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Five structural isomers have the formula C6H14. Where needed, draw the structure and give the IUPAC...

Five structural isomers have the formula C6H14. Where needed, draw the structure and give the IUPAC name of all five isomers.
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Concepts and reason

A chemical formula of a molecule is the symbolic representation of the number of atoms present in that molecule. It gives the proportion of atoms present in the molecule and not the arrangement of atoms that are present in the chemical formula. Isomers are the molecules which have same chemical formula but different structural arrangement. The structure of the molecule can be obtained by verifying the presence of the number of unsaturation (double or triple bonds) and rings in the chemical formula through DBE calculation.

Nomenclature of organic molecules can be done by following IUPAC rules. IUPAC stands for International union of Pure and Applied chemistry. The IUPAC name of an organic molecule consists of three parts namely prefix, root word, and suffix.

Fundamentals

• Double bond Equivalence (DBE) is calculated by the given formula:

DBE=C+1(H+XN2){\rm{DBE}}\,\,{\rm{ = }}\,\,{\rm{C}}\,{\rm{ + 1}}\,{\rm{ - }}\,\,\left( {\frac{{{\rm{H}}\,{\rm{ + }}\,{\rm{X}}\,{\rm{ - }}\,{\rm{N}}}}{{\rm{2}}}} \right)

Where Number of carbon atoms is denoted by C

Number of hydrogen atoms is denoted by H

Number of heteroatoms (other than carbon and hydrogen) is denoted by X

Number of nitrogen atoms is denoted by N

In general, alkanes do not contain any double or triple bonds.

• In IUPAC nomenclature, prefix represents the substituents present in the molecule.

Example :( methyl, ethyl, chloro, bromo etc...).

• Suffix represents the functional group present in the molecule.

Example: (ane for alkane, ene for alkene, yne for alkyne ol for alcohol, al for aldehyde etc.)

• Root word represents the longest carbon skeleton in the molecule.

The given chemical formula is C6H14.{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{{\rm{14}}}}{\rm{.\;}} .

Number of C present is 6.

Number of H present is 14.

The given chemical formula is C6H14.{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{{\rm{14}}}}{\rm{.\;}} .

Double bond Equivalence is calculated as follows.

DBE=C+1(H+XN2)=6+1(14+002)=77=0\begin{array}{c}\\{\rm{DBE}}\,\, = \,\,{\rm{C + 1}}\, - \,\left( {\frac{{{\rm{H + X}} - {\rm{N}}}}{{\rm{2}}}} \right)\\\\ = \,6\, + \,1\, - \,\left( {\frac{{14 + 0 - 0}}{2}} \right)\\\\ = \,\,7 - \,7\\\\ = \,\,0\\\end{array}

The possible structural arrangement of atoms for the given molecular formula C6H14.{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{{\rm{14}}}}{\rm{.\;}} is shown below.

CHз
Н2
с.
CH
Н2
с.
Н2
CHз
с
На
CHз
с
Нас
с
Нас
CHз
CHз
CHз
CHз
CHз
.CH
Нас.
Hас
Н2
Н2
Н2
CHз
CHз
CH
Нас
CHз
Figure 1
от

The longest carbon skeleton for the isomers is shown below.

Н2
Н2
CHз
6
1
3
5
с.
с.
4
Hас
С
2
Н2
Н2
оf
CHз
H2
C
2
Нас
CH 5
3
4
с
1
CHз
Н2

CH3
1
CHз
5
CH 4
HаС.
3
C
2 H2
Н2
CH3
CH3
CHз
С.
1
3
4
2
Нас
с
На

CH3
.CH 3
2
CH3
4
1
Hас
CH3

H2
H2
C.
1
3
CH3
6
4
C
H3C
C
2
H2
H2
No substituition
శలీల్ల
CH3
Н2
CH 5
3
4
1
CH3
Hас
Н2
1 methyl substituition

CH3
1
CH3
5
H2
CH 4
H3C
Hас.
3
2 H2
1 methyl substituition
CHз
.CH3
CHз
3
2
4
Hас
На
2 methyl groups substituited

CH3
3
CH3
4
.CH 3
2
Hас
CH3
2 methyl substituited

The substituted isomers are assigned with numbers as follows.

CH3
Н2
с.
CH
3
С
H2
5
2
4
CHз
Hас
2 - methyl pentane
CH3
1
CH 4
CH3
H3C
Нас.
5
3
C
2 H2
H2
3 methyl pentane

CH3
CH3
.CH3
C.
3
1
4
2
Нас
C
Н2
2,2 dimethyl butane
CH3
3
CH
CH3
1
2 H
Hас
CH3
2,3 dimethyl butane

Ans:

The structural isomers for the formula C6H14.{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{{\rm{14}}}}{\rm{.\;}} is

Н2
Н2
CHз
1
с.
6
4
с
На
Hас
с
2
Н2
Нехаne
и о
CH3
Н2
с.
CH
3
С
H2
5
2
4
CHз
Hас
2 - methyl pentane

CH3
1
CH 4
CH3
H3C
Нас.
5
3
C
2 H2
H2
3 methyl pentane
CH3
CH3
.CH3
C.
3
1
4
2
Нас
C
Н2
2,2 dimethyl butane

CH3
3
CH
CH3
1
2 H
Hас
CH3
2,3 dimethyl butane

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