Alkyl, Alkenyl and Alkynyl Substituent Groups The naming of prefix for alkyl, alkenyl and alkynyl groups are pretty simple. It follows a simple formula. We have already explored the appropriate prefix based on the number of carbons in the table above labelled “Naming unbranched Alkanes”. E.g. Hydrocarbons (Alkanes or Alkenes or Alkynes) with one carbon has prefix “Meth” E.g. Hydrocarbons with two carbons has prefix “Eth”, three carbons has prefix “Prop”, etc. Prefix of alkyl substituent groups = Appropriate prefix based on number of carbon atoms + “yl”. Prefix of alkenyl substituent groups = Appropriate prefix based on number of carbon atoms + “enyl”. Yes, methylene is an exception as it doesn’t follow the same format of adding ‘enyl‘ after meth. Prefix of alkynyl substituent groups = Appropriate prefix based on number of carbon atoms + “ynyl”. NOTE: We can express the organic molecules’ molecular formula using stick structures as depicted using pink lines in diagram above. We deliberately drew yellow dots on the line structure to depict how, at every yellow dot, there is a carbon atom. You should NOT draw the yellow dots in your line structure as we only did it here to get the idea through to you. In line structures, we DO NOT show hydrogen atoms that are attached to the carbon atoms (yellow dots in diagram) as that’s the purpose of line structures to make it faster and easier to draw, read and communicate the chemical structure of organic molecules. Learning Objective #1 - Investigate the IUPAC nomenclature for alkenes Now let’s move onto naming alkenes! These are organic molecules with a carbon double bond, (C = C). Basically, to name alkenes, we use the same procedure that we have outlined above for alkanes. However, there are few exceptions or things that you need take into consideration when naming alkene molecules. These important exceptions are for alkenes are: The parent carbon chain must contain the double-bond(s) (C=C) functional group. Like all alkenes molecules, there is an ‘ene‘ after the root-name of the alkene molecule. The C=C functional group has highest priority in alkene molecules. Therefore you should assign “1” to the the first carbon you encounter that makes up the C=C bond from one end of the parent carbon chain. Example: But-2-ene, where the “2” indicates that the carbon double bond (C=C) is found on the second carbon atom in butene molecule. Notice that you include the position (“2”) of the carbon double bond (C=C) before you write the suffix for alkene “ene” and it separates using dash symbols “-” as per IUPAC conventions. NOTE: Keep in mind that you must assign the least carbon number as possible to all of your substituent groups. So, check your work once you are finished! There are special cases when you are doing with cyclic compounds but these exceptions NOT necessary HSC Chemistry. You will learn in Higher Chemistry 1A if you decide to take Chemistry university. Learning Objective #1 - Investigate the IUPAC nomenclature for alkynes Moving on to the next station, alkynes! These are organic molecules with a triple carbon bond, (C ≡ C). To name alkynes, we use the same method that we used to name alkanes. However, again, there are some exceptions that you will need to evaluate. The parent carbon chain must contain the triple bond(s) (C ≡ C) functional group. Like all alkynes molecules, there is an ‘yne‘ after the root-name of the alkyne molecule. The C≡C functional group has the highest priority in alkyne molecules. Therefore, you should assign “1” to the first carbon you encounter that makes up the C ≡ C bond from one end of the parent carbon chain. In an organic molecule that contains both double and triple carbon bonds, they have equal priority. This means whichever of the two functional group appears on first from one side of the parent carbon chain, it will have the lower assigned carbon number. As mentioned in the alkane section, if the double and triple carbon bonds are in at equal carbon atom distance from each side of the parent carbon chain, we use assign the lowest carbon number according to alphabetical order, the carbon attached to the double bond (-ene-) will be assigned a lower carbon number than the carbon atom attached to the triple bond (-yne-) as e comes before y in the alphabet. NOTE: Keep in mind that you must assign the least carbon number as possible to all your substituent groups. So, check your work once you are finished! Moving into alcohols, aldehydes, ketones, carboxylic acids, amine, amides & halogenated compounds For alkanes, alkenes, alkynes and halogenated compounds, the substituent groups, i.e. alkyl, alkenyl, alkynyl and halogen groups, have equal priority and thus are named according to the alphabetical order of their substituent group’s prefix. However, for the following compounds, the highest priority substituent (or functional group to be more specific) are based on the following functional group priority naming order: Highest priority Side Carboxylic acid group Amide group Aldehyde group Ketone group Alcohol group Amine group Lowest priority Side Alkene (C=C) and alkynes (C≡C) have equal priority as we have mentioned before and they both have lower priority than amine groups. This is, of course, followed by the following substituent groups have equal priority and thus are named in the order in an alphabetical order which we have mentioned: Alkyl, Alkenyl, Alkynyl groups and halogen groups NOTE: Halogens are not treated as functional group in organic chemistry nomenclature. However, when the question is not concerning nomenclature, halogen are treated as functional group as they give chemical properties to the organic molecule. NOTE: There are many more other functional groups that fits between these groups but they are not relevant for HSC Chemistry purposes as they’re outside the scope of what the syllabus required us to know. Learning Objective #1 - Investigate the IUPAC nomenclature for alcohols A parent carbon chain with hydroxyl group as the highest priority substituent group is known as an alcohol. To name an alcohol, we can use the following the steps: Step 1: Identify the parent carbon chain, i.e. the longest, continuous carbon chain, that comprises of the highest priority functional group. In this case, the hydroxyl group (- OH) is the highest priority group. Step 2: Assign carbon numbers to every atom in the parent carbon chain with the carbon atom that is bonded to the highest priority group where the highest priority functional group have the the lowest carbon number. Step 3: Name any double and triple bonds in a similar fashion as we have discussed previously. Step 4: Since the parent carbon chain has hydroxyl group as the highest priority group, it is an alcohol. The organic molecule that is an alcohol has the suffix ‘ol’. Therefore, we need to replace the ‘e‘ at the end of the root-name of the organic molecule with the relevant suffix. In this case of alcohols, the suffix after the root-name would be ‘ol’. Again, like other cases, there’s no space between root-name and suffix. Include the carbon number too before the suffix using dash symbols “-” to signal where the main functional group is located. Finally, use the relevant prefix based on carbon number. E.g. Two carbons = ‘eth’. Three carbons = ‘Prop’. Step 5: Name any other other substituent groups (disregarding priority order now) that are present in the organic molecule in alphabetical order, according to their prefix names, and assigning lowest carbon position number to every substituent group as possible. Learning Objective #2 - Explore and distinguish the difference types of structural isomers, including saturated and unsaturated hydrocarbons including: Chain Isomers Alright, let’s explore the first type of structural isomers – chain isomers! We can say that two or more molecules are chain isomers if they satisfy the following conditions: They must have the same molecular formula. They must differ in their connectivity of carbon atoms to the parent carbon chain.* *This means that these isomers have different parent carbon chain lengths, hence they have the name being chain isomers. We are going to the next stop to explore another type of structural isomers, position isomers In order for two or more molecules to be considered position isomers, they need to meet the following criteria: Like all structural isomers, position isomers must have the same molecular formula. Unlike chain isomers, position isomers have the same carbon structure in terms of parent carbon chain length. One or more substituent and/or functional groups must differ in their connectivity to the carbon atoms along the parent carbon chain.* *This effectively means that these isomers needs to have one or more of their substituent and/or functional groups being in a different position along the carbon structure. Hence, they have the name being position isomers. Recall that from the table that we examined in Learning Objective #4, where we showed how double and triple bonds between carbon atoms are functional groups. So, isomers that differ in their position of double and triple carbon bonds in their carbon structure can be considered position isomers (provided that the molecules satisfy all of other criteria for position isomers outlined above). Functional isomers: Again, there are a set of conditions that two or more molecules must satisfy in order for them to be considered functional group isomers. These are: Like all structural isomers, they must have the same molecular formula. They may or may not have the same parent carbon chain length, depending on arrangement of atoms. Due to the isomers differing in their arrangement of atoms, (thus, NOT limited to only substituent and functional group like for positional isomers), this results in the isomers having different functional groups.* *Hence their name, functional group isomers. Variation in physical and structural properties between different types of structural isomers Chemical properties Structural isomers would have the same chemical property if they have the same functional group(s) in their structural formula.* Physical properties The following guidelines will help you compare the melting and boiling points between different isomer molecules. As you will explore with polymers which is part of the last Inquiry Question in Module 7, you learn that molecules (e.g. isomers) that exhibit a less branched (or linear) structural formula have a higher melting and boiling point. Therefore, you would expect the more chain branching that a molecule exhibits, the lower its melting and boiling point will be. This is because chain branching prevents the isomers from packing their carbon chains closely together and, thus, the intermolecular forces (e.g. dispersion forces) between them would be weaker. As we have explored in Preliminary HSC Chemistry, intermolecular forces play an important role in governing the melting and boiling point of molecules. As for functional group isomers, the isomers have different functional groups allowed differing extent of chain branching (thus affecting dispersion forces) and maybe different degree of dipoledipole and hydrogen bonding. Therefore, their melting and boiling point (physical property) may vary. Example: Alcohols (- OH) and carboxylic acids (- COOH) have different melting and boiling points due to different functional groups. Learning Objective #2 - Conduct an investigation to compare the properties of organic chemical compounds within a homologous series, and explain these differences in terms of bonding. General formula for: Alcohol Homologous Series is: Cn(2n+1)OH General formula for Aldehyde Homologous Series is: CnH(2n+1)CHO General formula for Ketone Homologous Series is: CnH(2n+2)CO General formula for Carboxylic acids Homologous Series is: CnH(2n+1)COOH General formula for organic amide homologous series: R-CO-NH2 General formula for organic amine homologous series: R-NH2 Homologues in the Alkane Homologous Series (without substituent groups) As we have learnt in the HSC Preliminary Chemistry Course, the type (and thus strength) of intermolecular forces plays a major role in determining the melting and boiling point of molecules. This is not limited to just alkanes but for all the compounds in each class of homologous series which we will explore soon. Here, we will talk about alkanes without substituent groups. We have already explored the structure of unbranched alkanes without substituent groups in last week’s notes as well as in this week’s learning objective #1. They only have of C-C and C-H bonds which are non-polar bonds due to the electronegativity difference between C-C is basically zero and the electronegativity difference between carbon and hydrogen in the C-H bond not being high enough to be considered a polar covalent bond. Recall that we have already learnt the electronegativity difference ranges that allows us to classify non-polar, polar and ionic bonds in Preliminary HSC Chemistry. For such reason, as both C-C and C-H bonds are non-polar in addition to the fact that unbranched alkanes are linear, it allows uniform distribute of any dipole moments, it is justifies why unbranched alkane molecules without substituent groups are non-polar. Since they are non-polar, it is only dispersion forces being the intermolecular force that is determining the melting and boiling points of unbranched alkanes without substituents. NOTE: We have already talked about dispersion forces in Preliminary HSC Chemistry, so we expect you to know how they are created. But, just for a quick recap on what they are, dispersion forces are intermolecular forces created due to the uneven distribution of electrons, thus even electron density, around the nucleus of an atom. This therefore results in a small dipole that is formed where one side of the atom is slightly positive and the other side being slightly negative. This dipole induces a similar dipole in a neighbour atom, resulting in a dispersion force being created between the two atoms. This means alkanes with larger molecular mass will have more atoms (carbon and hydrogen) and hence will have a greater number of electrons. Thus, the degree (and thus strength) of dispersion force formed between larger alkane molecules would therefore be greater than between smaller alkane molecules. So, a higher amount of energy is required to break the intermolecular forces (dispersion forces only between alkane molecules) in alkanes with higher molecular mass. Therefore, as the number of carbon and hydrogen atoms increase (i.e. alkane molecule increases in molecular mass – such as going from methane to octane), the melting and boiling point will increase. As you are aware from Preliminary HSC Chemistry, hydrocarbons from C1 to (and including) C4 are gases at room temperature. Hydrocarbons between C4 – C8 are liquids at room temperature. Branched vs Unbranched Alkanes For two alkanes, one being branched and the other unbranched, both with the same molecular formula and thus molecular mass. The unbranched alkane will exhibit higher melting & boiling point. The reason for this is because unbranched alkanes are more linear whereas branched alkenes are less linear (or more spherical or more bulky) in shape. For such reasons, the carbon chains between unbranched alkane molecules are able to pack more closely together than the carbon chains between branched alkanes. Therefore, a higher surface contact area means that the extent of dispersion force is greater between unbranched alkanes as more atoms’ electrons are able to interact with each other to induce dipoles. This results in unbranched alkanes having a higher melting & boiling point. Another differing physical property other than melting & boiling point between alkanes of different molecular weight is their density. Their density increase as the molecular weight increases, however, it doesn’t exceeds the density of water. Obviously, another density trend is that, since branched alkanes have their carbon chains packed more closely with each other, they have higher density than unbranched alkanes. With that, we conclude the physical property differences between homologues in the alkane homologous series! General physical properties of all alkanes share are: Alkanes are not soluble in polar substances but soluble in non-polar substances (the reason for this is already covered in HSC Preliminary Course, so we won’t re-explain here). Alkanes are colourless and odourless. NOTE: As we have mentioned previously, the chemical properties shared between homologues in a homologous series such as alkane are the same since there is no different atoms or functional groups. Homologues in the Alkene Homologous Series (without substituent groups) The differences between alkene molecules of increasing molecular mass is similar to the case of alkanes that we described above already. So, we will only do a very brief summary. Very brief summary (as the case of alkenes is similar to alkanes which we explained in detailed above): As the molecular mass of alkenes increases, the strength of dispersion forces increases and thus results in a trend of increasing melting & boiling point with increased molecular mass. It is important to note that alkenes have a lower melting and boiling point than their corresponding alkanewith the same number of carbon atoms. This is because alkenes have a lower molecular mass, lowering the strength of dispersion force. For the same reason, alkenes have a lower density than corresponding alkanes with the same number of carbon atoms as alkenes have a lower molecular mass. However, both alkanes and alkenes have less density than water as well as being insoluble. Similar to alkanes, the alkenes from C1 up to and including C4 exist in the gas state at room temperature whereas alkenes that is C5 – C8 are liquids. In general, apart from melting & boiling point, alkenes share similar physical properties to alkanes are they are comprised of the same types of atoms, carbon and hydrogen only. The chemical properties shared between alkene homologues are same as there is no difference in functional groups. Homologues in the Alkynes Homologous Series (without substituent groups) The physical properties are the same as alkenes and alkanes. However, alkynes have lowest melting & boiling point out of the three due to lower molecular mass than corresponding alkane and alkene hydrocarbons with the same amount of carbons. Alkyne homologues share similar with each other chemical properties as they all have a triple bond between carbon atoms. Difference in chemical properties between Alkanes, Alkenes & Alkynes withoutsubstituent groups As alkenes and alkynes have double and triple bonds between carbon atoms, they are more reactive than alkanes. This is because there is higher electron density in a double and triple bond region compared to single bonds meaning that breaking a C-C bond from double bond and triple bond between carbon atoms require less energy to break compared to breaking the one and only single carbon-carbon bond between carbon atoms. This therefore allows alkenes and alkynes to undergo addition reaction whereby a carbon-carbon bond is broken so that a substituent group or atom can be added to each carbon. Alkanes cannot undergo addition reactions but instead undergoes substitution reactions as they only have a single bonds, i.e. C-C and C-H. A common test for alkanes and alkenes is bromination (a type of halogenation reaction whereby Br2 is added to the organic compound). What happens is that alkanes will not react with the bromine molecule unless under the presence of high temperature or U.V light as discussed last week under substitution, halogenation reaction conditions. Comparatively, addition reaction occurring between alkenes and alkynes with bromine does not need high energy due to lower bond energy of breaking of one of the several C-C in a double bond or triple bond between carbons atoms. Therefore, the bromine molecule will be incorporated into the alkenes and alkyne molecules via addition reaction at room temperature. Since bromine has a red-brown colour and the hydrocarbons (e.g. alkane, alkene, alkynes) are colourless, the effect is that when bromine undergoes addition reaction or substitution reaction, there will no longer be bromine molecules in solution and thus the solution will become colourless after reaction. For alkynes, it basically is the same as alkenes in terms of chemical properties but they are more reactive than alkenes (and alkanes) as two C-C bonds can be broken with lower energy in its triple carbon bonds between carbon atoms. Therefore, alkynes can undergo addition reaction twice to be compared into an alkene which can only undergo only one addition reaction. In general, this means that alkynes are less stable than alkenes which are less stable than alkanes. In order of acidity, alkynes are the most acidic: i.e. alkynes > alkenes > alkanes. That is alkynes are the strongest acid as they are readily donate their hydrogen atom in the form of H+. The reason towards this trend of acidity is because the resulting lone electron pair on the carbon atom, after a hydrogen atom is donated is situated in the sp orbital compared to the sp2 in alkenes and sp3 in alkanes. Recall from Preliminary HSC Chemsitry: Sp orbital have 50% s character and 50% p character. Sp2 has 1/3 s character and 2/3 p character. Finally Sp3 orbitals have 25% s character and 75% p character. As electrons that are closer to the nucleus (more s character) are more stable than electrons found further away from the nucleus (less s character) due to higher effective nuclear charge, the reaction of alkynes acting as a Bronsted-Lowry Acid by donating a H atom occurs more readily compared to alkenes and alkanes as the conjugate base formed is more stable. This is because the conjugate base’s lone pair of electron have greater s character in alkynes than alkenes and alkanes which experiences a higher effective nuclear charge and thus is more stable as the electrons are closer to the nucleus which is positively charged. Finally, alkanes have higher chance of undergoing complete combustion than alkenes which have higher chance than alkynes. This is because the ratio between carbon to hydrogen is higher in alkynes than alkenes which is higher than alkanes so more oxygen is required to fully oxidise the carbon atoms. Most of the oxygen are used to form water and any remaining is used to form either carbon dioxide (two oxygen per carbon) or carbon monoxide (one oxygen per carbon), where carbon monoxide is favoured if there is limited availability of oxygen. Homologues in the Alkyl halides (or Haloalkane) Series The electronegativity difference between the halogen and the carbon atoms makes the C-X bond polar, where X represents a halogen atom. Due to this, there are dipole-dipole forces between haloalkane molecules as well as dispersion forces compared to only dispersion forces present between molecules in the three hydrocarbon homologous series (alkane, alkenes and alkynes). For such reasons, haloalkanes have stronger intermolecular forces than corresponding alkanes/alkenes or alkynes with the same number of carbon atoms. Again, similar to other homologous series, as haloalkanes increases in alkyl group chain length, the molecular mass will increase. This results in higher extent of dispersion forces due to greater number of electrons, resulting in stronger intermolecular forces. Another similar trend is that, as the length of the non-polar alkyl group increases, the solubility of haloalkanes in polar substances decreases due to increasing non-polar nature of the overall alkyl halide. Do note alkyl halides cannot perform hydrogen bonding with water molecules so they are not readily soluble in water. So, alkyl halides are less soluble than corresponding ketones, aldehydes, alcohols, amides and amines with the same number of carbon atoms (similar molecular mass) that can perform hydrogen bonding with water molecules. A thing to note about alkyl halides is that as its molecular mass increases (i.e. moving from C1 to C8), the density of the homologues in the series decreases (measured at a fixed temperature) as the ratio between mass to volume for density decreases. However, if the alkyl chain is the same length between two haloalkanes, the one with the heavier or higher molecular mass halogen will result in the haloalkane being more denser. Most alkyl halides exists as a liquid at room temperature except for few of those with low molecular mass such as fluoromethane and chloromethane. Homologues in the Ketone Homologous Series Compared to aldehydes, ketone molecules have a pleasant and sweet odour. Due to the electronegativity difference between the oxygen and carbon atom in the C=O group, the carbonyl group is polar. This essentially gives ketone molecules their polar properties. The main intermolecular force that determines the melting & boiling point of ketone molecules are dipole-dipole forces. Obviously, dispersion forces are present amongst the molecules too. Ketones have a higher boiling point than aldehydes because ketones have two alkyl (R) groups attached a central carbon atom where each alkyl group has a greater electron density than a hydrogen atom. Comparatively, aldehyde only has one alkyl (R) group and one hydrogen atom attached to a central carbon atom. Recall that since the C=O bond is polar due to the oxygen atom being more electronegative than carbon atom. This means that oxygen is partially positive and the carbon atom is partially negative. This means that two alkyl groups (each has higher electron density than a hydrogen atom) will attracted and be located closer to the central carbon atom. This effectively makes the central carbon atom in ketones more negatively charged, resulting ketones having a more polarised C=O bond compared to that C=O bond in aldehydes. A more polarised C=O bond in ketones means that the dipole-dipole forces between ketone molecules are stronger than those between aldehyde molecules. This means that more energy is required to break the dipole-dipole intermolecular forces between ketone molecules than between aldehydes. For such reason, ketones have a higher boiling point than aldehydes. Although the ketones with one to three carbons (C1 to C3) are very soluble in water, as the molecular mass increases (i.e. moving into C4 and beyond), the length of the non-polar hydrocarbon chain increases. This results in ketone molecules experiencing a gradual decrease in water solubility as molecular mass increases, resulting in the non-polar alkyl chain/group increasing in length. Like all of the previous cases, as the number of carbon and hydrogen atoms increases (i.e. molecular mass) for homologues in the ketone homologous series, the strength of dispersion force increases. Thus, the melting & boiling point increases. Homologues in the Aldehyde Homologous Series Compared to ketones, aldehydes have an unpleasant and strong/sharp odour. Aldehydes are polar for the same reason (for the most part) as Ketone as explained in the ketone section. The main intermolecular force between aldehyde molecules are dipole-dipole forces, followed by dispersion forces. If you pursue higher chemistry at university, you will learn that aldehydes can in fact hydrogen bond with themselves at certain configurations but that’s outside the scope of HSC Chemistry. To limit to the scope of HSC Chemistry scope, aldehydes and ketones cannot form hydrogen bonds with each other. They can form hydrogen bonds with other molecules such as water though! You can try draw out the structures to see how ketones and aldehyde can hydrogen bond with water as we have learnt about how hydrogen bonding can occur in different situations in Preliminary HSC Chemistry. Aldehydes have a lower melting and boiling point than ketones where the reason why has been explained in the previous section on ketones. Although the aldehydes with one to three carbons are very soluble in water, as the molecular mass increases (i.e. moving into C4 and beyond), the length of the non-polar hydrocarbon chain increases. This results in aldehydes experiencing a gradual decrease in water solubility as molecular mass increases, resulting in the non-polar alkyl chain/group increasing in length. Like all of the previous cases, as the number of carbon and hydrogen atoms increases (i.e. molecular mass) for homologues in the aldehyde homologous series, the strength of dispersion force increases. Thus, the melting & boiling point increases. Homologues in the Alcohol Homologous Series Like homologues in any homologous series, the homologues in the the alcohol homologous series differ by one carbon and one CH2 unit. This difference in molecular size that gives homologues in the alcohol homologous series their different physical properties. Since all homologues in the alcohol homologous series have one OH functional group, their chemical properties are the same. The main and strongest intermolecular force between homologues in the alcohol homologous series is hydrogen bonding. Of course, as you have learnt in Preliminary HSC Chemistry, molecules of a compound that are capable of hydrogen bonding amongst each other means that they will also have dipole-dipole as well as dispersion forces. Here, the learning objective is concerned about the properties between homologues in the alcohol homologous series. We know that the intermolecular forces between molecules of a compound plays a major role in determining the physical compound’s properties from Preliminary HSC Chemistry. Hydrogen bonding is stronger than dispersion forces as the electronegative difference between the atoms involved in creating the intermolecular force is greater than in generic dipole-dipole forces. Quick recap from Preliminary HSC Chemistry on hydrogen bonding: For a hydrogen bond to be formed, a highly electronegative atom must be added to a hydrogen, creating a slightly negative charge on the more electronegative atom and slightly positive charge on the hydrogen. The slightly positive charged hydrogen atom can perform hydrogen bonding with another electronegative atom (slightly charged) from another molecule as long as the electronegativity difference is strong enough. For such reason, alcohols have higher melting & boiling point than the hydrocarbon homologous series (i.e. alkanes, alkenes and alkynes) that do not have hydrogen bonds and only dispersion forces. Another factor that contributes towards alcohol’s higher boiling point is due to the existence of a hydroxyl group (OH) that replaced a hydrogen in the hydrocarbon homologous series (e.g. in alkanes). As you have learnt in Preliminary HSC Chemistry, the strength of dispersion force is proportional to electron number, an hydroxyl group (OH) has an extra oxygen atom (thus more electrons) compared to a hydrogen atom. Therefore, the strength of dispersion forces is also greater in alcohols than in homologues in hydrocarbon homologous series like alkanes. NOTE: Since homologues in the Aldehyde and Ketone categories only has dipole-dipole forces and not hydrogen bonding, corresponding alcohol molecules with the same number of carbon atoms (i.e similar molecular mass) would therefore have higher melting & boiling points. Although the alcohol molecules with one to three carbons (C1 to C3) are soluble in water, as the molecular mass increases (i.e. moving into C4 and beyond), the length of the non-polar hydrocarbon chain increases. This results in alcohols experiencing a gradual decrease in water solubility as molecular mass increases due to the increasing non-polar alkyl chain/group. Like all of the previous cases, as the number of carbon and hydrogen atoms increases (i.e. molecular mass) for homologues in the alcohol homologous series, the strength of dispersion force increases. Thus, the melting & boiling point increases. Lastly, as the molecular mass increases (due to increase carbon and hydrogen atoms) between alcohol homologues, the viscosity of the homologues increases. This is because the the larger molecular size of the alcohol molecules increases the extent of dispersion forces as well as the large size provide resistance in hindering alcohol molecules from escaping the liquid surface. This trend of viscosity can be applied to other applicable organic molecules that are in the liquid state in different homologous series. Amide Homologous Series Apart from the amide that has only one carbon (C1) in the homologous series existing as a liquid at room temperature, i.e. methanamide, all the rest of the other homologues in the amide homologous series are solids at room temperature. The main intermolecular force that determines the melting & boiling point of amides is hydrogen bonding. Compared to alcohols and carboxylic acids with the same number carbon of atoms, the corresponding amide has a higher melting & boiling point. This is because amides have the C=O bond as well as the two N-H polar bonds that can hydrogen bonds. As you will learn soon, all of the homologous series that we will talk and have talked about in this learning objective will have a lower density than water except for amides! This isn’t surprising considering amides are solids at room temperature except the amide with one carbon atom (C1), i.e. methanamide. An identical trend to homologues in other series, the density of amides will increasing molecular mass. In terms of solubility, this is similar to other cases we have talked about. As the molecular mass increases, the non-polar alkyl group will increase resulting in a gradual decline in solubility in polar substances. Like all of the previous cases, as the number of carbon and hydrogen atoms increases (i.e. molecular mass) for homologues in the amide homologous series, the strength of dispersion force increases. Thus, the melting & boiling point increases. Amine Homologous Series Amines in the gas phase have an ammonia odour. Comparatively, liquid amines have a “rotting fish” odour. Amines with one and two carbons (C1 and C2) are gas at room temperature – i.e. methanamine and ethanamine. The other amines from C2 to C8 are liquids at room temperature. The primary intermolecular force that governs the melting and boiling point of amines is hydrogen bonding. This is because the N-H bond is a polar covalent bond as the nitrogen atom is much more electronegative than the hydrogen atom. For such reasons, hydrogen bonding exists between amine molecules. If you draw out an amine molecule, you can see that one amine molecule has the capacity to perform hydrogen bonding with three water molecules. In terms of melting and boiling point, alcohols and carboxylic acids have a higher melting & boiling point than amines. This is because the O-H bond is more polar than the N-H bond as oxygen is more electronegative than nitrogen. Similar to alcohols, as the molecular mass of the amine molecules increases, the length of the nonpolar alkyl chain increases. Thus, the solubility of amines molecules in polar substances (or solvents) such as water decreases with increasing molecular size. Like all of the previous cases, as the number of carbon and hydrogen atoms increases (i.e. molecular mass) for homologues in the amine homologous series, the strength of dispersion force increases. Thus, the melting & boiling point increases. Carboxylic Acid Homologous Series The main intermolecular force that determines the melting & boiling point between homologues in the carboxylic acid homologous series is hydrogen bonding. As seen from the carboxylic acid functional group (COOH) structure we have explored in last week’s notes, we can see that there is a carbonyl group (C=O) and a hydroxyl group (O-H). So, each carboxylic acid group has the capacity to form two hydrogen bonds with another carboxylic acid molecule. Comparatively, when we compare carboxylic acids to alcohols, ketone and aldehydes, they can only form one hydrogen bond. Therefore, carboxylic acids have higher melting & boiling point compared to their corresponding counterparts in other homologous series with the same number of carbon atoms except for amide (which has the highest). Similar to alcohol, although the carboxylic acids with one to three carbons are soluble in water, as the molecular mass increases (i.e. moving into C4 and beyond), the length of the non-polar hydrocarbon chain increases. This results in carboxylic acids experiencing a gradual decrease in water solubility as molecular mass increases due to the increasing non-polar alkyl chain/group. That being said, when comparing carboxylic acid to their corresponding alcohol with the same number of carbon, the carboxylic acid is more soluble in water due to the capacity to form two hydrogen bonds compared to only one for alcohol. Like all of the previous cases, as the number of carbon and hydrogen atoms increases (i.e. molecular mass) for homologues in the carboxylic acid homologous series, the strength of dispersion force increases. Thus, the melting & boiling point increases. Summary of common trends amongst homologues in various homologous series We talked about the trend that melting & boiling point increases with the size of homologues (C1 to C8) within and between homologous series. General trend of decreasing melting & boiling point: Amide > Carboxylic Acid > Alcohol > Ketone > Aldehyde > Amine > Haloalkane > Alkyne > Alkane > Alkene. We also talked about solubility decreasing with increasing molecular mass. We also talked about density increasing with increasing molecular mass (except for alkyl halides). We also differentiated amides against amines in terms of odour. Likewise, aldehydes with ketone with odour. We differentiated some homologues in homologous series by their states at room temperature. Learning Objective #3 - Analyse the shape of molecules formed between carbon atoms when a single, double or triple bond is formed between them Single carbon-carbon bond An example of a molecule that comprises the single carbon-carbon (C-C) bond can be found in ethane, C2H6. All alkanes have single carbon-carbon bonds except for methane. The shape of carbon atoms in ethane have a tetrahedral shape according to VESPR theory since each carbon in ethane has four chemical bonds. Each carbon makes three C-H bonds and one C-C each with a 109.5 degree bond angle. NOTE: Drawing lewis structures is important to understanding and predicting molecular shapes which we have touched on when exploring the VESPR theory in the HSC Preliminary Course. NOTE: Since we are dealing with molecules with more than one central atom (two or more atoms bonded to it) such as ethane, sometimes there may not be no single shape that can be used to describe the shape of the full molecule if the central atoms have differing amount of atoms attached to them. In our case, we only have two central carbon atoms in ethane and both carbon atoms have the same amount (4) of atoms attached to it so they have the same shape. So, ethane can be described using one shape, it has a tetrahedral shape. NOTE: If you decide to take a first year chemistry course at university, you will realise that there is a difference between geometry and shape when describing molecules where geometry takes into account of lone electron pairs which we won’t and don’t have here. Double carbon-carbon bond Ethylene is an example of a molecule that consists of a double carbon-carbon (C=C) bond, C2H4. All alkenes have double carbon-carbon bonds. The shape of the central carbon atoms in ethylene have a trigonal planar shape, according to VESPR theory, since each carbon atom makes two C-H bonds and one C-C, each with are approximately 120 degree bond angle. Since both central carbon atoms have the same shape, ethylene has a trigonal planar shape. Triple carbon-carbon bond The molecule Ethyne is an example of a molecule that consists of a triple carbon-carbon bond, C2H2. All alkynes have triple carbon-carbon bonds. The shape of the central carbons ethyne is linear according to the VESPR theory since each carbon atom has one C-H and one C-C bond which each has 180 degrees bond angle. Since both central carbon atoms have the shape, ethyne has a linear shape. Learning Objective #5 - Describe the procedures required to safely handle and dispose of organic substances. Below are some common laboratory practices when dealing with organic substances. We will go through ten laboratory practices which you probably only need to write down three for HSC purposes. Handling – i.e. storing, using and treating spills of organic substances I. Store “like material with like” – so that flammable materials should be stored together and away (in a different cabinet) from materials that are corrosive, toxic, etc to prevent hazardous reactions. II. Oxidising agents must be stored separately from organic substances as they can ignite organic solvents or acids. III. Organic nitrates and peroxidies are shock-sensitive and should be handled with care as shock can result them in generating large volume of gases in a short period of time, resulting in a high increasing pressure. This would result in an explosion. IV. Some organic acids such as acetic and propanoic acid are corrosive which can cause burns upon contact with skin where severity depends on concentration of the acid. To minimise this risk, a laboratory coat is worn to as an additional protective layer against corrosive substances for the skin (arms, body & thighs). Safety gloves are worn to protect skin from being in physical contact with corrosive substances in case of spill. Safety googles are worn to prevent corrosive organic acids from being in contact with eyes in case of splashes. Enclosed leather shoes are worn to protect feet in case of accidental spills rather than open shoes. V. Ensure that a well-ventilated area such as a fume hood is used when handling flammable, volatile organic substances (e.g. alcohols). This is because the inhalation of alcohol can cause dizziness and could result in brain damage depending on duration of exposure. Ensure that the fume hood is closed when not in used to prevent escape of volatile, flammable vapour from the organic substance. Ensure that there is no open naked flame near the flammable organic substances like alcohols. VI. When you wish to diluting concentrated acids, you should always slowly add acid to water (of larger volume) and not the other way around. That is, never attempt to dilute acid by adding water to the acid. This is because adding acid to water (of larger volume) allows the water molecules to absorb the heat generated in the dissolution reaction (water also has a higher specific heat capacity than acids). So, it minimises the concentration of heat in a particular area which would otherwise result in splashing of water that contains the acids. The acid may end up in your eye, skin, etc which is undesirable and dangerous. If you do it the other way around (adding water to acid), the first water molecule that makes contact with the acid will react form hydronium ions which generates a big release of heat. This would cause the water to vapourise or splash which contains the dissolved acid coming at you which you don’t want. Therefore, never add water to acid when diluting but the other way around adding acid to water. “A comes before W” VII. To clean up an organic substance spill (e.g. organic vapour from, a half mask air-purifying respirator is used. A full-face air-purifying respirator is required if the spilled organic substance is a known eye irritant. VIII. Wash your hands before leaving the laboratory, even if gloves are worn as a safety precaution especiallyafter cleaning up a spill. This is because organic solvents can strip off your skin’s natural oil layer resulting in skin irritation whereby your skin would be more susceptible to absorbing toxic chemicals. Disposing organic substances IX. Ensure that all flasks containing organic substances are labelled such that they can be disposed into the right waste container. All waste containers must also be labelled appropriately such that the correct disposable method can be used. Liquid and solid organic waste are required to be segregated into different waste containers. Also, inorganic and organic liquid wastes are required to be disposed in separate waste containers. X. Do not dispose organic substances through the sink as they will contaminate in the waters in which aquatic organisms reside. Organic substances are toxic when ingested by aquatic organism and potentially humans or any terrestrial mammals that consume the water. Ideally, separate organic and aqueous wastes (mainly water) when disposing organic liquid wastes to reduce the cost of disposal. Learning Objective #6 - Examine the environmental, economic and sociocultural implications of obtaining and using hydrocarbons from the Earth Source of hydrocarbons Hydrocarbons, as we have defined already, as compounds that consists of only carbon and hydrogen atoms. There are many two classes of hydrocarbons. These are acyclic and cyclic hydrocarbons. Acyclic hydrocarbons are primarily derived from petroleum and natural gas. These hydrocarbons have a linear structure. Alkanes, Alkene, Alkynes that we have explored in last week’s and this week’s notes are examples of acyclic hydrocarbons. Comparatively, cyclic hydrocarbons are generally sourced from coal. These hydrocarbon have a cyclic or circular structure as they have a hexagonal-shaped phenyl ring, C6H5, which has three alternating C=C bonds. Petroleum and Natural Gas can be found at different levels deep in the Earth’s crust. Petroleum and Natural Gas formed as dead aquatic organism are covered in sand, clay and mud at the bottom of the sea. Due to the environment consisting of the lack of oxygen, high heat and high pressure, the dead matter from the organisms are converted into petroleum and natural gas. As petroleum is lighter than water, it is situated above water. Comparatively, the natural gas is formed above the petroleum layer but then becomes trapped between the impervious rocks that is above it. Therefore, natural gas can be sourced from the impervious rocks that are located deep in the Earth’s crust. Coal is formed by the decomposition of dead plants and trees on land covered in soil and mud. Over time, the dead matter is trapped and subjected to an environment that has no oxygen (anaerobic), high heat and high pressure. Thus, coal is sourced deep within the Earth’s curst. Now with that introduction out of the way, let’s address the hydrocarbons that are derived from in petroleum, natural gas and coal! Petroleum mainly consists of hydrocarbons from C1 to C30 that are used in everyday life or industrial processes. For example: Petrol used in motor vehicles have hydrocarbons mainly consisting of 8 carbons (octane) Paraffin wax used in candles & machine lubricants consists of hydrocarbons with 20 carbons or higher. Natural gas mainly consists light weight hydrocarbon such as methane (~80%) which has the molecular formula, CH4. Other light weight hydrocarbons include ethane, propane and butane. Yes, these unbranched hydrocarbons from C1 to and including C4 (e.g. butane) are all gases at room temperature as we have pointed out in Preliminary HSC Chemistry. Coal mainly consists of carbon rich compounds which can be burnt in an anaerobic environment to produce methane. Environmental Implications As we have mentioned previously, petroleum has a wide range of hydrocarbons with differing molecular weights from C1 to C30 (or even higher). Depending on the hydrocarbon’s physical & chemical properties, amount of hydrocarbon and time of contact, the hydrocarbon’s effect of the environment can vary upon contact. It is important to note that only a small portion of the wide range of hydrocarbons in petroleum actually do yield harmful effects (e.g. toxic) to the environment. Let’s first explore how obtaining hydrocarbons can result in environmental implications! Obtaining petroleum and natural gas from the Earth results drilling through rocks deep in the Earth’s crust whereby hydrocarbons from the drill machine’s lubricants can be dispersed into surrounding water pollutes the surrounding seawater or ocean. These hydrocarbons are toxic to aquatic organisms that reside in the sea. These rocks are also returned into the ocean which often contain barium ions from the traces of lubricant that remain. These barium ions are toxic as they interfere with enzyme activities which can result in death of living organisms. Furthermore, there are potassium ions in the machine lubricants in extracting the hydrocarbons (e.g. petroleum & natural gas) where high levels of potassium can result in uncontrollable algae growth leading to a state of eutrophication. This essentially allows algae to grow on the water surface which blocks the sunlight reaching to the plants beneath the water as well as oxygen gas that is dissolved in the water. This will reduce in the death of plants which decreases the oxygen availability in the water for aquatic animals like fish as well as bacteria that further uses more oxygen to decomposes dead matter of demised plants. Overall, it destroys the aquatic ecosystem turning a habitable environment into a toxic one for the original species. Other than eutrophication, an increase in ion concentration (e.g. K+ ions) in the surrounding water can disrupt the osmotic balance in living organisms, resulting in the deformations in aquatic fauna or flora. Another environmental implication would involve the noise pollution through the sending of sound waves to detect potential hydrocarbon deposits for drilling. This would disturb local aquatic organisms, such as whales, as well as any humans that near the area. The sound waves can disorientate whales which can result in large scale whale stranding on beaches. This results in their death through multiple means on is due to dehydration. To obtain hydrocarbons for use, it must be transported from the sea to land to oil refinery so that petroleum can be split into their components. The mode of transportation is through the use of ships. Oil spillages due to various accidents such as collisions with rocks, other ships, etc have resulted in environmental damages similar to those which we have mentioned already. These hydrocarbons can be washed ashore which can pollute beaches whereby humans have contact with. Hydrocarbons that enter the human body could cause severe respiratory irritation. The long term effects of hydrocarbon exposure is also currently not fully understood. Economic Implications Through potential implications such as leaking toxic hydrocarbons, potassium ions and eutrophication as we have discussed earlier as an environmental implication, they could all result in the death and reduction in biodiversity of aquatic organisms. It is important to preserve biodiversity for many countries as aquatic organisms serves a major source of economic revenue. For example, aquatic organisms sold from Philippines accounts for over $550 billion USD dollars annually. The reduction of Philippines’ aquatic organisms would result in a major hit to the country’s economy and have many social implications such as poverty (which is relevant for the following section – Sociocultural Implications of obtaining hydrocarbons). Philippines recognises the importance of preserving biodiversity of its aquatic organisms as it is first south eastern asian country to regulate the risk of biotechnology as the technology could reduce the country’s rich aquatic organisms biodiversity. Sociocultural Implications Workers involved in the drilling process to obtain hydrocarbons will be exposed to drilled rocks that are covered in toxic hydrocarbon lubricants, as well as the lubricants in the machinery itself, which can be accidentally or voluntarily inhaled as oil mists in the air. We have already mentioned how the leaking of toxic hydrocarbons, potassium ions and eutrophication during the extraction of hydrocarbons can result in a decline in aquatic organism’s biodiversity. This has many implications in removing or limiting the diverse range of food in which humans can enjoy in our everyday lives. Also, as the supply of aquatic organisms decreases due to hydrocarbon pollution, the price of seafood would increase which means that it would be less affordable for the global population in general. Furthermore, the presence of toxic hydrocarbons would result in our everyday potable (drinkable) water derived from the sea being toxic. The treatment of water would thus be more thorough which would incur additional cost to consumers. Learning Objective #1 - Write equations and construct models to represent the reactions of unsaturated hydrocarbons when added to a range of chemicals Hydrogenation: Adding hydrogen (H2) to unsaturated hydrocarbons Hydrogenation is an example of an addition reaction whereby, in our case, hydrogen gas reacts with unsaturated hydrocarbons such as alkenes and alkynes to form one combined product. In hydrogenation specifically, the hydrogen reacts with and breaks the C=C bond such that one hydrogen atom is added (i.e. covalently bonded) to each carbon atom whereby the alkene is converted into an alkane product. That is, the carbon that was originally unsaturated became saturated after hydrogenation. Halogenation: Adding Halogens (X2) to unsaturated hydrocarbons Another type of addition reaction is halogenation whereby diatomic halogen molecules reacts with the unsaturated carbon bond, e.g. C=C bond, resulting in one halogen atom being added to each carbon atom. The resulting product is a haloalkane (alkane containing halogen atom(s)) This effectively converts the unsaturated carbon in an alkene into an saturated carbons in an alkane. Hydrohalogenation: Adding Hydrogen Halides (HX) to unsaturated hydrocarbons Hydrohalogenation is the third type of addition reaction in which a hydrogen halide (HX) molecule breaks the C=C bond so that a hydrogen is added to one carbon atom and a halogen atom is added to the other carbon. The result is a haloalkane being produced. In HSC Chemistry, we do not need to need to understand which carbon atom the hydrogen is added to and which carbon atom the halogen is added to. It is because it is outside the HSC Chemistry Syllabus. However, we will explore the Markovnikov’s rule in this week’s Youtube Video so that we can create a more accurate model for haloalkane that is produced via hydrohalogenation reaction. Again, you can simply add to any of the two unsaturated carbon as you wish for HSC Chemistry purpose if you decide to not learn the Markovikov rule as it’s not part of HSC Chemistry syllabus. This is because Markovnikov’s Rule is outside the scope of HSC Chemistry. Hydration: Adding Water (H2O) to unsaturated hydrocarbons The final addition reaction is hydration. In hydration, there is a hydroxyl group (-OH) that is added to the unsaturated carbon atom (i.e. part of the C=C bond) and a hydrogen atom is added the other carbon atom which was part of the original C=C bond. The resulting product is a saturated alcohol. The second type of unsaturated hydrocarbons is alkyne. So let’s explore how alkynes can undergo additional reaction. NOTE: Alkynes can double the number of chemical reaction with halogens compared to alkenes. This is because alkynes have triple carbon bonds while alkenes have double carbon bonds. We will only explore halogenation and hydrohalogenation addition reactions for alkynes as the other reactions we have explored for alkenes will follow the same principle. That is, alkynes will have double the number of addition chemical reactions that alkenes can perform. Halogenation of unsaturated hydrocarbons - Alkynes Hydrohalogenation of unsaturated hydrocarbons - Alkynes Learning Objective #1 - Investigate the structural formulae, properties and functional group including: Primary Alcohols Secondary Alcohols Tertiary Alcohols We have already talked about the physical shared by all alcohols in general in last week’s notes, except for the boiling points between primary, secondary and tertiary alcohols which we will explore in this week’s note. So please revisit last week’s notes, if you need to revise on the other physical properties of alcohols homologues. That being said, we will explain some differences in the chemical properties between primary, secondary and tertiary alcohols towards the end of this learning objective as well as in Learning Objective #4 and #6 which we haven’t explored in last week’s notes. Anyways, returning to this learning objective. This question we want to address is what are the differences between primary, secondary and tertiary alcohols? Well, primary alcohols are alcohols whereby the carbon atom that is bonded to the hydroxyl group (OH) is only bonded to one R group. On the other hand, secondary alcohols are organic molecules whereby the two R groups (can same or different) are attached the carbon atom that is bonded to the hydroxyl group. Lastly, in the case of tertiary alcohols, they have three R groups (can be same or different) are bonded to the carbon atom which is also bonded to the hydroxyl group. NOTE: The ‘R’ groups used here to differentiate primary, secondary and tertiary molecules are groups that contain at least one carbon atom attached to hydrogen atoms. The R groups here cannot be hydrogen atoms. Check out the following diagram that compares the differences between the three categories of alcohols. NOTE: The direction in which you draw the R-C bond(s) does not matter. As you can see in the diagram, the red arrows are drawn to remind and show you that the chemical bonds can be rotated (clockwise or anti-clockwise) NOTE: Some sources include methanol as a primary alcohol. Some do not. It is up to you to decide. It is probably more accurate to exclude it from the primary alcohol class. So let’s explore in terms of melting & boiling point shared between primary, secondary and tertiary alcohol with similar molecular masses (i.e. same number of carbon atoms). So, in the order of increasing melting & boiling points: Tertiary, secondary, primary alcohols and finally methanol. An important factor that is contributing towards this trend is that an alkyl group is bigger in size than a hydrogen atom. But let’s explore why this is important in dictating the trend above. As tertiary alcohol has more alkyl groups than secondary alcohols which has more alkyl groups than primary, the hydroxyl group is less exposed due to more alkyl groups. This hinders the extent at which hydrogen bonding can occur. This means that there would be less hydrogen bonding occurring in tertiary alcohol compared to secondary alcohols due to the hinderance of alkyl groups. Therefore, tertiary alcohols have lower melting & boiling point than secondary alcohols and also the reason for why secondary alcohols have lower M.P and B.P than primary alcohols. Needlessly, to say by now, methanol does not have any alkyl groups, hence, it has higher melting and boiling point than corresponding primary, secondary and tertiary alcohols. As promised at the start of this learning objective, we will discuss about a difference in chemical property between primary, secondary and tertiary alcohols here. This difference in chemical property amongst the three different types of alcohol is their acidity. In order of increasing acidity (from low to high): tertiary alcohol -> secondary alcohol -> primary alcohol -> methanol For HSC Chemistry purposes, we can explain this trend by understanding that alkyl groups are better electron donators than hydrogen atoms. Therefore, as the number of alkyl groups increases from primary to tertiary alcohol, more electron density is pushed towards the central carbon atom which passes on some of these electron towards the oxygen atom in the hydroxyl group. As a result, the electrons that oxygen attracts from hydrogen atom in the OH group is less extensive. Therefore, the energy required to break the OH bond increases. This makes it harder to break the hydroxyl bond and release a H+ ion. Another way to explain this is that as the number of electron donating alkyl groups increases, the more unstable the conjugate base will be (due to increased number of electrons which the conjugate base carries) after the alcohol donates its hydrogen atom. Therefore, tertiary alcohols form the least stable conjugate base, thus, making it the weakest acid of the three. Lastly, in terms of chemical properties differences, due to their different number of alkyl groups present in the different types of alcohols, primary alcohol be oxidised to aldehyde and carboxylic acid but not to ketone. In the case secondary alcohols, they can be oxidised to ketones but not to aldehyde or carboxylic acid (unless in extreme oxidative conditions). Comparatively,, tertiary alcohol does not readily oxidise unless in extreme oxidative conditions involving the breaking of C-C bonds such as in combustion. Learning Objective #4 - Write equations, state conditions and predict products to represent the reaction of alcohol, including but not limited to: - Combustion - Dehydration - Substitution with HX - Oxidation Combustion of alcohols Learning Objective #4 - Write equations, state conditions and predict products to represent the reaction of alcohol, including but not limited to: - Combustion - Dehydration - Substitution with HX - Oxidation Combustion of alcohols Dehydration of alcohols Reaction condition: Heating the alcohol in the presence of a concentrated acid catalyst (e.g. H2SO4 or H3PO4) at 170 degrees celsius. The product as a result of the dehydration of alcohol is an alkene as well as water as a byproduct. Since alkene is formed as a product, the dehydration of alcohol can only occur if the alcohol has two or morecarbon atoms. NOTE: If the temperature is not suffice then ethers would be produced rather than alkenes. This is a condensation reaction as water is released as a by-product and the product is larger than the reactants, i.e. alcohol. Dehydration is a type of condensation reaction. Substitution reaction involving alcohols with HX (hydrogen halide) What happens in substitution reaction is that the HX ‘attack’s (interacts) and react with the C – OH bond to free the hydroxyl group and substitute the original alcohol molecule with the halogen atom “X”. The free, negatively charged hydroxide ion can bond with the positively charged hydrogen ion (from the HX molecule) to form water as a by-product. If we wish to determine the alcohol is primary, secondary or tertiary, we can use the Lucas Test which involves the use of a Lucas’ reagent which is essentially a solution comprised of an anhydrous zinc chloride catalyst dissolved in concentrated hydrochloric acid solution. The products produced is a chloroalkane (R-Cl) and water. To determine whether the alcohol is primary, secondary or tertiary, the rate of reaction can be used. Tertiary alcohol is able to completely react with the hydrochloric acid via substitution reaction in less than one minute to form the cloudy, white haloalkane precipitate and water as a by-product. This means that two layers are formed (originally there is only one layer of alcohol solution at start of experiment). The bottom layer is the precipitate that is more dense and the top layer is the water alongside any excess alcohol and the zinc chloride catalyst. For the case of secondary alcohol, the reaction typically requires 2-5 minutes. Again, the products are precipitate haloalkane and water. Lastly, it would take more than 6 minutes for primary alcohol to react with the hydrochloric acid in the case of primary alcohols. When the reaction is complete, the haloalkane product is formed as a precipitate as well as water. You may need to heat the solution containing primary alcohol and hydrogen halide (HX) to speed up the rate of reaction.