18C.18-18C.21 - Organic Synthesis And Chain Extension

Identifying Organic Structures

An unknown organic liquid may have no obvious colour, no useful smell and no single test that proves its structure. Synthesis questions often start by making you deduce the target first: a formula, a functional group, a structural formula, or a set of possible structures. A synthesis route is only as good as the structure it is trying to make.

Use evidence in a layered order:

The key habit is to make each conclusion proportional to its evidence. A broad acid $\ce{O-H}$ absorption and a carbonyl absorption support a carboxylic acid group; they do not automatically identify the exact acid unless the formula and environment data also fit.

Formula routes

For percentage composition: percentage mass → divide by $A_{\mathrm r}$ → mole ratio → empirical formula.

For combustion analysis: moles of $\ce{CO2}$ → moles of C; moles of $\ce{H2O}$ → moles of H atoms ($\times 2$); mass of C and H → mass of O by difference, if oxygen is present; mole ratio → empirical formula.

Do not round the ratio too early. If the ratio is near $1 : 2 : 1$, write $\ce{CH2O}$; if it is near $1 : 2.5 : 1$, multiply all parts by 2 before writing the empirical formula.

Worked example: from data to structure

An unknown compound $\ce{X}$ contains only carbon, hydrogen and oxygen. Its percentage composition is 54.5% C, 9.1% H and 36.4% O. Its molecular ion peak is at m/z = 88. It has a strong IR absorption in the carbonyl region, no broad $\ce{O-H}$ absorption, no effervescence with sodium carbonate solution, and the following supplied proton NMR summary: three proton environments in the ratio $3 : 2 : 3$; a singlet at $2.05\,\mathrm{ppm}$ for $\ce{3H}$, a quartet at $4.10\,\mathrm{ppm}$ for $\ce{2H}$, and a triplet at $1.25\,\mathrm{ppm}$ for $\ce{3H}$.

Formula first:

The empirical formula is $\ce{C2H4O}$, with empirical formula mass $44$.

$88 \div 44 = 2$, so the molecular formula is $\ce{C4H8O2}$.

Functional-group evidence: the strong carbonyl absorption shows a $\ce{C=O}$ group. The absence of a broad $\ce{O-H}$ absorption and the lack of effervescence with sodium carbonate argue against a carboxylic acid. With formula $\ce{C4H8O2}$, an ester is a strong possibility.

NMR evidence: the quartet and triplet in a $2\mathrm{H} : 3\mathrm{H}$ pattern support an ethyl group, $\ce{-CH2CH3}$. The quartet at $4.10\,\mathrm{ppm}$ places the $\ce{CH2}$ next to oxygen. The $\ce{3H}$ singlet at $2.05\,\mathrm{ppm}$ is consistent with a methyl group next to a carbonyl carbon, $\ce{CH3CO -}$, with no adjacent hydrogens on the carbonyl carbon.

Deduction: the structural formula is $\ce{CH3COOCH2CH3}$, ethyl ethanoate.

Common-error contrast

Do not write "carbonyl peak, therefore ketone" when the molecular formula contains two oxygen atoms and other data suggest an ester or acid. A better answer says what the carbonyl proves, then uses the missing acid $\ce{O-H}$, reaction data and NMR evidence to narrow the class and structure.

Worked combustion analysis

A $0.740\,\mathrm{g}$ compound contains C, H and O only and gives $1.760\,\mathrm{g}$ of $\ce{CO2}$ and $0.900\,\mathrm{g}$ of $\ce{H2O}$ on complete combustion. Using $A_r(\ce C)=12.0$, $A_r(\ce H)=1.0$, $A_r(\ce O)=16.0$:

The carbon and hydrogen masses are $0.480$ and $0.100\,\mathrm g$, leaving $0.160\,\mathrm g$ of oxygen, or $0.0100\,\mathrm{mol}$. The atom ratio $0.0400:0.100:0.0100=4:10:1$ gives $\ce{C4H10O}$. The factor 2 belongs to hydrogen atoms because each water molecule contains two H atoms. Oxygen by difference is valid only because no other elements are present.

Planning Routes Backwards

Synthesis planning is a controlled search through functional-group conversions. Start with the target, identify the functional group that must be made, then ask what precursor could give that group in one known step. Work backwards until the starting material is available, then write the route forwards with reagents and conditions.

target functional group ← nearest precursor ← earlier precursor ← starting compound

Routes may have up to four steps and may involve unfamiliar compounds. That does not mean unfamiliar reactions: it means familiar transformations applied to a new carbon skeleton.

The route must preserve the carbon count unless a step deliberately changes it. Carbon-chain extension is a decision, not an accident.

Worked route: making an ester from an alcohol

Task: plan a route from propan-1-ol to ethyl propanoate. Ethanol is also available.

Work backwards:

ethyl propanoate ← propanoic acid + ethanol
propanoic acid ← propan-1-ol

Write forwards:

1. Oxidise propan-1-ol to propanoic acid using acidified potassium dichromate(VI) under reflux.
2. React propanoic acid with ethanol using a concentrated acid catalyst under reflux to form ethyl propanoate.

Why this route works: the carbon skeleton from propan-1-ol becomes the acyl part of the ester, $\ce{CH3CH2CO -}$. Ethanol supplies the alkoxy part, $\ce{-OCH2CH3}$. The route uses two familiar transformations rather than trying to replace the $\ce{-OH}$ group directly with an ester group.

Guided practice

Plan a route from propan-1-ol to propanamide.

Expected route: oxidise propan-1-ol to propanoic acid using acidified potassium dichromate(VI) under reflux; convert propanoic acid to propanoyl chloride using phosphorus(V) chloride, $\ce{PCl5}$; react propanoyl chloride with ammonia to form propanamide. A strong answer names each intermediate and pairs each reagent with the relevant condition.

A complete four-transformation example is ethanol → bromoethane → propanenitrile → propanoic acid → methyl propanoate. Use HBr (or a suitable bromide/acid preparation) for the alcohol-to-halide conversion; ethanolic KCN and heat under reflux for the nitrile; dilute acid and reflux for hydrolysis; then methanol with an acid catalyst and heat for esterification. Isolate the needed intermediate between incompatible stages. The cyanide step alone extends the original two-carbon chain; methanol later supplies the separate ester alkyl group.

Forming Grignard Reagents

Grignard reagents give a way to extend carbon chains. They are made by reacting an organic halide with magnesium in dry ether:

$\ce{R-X + Mg -> R-MgX}$

In many A level routes, $\ce{X}$ is bromine or iodine because those compounds usually form Grignard reagents more readily than chlorides.

The important electronic model is that the carbon attached to magnesium behaves as a powerful carbon nucleophile. It attacks electron-poor carbon atoms, especially the carbon in carbon dioxide or in a carbonyl group. This is why Grignard chemistry can form a new $\ce{C-C}$ bond.

Grignard reagents are destroyed by water, alcohols, carboxylic acids and other proton sources:

$\ce{R-MgX + H2O -> R-H + Mg(OH)X}$

That reaction wastes the carbon nucleophile before it can extend the chain. Dry ether is used because it dissolves or supports the organomagnesium reagent while excluding water. The glassware, solvent and reagents must be dry for the chemistry to work.

Mini worked example

Question: name the Grignard reagent formed from bromoethane, and explain why water must be absent.