Chapter 12, Problem 12.39UKC

Answer the following questions about alcohol A.

1. a. Give the IUPAC name.
2. b. Classify the alcohol as 1°, 2°, or 3°.
3. c. Draw the products formed when A is dehydrated with H₂SO₄.
4. d. What product is formed when A is oxidized with K₂Cr₂O₇?
5. e. Draw a constitutional isomer of A that contains an OH group.
6. f. Draw a constitutional isomer of A that contains an ether.

Interpretation Introduction

Interpretation:

For the below compound IUPAC name has to be determined.

Concept introduction:

Nomenclature of alcohol:

Firstly, find the longest carbon chain, which is bonded with the $\text{-OH}$ group. The alcohol has to be identified and named as –ol, in the suffix. In case of cyclic compounds, numbering should be given from $\text{-OH}$ group-attached carbon, which is considered as the lowest carbon. If any substituents other than $\text{-OH}$ group are present, it is necessary to mention their name in prefix and if more than one substituents are present, then naming should be in alphabetical order. In the name of the parent carbon chain ‘e’ is replaced by –ol. If more than, one alcohol is present then it is named as di, tri, tetra, etc after the parent name but before the –ol.

Explanation of Solution

Given compound A is,

Step 1: In the given compound, black, white and red color balls indicates carbon, hydrogen and oxygen atoms respectively. Firstly, the ring which, is attached to the $\text{-OH}$ group, is found and numbering is given from $\text{-OH}$ group-attached carbon, which is considered as lowest numbering carbon.

Step 2: For naming the compound, firstly, the parent carbon chain should be checked, and as it possess five membered ring, it is named as cyclopentane. Due to $\text{-OH}$ group present in the compound the last letter from the parent carbon naming ‘e’ is replaced with -ol. As one ethyl group is present in the third position of ring, it is named as 3-ethyl. Finally, the compound name is 3-ethylcyclopentanol.

Interpretation Introduction

Interpretation:

From the compound A, the type of alcohol has to be determined.

Concept introduction:

Classification of alcohol:

Generally, the alcohol group is bonded with minimum one alkyl group. The alcohol group is classified into three types, such as primary $(1^{\circ })$, secondary $(2^{\circ })$ and tertiary $(3^{\circ })$. It depends on the number of carbons attached to the single carbon atom bonded with the alcohol group. Primary $(1^{\circ })$ alcohol contains one carbon bonded with alcohol attached carbon atom. Secondary $(2^{\circ })$ alcohol contains two carbon atoms bonded with alcohol attached carbon atom. Tertiary $(3^{\circ })$ alcohol contains three carbon atoms bonded with alcohol attached carbon atom.

Explanation of Solution

In the below compound A, $\text{OH}$ group attached carbon atom is bonded with two carbon atoms, therefore it is secondary $(2^{\circ })$ alcohol.

Interpretation Introduction

Interpretation:

The product obtained when the compound A reacts with H₂SO₄ reagent has to be determined.

Concept introduction:

Dehydration of alcohol:

When alcohol reacts with ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$, it gives alkene as a product. This reaction involves elimination of water moleculeby breaking of bonds between two adjacent atoms (C-OH and C-H), which is called as a dehydration reaction.

Explanation of Solution

Given stating material A is 3-ethylcyclopentanol ${\text{(2}}^{\text{o}}\text{alcohol)}$,

When 3-ethylcyclopentanol reacts with ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$, elimination of water molecule from reactant and double bond formation on product side takes place resulting into 3-ethylcyclopent-1-ene and 4-ethylcyclopent-1-ene as products as shown below.

Interpretation Introduction

Interpretation:

The product obtained when the compound A reacts with K₂Cr₂O₇ reagent has to be determined.

Concept introduction:

Oxidation of alcohol:

During oxidation of alcohol, the number of $\text{C-O}$ bonds gets increased on product side and $\text{C-H}$ bond gets decreased on reactant side. In the oxidation reaction, reaction takes place between carbon attached to hydrogen atom and carbon attached to $\text{OH}$ group and finally C-O bond is formed leading to ${\text{H}}_{\text{2}}$ removal from the reactant. Oxidation of an alcohol gives different products depending upon the type of alcohols and reagents take part in a reaction. Oxidation of alcohols give carbonyl group in the product side. In a primary alcohol, among the two $\text{C-H}$ bonds present, the first $\text{C-H}$ bond gets oxidized into aldehyde group and then further it gets oxidized into carboxylic acid group. In case of the secondary alcohol, only the one $\text{C-H}$ bond present is oxidized into ketones. For tertiary alcohol, there are no hydrogen atoms present and therefore, tertiary alcohols are not been able to be oxidized.

Explanation of Solution

Given stating material A is 3-ethylcyclopentanol ${\text{(2}}^{\text{o}}\text{alcohol)}$,

When 3-ethylcyclopentanol, the secondary alcohol containing one hydrogen atom on the carbon attached with OH group reacts with K₂Cr₂O₇, it gets oxidized to 3-ethylcyclopentanone as shown below.

Interpretation Introduction

Interpretation:

The constitutional isomer of compound A that contain $\text{OH}$ group has to be determined.

Concept introduction:

Constitutional isomers:

The compounds, that contain same molecular formula but different with respect to molecular orientation. In other words compounds having similar molecular formula and different structure are called as Constitutional isomers.

Explanation of Solution

In the given compound A molecular formula is ${\text{C}}_{\text{7}}{\text{H}}_{\text{14}}\text{O}$, it contains one $\text{OH}$ group. This compound has one constitutional isomer as given below.

Interpretation Introduction

Interpretation:

The constitutional isomer of compound A that contain an ether group has to be determined.

Concept introduction:

Refer to part ‘e.’.

Explanation of Solution

In the given compound A molecular formula is ${\text{C}}_{\text{7}}{\text{H}}_{\text{14}}\text{O}$, it contains one $\text{O}$ atom. Therefore, only one constitutional isomer of ether containing compound is possible and it as given below.