<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>
Hey There!
Here is your answer:
1.8
2.11
3.9
4. C
5. True.
Hope this helps!
Answer: 5.47x10^24 molecules
Explanation:
6.02214076x10^23 molecules CO2 = 44g
400g = (400/44)*6.02214076*10^23 = 5.47x10^24
The temperature for the reaction is calculated as follows
delta H/ delta S
delta H=-70 Kj
delta S =-0.300 kj/k
temperature is therefore = -70kj/-0.300kj/k =233.33 K