Unit 9 - Organic Chem basics

Question 1

Hydrocarbons basics

Answer 1

• carbons can bond to itself to form rings, single/double/triple bonds, or any number of other atoms
• smaller, branched chains of carbon with no clear dipoles/hydrogen bonds have weakest IMF and lowest boiling points due to less interaction and less VdW
• long, straight chain molecules have higher boiling points for opposite reasons
• chains with very electronegative atoms e.g halogenoalkenes have stronger dip-dip forces
• if molecule has O-H, N-H or F-H there are very strong Hydrogen bonds

Question 2

nomenclature rules

Answer 2

• stem = no. carbons
• prefix = before stem (additional atoms, alkyl branches)
• suffix = after stem (homologous series)
• order of group priority for names = Halogens, alkyl groups, alkenes, everything else

Question 3

Homologous series and functional groups

Answer 3

• Alcohol functional group -OH
• Ether func group -O-
• carboxylic acid group -C(=0)-OH
• Ester func group -C(=O)-O
• Aldehyde func group -C(=O)-H
• Ketone func group -C(=O)-
• Amine func group -NH2
• Amide func group -C(=O)-NH2

Question 4

isomerism

Answer 4

• When other molecules have different structure but same atoms
• stereoisomers have same atoms located in different positions
• positional isomers have func groups on different atoms
• Func group isomers have different functional groups
• chain isomers have different lengths
• E/Z isomers have different rotations around a double bond, with E isomers having pairs of atoms with higher Mr diagonal across bond, and Z isomers have them opposite each other

Question 5

Electrophyllic addition

Answer 5

• one atom in dipole acts as electrophile and takes electron from double bond, bonding to one carbon and forming a carbocation on the other one
• other atom in dipole attracted to cation and bonds to it
• cation can form either side of the double bond, in some cases forming multiple products
• carbon with more surrounding carbons bonded to it will be more likely to form a cation as it is more stable, skewing the ratio of the two products formed

Question 6

free radical substitution

Answer 6

• initiation step - UV radiation splits a bond in half
• propagation steps - radicals react with atoms to form other radicals and will continue until termination
• termination - two radicals bond and cancel out to form a normal molecule

Question 7

CFC’s and ozone

Answer 7

• intiation step forms chlorine radicals when CFC’s get into atmosphere
• Cl radicals react with ozone to form oxygen and ClO radicals ( Cl. + O3 -> ClO. + O2)
• ClO radicals react with more Ozone to form more oxygen and Cl radicals, making Cl radicals a catalyst for decomposition of Ozone into oxygen (ClO. + O3 -> Cl. + 2O2)
• CFC’s used to be used in loads of things like fridges and aerosols, now there are CFC free alternatives after they were banned
• ozone is useful because it absorbs UV radiation from the sun

Question 8

Nucleophillic substitution

Answer 8

• as halogenoalkanes are polar, they react with nucleophiles
• requires warm aqueous conditions for substitution with OH-, and Ethanolic warm conditions with CN- and NH3
• nucleophillic reactant donates electrons to a positive dipole in a hydrocarbon, and the negative dipole takes a pair of electrons off with it when it is substituted for the other reactant
• NH3 will then also form a cation, so another NH3 takes a hydrogen and becomes NH4+ and an amine

Question 9

Elimination reactions

Answer 9

• under hot ethanolic, reflux conditions with -OH
• OH takes a proton, carbon then adds its spare pair to the central bond forming a double,other carbon must let go of its negative dipole to form this
• creates a negative ion, water and an alkene