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4

Basically, you are on the right track and answered the question. First, NaNH$_2$ is used as a base to deprotonate the terminal alkyne (see this Reagent Friday). Second, the generated nucleophile opens the epoxide in an S$_\mathrm{N}$2 fashion, usually adding to the lesser substituted carbon which in your case yields a quaternary alcohol after acidic workup (...

3

The word "Contact process" has an historical origin. Before the $18$th century, sulfuric acid was difficult to obtain. It was synthetized by first oxidizing pyrite $\ce{FeS2}$ in wet air, in order to obtain iron sulfate $\ce{Fe2(SO4)3)}$, which had to be later pyrolized according to : $$\ce{Fe2(SO4)3 + 3 H2O -> Fe2O3 + 3 H2SO4}$$ Unfortunately, ...

3

Chemical bonds aren't just electrostatic attractions between some cation $\ce{H+}$ and anion $\ce{X-}$. Well, purely ionic bonds are; but if you don't have a purely ionic bond, then there's some covalent part that also matters, which is the part that you're neglecting. You can't treat a covalent bond as if it were purely ionic—which is what you're implicitly ...

3

Yes, $\ce{FeCl3}$ is a well known Lewis acid that is used in place of $\ce{AlCl3}$ for FC reaction. See the paper linked by user55119 and also this question. Other $\ce{MX_n}$ types of reagents used in FC reactions are $\ce{BF3}$, $\ce{BeCl2}$, $\ce{TiCl4}$, $\ce{SbCl5}$, $\ce{SnCl4}$, $\ce{SeCl4}$, $\ce{TeCl4}$, $\ce{InCl3}$, $\ce{NbCl5}$, $\ce{IrCl3}$ or $\... 3 In neutral rhodizonic acid water solution, not all carbonyl groups are available for delocalization. Some are hydrated i.e. there are gem-diol ($\ce{C(OH)2}$) groups instead of carbonyls (1). This may be surprising, but it's generic behavior for compounds with conjugated carbonyl groups and electron-withdrawing groups bound to$\ce{C=O}$in general. Because ... 2 Your second equation is an equilibrium. Both acetic acid and acetate ions are always present in a solution, although their proportion may vary. When$\ce{CH3COOH}$is dissolved into water, a fraction of the molecules are transformed by reaction with water, producing some$\ce{H3O+}$But if these ions are destroyed by adding$\ce{OH-}$ions, more molecules$\...

2

$\ce{NaHCO3}$ and $\ce{Na2CO3}$ are not molecules, but ionic compounds consisting of $\ce{Na+},$ $\ce{HCO3-},$ respectively $\ce{CO3^2-}$ ions. . There is no ion like $\ce{NaCO3-}$, unless in extreme conditions like kryogenics or interstellar space. $\ce{CO3^2-}$ is the conjugate base to $\ce{HCO3-}$, because of : $$\ce{NaHCO3(s) ->[H2O] Na+(aq) + HCO3-(... 1 Carbonates, \ce{CO^{2-}_3}, formally derive from \ce{H2CO3}. You may postulate two deprotonations:$$\ce{H2CO3 + H2O <=> HCO^-_3 + H3O+}$$and$$\ce{HCO^-_3 + H2O <=> H3O+ + CO^{2-}_3}$$Thus, hydrogencarbonates are monobasic; and carbonates are dibasic. However, note that \ce{CO2} gas may solute physically in water (like, for example, ... 1 In such cases, you cannot afford to involve implicit simplifications. In this particular case, the equilibrium ratio \frac {[A-]}{[\ce{HA}]} is not the same as the nominal \frac {[A-]}{[\ce{HA}]} ratio, but reaching acido-basic equilibrium is to be considered. One has to count also with water auto-ionization equilibrium. If the added \ce{HCl} ... 1 In contrary to acid-base reactions(1), a redox reaction is such a reaction, which can be formally divided into two half-reactions. At least one reactant in such a half-reaction explicitly formally acts as a donor or acceptor of one or more electrons.$$\ce{Cl2(aq) + 2 Fe^2+(aq) -> 2 Cl-(aq) + 2 Fe^3+(aq)}$$can be formally divided to:$$\ce{Cl2(aq) + 2 e- ...

1

A rough calculation for the pH of 0.2-ᴍ formic acid gives: $$\mathrm{pH} = 1/2 (\mathrm{p}K_\mathrm{a} - \log(0.2)) = 1/2 (3.74 + 0.70) = 2.22$$ You can check by calculating the equilibrium constant from the concentrations of all the species, and the estimate is pretty good. The pH of the buffer is 3.44 using the Henderson Hasselbalch expression (see other ...

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