<span>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</span>
CO + 2 H2 → CH3OH
<span> find # of mols in each reactants, </span>
<span>152500 g CO x 1 mol CO / 28.01g CO = 5444 mol CO </span>
<span>24500 g H2 x 1 mol H2 / 2.02 g H2 = 12129 mol H2 </span>
<span>mol ratio between CO and H2 is 1:2, which means each mol of production of CH3OH need 1 mol of CO and 2 mol of H. </span>
<span>H2 is enough to produce 6064 mols of CH3OH but there are only 5444mol of CO. </span>
<span>5444 mol CH3OH x molar mass of CH3OH / 1 mol CH3OH </span>
<span>= 174371 g = 174.4 kg</span>
Answer:
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Explanation:
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