Why does Birch reduction of naphthalene with isoamyl alcohol form tetralin instead of 1,4-dihydrodialin (as formed with C2H5OH)?
As I noted in a Comment that there should be no significant difference between the use of ethanol or isoamyl alcohol in the reduction of naphthalene with sodium in that both alcohols are primary. The difference lies in the conditions of the reaction. The sequence is a classic case of kinetic vs. thermodynamic conditions.
The reaction conditions in the diagram are from this paper.
Naphthalene (1) reversibly adds an electron from sodium under ambient conditions to form the resonance stablized radical anion 2 that is shown in a canonical form. This species is readily protonated by an alcohol, be it primary or tertiary as in this instance, at the site of highest electron density to afford the resonance stabilized radical 3. The addition of a second electron to 3 in this 2-electron reduction gives rise to the resonance stabilized anion 4, which, in turn, is protonated to form the deconjugated 1,4-dihydronaphthalene 5, the kinetically formed product. Just like cyclohexene is unable to reduce under these reaction conditions, neither can alkene 5.  During the course of this reduction, something else is formed -- namely -- a base, sodium tert-butoxide.
To achieve reduction of the double bond in 5 it is not sufficient to conduct the reaction at reflux but rather to conjugate the double bond with the aromatic ring. This is the role of the base at elevated temperature. Deprotonation of 5 at the benzylic site provides resonance stabilized anion 4 again, which can reversibly protonate to form 1,2-dihydronaphthalene (dialin) 6. This styrene-like double bond has a low reduction potential. Addition of an electron produces the radical anion 7 that protonates to form stabilized benzyl radical 8. Subsequent addition of a second electron affords the benzylic anion 9. This anion is readily protonated by the alcohol to form 1,2,3,4-tetrahydronaphthalene 10 (tetralin).