Answer:
See below.
Step-by-step explanation:
Ethers react with HI at high temperature to produce an alky halide and an alcohol.
R-OR' + HI ⟶ R-I + H-OR'
<em>Benzylic ethers</em> react by an Sₙ1 mechanism by forming the stable benzyl cation.
- PhCH₂-OR + HI ⟶ PhCH₂-O⁺(H)R + I⁻ Protonation of the ether
- PhCH₂-O⁺(H)R ⟶ PhCH₂⁺ + HOR Sₙ1 ionization of oxonium ion
- PhCH₂⁺ + I⁻ ⟶ PhCH₂-I Nucleophilic attack by I⁻
If there is excess HI, the alcohol formed in Step 2 is also converted to an alkyl iodide:
ROH +HI ⟶ R-I + H-OH
Thus, benzyl ethyl ether reacts to form benzyl iodide (a) and ethanol (b).
The ethanol reacts with excess HI in an Sₙ2 reaction to form ethyl iodide (c).
8 carbons is "oct" prefix (like polygons)
No pi bonds is "ane" suffix.
put it together to get.... Octane
PV=nRT n=PV/RT
R=.0821 atm x L
Mol x K
T=318K
P=.55atm
V=37.4L
n= (.55atm) (37.4L) /
.0821atm x L (318k)
mol x K
n= .79 mol of sulfur dioxide
Answer:
<h2>put the bigger number on top then add </h2>
Explanation:
41.369