Answer:
D) sodium t-butoxide + bromomethane
Explanation:
The alkoxide ion is a strong nucleophile, that unlike alcohols, will react with primary alkyl halides to form ether. This general reaction is known as <em>the Williamson synthesis</em>, and is a SN₂ displacement. The alkyl halide must be primary so the back side attack is not hindered, and the alkoxide ion must be formed with the most hindered group.
The mechanism can be seen in the attachment.
Answer:
(a) Pair 1: H₂S and HS⁻
Pair 2: NH₃ and NH₄⁺
(b) Pair 1: HSO₄⁻ and SO₄⁻
Pair 2: NH₃ and NH₄⁺
(c) Pair 1: HBr and Br⁻
Pair 2: CH₃O⁻ and CH₃OH
(d) Pair 1: HNO₃ and NO₃⁻
Pair 2: H₃O⁺
Explanation:
When an acid loses its proton (H⁺), a conjugate base is produced.
When a base accepts a proton (H⁺), it forms a conjugate acid.
(a) H₂S is an acid. When it loses a proton, it forms the conjugate base HS⁻.
NH₃ is a base. When NH₃ gains a proton, it forms the conjugate acid NH₄⁺
(b) The acid HSO₄⁻ loses a H⁺ ion and forms the conjugate base SO₄²⁻.
The base NH₃ accepts a H⁺ ion to form the conjugate acid NH₄⁺.
(c) HBr is an acid. When loses the H⁺ ion, it forms the conjugate base Br⁻.
CH₃O⁻ accepts a H⁺ ion to form the conjugate acid CH₃OH.
(d) HNO₃ loses a proton to form the conjugate base NO₃⁻.
H₂O gains a proton to form the conjugate acid H₃O⁺.
After crossbreeding, the seeds will be in the form below
Rr, Rr, rr and rr
which is two for each in four possible outcomes bringing the ratio to 1:1
which is equivalent to 1/2
so 1/2 as a percentage is
=1/2*100
=50%
Chemical reaction: C₃H₇COOH → C₃H₇COO⁻ + H⁺.
c(<span>butanoic acid) = 0,100 M.
</span>α = 1,23% = 0,0123.
Ka = α² · c / 1 - α.
Ka = 0,0123² · 0,1 M / 1 - 0,0123.
Ka = 0,0000153 M.
Ka = c(C₃H₇COO⁻) · c(H⁺) / c(C₃H₇COOH).
c(H⁺) = α · c(C₃H₇COOH).
c(H⁺) = 0,0123 · 0,1 M = 0,00123 mol/L.
pH = -log c(H⁺).
pH = 2,91.
Given -
T = 450K
∆H = 3000J
∆S = 4J/K
Using formula
∆G = ∆H - T∆S
∆G = 3000- 4*450
∆G = 1200 Joule
∆G is positive so the reaction is non spontaneous!
