Answer:
A. Methanol
B. 2-chloropropan-1-ol
C. 2,2-dichloroethanol
D. 2,2-difluoropropan-1-ol
Explanation:
Primary alcohols are stronger acids than secondary alcohols which are stronger than tertiary alcohols.
This trend is so because of the stability of the alkoxide ion formed(stabilising the base, increases the acidity). A more stabilised alkoxide ion is a weaker conjugate base (dissociation of an acid in water).
By electronic factors, When there are alkyl groups donating electrons, the density of electrons on th O- will increase a d thereby make it less stable.
By stearic factors, More alkyl group bonded to the -OH would mean the bulkier the alkoxide ion which would be harder to stabilise.
Down the group of the periodic table, basicity (metallic character) decreases as we go from F– to Cl– to Br– to I– because that negative charge is being spread out over a larger volume that is electronegativity decreases down the group.
Electronegative atoms give rise to inductive effect and a decrease in indutive effects leads to a decrease in acidity. Therefore an Increasing distance from the -OH group lsads to a decrease in acidity.
From above,
A. Methanol
B. 2-chloropropan-1-ol
C. 2,2-dichloroethanol
D. 2,2-difluoropropan-1-ol
Answer:
<h2>ignore your body being cold</h2>
Hope it helps
CuSO4 *5H20
In the formula of the hydrate, on the first place is the salt and on the second of water. so in ration on the first place should be salt also.
Answer:
NaOH(aq) + HNO3(aq)------>NaNO3(aq) + H2O(l)
Explanation:
A thing to note is that an acid and a base will react to form a metal salt + H2O.
~Hope it helps:).
I'm pretty sure it's B)constructive
