4.7 Organic chemistry — coverage pack

Answer

• It contains carbon and hydrogen only.
• For $n=3$, $2n+2=8$, so it fits $\mathrm{C_nH_{2n+2}}$.

Method: First inspect the elements: C₃H₈ contains only carbon and hydrogen, so it is a hydrocarbon. Then substitute $n=3$ into the alkane formula: $2(3)+2=8$, matching the eight hydrogen atoms.

Answer

• C₇H₁₆
• Using $\mathrm{C_nH_{2n+2}}$ with $n=7$ gives $2n+2=16$ hydrogen atoms.

Method: Use $\mathrm{C_nH_{2n+2}}$ with $n=7$. The number of hydrogen atoms is $2(7)+2=16$, so the molecular formula is C₇H₁₆.

Answer

• C₄H₁₀
• Butane

Method: Write the alkane as $\mathrm{C_nH_{2n+2}}$. Its relative molecular mass is $12n+(2n+2)=14n+2$. Solve $14n+2=58$ to obtain $14n=56$ and $n=4$. The formula is therefore C₄H₁₀, which is butane.

Answer

• A mixture of hydrocarbons with similar numbers of carbon atoms.
• Their different boiling points allow separation.

Method: A fraction is not a single pure hydrocarbon; it is a group of hydrocarbons of similar molecular size. Their boiling points differ from those in other groups, enabling fractional distillation.

Answer

• Hydrocarbon B condenses lower in the column.
• B has the higher boiling point, so it condenses at a higher temperature.
• The column is hotter at the base, so vapours with higher boiling points condense lower down.

Method: The bottom of the column is hotter than the top. B condenses when the vapour cools below $210\,^\circ\mathrm{C}$, which happens relatively low in the column. A remains a vapour until it reaches a cooler, higher region near $75\,^\circ\mathrm{C}$.

Answer

• The hydrocarbons vaporise, cool as they rise, and condense at heights determined by their boiling points.
• Fractions are feedstock for making useful petrochemicals such as polymers, solvents, lubricants or detergents.

Method: Heat the crude oil so that its hydrocarbons enter the column mainly as vapours. The temperature falls up the column. Each hydrocarbon condenses where the temperature becomes lower than its boiling point, so similar-sized molecules are collected together. Some fractions are processed as petrochemical feedstock, supplying carbon compounds used to manufacture materials such as polymers and solvents.

Answer

• Boiling point increases.
• Viscosity increases.

Method: Recall the molecular-size trends: larger hydrocarbons have stronger intermolecular attractions, so they boil at higher temperatures and flow less readily.

Answer

• $\mathrm{C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O}$

Method: Balance the three carbon atoms with $3\mathrm{CO_2}$ and the eight hydrogen atoms with $4\mathrm{H_2O}$. The products then contain $6+4=10$ oxygen atoms, so use $5\mathrm{O_2}$.

Answer

• Q has the higher boiling point and viscosity but is less flammable than P.
• $\mathrm{C_{15}H_{32} + 23O_2 \rightarrow 15CO_2 + 16H_2O}$

Method: Q is the larger molecule, so it has stronger intermolecular forces: its boiling point and viscosity are higher, while its flammability is lower. For combustion, use $15\mathrm{CO_2}$ for carbon and $16\mathrm{H_2O}$ for hydrogen. These products contain $30+16=46$ oxygen atoms, requiring $23\mathrm{O_2}$.

Answer

• An alkene.

Method: The orange-to-colourless bromine-water result is the characteristic test for the carbon-carbon double bond in an alkene.

Answer

• X is C₄H₈.
• X is an alkene.

Method: Conserve atoms. After forming C₈H₁₈, four carbon atoms and eight hydrogen atoms remain, so X is C₄H₈. This fits $\mathrm{C_nH_{2n}}$, so it is an alkene.

Answer

• The other product is C₆H₁₂.
• Vaporise the hydrocarbon and pass the vapour over a hot catalyst.
• It supplies high-demand short-chain fuels and alkenes used for polymers or other chemicals.

Method: Subtract the atoms in C₉H₂₀ from C₁₅H₃₂, leaving C₆H₁₂, an alkene. For catalytic cracking, vaporise the feed and contact it with a hot catalyst. The reaction turns surplus long molecules into more flammable short-chain fuels and valuable alkene feedstock.

Answer

• C₅H₁₀

Method: Pentene contains five carbon atoms, so set $n=5$ in $\mathrm{C_nH_{2n}}$. This gives ten hydrogen atoms and the formula C₅H₁₀.

Answer

• Propene.
• It is unsaturated because it contains a carbon-carbon double bond.

Method: Three carbon atoms give the prefix prop-. A C=C bond identifies the alkene ending -ene, so the name is propene. The double bond is also the structural reason it is described as unsaturated.

Answer

• The alkene is C₄H₈, butene.
• The corresponding alkane is C₄H₁₀.

Method: For $\mathrm{C_nH_{2n}}$, the relative molecular mass is $12n+2n=14n$. Solve $14n=56$ to get $n=4$, giving C₄H₈, butene. The alkane formula is $\mathrm{C_nH_{2n+2}}$, so with four carbons it is C₄H₁₀.

Answer

• Orange to colourless.
• Addition reaction.

Method: Bromine adds across ethene's C=C bond. Its orange colour disappears as bromine is consumed, so the reaction is addition.

Answer

• $\begin{array}{c}\quad \mathrm{H}\qquad\mathrm{H}\\ \quad\vert\qquad\vert\\ \mathrm{H-C-C-H}\\ \quad\vert\qquad\vert\\ \quad \mathrm{H}\qquad\mathrm{H}\end{array}$
• Ethane.
• Nickel catalyst.

Method: Add one hydrogen atom to each carbon of C=C and replace the double bond with a single C-C bond. Draw every C-H bond: each carbon in ethane has three hydrogen atoms and one bond to the other carbon. The hydrogenation uses a nickel catalyst.

Answer

• It is an alkene: incomplete combustion can make a smoky flame, and bromine adds across C=C.
• React it with steam using a phosphoric acid catalyst; ethene forms ethanol.

Method: A smoky flame indicates incomplete combustion, which alkenes tend to show in air. Decolourising bromine water confirms a reactive C=C bond because bromine adds across it. Hydration adds water across the double bond: pass ethene with steam over a phosphoric acid catalyst to produce ethanol.

Answer

• Ethanol.
• The -OH functional group.

Method: The molecule contains two carbon atoms, giving the prefix eth-. The -OH group gives the alcohol ending -anol, so the compound is ethanol.

Answer

• Yeast.
• Warm conditions and absence of oxygen.
• Carbon dioxide.

Method: Add yeast to the glucose solution and keep it warm so its enzymes work. Exclude oxygen so fermentation rather than aerobic respiration occurs. The sugar is converted into ethanol and carbon dioxide.

Answer

• $\mathrm{C_2H_5OH + 3O_2 \rightarrow 2CO_2 + 3H_2O}$
• Ethanoic acid.
• The oxidising agent supplies oxygen or removes hydrogen from ethanol.

Method: For combustion, balance carbon with $2\mathrm{CO_2}$ and hydrogen with $3\mathrm{H_2O}$. The products then contain seven oxygen atoms, one already in ethanol, so $3\mathrm{O_2}$ supplies the other six. Oxidation converts ethanol into ethanoic acid; the oxidising agent causes this gain of oxygen or loss of hydrogen.

Answer

• Ethanoic acid.
• The -COOH functional group.

Method: Count both carbon atoms, including the one in -COOH, to obtain the eth- prefix. The -COOH group gives the name ethanoic acid.

Answer

• Fizzing or effervescence is seen.
• A calcium ethanoate salt, water and carbon dioxide are formed.

Method: Use the general acid-carbonate reaction: acid plus carbonate produces a salt, water and carbon dioxide. The escaping carbon dioxide causes fizzing; ethanoic acid forms an ethanoate salt.

Answer

• Add a carbonate: ethanoic acid fizzes and releases carbon dioxide, whereas ethanol does not.
• Ethyl ethanoate is formed, together with water.

Method: Add the same small amount of carbonate to separate samples. Only ethanoic acid gives effervescence because an acid-carbonate reaction releases carbon dioxide. When ethanol and ethanoic acid react, the alcohol and acid combine in a condensation reaction to make the ester ethyl ethanoate and water.

Answer

• Ethene.
• Addition polymerisation.

Method: The name inside poly(ethene) identifies the alkene monomer as ethene. Alkene monomers join by opening C=C bonds, which is addition polymerisation.

Answer

• $[\mathrm{-CH_2-CH(CH_3)-}]_n$
• C=C becomes C-C as new bonds link the monomers.

Method: Use the two double-bonded carbon atoms as the polymer backbone, retain the CH₃ side group, and replace C=C with C-C. Put the unit in brackets with bonds passing through the brackets: $[\mathrm{-CH_2-CH(CH_3)-}]_n$.

Answer

• $\mathrm{CH_2{=}CHCl}$
• The double bond opens and all monomer atoms remain in the polymer, so no small molecule is eliminated.

Method: Take one pair of backbone carbons and restore a double bond between them, keeping every attached atom in place. This gives $\mathrm{CH_2{=}CHCl}$. Addition polymerisation only rearranges the bonding around C=C, so the repeating unit has the same atoms as the monomer and there is no by-product.

Answer

• Each monomer has two functional groups.
• The monomers join repeatedly while a small molecule, commonly water, is released.

Method: Two reactive ends are needed so a monomer can connect on both sides and extend the chain. Each condensation link usually forms with the elimination of water.

Answer

• Ethanediol has two -OH groups and butanedioic acid has two -COOH groups.
• A polyester forms and water is eliminated.

Method: The diol supplies an -OH group at each end, while the dicarboxylic acid supplies two -COOH groups. Alcohol and carboxylic-acid groups form ester links, so the product is a polyester and each new link eliminates water.

Answer

• $[\mathrm{-O-CH_2-CH_2-O-CO-CH_2-CH_2-CO-}]_n$
• -OH and -COOH groups react to form ester links, eliminating water.

Method: Remove H from an alcohol -OH and OH from a carboxyl -COOH at each joining point; these form water. The remaining oxygen connects to the carbonyl carbon, producing ester links. Keeping both carbon chains gives $[\mathrm{-O-CH_2-CH_2-O-CO-CH_2-CH_2-CO-}]_n$.

Answer

• An amino group, -NH₂.
• A carboxyl group, -COOH.
• Glycine.

Method: Read the end groups separately: H₂N- is the amino group and -COOH is the carboxyl group. The specified amino acid with formula H₂NCH₂COOH is glycine.

Answer

• H₂NCH₂CONHCH₂COOH
• A peptide link, -CONH-, and water.

Method: Join the -COOH group of one glycine to the -NH₂ group of the other. Remove OH from -COOH and H from -NH₂ to make water. The remaining groups form H₂NCH₂CONHCH₂COOH with a -CONH- peptide link.

Answer

• Three peptide links and three water molecules.
• Different amino acids can be arranged in different sequences, producing different polypeptides or proteins.

Method: Joining four separate molecules into one chain requires three joining reactions. Each condensation reaction makes one peptide link and releases one water molecule, so there are three of each. Changing the amino acids or their order changes the sequence of groups along the polypeptide, giving a different protein.

Answer

• Nucleotides.
• Four different nucleotides.

Method: Recall that each DNA chain is a nucleotide polymer. The specification identifies four different nucleotide monomers.

Answer

• Protein - amino acids.
• Starch - glucose.
• Cellulose - glucose.
• DNA - nucleotides.

Method: Proteins are polypeptides, so their monomers are amino acids. Starch and cellulose are both glucose polymers. DNA is formed from nucleotide monomers.

Answer

• The claim is incorrect: DNA uses four different nucleotides, proteins use amino acids, and starch uses glucose.
• Most DNA has two polymer chains arranged in a double helix.

Method: Being naturally occurring does not determine a polymer's monomer. Identify each separately: nucleotides build DNA, amino acids build proteins, and glucose builds starch. DNA is further distinguished because most DNA molecules contain two polymer chains coiled into a double helix.