Chemistry NCERT Exemplar Solutions Class 12th Chapter Eleven

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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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(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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(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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(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) 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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(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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(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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(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