10. Organic SL

Welcome to Organic Chemistry!

Organic chemistry is the study of carbon-based compounds

Elements most commonly worked with: hydrogen and carbons (which form hydrocarbons), oxygen, nitrogen, and the halogens

Describe the features of a homologous series

Homologous series- Long chains of hydrocarbons with the same functional group but differ by a different number of CH₂. Have similar chemical properties and gradiation in physical properties.

Predict and explain the trends in boiling points of members of a homologous series

As the hydrocarbon gets longer, so there are more van der waals forces, and the melting/boiling point is much higher

Distinguish between empirical, molecular, and structural formula

(Example: Butane)

Molecular: the full formula written out with all numbers (C₄H₁₀)

Empirical: the simplest ratio of the elements to one another (C₂H₅)

Condensed Structural: Shows overall structure in one line (CH₃CH₂CH₂CH₃)

Full structural: every atom and/or bond is shown

Describe structural isomers as compounds with the same molecular formula but with different arrangements of atoms

Structural Isomers-compounds with the same molecular formula but different arrangements of bonds (structural formulae different)

            Positional- Have the same chain, but different functional group location (position)

                        Example: Propan-1-ol and Propan-2-ol

            Hydrocarbon- Have same functional group but different length chains

                        Example: bromobutane and methyl-bromopropane

            Functional group-Have the same molecular formula but different functional group

                        Example: Ethanol and Ethanone

Deduce structural formulas for the isomers of the non-cyclic alkanes up to C₆ AND apply IUPAC rules for naming the isomers of the non-cyclic alkane up to C₆

ORGANIC NAMING TABLE OF DOOM!!

Number of Carbons	Stem	Side Chain
1	Meth-	Methyl
2	Eth-	Ethyl
3	Prop-	Propyl
4	But-	Butyl
5	Pent-	Penthyl
6	Hex-	Hexyl
Benzene Ring	Benz-	Phenyl

Alkanes- Homologous series with all single-bonds. Noted by the suffix -ane

Example: A two-carbon chain would be ethane, and a  six-carbon chain would be hexane

Side chains- “Branches” of hydrocarbons which come off of the main chain

Example: A five-carbon chain with a CH₃ group hanging off the 2^nd carbon will be 2-methylpentate

Alkenes- Homologous series with at least one double-bond

Alkynes- Homologous series with at least one triple-bond

Saturated compounds are composed of all single bonds.

Unsaturated compounds have at least one double or triple bond.

Rules for double and triple bonds-Always put the number of the Carbon after which the double-bond occurs Imagine if the Carbons were numbered, the number would be “2” because the double-bond comes up right after C²

C¹H₃-C²H=C³H-C⁴H₂-C⁵H₃

CH₃-CH=CH-CH­₂­-CH­₃ is called Pent 2-ene

CH₃-CH₂-CH=CH-CH­₃ is still called Pent 2-ene because molecules have no set front or back. It will always be after the smallest numbered carbon

Rules for naming compounds with side chains- Write the number of the carbon from which it branches off, a dash, and then the name of the side chain (based off of the column from the ORGANING NAMING TABLE OF DOOM, above. If you have more than one  of each group, add prefixes (di-, tri-, tetra-). Make sure the compounds are in alphabetical order (Example: 2- Ethyl 3,4-diMethyloctane)

Deduce structural formulas AND Apply IUPAC rules for naming compounds containing up to six carbon atoms with one of the following functional groups: alcohol, aldehyde, ketone, carboxylic acid, and halide AND Identify the following functional groups when present in structural formulas: amino (NH₂), benzene ring, and esters.

Functional groups are groups of atoms that bond to carbons, creating a series of compounds with similar chemical and physical properties

FUNCTIONAL GROUP TABLE OF DOOM!

Name	Functional	Prefix/Suffix	Examples
Alkane	None	-ane	CH3-CH2-CH­3 (Propane­)
Alkene	C=C	-ene	CH3-CH­=CH2 (Propene)
Alkyne	C=C	-yne	CH=CH-CH­3­ (Propyne)
Alcohol	-OH	-anol (hydroxyl-)	CH3CH2OH (Ethanol)
Aldehyde	-CHO	-anal	CH3CH2CHO (Propanal)
Ketone	-CO	-anone	CH3CH­2COCH­3 (Butanone)
Carboxyl Acid	-COOH	-anoic acid	CH­3CH2CH­2COOH (Butanoic Acid)­
Halide	-F, -Cl, -Br, -I	Flouro-, Chloro-, Bromo-, Iodo-	CH3CH2CH2CH3F (Flouro-butane)

Identify primary, secondary, and tertiary carbon atoms in alcohols and halogenoalkanes

1º (Primary) Carbon- The carbon bound to the functional group (C₁) is bound to ONE other carbon

2º (Secondary) Carbon- C₁ binds to TWO other carbons

3º (Tertiary)  Carbon- C­­₁ binds to THREE other carbons

Primary alcohols/halogenoalkanes have functional group bound to primary carbon, etc…

Explain the low reactivity of alkanes in terms of bond enthalpies and bond polarity

• Alkanes have low bond enthalpy and are fairly stable, so they are unwilling to react under many conditions

• Alkanes are symmetrical and have non-polar (C-H) single bonds. These add to the unreactive nature

Descrube using equations, the complete and incomplete combustion of alkanes

Complete combustion always produces water and CO₂

2C₈H₁₈ + 25O₂ à 16CO₂ + 18H₂O

Incomplete combustion can produce carbon monoxide, hydrogen gas [H_{2 (g)}], [CO_(g)], and/or soot [C_(s)]

C₃H₈ + 4O₂ à CO₂ + 2CO + 4H₂O

***this occurs in the absence of sufficient O₂

Describe, using equations, the reactions of methane and ethane with chlorine and bromine

Alkane + Hal₂ à Halogenoalkane + HHal                 (With Heat/UV)

            Example: C₂H₆ + Br₂ à C₂H₅Br + HBr

Explain the reactions of methane and ethane with chlorine and bromine in terms of a free-radical mechanism

Free Radical Chain Reaction- The mechanism by which alkanes form halgenoalkanes. Involves three steps.

1. Initiation- Free radicals (atoms with unpaired electrons) are produced when heat/UV light splits Cl₂ into 2 Cl*
2. Propagation- Most of the product is made and radicals are reformed
1. Cl* + HCH₃ à HCl + *CH₃
2. CH₃ + Cl₂ à CH₃Cl + Cl*
3. Termination: The radicals are consumed and the reaction ends
1. Cl* + Cl* à Cl₂
2. Cl* + *CH₃ à CH₃Cl
3. CH₃ + CH₃ à CH₃CH₃

Describe, using equation, the reactions of alkenes with hydrogen halides and water

Alkene + HHal à Halogenoalkane                            (Room Temperature)

            Example: C₂H₄ + HCl à C₂H₅Cl

Alkene + H₂O à Alcohol                              (300^OC , 7 ATM, with H₂SO₄/H₃PO₄/Al₂O₃)

            Example: C₂H₄ + H₂O à C₂H₅OH

Alkene + H₂ à Alkane                                               (With HEAT + Ni catalyst)

            Example: C₂H₄ + H₂à C₂H₆

Distinguish between alkanes and alkenes using bromine water

When alkenes are mixed with bromine water, it is decolorized. The HBr will break the C=C double bond and convert the substance into a halogenoalkane, thus causing the solution to shift from brown à colorless.

When alkanes are mixed with bromine water, there is no change, because no reaction occurs

Outline the polymerization of alkenes

Polyelkene- Long chain of repeating monomers

Naming Polyelkenes- Prefix “poly-“ and the ORIGINAL COMPOUND as the suffix. So

            Example- propene would be polymerized into polypropene.

Repeating unit- One monomer of the original compound

Structural formula of polyelkenes- Show the repeating units, the bonds on either ends, parenthesis to show that the unit repeats, and n to show that it repeats any # of times

            Example: Structural formula of polyethene is  (-CH₂CH₂-)_n

            ***NOTE: Although this is polyethene, there are no double bonds

            Example: Structural formula of  polychloroethene is (-CHCl-CHCl-)_n

Outline the economic importance of the reactions of alkenes

• Hydrogenation of vegetable oils à margarine
• Hydration of Ethene à Ethanol
• Polymerization à manufacture of plastics

Describe, using equations, the complete combustion of alcohols

C₂H₅OH + 3O₂ à 2CO₂ + 3H₂O

C₄H₉OH + 6O₂ à 4CO₂ + 5H₂O

Etc…

Describe, using equations, the oxidation reactions of alcohols AND determine the products of formed by the oxidation of primary and secondary alcohols

1^o Alcohol à Aldehyde à Carboxylic Acid

***Oxidizing agent (Catalyst) is acidified potassium dichromate (H⁺/ K₂Cr₂O₇)

Alcohol + O – (H⁺/K₂Cr₂O₇)à Aldehyde + 2H⁺                   (Heat + Distillation)

C₂H₅OH + O -- (H⁺/K₂Cr₂O₇) à C­H₃CHO + H₂O

Aldehyde + O --(H⁺/K₂Cr₂O₇)à Carboxylic Acid               (Heat + Reflux)

CH₃CHO à O --(H⁺/K₂Cr₂O₇)à CH₃COOH

Alcohols have higher boiling points than aldehydes because while both have dipole-dipole interactions, alcohols can hydrogen bond, so they are stronger and need more energy to boil.

Aldehydes have higher boiling points than carboxylic acids because, although the carboxylic acids have a OH group, they do not hydrogen bond due to interference from the =O group.

2^o  Alcohol à Ketone

***Oxidizing agent (Catalyst) is acidified potassium dichromate and/or magnesium (IV) oxide (H⁺/K₂Cr₂O₇/MnO₄)

Alcohol + O --(H⁺/K₂Cr₂O₇/MnO₄)à Ketone + H₂O        (Heat + Distillation OR Reflux)

CH₃OHCH₃ + O à CH₃OCH₃ + H₂O

Describe, using equations, the substitution reactions of halogenoalkanes with NaOH

Halogenoalkane + NaOH à Alcohol + NaHal                     (Warm)

C₂H₅Cl + NaOH à C₂H₅OH + NaCl

Describe the substitution reactions of halogenoalkanes with sodium hydroxide in terms of SN₁ and SN₂ pathways

Nucleophilic Substitution-The mechanism by which halgenoalkanes form alcohols, nitriles, and amines. In nucleophilic substitution a Lewis base called a nucleophile donates a lone pair of electrons to the central atom, forming a dative covalent bond.

There are two potential pathways:

1.      S_N1 Pathway- Two step process. A tertiary halgenoalkane forms a carbocation intermediate before forming the end product.

Example: (CH₃)₃CCl + NaOH à

     (CH₃)₃C⁺ + Cl^- Na⁺ + OH^- à

     (CH₃)₃COH + NaCl

2.      S_N2 Pathway- One step process. The primary or secondary halogenoalkane immediately forms the product, with an incredibly brief transition state in between (where the electrons are transferring from the C-Hal bond to the C-Nucleophile bond).

Example: CH₃Cl + NaOH à

    CH₃ClOH (trans state) + Na⁺ à

    CH₃OH + NaCl

Examples of nucleophiles- H₂O, OH^-, NH₃, CN^-

Heterolytic Fission- The division of a band that causes an unequal distribution of electrons (i.e. does form ions). This occurs in the S_N1 Pathway

Homolytic Fission- The division of a bond that causes an equal distribution of electrons (i.e. does not form ions). This occurs in the S_N2 Pathway

Carbocation intermediate- A positively charged atom formed after the halide ion forms, but before the nucleophile bonds

Deduce reaction pathways given the starting materials and the products

THIS OBJECTIVE SUMMARIZES A LOT OF THE PREVIOUS OBJECTIVES