid occur when it is treated with acetic anhydride

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The nitration of aromatic compounds occurs by using HSO₄ and HNO₃ which leads to the formation of  NO₂⁺ (nitronium ion) electrophile.

Due to the formation of  NO₂⁺ electrophile, we can say that nitration is an example of aromatic electrophilic substitution. The electron releasing group on the benzene ring increases the rate of nitration and vice-versa.

Phenol is more easily nitrated than benzene as the hydroxyl (-OH) group on the benzene ring is an electron releasing group which increases the electron density at the ortho and para position due to +R effect of -OH group. The nitration occurs at these electron rich sites by the attack of NO₂⁺ electrophile. 

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Phenoxide ion is more reactive than phenol to electrophilic aromatic substitution reaction  due to the more in electron density of  phenoxide ion and hence, phenoxide ion easily undergoes the electrophilic substitution reaction than phenol by the weak CO₂ electrophile.

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In phenol, the electron pairs on oxygen atom of -OH group are in conjugation (or resonance) with phenyl ring which decreases the polarity of the C-OH bond whereas in methanol the polarity of C-OH bond is more due to electron releasing group of -CH₃.

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Ethers are prepared by Williamson synthesis by the reaction of alkyl halide with the stadium alkoxide. But this is not possible in case of di-tert-butyl ether. In case of preparation of di-tert-butyl ether, tert-butyl halide must react with sodium tert-butoxide. Alkoxides are nucleophiles but they are strong bases as well, due to which elimination is favoured instead of substitution and leads to the formation of 2-methylprop-1-ene.

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The bond angle C—O—H in alcohols is less than tetrahedral angle because of the repulsion between the lone pairs present on the oxygen atom, which pushes the bond C—O—H closer. Therefore,   C—O—H bond angle in alcohols is slightly less than the tetrahedral angle.

In  ethers, the lone pairs on oxygen atoms are also present but the two bulky alkyl groups also have repulsive force which increases the C—O—C bond angle. Therefore,   the C—O—C bond angle in ether is slightly greater than tetrahedral angle .

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The occurrence of intermolecular hydrogen bonds due to the H- bonds in alcohol compounds play a major role . With the increasing number of alkyl groups the molecular mass increases due to which the polar nature of these compounds get suppressed . Hence, the solubility factor of alcohols is indirectly proportional to the molecular mass.

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The electron withdrawing group (-NO₂), withdraws electrons and disperses the negative charge. Therefore, -NO₂ group stabilizes the phenoxide ion. Hence p-nitrophenol is more acidic than phenol.

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Due to the presence of intermolecular H- bonds in alcohol the energy required to break H – bonds is more than the energy required to break simple dipole bonds present in ethers . Thus, the boiling point of alcohol is higher than the boiling point of ethers.

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This can be explained as under:

(i) In phenol, the conjugation of unshared electron pairs over oxygen with aromatic ring results in partial double bond character in C – O bond.

In methanol, no such conjugation (resonance) is possible.

(ii) In phenol, oxygen is attached to sp² hybridised carbon while in methanol, oxygen attached to sp² hybridised carbon. An sp² hybridised carbon is more electronegative (because of greater 5-character) than sp³ hybridised carbon atom. Therefore, the bond between oxygen and sp² hybridised carbon is more stable than the bond between oxygen and sp², hybridised orbital.

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Increasing order of acidity is: ethanol < water < phenol.
This is because the phenoxide ion obtained after the removal of H⁺ is resonance stabilised, while the ethoxide ion obtained after the removal of H⁺ is destabilised by +1 effect of ethyl group. Thus phenol is a stronger acid than ethanol.
Now, ethanol is a weaker acid than water because the electron releasing ethyl group increases the ethanol density on oxygen and consequently the proton will not be released easily. There is no such effect is water.

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(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

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

(a) CH₃-CH₂CH₂-CH₂OH   (Butane-1-ol)

(b) CH₃-C^*H-CH₂-CH₃  (Butane-2-ol)

              |

            OH

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

               |

            CH₃

            CH3

              | 

(d) 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.

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(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° 

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(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

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(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

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(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.

So,   compound B and C have -OH groups attached to sp³ hybridized  -CH₃ groups and hence are aromatic alcohols and compounds  A and D have -OH groups attached to sp² benzene rings, hence are non-aromatic compounds.

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(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 the 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.

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(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

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(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. 

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(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.

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(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. 

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(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.

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(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.

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(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).

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(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.

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(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

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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). 

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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.

The nucleophilic substitution of haloarenes such as chlorobenzene requires drastic conditions for conversion to phenol so chlorobenzene will not be converted to phenol on heating with NaOH at 298K / 1atm.

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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.

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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.

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II. 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

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(i)- (d);    (ii)- (c);  (iii)- (f);  (iv)- (a); (v)- (g);  (vi)- (b)

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  (i)- (d)     (ii)- (e);     (iii)- (b);     (iv)- (a)

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(i)- (e) ;    (ii)- (f) ;     (iii)- (d) ;    (iv)- (c);     (v)- (a);     (vi)- (b)

This is a matching answer type question as classified in NCERT Exemplar

(i)- (d);    (ii)- (a);     (iii)- (f);     (iv)- (e);    (v)- (c);     (vi)- (b)

This is a assertion and reason answer type question as classified in NCERT Exemplar

(ii) Assertion and reason both are wrong statements.

Correct assertion: Addition reaction of water to but-1-ene in acidic medium yields butan-2-o1.

Correct reason: Addition of water in acidic medium proceeds through the formation of secondary carbocation.

This is a assertion and reason answer type question as classified in NCERT Exemplar

(i) Assertion and reason both are correct and reason is correct explanation of assertion.

Explanation:  P-nitrophenol is more acidic as nitro group helps in the stabilization of the phenoxide ion by dispersal of negative charge due to resonance.

This is a assertion and reason answer type question as classified in NCERT Exemplar

(iv) Assertion is a wrong statement but reason is correct.

Explanation:  Correct assertion is the IUPAC name of the compound is 1- (2-propoxy) propane.

This is a assertion and reason answer type question as classified in NCERT Exemplar

(iv) Assertion is a wrong statement but reason is correct. 

Explanation: 141 pm

This is a assertion and reason answer type question as classified in NCERT Exemplar

(ii) Assertion and reason both are wrong statements.

Explanation: Correct assertion: Boiling points of alcohol  are higher than that of ethers of comparable molecular mass.

Correct reason: Alcohols can form intermolecular hydrogen bonding while ethers cannot.

This is a assertion and reason answer type question as classified in NCERT Exemplar

(iv) Assertion is a wrong statement but reason is correct.  

Explanation:  Bromination of phenol cannot be carried out in presence of Lewis acid.

This is a assertion and reason answer type question as classified in NCERT Exemplar

(v) Both assertion and reason are correct statements but reason is not correct explanation  of assertion.

Explanation:  Due to intramolecular hydrogen bonding o-Nitrophenol does not form hydrogen bond with  water

This is a assertion and reason answer type question as classified in NCERT Exemplar

(iii) Assertion is a correct statement but reason is wrong.

Explanation:  Phenoxide ion is stabilized by resonance which is not possible in alkoxide ion.

This is a assertion and reason answer type question as classified in NCERT Exemplar

(ii) Assertion and reason both are wrong statements.

Explanation: Correct assertion: Phenol on treatment with bromine water can form 2,4,6- tribromophenol

Correct reason: In water, phenol gives phenoxide ions which activate the ring towards electrophilic substitution reaction.

This is a assertion and reason answer type question as classified in NCERT Exemplar

(iv) Assertion is a wrong statement but reason is correct. 

Explanation:  Phenol on treatment with dil. HNO₃ form o-nitrophenol and p-nitrophenol.