Organic Chemistry
Organic Chemistry
3rd Edition
ISBN: 9781119338352
Author: Klein
Publisher: WILEY
Question
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Chapter 1, Problem 66IP

(a)

Interpretation Introduction

Interpretation:

To find which two compounds are constitutional isomers of the given three structures

Concept introduction:

Constitutional isomers are the compounds which have same molecular formula but differ in connectivity between the atoms present in it.

The atomic orbitals mix together to form a hybrid orbital which is suitable for forming bonds between the atoms to form a compound.  This mixing of hybrid orbitals is known as hybridization.  The hybridization determines the geometry of the particular atom in the compound.  Shortly to say

sp3hybridized =tetrahedral

trigonal pyramidal(if one lone pair of electron is present)

bent geometry (if two lone pair of electrons are present)

sp2hybridized =trigonal planar

sp hybridized =linear

To find:

Constitutional isomers in the given three compounds.

(b)

Interpretation Introduction

Interpretation:

To find which compound has nitrogen in trigonal pyramidal geometry

Concept introduction:

A bond is formed by sharing of valence electrons.  A σ  bond is single bond that is formed between the atoms.  Double bond contains one σ  bond and one π bond.  Triple bond contains one σ  bond and two π bonds

To identify: Which compound has nitrogen atom with trigonal pyramidal geometry.

(c)

Interpretation Introduction

Interpretation:

To find which two compounds are constitutional isomers of the given three structures

Concept introduction:

A bond is formed by sharing of valence electrons.  A σ  bond is single bond that is formed between the atoms.  Double bond contains one σ  bond and one π bond.  Triple bond contains one σ  bond and two π bonds

To identify: Which compound has larger number of σ bonds.

(d)

Interpretation Introduction

Interpretation:

To find which two compounds are constitutional isomers of the given three structures

Concept introduction:

A bond is formed by sharing of valence electrons.  A σ  bond is single bond that is formed between the atoms.  Double bond contains one σ  bond and one π bond.  Triple bond contains one σ  bond and two π bonds

To identify: Which compound has lesser number of σ bonds.

(e)

Interpretation Introduction

Interpretation:

To find which two compounds are constitutional isomers of the given three structures

Concept introduction:

The atomic orbitals mix together to form a hybrid orbital which is suitable for forming bonds between the atoms to form a compound.  This mixing of hybrid orbitals is known as hybridization.  The hybridization determines the geometry of the particular atom in the compound.  Shortly to say

sp3hybridized =tetrahedral

trigonal pyramidal(if one lone pair of electron is present)

bent geometry (if two lone pair of electrons are present)

sp2hybridized =trigonal planar

sp hybridized =linear

To identify: Which compound has more number of π bonds.

(f)

Interpretation Introduction

Interpretation:

To find which two compounds are constitutional isomers of the given three structures

Concept introduction:

The atomic orbitals mix together to form a hybrid orbital which is suitable for forming bonds between the atoms to form a compound.  This mixing of hybrid orbitals is known as hybridization.  The hybridization determines the geometry of the particular atom in the compound.  Shortly to say

sp3hybridized =tetrahedral

trigonal pyramidal(if one lone pair of electron is present)

bent geometry (if two lone pair of electrons are present)

sp2hybridized =trigonal planar

sp hybridized =linear

To identify: Which compound has carbon atom with sp2hybridization .

(g)

Interpretation Introduction

Interpretation:

To find which two compounds are constitutional isomers of the given three structures

Concept introduction:

Boiling point of a compound is the temperature at which the compound start to escape into vapor phase from liquid state when temperature is raised.  The total number of carbon atoms determines the boiling point of the compound.

To identify: Which compound has all atoms with sp3hybridization except hydrogen atoms.

(h)

Interpretation Introduction

Interpretation:

To find which two compounds are constitutional isomers of the given three structures

To identify: Which compound has high boiling point.

Blurred answer

Chapter 1 Solutions

Organic Chemistry

Ch. 1.3 - Prob. 8PTSCh. 1.3 - Prob. 9PTSCh. 1.3 - Prob. 10PTSCh. 1.3 - Prob. 11ATSCh. 1.4 - Prob. 4LTSCh. 1.4 - Prob. 12PTSCh. 1.4 - Prob. 13PTSCh. 1.4 - Prob. 14ATSCh. 1.5 - Prob. 5LTSCh. 1.5 - Prob. 15PTSCh. 1.5 - Prob. 16ATSCh. 1.5 - Prob. 17ATSCh. 1.6 - Prob. 6LTSCh. 1.6 - Prob. 18PTSCh. 1.6 - Prob. 19PTSCh. 1.6 - Prob. 20ATSCh. 1.9 - Prob. 21CCCh. 1.9 - Prob. 22CCCh. 1.9 - Prob. 23CCCh. 1.9 - Prob. 7LTSCh. 1.9 - PRACTICE the skill Determine the hybridization...Ch. 1.9 - APPLY the skill Nemotin is a compound that was...Ch. 1.9 - Prob. 26CCCh. 1.10 - 1.8 PREDICTING GEOMETRY LEARN the skill Using...Ch. 1.10 - PRACTICE the skill Use VSEPR theory to Predict the...Ch. 1.10 - Prob. 28PTSCh. 1.10 - Ammonia (NH3) will react with a strong acid, such...Ch. 1.10 - Volatile organic compounds (VOCs) contribute to...Ch. 1.11 - Prob. 9LTSCh. 1.11 - Prob. 31PTSCh. 1.11 - Volatile organic compounds (VOCs) contribute to...Ch. 1.12 - Prob. 10LTSCh. 1.12 - Prob. 33PTSCh. 1.12 - Epichlorohydrin (1) is an epoxide used in the...Ch. 1 - Prob. 35PPCh. 1 - Prob. 36PPCh. 1 - Prob. 37PPCh. 1 - Prob. 38PPCh. 1 - Prob. 39PPCh. 1 - Prob. 40PPCh. 1 - Prob. 41PPCh. 1 - Prob. 42PPCh. 1 - Prob. 43PPCh. 1 - Prob. 44PPCh. 1 - Prob. 45PPCh. 1 - Prob. 46PPCh. 1 - Prob. 47PPCh. 1 - Prob. 48PPCh. 1 - Prob. 49PPCh. 1 - Prob. 50PPCh. 1 - Prob. 51PPCh. 1 - Prob. 52PPCh. 1 - Prob. 53PPCh. 1 - Prob. 54PPCh. 1 - Prob. 55PPCh. 1 - Prob. 56PPCh. 1 - Prob. 57PPCh. 1 - Prob. 58PPCh. 1 - Prob. 59PPCh. 1 - Prob. 60PPCh. 1 - Prob. 61PPCh. 1 - Prob. 62PPCh. 1 - Prob. 63PPCh. 1 - Prob. 64PPCh. 1 - Prob. 65PPCh. 1 - Prob. 66IPCh. 1 - Propose at least two different structures for a...Ch. 1 - Prob. 68IPCh. 1 - Prob. 69IPCh. 1 - Prob. 70IPCh. 1 - Prob. 71IPCh. 1 - Prob. 72IPCh. 1 - Prob. 73IPCh. 1 - Prob. 74IPCh. 1 - Prob. 75IPCh. 1 - Prob. 76IPCh. 1 - Prob. 77IPCh. 1 - Prob. 78CPCh. 1 - Prob. 79CPCh. 1 - Prob. 80CPCh. 1 - Prob. 81CP
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