Topic 4: Introductory Organic Chemistry AS and Alkanes

Concentrated hydrochloric acid is corrosive: that is its hazard. Using a large open beaker near eye level creates a higher risk than handling a few drops in a tray while wearing eye protection, because exposure is more likely or consequential.

Risk assessment therefore asks: what harm can occur, who or what could be exposed, how likely and severe is it, and which controls reduce that risk to an acceptable level?

A risk assessment identifies the substances, quantities, concentrations, temperatures, apparatus, exposure routes and people involved. It then selects controls, records emergency action and considers disposal before work begins.

Organic compounds are often volatile and flammable, so vapour can spread beyond the vessel and meet an ignition source. The same chemical can present very different risks at microscale and at bulk scale.

Prefer controls that remove or contain the hazard before relying only on personal protective equipment. Check ventilation, heating method, secure apparatus, spill response and waste route for the actual procedure.

For carbon monoxide, improve containment and ventilation, use a fume cupboard or suitable extraction, monitor where necessary, and limit exposure time. Gloves alone do not control an inhalation hazard.

A functional group is the atom or group of atoms responsible for the characteristic reactions of an organic compound. Examples include C=C in alkenes, –OH in alcohols and –COOH in carboxylic acids.

Choose the longest parent chain containing the principal functional group and any required multiple bond. Number from the end giving the lowest relevant locants. Name and alphabetise substituents, use di-, tri- or tetra- for repeats, then add the suffix and locant for the functional group.

CH$_3$C(CH$_3$)$_2$CH$_2$CH(CH$_3$)CH$_3$ has a five-carbon parent chain and methyl groups at 2, 2 and 4, so its name is 2,2,4-trimethylpentane. Commas separate numbers and hyphens separate numbers from words.

Compare bonds and functional groups in reactants and products. Classify the actual transformation, not the reagent name: converting C=C to C–C by adding atoms is addition, while replacing C–H by C–Cl is substitution.

Under ultraviolet light, Cl–Cl can split homolytically: $\mathrm{Cl_2 \rightarrow 2Cl\boldsymbol{\cdot}}$. H–Cl can be represented as breaking heterolytically to H$^+$ and Cl$^-$ when both electrons go to chlorine.

After drawing arrows, count electrons and charges in every product. A radical has an unpaired electron; heterolysis produces opposite charges whose total equals the reactant charge.

The unpaired electron makes many radicals highly reactive. An electrophile is electron-deficient and is attracted to an electron-rich region such as a lone pair or a $\pi$ bond.

A hydrocarbon contains carbon and hydrogen only. Saturated means that it contains no carbon–carbon multiple bond, so each carbon has the maximum number of hydrogen atoms allowed by its C–C connectivity.

Closing an alkane chain to make one ring removes two hydrogen atoms, which explains the change from C$_n$H$_{2n+2}$ to C$_n$H$_{2n}$. For example, propane is C$_3$H$_8$ and cyclopropane is C$_3$H$_6$.

C$_4$H$_{10}$ can be CH$_3$CH$_2$CH$_2$CH$_3$ (butane) or CH$_3$CH(CH$_3$)CH$_3$ (2-methylpropane). Both contain four carbons and ten hydrogens, but the carbon skeletons differ.

To find isomers, vary the carbon skeleton, functional-group position or functional group where the formula permits it. After each drawing, recount atoms, check valencies, assign an IUPAC name and reject any structure that is just a rotated or renumbered duplicate.

For a cycloalkane, choose a ring size from three up to the total carbon count, then distribute the remaining carbons as substituents. Number substituted rings to give the lowest set of locants and reject rotations, reflections and alternative numbering of the same connectivity.

C$_4$H$_8$ gives cyclobutane and methylcyclopropane. For C$_5$H$_{10}$, valid connectivities include cyclopentane, methylcyclobutane, ethylcyclopropane, 1,1-dimethylcyclopropane and 1,2-dimethylcyclopropane. This same ring-size method extends to six carbons.

Use displayed, structural or skeletal formulae consistently. In a skeletal ring, every unlabelled vertex is carbon and enough hydrogens are implied to give carbon four bonds.

Alkanes burn exothermically, so fractions containing them are used as fuels. Supply and demand rarely match crude-oil composition, so cracking increases smaller fuels and alkene feedstocks, while reforming improves combustion quality.

For every cracking or reforming equation, conserve the number of carbon and hydrogen atoms. Fractional distillation has no reaction equation because no covalent bonds change.

Pollutant formation depends on fuel composition and combustion conditions. Improving oxygen mixing can reduce CO and soot, but high combustion temperature can favour nitrogen-oxide formation.

Acid deposition lowers the pH of soils and surface waters and can damage carbonate stone and living systems. In an enclosed space, incomplete combustion makes CO especially dangerous because it is colourless and toxic.

Alternative fuels are developed to reduce dependence on finite crude oil, improve long-term security of supply and reduce harmful or greenhouse-gas emissions over the fuel's life cycle.

Combustion of fossil alkanes transfers geologically stored carbon to atmospheric CO$_2$. CO$_2$ absorbs outgoing infrared radiation, so increasing its concentration strengthens the greenhouse effect and contributes to climate change.

State the system boundary, identify every carbon or energy input, compare atmospheric uptake with emissions, and give a conditional conclusion rather than relying on a label such as 'bio' or 'hydrogen'.

To balance complete combustion of C$_x$H$_y$, form $x$CO$_2$ and $y/2$H$_2$O, then balance O$_2$. Include state symbols when requested.

The carbon skeleton remains while a C–H bond and X–X bond are replaced by C–X and H–X bonds. Ultraviolet radiation initiates radical formation.

A curly half-arrow moves one electron. In each propagation step, use half-arrows so bond breaking and bond formation account for the unpaired electron without inventing charge.

The product can undergo further substitution because it still contains C–H bonds. Longer alkanes can also substitute at different carbon positions. The resulting mixture of products and isomers makes the reaction poorly selective for synthesis.