IB Chemistry S2.4: From models to materials worked solutions

Take one electronegativity for each element, not one per atom in the formula. The coordinate is (0.9+2.7)/2=1.8. Its value has no unit. Weighting iodine twice instead gives the distracting value 2.1.

Unequal site sizes interrupt the uniform arrangement of the starting metal. This makes permanent slip harder and can increase deformation resistance. Thermal vibration continues, and mobile electrons can still carry current.

Trace the two carbon sites joined by the double bond. They become the two backbone sites. Both methyl and ethyl started on the substituted site, so both remain attached there. The other three carbon atoms are side-group atoms.

With repeat chemistry and chain length held comparable, fewer branches allow neighbouring chains to approach and pack more efficiently. The expected larger mass per bulk volume belongs to V. No change of repeat molar mass or ionic bonding is supplied.

A chain must continue from each end of the partner molecule. The four-carbon diol has a reactive hydroxyl at both ends. Ethanol can make one ester but then caps that end; the amine and hydrocarbon are not the required two-ended alcohol.

Restore each ester to an alcohol end and a carboxyl end, including the link completed by the next repeat. The alcohol-derived spacer contains four methylene groups; the acid-derived spacer contains three methylene groups plus its two carbonyl carbons. The recovered partners are \mathrm{HO(CH_2)_4OH} and \mathrm{HOOC(CH_2)_3COOH}.

The supplied absorptions distinguish bond types at this level of evidence. Different lengths and arrangements can share those bonds. Similar absorptions therefore leave chain architecture unresolved; they cannot by themselves prove equal chain lengths or assign a polymerisation route.

The backbone pair is CH2 and CH, with propyl attached to the latter. Remove the continuing bonds and put C=C between this pair. The recovered structure is \mathrm{CH_2=CHCH_2CH_2CH_3}; moving the double bond into the propyl group would change the attachments.

(a) Retain F and ethyl on the substituted carbon. Replace C=C by a single backbone bond, extend a bond from both carbon sites, and place brackets with n around one repeat. This gives [-\mathrm{CH_2-C(F)(CH_2CH_3)-}]_n. [3] (b) The alternative gives [-\mathrm{CH(F)-CH(CH_2CH_3)-}]_n. Its F and ethyl groups lie on different backbone sites, so the connectivity is different. The common atom inventory \mathrm{C_4H_7F} is insufficient to identify a polymer. [2]

(a) Average the two element values once each: (1.3+3.2)/2=2.25. Their absolute difference is 1.9. Both coordinates are dimensionless. [2] (b) The relatively large difference supports predominantly ionic bonding. In this model, ions in the solid are constrained to lattice sites and cannot transport charge through the sample. Melting allows ions of both signs to move, so the melt has mobile carriers. [3]

(a) Differently sized atoms disturb the regular array. The disturbed arrangement impedes permanent layer displacement, increasing resistance to indentation. [2] (b) Deformation resistance helps the clip retain its shape; lower corrosion helps it survive chemical exposure. Mechanical slip and surface reactions differ, so the indentation result alone would not establish the corrosion benefit. [2]

(a) Open the C=C in each molecule. Each contributes two backbone carbons, so three molecules give six backbone sites. The three CH2Cl groups remain on alternating CH sites. The section is -\mathrm{CH_2-CH(CH_2Cl)-CH_2-CH(CH_2Cl)-CH_2-CH(CH_2Cl)}-, with a continuing bond at each end. [3] (b) This is addition polymerisation, without loss of a small molecule. [1]

(a) London attractions act between separate hydrocarbon chains. Heating permits more relative motion as these attractions are overcome; ordinary softening does not require severing the backbone. [2] (b) Resistance to reaction with the specified acid is a chemical property that supports durability in that setting. Flexibility describes mechanical response without changing chemical identity and is physical. The observation does not prove resistance to every reagent. [2]

(a) Molecules inserted between chains increase their separation, reducing effective interchain attractions. Chains can move past one another more readily, increasing flexibility. [2] (b) The additive is not covalently fixed and may migrate to a surface or into a contacting material. Losing it can restore closer chain contact, so the coating may become stiffer. Neither toxicity nor the rate of loss follows from this structure alone. [2]

(a) Remove OH from the A carboxyl and H from the B amino group. Join the carbonyl C directly to N, giving \mathrm{H_2NCH(CH_3)C(=O)NHCH_2COOH}. The new C(=O)-NH group is an amide (peptide) linkage; one water is removed. The free amino and acid ends remain. [3] (b) B–A places CH2 next to the free amino end and CH(CH3) next to the free acid end. The residue order and connectivity differ. [1]

(a) Break the C(=O)-N bond within the amide; the free COOH group is not an additional peptide linkage. [1] (b) Add OH to the carbonyl carbon to restore COOH, and H to N to restore NH2. Both fragments are \mathrm{H_2NCH_2COOH}. One bond is hydrolysed, so one H2O is consumed. Acidic or alkaline reaction media would change protonation; the requested structures are neutral. [3]

(a) Using \rho=m/V gives \rho_X=9.20/10.00=0.920\,\mathrm{g\,cm^{-3}} and \rho_Y=9.60/10.00=0.960\,\mathrm{g\,cm^{-3}}. [2] (b) Use the requested first-order sum of relative absolute limits: 100(0.02/9.20+0.05/10.00)=0.71739\ldots\%\approx0.72\%. The absolute result is 0.920(0.02/9.20+0.05/10.00)=0.0066\,\mathrm{g\,cm^{-3}}, or 0.007 at one uncertainty digit. This is a limits convention, not a statistical confidence interval. [2] (c) Less-branched Y has greater mass per volume. More efficient packing leaves less volume for a given mass, consistent with the result. [2] (d) Repeating specimens shows whether the difference persists beyond random specimen variation. Chemistry, preparation and temperature should also be comparable before assigning the whole effect to branching. [1]

(a) The paired repeat is [-\mathrm{OCH_2CH(F)CH_2OC(=O)(CH_2)_4C(=O)-}]_n. The diol retains F and its alcohol oxygens; the diacid retains both carbonyls and the four-methylene spacer. Continue both terminal bonds through the brackets and write n outside. Under the supplied no-exchange rule, acid hydroxyl oxygens leave at the internal joins, so the periodic repeat has no label. [4] (b) Fourteen starting molecules require thirteen joins to form a single open chain. Each join removes one labelled acid OH in water, so thirteen labelled waters form. Fourteen initial labels minus thirteen lost leaves one label on the hydroxyl oxygen of the terminal carboxyl group. The terminal alcohol and carbonyl oxygens are unlabelled. [2]

(a) Let the values be x and y, with y greater. Then x+y=5.2 and y-x=1.2. Adding gives 2y=6.4, hence y=3.2 and x=2.0. [2] (b) The coordinates support a bonding description but do not specify whether the solid contains separate molecules or an extended network. A covalent network cannot melt by merely separating whole chains: its continuous covalent connections must be disrupted. A thermoplastic can soften through increased motion between chains as interchain attractions are overcome. A high mean alone therefore does not justify either melting prediction. [3]

(a) Cut the amide connections and restore H on each N and OH on each carbonyl carbon. This gives \mathrm{H_2N(CH_2)_5NH_2} and \mathrm{HOOCCH(CH_3)COOH}. The last carbonyl in one repeat bonds directly to the first NH of the next: C(=O)-NH. [3] (b) Amino ends require a diamine residue at each end. Alternating nine residues must therefore contain five diamines and four diacids. Nine initially separate molecules need eight joining events to form one open chain, so eight waters are eliminated. All eight acid groups have reacted while two amino groups remain. [3]

(a) Select P for the stated water exposure because it retains strength in the trial. Water can hydrolyse the ester links in the backbone of Q, cutting its chains and reducing their ability to maintain a continuous pouch. P has no corresponding ester bond in its C–C backbone, so that cleavage route is absent. This comparison does not establish resistance to every temperature or reagent. [3] (b) The different outcomes in warm water and cold, dry storage show that breakdown depends on conditions. Fragmentation into smaller pieces does not by itself prove complete biological conversion to harmless small molecules. The trial gives no information about ecological effects or all disposal settings, so the universal claim is unsupported; an appropriate collection or treatment route still matters. [3]

(a) The two alkene carbons become a single-bonded backbone pair, with ethyl retained on CH: [-\mathrm{CH_2-CH(CH_2CH_3)-}]_n. Show both continuing bonds. [2] (b) Initial amount =11.22/56.1=0.200\,\mathrm{mol}. Addition removes no small product, so incorporated mass =0.750(11.22)=8.415\,\mathrm{g}\approx8.42\,\mathrm{g} and amount of repeat units =0.750(0.200)=0.150\,\mathrm{mol}. [2] (c) Amount of chains =0.150/1000=1.50\times10^{-4}\,\mathrm{mol}. The backbone consumes alkene C=C, so diminished C=C-related IR evidence can support conversion. Such a change alone does not determine chain length or prove all starting monomer reacted. [2]

(a) n(\mathrm{Cu})=40.0/63.55=0.629426\,\mathrm{mol} and n(\mathrm{Zn})=10.0/65.38=0.152952\,\mathrm{mol}. The Zn atom fraction is 0.152952/(0.629426+0.152952)=0.195496\approx0.195. Its mass fraction is 10.0/50.0=0.200; the slightly larger Zn molar mass makes its atom fraction smaller. [2] (b) Unequal sizes disturb the regular array and impede permanent layer displacement, increasing indentation resistance. [2] (c) Delocalised electrons remain mobile through the metal-based solid and carry current; the metal cores need not migrate. Corrosion resistance under the intended outdoor exposure needs a separate test, since hardness does not establish chemical durability. [3]

(a) Ester cleavage restores the alcohol and acid ends: \mathrm{HO(CH_2)_3OH} and \mathrm{HOOC(CH_2)_4COOH}. The four acid methylenes exclude its two carbonyl carbons. [2] (b) Six monomer residues in one open chain require five joining links. Hydrolysing each link consumes one water, so complete cleavage requires five waters per chain. The periodic-repeat drawing does not set the finite-chain end count. [2] (c) Two COOH groups require two OH- per diacid. Thus n(\mathrm{NaOH})=2(0.0100)=0.0200\,\mathrm{mol} and V=n/c=0.200\,\mathrm{dm^3}=200\,\mathrm{cm^3}. The product is \mathrm{{^-}OOC(CH_2)_4COO^-} with total charge 2-. Each COOH donates a proton to hydroxide, forming water and carboxylate; the diol is not counted as another carboxylic acid. [4]