When 3-iodo-3-ethylpentane is heated in methanol, the major organic product is an __________________ that is generated through a
1 answer:
Answer:
Ether
SN1 mechanism
Explanation:
The nucleophile in this reaction is CH3OH. It is a poor nucleopile. We already know that a poor nucleophile reacting with a tertiary alkyl halide often leads to the substitution product as the major product.
Also, the iodide ion is a good leaving group. This makes the SN1 substitution more likely yielding the ether as the major product as shown in the image attached.
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Answer:
(a) Δ = 2.881 J/K; Δ
= -2.881 J/K; total change in entropy = 0
(b)Δ = 2.881 J/K; Δ
= 0 ; total change in entropy = 2.881 J/K
(c) Δ = 0 ; Δ
= 0 ; total change in entropy = 0
Explanation:
In the given problem, we need to calculate the change in the entropy of the system and also the change in the entropy of the surroundings, and the resulting total change in entropy, when a sample of nitrogen gas of mass 14 g at 298 K and 1.00 bar doubles its volume. We have the following variable:
mass (m) = 14 g
Temperature = 298 K
Pressure = 1.00 bar
Initial volume =
Final volume = =
(a) Change in entropy of the system Δ =
where R = 8.314 J/(mol*K)
n = number of moles = mass/molar mass = 14/ 28 = 0.5 moles
Δ = 0.5*8.314*ln2 = 2.881 J/K
Change in entropy of the surrounding Δ = -2.881 J/K
Therefore, for a reversible process, the total change in entropy = Δ+Δ
= 2.881 - 2.881 = 0
(b) Because entropy is a state function, we use the same procedure as in part (a). Thus, Δ = 2.881 J/K
Since surrounding does not change in this process Δ = 0.
total change in entropy = Δ+Δ
= 2.881 - 0 = 2.88 J/K
(c) For an adiabatic reversible expansion, q(rev) = 0, thus:
Δ = 0
Since heat energy is not transferred from the system to the surrounding
Δ = 0
total change in entropy = Δ+Δ
= 0