What language is even this Oop i DONT know good luck
Answer:
Explanation:
Step 1. Calculate the pOH
pOH =-log[OH⁻]
pOH =-log(1.0 × 10⁻⁹)
pOH = 9.00
Step 2. Calculate the pH
pH + pOH = 14.00
pH + 9.00 = 14.00
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).
1s^2 2s^2 2p^6 3s^2 2p^3 or the shortcut way is [Ne] 3s^2 2p^3