Chapter 12, Problem 12.47AP

a.

Interpretation Introduction

Interpretation:

A secondary alcohol of molecular formula of ${\text{C}}_{\text{6}}{\text{H}}_{\text{14}}\text{O}$ 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 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 bonded with alcohol attached carbon atom. Tertiary $(3^{\circ })$ alcohol contains three carbon bonded with alcohol attached carbon atom.

b.

Interpretation Introduction

Interpretation:

An ether with molecular formula of ${\text{C}}_{\text{6}}{\text{H}}_{\text{14}}\text{O}$ that has a methyl group bonded to oxygen has to be determined.

Concept introduction:

Structure of ether:

Ether is an organic compound, it contains one oxygen atom bonded with two alkyl groups $\text{(R-O-R)}$. This oxygen atom contains two lone pairs of electrons. The ether structure is look like bent like structure shapes, same as shape of water molecule. The naming of ether structure is very simple. First need to identify the two alkyl groups attached with the oxygen atom, and the names are given in alphabetical order, finally the word ether is included. If two alkyl groups are same then di- should be added on the prefix.

c.

Interpretation Introduction

Interpretation:

A thiol that contain four carbons has to be determined.

Concept introduction:

Thiols:

Thiols contains $\text{-SH}$ group, it is called sulfhydryl group, and also it is a organic compound attached with tetrahedral carbon. This thiols are varied from alcohols, because alcohol contains $\text{O-H}$ bond, so it can make intermolecular hydrogen bonding.

d.

Interpretation Introduction

Interpretation:

Secondary alkyl bromide that contains five carbons has to be determined.

Concept introduction:

Classification of alkyl halides:

The organic molecules contains halogen atoms is called alkyl halides. Generally, the halogen atom is denoted as $\text{X}$, which includes $\text{X=}\text{F, Cl, Br, I}$. This halogen atom is attached with tetrahedral carbon atom. Depends on the number of carbon atoms attached with carbon bonded halogen atom it is classified into primary ${\text{(1}}^{\text{o}}\text{)}$, secondary ${\text{(2}}^{\text{o}}\text{)}$ and tertiary ${\text{(3}}^{\text{o}}\text{)}$ halides. For primary halides ${\text{(1}}^{\text{o}}\text{)}$, the halogen bonded carbon atom is attached with one carbon atom For secondary halides ${\text{(2}}^{\text{o}}\text{)}$, the halogen bonded carbon atom is attached with two carbon atoms. For tertiary halides ${\text{(3}}^{\text{o}}\text{)}$, the halogen bonded carbon atom is attached with three carbon atoms.