What would be the major product if 1,4-dibromo-4-methylpentane was allowed to react with one equivalent of silver nitrate in ethanol?

a) Iodide is a strong nucleophile but a weak base, so SN2 is the preferred reaction. Only the bromine on C1 is eligible to undergo SN2, so that one will be replaced by iodide. b) Silver ion tends to suck off a halide ion and leave a carbocation, which means E1 and SN1. If there's only one equiv, then the tertiary bromide on C4 is the one that will go. The resulting carbocation can give: E1 products 5-Br-2-Me-2-pentene (major, trisub) and 5-Br-2-Me-1-pentene (disub, minor). SN1 product 5-bromo-2-ethoxy-2-methylpentane

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Hope that helps:)