B. Decomposition is the correct answer~
Not all acid-catalyzed conversions of alcohols to alkyl halides proceed through the formation of carbocations. Primary alcohols and methanol react to form alkyl halides under acidic conditions by an SN2 mechanism.
Not all acid-catalyzed conversions of alcohols to alkyl halides proceed through the formation of carbocations. Primary alcohols and methanol react to form alkyl halides under acidic conditions by an SN2 mechanism.
In these reactions the function of the acid is to produce a protonated alcohol. The halide ion then displaces a molecule of water (a good leaving group) from carbon; this produces an alkyl halide:
Again, acid is required. Although halide ions (particularly iodide and bromide ions) are strong nucleophiles, they are not strong enough to carry out substitution reactions with alcohols themselves. Direct displacement of the hydroxyl group does not occur because the leaving group would have to be a strongly basic hydroxide ion:
We can see now why the reactions of alcohols with hydrogen halides are acid-promoted.
Carbocation rearrangements are extremely common in organic chemistry reactions are are defined as the movement of a carbocation from an unstable state to a more stable state through the use of various structural reorganizational "shifts" within the molecule. Once the carbocation has shifted over to a different carbon, we can say that there is a structural isomer of the initial molecule. However, this phenomenon is not as simple as it sounds.
<em>-</em><em> </em><em>BRAINLIEST</em><em> answerer</em>
Answer is: 3,4
Chemical reaction: HNO₂ + NaOH → NaNO₂ + H₂O.
c₀(HNO₂) = 1,2 M = 1,2 mol/dm³.
c₀(NaNO₂) = 0,8 M = 0,8 mol/dm³.
V₀(HNO₂) = V₀(NaNO₂) = 1 dm³ = 1 L.
c₀(NaOH) = 0,5 M = 0,5 mol/dm³.
n₀(HNO₂)= 1,2 mol/dm³ · 1 dm³ = 1,2 mol.
n₀(NaNO₂) = 0,8 mol/dm³ · 1 dm³ = 0,8 mol.
V(NaOH) = 400 mL · 0,001 dm³/mL = 0,4 dm³.
n₀(NaOH) = c₀(NaOH) · V₀(NaOH).
n₀(NaOH) = 0,5 mol/dm³ · 0,4 dm³ = 0,2 mol.
n(HNO₂) = 1,2 mol - 0,2 mol = 1 mol.
n(NaNO₂) = 0,8 mol + 0,2 mol = 1 mol.
c(HNO₂) = 1 mol ÷ 1,4 dm³ = 0,714 mol/dm³.
c(NaNO₂) = 1 mol ÷ 1,4 dm³ = 0,714 mol/dm³.
pH = pKa + log (c(HNO₂) / c(NaNO₂)).
pH = 3,4 + log (0,714 mol/dm³ / 0,714 mol/dm³) = 3,4.
The equation that shows the formation of chromium (ii) ion from neutral chromium atom is as follow
Cr ---> cr^2+ + 2e-
Cr^2+ is the chromium ion with oxidation state of two which is one of the common ion of chromium. Other common ion of chromium include chromium of oxidation state 6 and 3
Answer:
Atoms are neither created, nor distroyed, during any chemical reaction ... Chemical reactions are represented on paper by chemical equations. For example, hydrogen gas (H2) can react (burn) with oxygen gas (O2) to form water (H20). ... Write 'balanced' equation by determining coefficients that provide equal numbers of ...
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