Two stereoisomeric 2-isopropyl-5-methylcyclohexyl chlorides (Cl on C-1, isopropyl on C-2, methyl on C-5) undergo E2 with $\mathrm{NaOEt}$/ethanol. In the more stable chair of chloride P all three substituents are equatorial. In the more stable chair of chloride Q the isopropyl and methyl groups are equatorial and Cl is axial. Which statement is correct?

1. AQ reacts faster than P but gives only the disubstituted alkene, because the hydrogen on C-2 is equatorial in Q
2. BQ reacts faster than P and gives mainly the trisubstituted alkene (double bond C-1=C-2); P gives only the disubstituted alkene (double bond C-1=C-6) because only one axial hydrogen is anti to Cl in its reactive chair
3. CP and Q give the same alkene mixture at the same rate, since the Saytzeff rule always governs
4. DP reacts faster than Q because all-equatorial is more stable, and P gives mainly the trisubstituted alkene