Hydrocarbons are organic molecules made up of only carbon and hydrogen. If the hydrocarbon has only single bonds between the carbon atoms , it is said to be an alkane. If there is a double bond between two carbon atoms, then the molecule is said to be an alkene. When there is a triple bond between two carbon atoms, that molecule is called an alkyne.
Benzene is a special case because there are three alternating double bonds within the six carbon ring, presenting chemical and physical properties that are not possessed by either alkanes or alkenes. Hydrocarbons with a benzene like ring within them are said to be aromatic hydrocarbons
The forces that hold molecules together in a liquid, solid and solution phases are very weak. They are generally called London dispersion forces.
The electrons in the orbitals of a molecule are free to move around. If you could compare a "snapshot" of the molecule at an instant in time, you would see that there would be slightly different charge distributions caused by the different positions of the electrons in the orbitals. The amount of difference is based on the polarizability of the molecule, which is a measure of how well electrons can move around in their orbitals. In general, the polarizability increases as the size of the orbital increases; since the electrons are further out from the nucleus they are less strongly bound and can move about the molecule more easily.
When two molecules come together, these variations in charge can create a situation where one end of a molecule might be slightly negative and the other end of that molecule could be slightly positive. This would result in a slight attraction of the two molecules (until the charges moved around again) but is responsible for the attractive London dispersion forces all molecules have.
These London dispersion forces are weak, the weakest of all the intermolecular forces. Their strength increases with increasing size and polarizability of the molecule.
The rule to use when determining hydrocarbon solubility is: Like dissolves like.
This means that polar compounds (water, alcohols, and carboxylic acids) dissolve other polar compounds. Water can be broken down into H - OH, thus it has the -OH group which identifies alcohol and carboxylic acids. Nonpolar compounds dissolve other nonpolar compounds but tend not to dissolve polar compounds.
When you test for solubility you are looking for either a homogeneous solution or a heterogenous solution. Homogeneous solutions have no layers evident indicating the hydrocarbon being tested is soluble. These hydrocarbons are miscible. Heterogenous solutions have layers evident indicating the hydrocarbons are insoluble. These hydrocarbons are immiscible.
When you are trying to evaluate the solubility properties of alcohols and carboxylic acids, it becomes necessary to consider the relative sizes of the hydrocarbon and water-like portions of the molecule. The hydrocarbon portion is said to be hydrophobic (water hating) because it will not hydrogen bond with water but does tend to dissolve in hydrocarbon liquids. The water-like alcohol and carboxylic acid groups hydrogen bond with water and are said to be hydrophilic (water-loving).
If the ratio of the size of the hydrophilic portion to the hydrophobic portion is small, the hydrophilic portion is too small to carry the molecule into solution with water. If the ratio is large, it can carry the molecule into solution.
The solubility of alcohols and carboxylic acids in water is made smaller when the hydrophobic portion of the molecule is made larger.
When determining the solubility of a molecule there is one final rule.
The solubility of hydrogen bonding molecules is improved if either the positive charge on the hydrogen is made larger or the negative charge on the electronegative atom (oxygen or nitrogen) is increased.
Alkanes are considered saturated because they have only single C-C bonds and cannot add a hydrogen. Alkenes are unsaturated because they are capable of adding a hydrogen when the C=C double bond is broken.
The bromine test is used to determine if the colorless organic compound contains any double C=C bonds (the alkene functional group). Bromine does not react with an alkane because the alkane contains only single C-C bonds which cannot add the bromine. Alkanes merely dilute the red-brown bromine color to an orange or yellow color in the absence of a strong catalyst.
Due to their C=C double bonds which can be broken, alkenes react readily with bromine to produce saturated dibromoalkanes. When an alkene is reacted with bromine, the red-brown color of the bromine is immediately lost due to the reaction of the bromine.
Alkane + bromine (No strong light or heat) results in diluted solution colored orange or yellow indicating no reaction.
Alkane + bromine (heat or strong light acting as a catalyst) results in the brown red color of bromine slowly disappearing.
Alkene + bromine results in the red brown color of bromine rapidly or immediately disappearing giving a colorless solution
Aromatic hydrocarbon + bromine (no heat or light acting as a catalyst) results in no reaction and the red brown color of bromine is diluted to orange or yellow.
Aromatic hydrocarbons are too stable to react without a catalyst so they act like alkanes.
Red brown color of bromine
Bromine Test (left) No reaction, saturated, (right) Reaction, unsaturated
If an alkene is present, the π bond breaks and bromine is added in two places.
Cholesterol has only one -OH group which is polar. This single alcohol functional group does not make it polar enough to dissolve in water or blood (made up of mostly water). Since cholesterol is not polar enough to dissolve in water it will dissolve in nonpolar solvents. Cholesterol when tested with the bromine test will react, losing the red-brown color of bromine and turning colorless. This is due to the one double (π) bond. This double bond can break allowing bromine to be added, indicating an alkene is present and cholesterol is unsaturated allowing the red brown color of bromine to slowly disappear.
Red brown color of bromine slowly disappears = unsaturated.
Red brown color of bromine dilutes to yellow or orange = saturated.
Again bromine is used to determine whether an alkane, alkene, or aromatic hydrocarbon is present. If the substance is an alkene it will react with the bromine, indicating the compound is unsaturated. When the red-brown color of bromine disappears it indicates the substance is unsaturated and a reaction has occurred.
ALKANES contain only σ (single C-C) bonds and cannot add any Br so red brown color of bromine dilutes to yellow or orange = saturated.
If a catalyst (stong or intense light or heat) is used with an ALKANE one H is removed and one Br is added. The color of the bromine will SLOWLY disappear.
ALKENES contains a π bond making it possible to add Br so the red brown color of bromine disappears rapidly or immediately = unsaturated.
Alkenes add two bromines.
AROMATIC HYDROCARBONS are extremely stable and will not react without a catalyst = saturated.
If a catalyst is used with an AROMATIC HYDROCARBON, no C=C (double bonds) are broken. The aromatic hydocarbon will simply replace one H with one Br.
Bromine is a red-brown color. To determine unsaturation or saturation, ask yourself "What does the bromine do?". If the red-brown color goes colorless, then you have UNSATURATION.
Red brown color of bromine
Red-brown color of bromine is diluted to yellow or orange (no reaction), the substance is SATURATED.
Red brown color of bromine disappears (reaction), the substance is UNSATURATED.
Bromine Test (left) No reaction, saturated, (right) Reaction, unsaturatedDigital Video from DVAction of the Bromine Test for Alkenes
The Baeyer reagent is a cold dilute aqueous solution of potassium permanganate which is a deep purple color. Potassium permanganate does not react with alkanes because they are saturated (single bonds which are all taken). When it is added to alkanes the purple color does not change. However, when it is added to an alkene, the purple color slowly disappears and a brown MnO2 precipitate forms. The appearance of the brown precipitate indicates a positive test for unsaturation. The Baeyer test for unsaturation is used when the color of the organic compound may interfere with the result of the Bromine Test for Unsaturation.
Purple color dilutes to light purple = SATURATED.
Purple color disappears and brown ppt forms = UNSATURATED.
The purple color of KMnO4
Baeyer Test for Unsaturation (left) purple diluted to light purple: saturated, (right) purple disappears, brown preciptate forms: unsaturated.
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last updated: January 28, 2015