Q.Account for the following:

Step 1 (i). t-Butyl chloride, (CH3)3C-Cl, ionising at its C-Cl bond gives a TERTIARY carbocation, stabilised by hyperconjugation and the +I effect of three alkyl groups -- stable enough to form readily, so the reaction proceeds by the two-step SN1 pathway (ionise first, then the nucleophile attacks the planar cation from either face).

Step 2 (i, contd). n-Butyl chloride, CH3CH2CH2CH2-Cl, is PRIMARY; a primary carbocation would be far too unstable to form at any reasonable rate (almost no alkyl-group stabilisation), so the SN1 pathway is not viable here. Its C-Cl carbon is also sterically OPEN (only one other carbon attached), so the nucleophile can approach freely from the back -- making the single-step SN2 pathway both accessible and fast.

Step 3 (i, contd). So the same nucleophile (aqueous KOH) reacts with the two substrates by genuinely different mechanisms, purely because of how substituted (and therefore how carbocation-stabilising or sterically open) each C-X carbon is.

Step 4 (ii). p-Dichlorobenzene has both chlorines symmetrically placed (1,4-), giving the molecule an overall symmetric, 'straight' shape that packs very efficiently and regularly into a crystal lattice, maximising intermolecular contact.

Step 5 (ii, contd). The less symmetric ortho (1,2-) and meta (1,3-) isomers cannot pack as tightly or as regularly, so their crystals are held together less strongly. …

For (i), connect substrate substitution level to which mechanism its carbocation stability (or lack of it) and steric openness favours; for (ii), connect molecular SYMMETRY specifically to crystal packing efficienc …

• For (ii), reasoning from polarity or dipole moment instead of symmetry/packing -- boiling point (which is more sensitive to polarity in the liquid state) is in fact SIMILAR across the three dichlorobenzene isomers; it is specifically the …