2.5 organic

Functional group priority list

1. carboxylic acid

2. alcohol

3. amine

4. alkene

5. alyne

6. alkane/haloalkane

compounds with functional group number at the beginning of name

haloalkane

compounds with functional group number at the middle of name

alkene, alkyne, amine, alcohol

compounds with that need n after prefix

amine, alcohol, carboxylic

boiling point order

alkane < alkyne < alkene < haloalkane < amine < alcohol < carboxylic

but the longer carbon chain > higher up in list

• the only exception is carboxylic (only 8 and above carbon is higher)

Alkane properties

• saturated carbons, meaning they are very strong and stable due to single bonds

• insoluble in water (non-polar), has no attraction to water (polar)

• poor conductor of electricity and heat (no free-moving charged particles)

• boiling point increases with carbon chain length

Justifying boiling point

As the carbon chain length increases, there are more atoms, therefore there will be greater intermolecular forces due to greater surface area. This means that more energy is required to separate molecules, resulting in a higher boiling point.

Straight chain compounds have greater SA than branched compounds, creating more points of contact between molecules and therefore have stronger intermolecular attractions.

Substitution reaction example answer

[compound] reacts with [reagent] under [conditions] in a substitution reaction. The __ atom in [compound] is replaced by the __ group from [reagent], forming [products].

• don’t list conditions if none

• optional: the __ removed from the [compound] combines with __ from [reagent] to form __ as a byproduct

Cycloalkane properties

• Ring structure

• same properties as alkane

Haloalkane properties

• Same structure as alkane

• Heavier halogens increase in boiling point: I > Br > Cl > F (highest to lowest)

Elimination reaction example answer

[compound] reacts with [reagent] under [conditions] in an elimination reaction. The [reagent] removes a hydrogen atom from one carbon, and the [functional group] atom from the adjacent carbon, forming a C=C double bond between them and producing [alkene]. The __ removed from [compound] combines with __ from [reagent] to form __ as a byproduct.

2 degree haloalkane / alcohol compound - saytzeff’s rule elimination (full answer)

[compound] is an asymmetric/ 2 degree haloalkane/alcohol. It reacts with [KOH aq/ conc H₂SO₄] under heat by elimination. The [reagent] removes the [OH/X] along with a hydrogen from either adjacent carbon to the [OH/X group] (carbon __ or __), therefore 2 possible alkenes can be formed.

According to Saytzeff’s rule, the H atom will often be removed from the carbon with the lowest number of carbon atoms. Therefore, the H atom will be removed from carbon __ as it only has (lower no. H atoms) attached to it, whil carbon has (higher no. H atoms) attached to it. This will form the major product [alkene]. The minor product [alkene] can be formed if the H atom from carbon (lower H atom) is removed. The removed [OH/X] and K⁺/H forms KX/water as a byproduct.

Alkene properties

• same properties as alkane

Addition reaction example answer

[compound] reacts with [reagent] under [conditions] in an addition reaction. The C=C double bond breaks and each carbon gains one [reagent] atom, forming [compound].

asymmetrical alkene to haloalkane/alcohol markovnikov addition example answer

[alkene] is an asymmetrical alkene. It reacts with [HX / H₂O/H⁺] in an addition reaction. [alkene] has a double bond between carbon __ and carbon 2 __. Carbon __ has (higher H atom) while carbon __ has (lower H atom). During an addition reaction, H and X from [HX] / H and OH from [H₂O/H⁺] is added across the double bond, forming 2 different products (double bond is broken).

According to Markovnikov’s rule, the carbon in the C=C double bond with more H atoms will often have another H atom bonded to it. This means that the H atom from [HX/ H₂O/H⁺] will bond to carbon __ (more H atom), and the X/OH atom will bond to carbon __ (less H atom), forming the major product [haloalkane/alcohol]. The minor product [haloalkane/alcohol] can be formed if the H atom bonds to carbon __ (less H atom) instead, and the X/OH atom bonds to carbon __ (more H atom).

Minor/major product drawing

If major: the functional group is not at the end

If minor: the functional group is at the end

Alkene + H₂ —> alkane reaction explained

This is an addition reaction. [alkene] reacts with H₂ in the presence of a Ni/Pt catalyst and heat. The H₂ is added to the C=C double bond in the addition reaction, converting [alkene] into [alkane]

Alkene —> polymer / addition polymerisation

[alkene] undergoes addition polymerisation. The C=C double bond is broken down and new C–C single bonds are formed between monomers, making long chains of repeating units called poly[alkene]. Because poly[alkene] only contains single saturated C–C bonds, it is more stable and less reactive than the alkene monomer [alkene]. This make them safe and durable to use as they are chemically inert.

alkene oxidising reaction

Alkenes react with acidified dichromate or permanganate in an oxidation reaction. The oxidising agent adds across the C=C double bond, breaking it and forming a diol. Each carbon from the double bond gains an OH group.

Alcohol properties

• As the carbon length increases, the solubility of the alcohol decreases.

  • Small alcohols (C1 - C3) are soluble in water. C4 is slightly soluble in water. Anything above C4 is not.

  • This is because small alcohols have a small non-polar carbon chain and a very polar OH group, which forms an attraction to the polar water. As the non-polar hydrocarbon chain increases, the alcohol becomes less polar overall and the OH group becomes a smaller part of the molecule, therefore becoming less soluble with water.

• BP same as alkane

primary alcohol oxidising reaction to carboxylic

Primary alcohols are oxidised by acidified dichromate or permanganate. During oxidation, the first carbon (bearing the OH group) loses hydrogen atoms and gains an oxygen atom from the oxidising agent, forming a C=O double bond. This converts the alcohol into a carboxylic acid.

Amine litmus paper

Turns wet red litmus paper blue (indicates it is acting as a base).

Turns wet blue litmus paper red (indicates it is acting as an acid).

carboxylic properties

• COOH is very polar

• High melting and boiling points due to stronger intermolecular attractions. (They can form two hydrogen bonds, harder to separate).

• Small acids (C1–C4) are very soluble in water. As the carbon chain increases, the non‑polar hydrocarbon part dominates → solubility decreases

what does this mean

• number of points = carbon atoms

• because amine is attached to the second carbon, it would be butan-2-amine

Geometric isomerism existing

Geometric isomers contain a rigid double bond which prevents rotation. Because the atoms are not able to freely rotate, groups attached to the double bonded carbons stay locked in place.

For geometric isomerism to exist, each carbon in the C=C double bond must have 2 different groups attached to it. In [the one with geometric isomer], one alkene carbon is bonded to [x] and [y], while the other carbon is bonded to [x] and [y].

Geometric isomerism not existing

say everything except geometric isomers have rigid part until the ‘in the’ part

In [the one without geometric isomer], one carbon in the C=C double bond has two identical [x] groups, therefore it cannot form geometric isomers.

cis vs trans

• Cis = the groups are on the same side of the C=C double bond.

• Trans = the groups are on different sides of the C=C double bond. 

Classifying alcohol/haloalkane

• The carbon atom attached to the halogen is joined to only one/two/three carbon atom, it is a primary/secondary/tertiary haloalkane. Same thing for alcohol.

Drawing polymer

1. Draw the monomer so that the double bonds line up across the page (make sure to break it up into 2 carbons each)

2. Change the double bond to a single bond, line it up across a horizontal line

3. Put double brackets around the polymer and an n 

4. The name of the polymer would be poly - name of monomer

Drawing monomer

1. see where the repeating unit stops

2. Draw the double bond first, and see what part is attached to what

Distinguish between compounds tips

• You have to say what happens to the other compound no matter what.

• If nothing happens to something, state no reaction happens

• justify some things like BP or solubility

to distinguish between carboxlyic and amine using a metal carbonate

• state that it’s an acid base reaction

• state that carboxylic produces carbon dioxide. fizzing will be observed

• amine is a base and will not react with metal carbonate

to identify between alkene, alcohol, and alkane using br and h2o

• br can be used to identify alkane, it decolourises from red-brown colour

• hexene will form two layers when reacted with water, because it has no attraction to it (non-polar)

• ethanol is a polar solvent and is miscible in water (polar solvent as well)

chemical test for an alcohol and an alkene

• you can use warm acidified potassium dichromate for alcohol as it will oxidise into a carboxylic acid, the colour change observed will be orange to green

• you can use bromine water to identify alkene as it will decolourise bromine water from orange-brown colour to colourless rapidly.

• permanganate cannot be used as both alkene and alcohol will react with it, and the same purple to colourless observation will be seen. alcohol will oxidise into a carboxylic, alkene will oxidise into a diol.