All Advanced Organic Chemistry Notes

Case study 1b.1

(a) Positional isomers of C₂H₄X₂ and C₂H₃X₃ where X = halogen

In all cases there are differences in physical properties e.g. different boiling points and liquid densities.

The molecular formula C₂H₄X₂ will give rise to two positional isomers i.e. 1,1-di ... and 1,2-di ...

1,1-dichloroethane  and  1,2-dichloroethane

Boiling points: 57.3  and  83.7^oC,  densities: 1.178  and 1.253 g/cm³

1,1-dibromoethane  and  1,2-dibromoethane

Boiling points: 110  and  132^oC,  densities: 2.055  and  2.180 g/cm³

 

The molecular formula C₂H₃X₃ also gives rise to two positional structural isomers

e.g. 1,1,1-trichloroethane  and  1,1,2-trichloroethane  where X = chlorine Cl

Boiling points: 74  and  ~112^oC,  densities: 1.320  and 1.435 g/cm³

Once an alkane has at least 3 carbon atoms, substituent groups e.g. halogen or amino groups, can take up different positions on the carbon chain.

In the uv light catalysed reaction of bromine and propane gases, the free radical substitution reaction can produce two initial mono-substitution products. [full mechanism]

CH₃CH₂CH₃ + Br₂ {CH₃CH₂CH₂Br or CH₃CHBrCH₃} + HBr

(1) 1-bromopropane, , , bpt 71^oC, primary halogenoalkane,

(2) 2-bromopropane,  , , bpt 59^oC, secondary halogenoalkane, 

Only two isomers are possible. They are both low boiling colourless liquids, but the more compact molecule (2) has a lower boiling point - similar physically, but differences in boiling point.

Chemically they are very similar e.g. both undergoing all the nucleophilic substitution reactions with ammonia, cyanide ion, and hydroxide ion etc.

In the case of the latter, (1) would give the primary alcohol, propan-1-ol and (2) would give the secondary alcohol, propan-2-ol - these are important different, if similar outcomes, from the same reaction.

[lots of named halogenoalkane structures]

For higher bromoalkanes e.g. 1-bromobutane CH₃CH₂CH₂CH₂Br and 2-bromobutane CH₃CH₂CHBrCH₃, another chemical difference will show up on refluxing them with ethanolic potassium hydroxide, by which, following an elimination reaction,

1-bromobutane can only form but-1-ene CH₃CH₂CH=CH₂, but 2-bromobutane can form two isomeric elimination products:

CH₃CH₂CH₂CH₂Br  +  KOH  ===> CH₃CH₂CH=CH₂  +  KBr  +  H₂O

but-1-ene CH₃CH₂CH=CH₂, and but-2-ene CH₃CH=CHCH₃, i.e. you can eliminate either side of the C-Br bond.

CH₃CH₂CH₂CH₂Br  +  KOH  ===> {CH₃CH₂CH=CH₂  and  CH₃CH=CHCH₃} +  KBr  +  H₂O

 

See also other positional isomers of saturated mono-halogenated alkanes (haloalkanes)

Isomers of molecular formula C₄H₉X  (where X = F, Cl, Br or I)

Isomers of molecular formula C₅H₁₁X  (where X = F, Cl, Br or I)

Isomers of molecular formula C₆H₁₃X  (where X = F, Cl, Br or I)

These sets of isomers overlap with carbon chain isomerism.

In this case it is the different position of the C=C alkene functional group give rise to two structural isomers.

(1) , pent-1-ene, bpt 30^oC, 

(2) , pent-2-ene, cis bpt 37^oC, trans bpt 36^oC,

They are very similar physically e.g. relatively non-polar volatile colourless liquids, with similar low boiling points. Chemically similar e.g. all the usual electrophilic addition reactions of any alkene, though may, or may not be, some 'isomeric consequences' as regards both their formation or addition reaction products and some examples are outlined below.

Unlike pent-1-ene, pent-2-ene can also exist as E/Z (cis/trans isomers)

Both can be formed in cracking pentane or higher alkanes in which various isomers of C₅H₁₀ would be formed. In the laboratory they can be made by elimination reactions e.g.

(a) the 'dehydration' of isomeric pentanols  with conc. sulfuric acid or

(i) CH₃CH₂CH₂CH₂CH₂OH ==> CH₃CH₂CH₂CH=CH₂ + H₂O

Pentan-1-ol (above) can only give 1 isomer, pent-1-ene,

(ii) CH₃CH₂CH₂CHOHCH₃ ==> {CH₃CH₂CH₂CH=CH₂ or CH₃CH₂CH=CHCH₃} + H₂O

but pentan-2-ol (above) can give 2 isomers, pent-1-ene and pent-2-ene, because elimination of a -H (as well as the -OH) can take place either side of the >CH-OH group from an adjacent C-H.

This is not possible with pentan-1-ol with the -OH group on the end carbon.

(b) or by  refluxing with ethanolic potassium hydroxide to give an elimination of HBr reaction. The formation of more than one isomer of the pentenes depends on the position of the -OH in alcohols or the -Br in bromoalkanes e.g.

(i) CH₃CH₂CH₂CH₂CH₂Br + KOH ==> CH₃CH₂CH₂CH=CH₂ + H₂O + KBr

1-bromopentane can only give 1 isomer, pent-1-ene (above), but 2-bromopentane (below)

(ii) CH₃CH₂CH₂CHBrCH₃ + KOH ==> {CH₃CH₂CH₂CH=CH₂ or CH₃CH₂CH=CHCH₃} + H₂O + KBr

can give two isomers, pent-1-ene and pent-2-ene, because elimination of a -H (as well as the -Br) can take place either side of the >CH-Br group from an adjacent C-H.

This is not possible with 1-bromopentane with the -Br group on the end carbon.

 

You can also derive many other isomers from the molecular formula C₅H₁₀ e.g. methylbutenes (chain/positional isomers with respect to pentenes), methylcyclobutane and dimethylcyclopropanes (both chain/functional group isomers with respect to pentenes).

[lots of named alkene structures], [named cyclo-alkane structures] or [halogenoalkane structures]

Other alkene positional isomers e.g.

, but-1-ene

and  , but-2-ene (the latter can also exhibit E/Z isomerism (cis/trans)

 

See also other examples of positional isomers

Isomers of molecular formula C₄H₈

Isomers of molecular formula C₅H₁₀

Isomers of molecular formula C₆H₁₂

Isomers of molecular formula C₇H₁₄

which overlap with carbon chain and functional group isomerism.