Acidity and Basicity - Notes - Class 11

Carboxylic Acid and Acid Strength
An aliphatic carboxylic acid ionizes as
an electron withdrawing A will increase the acid  strength while, electron
donating A will decrease the acid strength

Compare the strength of following carboxylic acids
CH₁OOOH   CH₃CH₂OOOH  HCOOH (CH₃)₂CHCOOH
    (I)                      (II)                   (III)           (IV)
Solution: Alkyl groups are electron-donating, +I effect giving group, decreases acid strength. Methanoic acid (III) has no alkyl group bonded to -COOH, hence strongest acid. The smallest alkyl group (-CH₃) is in ethanoic acid, it is the strongest amongst I, II, IV. Between II and IV, IV has two alkyl groups bonded at C-2, greater +1 effect, weaker the acid. thus, the overall order of acid strength is

III > I > II > IV

Example:
Compare the strength of following carboxylic acids
CH3OOOH   CH₃CH₂OOOH  HCOOH (CH₃)₂CHCOOH
    (I)                      (II)                   (III)           (IV)

Solution: Alkyl groups are electron-donating, +I effect giving group, decreases acid strength. Methanoic acid (III) in has no alkyl group bonded to —COOH, hence, strongest acid. The smallest alkyI group ( -CH₃) is in ethanoic acid, it is the strongest amongst I, II, IV. Between II and IV, IV has two alkyl groups bonded at C-2, greater +1 effect, weaker the acid. Thus, the overall order of acid strength is

III > I > II > IV

Compare the acid strength of following acids.

Solution: Halogens are all -I effect giving groups, increases acid strength. In III and IV halogens are at C-2 while, in I & II, they are at C-3 and C-4. Therefore, considering distance of halogens first, II is weakest, I is 2nd weakest. IV is stronger than III due to an additional F at C-3. The overall order is weakest, I is 2nd weakest. IV is stronger than III due to an additional F at C-3. The overall order is

IV > III > I >II 

Example:
Arrange the following in decreasing order of their acid strength
Solution: IV is the strongest acid since, it has stronger -l effect giving group (-NO₂) at C-2 and additional -l effect giving group (F) at C-3. III is 2nd strongest since, it has stronger electron withdrawing group (-NO₂) at C-2 than F on C-2 in I. II is the weakest since, it has -l effect producing group but at greater distance C-3. Hence, the overall order is 

IV > III > I >II 

Example:
Arrange the following acids increasing order of their acid strength
(a) CH₃CH₂CH2COOH (I),
   
CH₃CH₂COOH (III),
H₂C = CH — COOH (IV)

(b) CH₃CH₂CH₂COOH (I).
CH₃CHFCHFCOOH (II),



(c)


Solution: (a) II < I < III < IV: sp² carbon is more electronegative than a sp³ 
(b) I < II < lV < I :-I ( —N0₂)>—(—F)
(c) I < II

Example:
Arrange the following carboxylic acids increasing order of their acid strength




Solution:
III strongest since, CI(-I effect) is at closest distance
II weaker since, —CH3(+I effect) is at closest distance
Also, I is stronger than IV since, —CH2Cl(-I effect) is at closer distance in I than in IV, hence< overall order is

II < IV < I < III

Basic Strength and its Determination by Inductive Effect
Ammonia and its alkyl derivatives amines, aryl derivative anilines, are all Lewis- Base because they all have a lone-pair of non bonded electron on nitrogen that can be donated to a Lewis acid. An amine can also act as a Bronsted- Lowry base by accepting a proton from a proton acid.
Reaction of amine as nucleophile

Reaction of amine as proton base


Compare the basic strength of following ,amimes.

Solution: 
I is the strongest base since, it has no electron withdrawing group bonded.
IV is the weakest since, it has most electronegative F bonded at closest distance to nitrogen.
II is stronger base since, it has two electron withdrawing groups but at larger distance than in III
Hence, overall order is

IV < III < II < I

Compare the basic strength of following
Solution:
II is the strongest since, it has an electron donating group and no electron withdrawing group.
IV is the weakest since, electron withdrawing group is at closer distance.
III is weaker than I since. - I effect of Cl is greater than the +I effect of —CH₃ from same position. Hence, the overall order is

IV < III < I < II

Effect of Resonance over Acid-Strength
As discussed earlier, an election withdrawing group increases acid strength by facilitating the release of H⁺ ion while an electron donating group decreases the acid strength an it opposes the ionization process and makes release of H⁺ difficult. A further better approach to compare the acid strength is to compare the stability of conjugate base - a strong acid is one that produce a stable conjugate base after ionization. To illustrate it let us consider the case of aliphatic and aromatic alcohols

Since, phenoxide ion as stabilized as shown below, a more stable conjugate base than alkoxide ion (RO)

Due to above resonance stabilization of conjugate base, phenol is stronger acid than aliphatic alcohols

Comparison of Acid Strength of Substituted Phenols
An election withdrawing group on ring further increases the stability of phenoxide ion and therefore, increases the acid strength. On the other hand, an electron donating group destabilizes the conjugate base (phenoxide ion) and lowers the acid strength of conjugate acid
(i) Nitrophenols: Three isomeric nitrophenols are possible. All isomers are stronger acid than phenol because nitro group is a strong election withdrawing group. Also acid strength of three isomeric phenols vary significantly. The relative acid strength of para and meta isomers can be explained either on the basis of electron withdrawing effect of substituents on the ease of ionization or stabilization of conjugate base as

(Here positive charge does not come on the carbon bearing - OH group, not stabilized and therefore, ionization  is not helped by resonance)

In the above resonance description, one of the resonance structure of p-nitrophenol has positive charge on oxygen making oxygen strongly electron withdrawing, withdraw electron pairs of O—H bond away from H, facilitating the release of H⁺ and makes stronger acid. On the other hand, in m-nitrophenol, no resonance structures has positive charge on oxygen atom of —OH, therefore, release of H⁺ is not facilitated by resonance, relatively weaker acid than para nitrophenol. Another approach is through stability of conjugate base. Let as draw resonance structures of conjugate base of above nitrophenols

Above resonance description of conjugate base of p-nitrophenol and m-nitrophenol indicates that conjugate base of p-nitrophenol is more stable than conjugate base of m-nitrophenol. Therfore, once again, p-nitrophenol is stronger acid than m-nitrophenol.
Here, resonance effect is only considered for comparing acid strengths of the above two isomers of nitrophenol. Had the inductive effect been considered, meta isomers would have been stronger acid which is not true. it is due to the stronger nature of resonance effect than inductive effect. However, it is the inductive effect only in meta nitrophenol that makes it stronger acid than and unsubstituted phenol.
 
Ortho nitrophenol: In case of ortho isomer of nitrophenol, the resonance effect favours ionization and hence, acid strength as
On the basis of resonance effect, ortho nitrophenol is stronger acid than meta nitrophenol as well as phenol. However, between ortho and para, the resonance effect is common and has no role in comparative acid strength. Nitro group is a strong electron withdrawing group by -I effect also and the magnitude of inductive effect depends on distance. On that basis ortho nitrophenol should have been stronger acid than its para isomer since, ortho position is closer. However, this is not true on the not of a new effect called ' Ortho effect". Nitro group from ortho position stabilize the acid by intramolecular H-bonding as shown below, and lowers the acid strength
Due to the above ortho effect, release of H⁺ becomes slightly difficult. Ortho effect wins over the inductive effect and makes ortho nitrophrno weaker acid than pare nitrophenol.
Ortho effect is stronger than inductive effort but weaker than resonance effect. Therefore, the acidity order of three nitrophenols and unsubstituted phenol is

Phenol < m-nitrophenol < ortho nitrophenol < p-nitrophenol.

Increases in number of nitro substituents on benzene ring of phenol increases the acid strenght and trinitrophenol (picric acid) is equivalent to common mineral acid in the acid strenght.

Example:
Rank the following substituted phenols in increasing order of their acid strength explaining reasons for your order.

Nitro group from ortho and para position increases the acid strength via resonance whereas, it increases the acid strength from meta position, via induction Since, resonance effect is more effective in increasing acid strength than inductive effect, A and C arc stronger acid than B. A also has additional ortho effect stabilization of acidic hydrogen. C is stronger acid than A. Hence, the overall order of acid strength is B < A < C.

Effect of Resonance Over Basic Strength
Basic strength depends on how easily the base can react with an acid. Let us consider the case of amines. Amines are all Lewis base which denote a long-pair of electron to a  Lewisacid . The strenght of a Lewis base depends on its ability to donate its base-past to a Lewis acid. In turn electron-donating ability of a Lewis base depends on availability of electron around the donor atom. Resonance effect can influence the basic strength by influencing the electron density around the donor atom.

Aliphatic Amines and Aromatic Amines
In aliphatic amine electron donating inductive effect of alkyl group increases the electron density around nitrogen hence, the basic strength. On the other hand, in aromatic amine, lone-pair of electron on nitrogen is involved in resonance, less available in nitrogen for donation to a Lewis acid. hence, a weaker base as-
Above resonance delocalization of base-pairs of nitrogens in aniline decreases its basic strength.

Relative Basicity of Substituted Aniline
If aniline substituted on the ring, electron donating substituents will increase basic strength while, electron withdrawing substituent will decrease the basic strength. Let us consider the three isomers of nitro-aniline. Nitro groups from para and ortho are in direct resonance with —NH₂ group as
Between ortho and para nitro aniline, influence of resonance is common but at the same time nitro group is also a very strong -I effect giving group, which depends on distance. Since, ortho position in closer, it is further weaker base due to greater -I effect from this position. Nitro group is not in resonanace with —NH₂ from meta position as described below
Therefore, the overall basic strength of these three nitro aniline is ortho nitroaniline < para nitroaniline < meta nitroaniline. Also all three isomers of nitroaniline is weaker base than unsubstituted aniline.

Steric Inhibition to Resonance and Acid-Base Strength
Resonance ability of an atom is lost if it loses planarity with the other part of system due to steric crowding by bulky group on adjacent position. To illustrate it, let us consider the following two examples -
The above two compounds A and B have everything identical except position of two methyl groups. It is expected that A should be stronger base than B due to closeness of two electron donating methyl groups to — NH₂. The fact is opposite to this. The significant factor affecting the basic strength of these two compounds is -R effect of -NO₂ group. In compound B, - NO₂ is surrounded by two bulky methyl group and they sterically repel the - NO₂ group. In order to minimize the steric repulsion by the two adjacent methyl group, the nitro group loses planarity with the benzene ring. So now - NO₂, due to lack of planarity with ring. not able to resonate. This is known as steric inhibition to resonance.Thus, in B, -NO₂ is not decreasing basic strength by resonance. In A -NO₂ lies in the plane of ring, it is in resonance with ring, decreases basic strength of - NH₂ by resonance, hence, weaker base. The above principle applies well in case of acid strength also. Let us consider the following two acids
A is stronger acid inspite of closeness of two electron donating methyl group to —COOH. It is once again due to the same reason that in A, -NO₂ is in resonance with the ring, increases the acid strength. In B, there is steric inhibition to resonance to -NO₂ group, therefore, a weaker acid.

Example: Compare the acidic strength of the following acids:
(a) C - C - C - COOH
(b) C = C - C - COOH
(c) C ≡ C - CCOOH
Solution: The acid whose conjugate base is most stable will be more acidic.

After forming conjugate base from the above acids.
(a)
(b)
(c)
It is clear that sp hybridised carbon being most electronegative will decrease e-density from O most effectively making the conjugate base most stable.
c > b > a (acidic strength)

Example: Which is more acidic between the two:
(a) CHF₃ 
(b) CHCl₃ 
Solution: CHF₃ > CHCl₃ 
If we consider the -I effect of F and Cl But this effect will not be considered here After the removal of proton

(a)
(b)
(vacant d-orbital available where C will coordinate its electron) (pπ -dπ bonding) → a < b (acidic strength)

Example: Compare the acidic strength of the following:
(a) CHF₃ 
(b) CHCl₃ 
(c) CHBr₃ 
(pπ - dπ bonding in Br is not as much as effective as in Cl due to large size of Br) Solution: CHCl₃ > CHBr₃ > CHF₃

Example: Compare the acidic strength of the following:
(a) CH (CN)₃ 
(b) CH (NO₂)₃ 
(c) CHCl₃ 
Solution: After removing H

(Resonance) In its resonating structure, -ve charge will be on 

N)

(Resonance) (- In its resonating structure -ve charge will reside on O)
→ more effective Resonance

(pπ - dπ)
b > a > c
* -ve charge on O is more
stable than -ve charge on N as O is more electronegative than N.
* pπ - dπ Resonance < Actual Resonance

Example: Compare the acidic strength of the following: 
(a) CH ≡ CH
(b) CH₂ = CH₂ 
(c) CH₃ - CH₃
Solution: 

(Stability of the conjugate base)
→ a > b > c (acidic strength)

Example: Compare the acidic strength of the following:
(a) CH₃ - CH₂ - CH₂ - COOH
(b)
(c)
(d)
Solution: d > c > b > a

Example: Compare the acidic strength of the following: 
(a) H₂O
(b) H₂S
(c) H₂Se
(d) H₂Te
Solution: Conjugate base has the stability order:

→ H₂O < H₂S < H₂Se < H₂Te (acidic strength)

Example: Compare the acidic strength of the following compound:
(a)
(b)
(c)
(d)
Solution: After forming conjugate base of the above:
c > d > b > a

Example: Compare the reactivity of the following compounds with 1 mole of AgNO₃:
(a)
(b)
(c)
(d)
Solution: After removing Cl-
( +ve charge is not on resonance least stable)
extent of +ve charge decreases stability increases.

Example: Compare the acidic strength:
(a)
(b)
(c)
(d)
Solution: After making conjugate base
c > b > a > d

BASIC STRENGTH

Basic strength directly depends on the availability of lone pair for H.

Example: Compare the basic strength of following:
Solution:


Example: Compare the basic strength of the following:
(a)
(b)
(c)
(d)
Solution:

CH₄ < NH₃ < H₂O < HF

(acidic strength)

* Strong Acids have weak conjugate base.

* For the same period
less electronegativity, more nucleophilicity as more electronegative element has less tendency to give its electron pair.

Example: Which is more basicor?
Solution:> 
Which is more basic NH₃ or forming conjugate acid


Comparison of Basicity of Ammonia and Alkyl Amines: 
Example: Compare the basic strength of the following:
NH₃, CH₃NH₂, (CH₃)₂NH, (CH₃)₃N
Solution: Factors which affect the basicity of Amines
(1) steric effect
(2) Inductive effect
(3) solvation effect.
The base whose conjugate acid is more stable will be more acidic forming conjugate acid of the given base.

Stability order of conjugate acid

Therefore basic strength,
(CH₃)₃N > (CH₃)₂NH > CH₃NH₂ > NH₃
(vapor phase or gaseous phase or in Non polar solvent)
In Aqueous solution or in polar solvent
(CH₃)₂NH > CH₃NH₂ > (CH₃)₃N > NH₃

In aqueous solution, the conjugate acids form H-bonds (intermolecular) with water molecules and stabilise themselves. Conjugate acid of 1° amine which has largest no. of H-atoms form maximum H-bond with water and is most stable. Consequently 1° amine is most basic.
Due to steric effect 1° amine is considered more basic as compared to 3° amine as lone pair is hindered by three alkyl group and less available for H .

Considering the combined effect of the three (Inductive, solvation and steric effect) we can conclude that
2° > 1° > 3° > NH₃

Aromatic amines are least basic as their lone pair is in conjugation and less available for protonation.

Example: Compare the basic strength of the following:
(a)

(b) 

(c)

Solution: 
(if L.P. will participate in Resonance, then molecule becomes aromatic)
Hence L.P. will have a greater tendency to take part in Resonance and will be less available for H+
This compound will be least basic.

Example: Compare the basic strength of the following:

Solution:
sp hybridised carbon being most electronegative will attract edensity from nitrogen and will make it less available for H+. Hence basicity decreases.
c > b > a

Example: Compare the basic strength:
(a)

(b)

Solution:
a < b

Example: Compare the basicity of the following compounds:
(a)
(b)
(c)
(d)
Solution: In part (a) the lone pair of nitrogen is in resonance therefore will be less available for H making it least basic among all followed by sp, sp², sp³ hybridised carbon atoms.
b > c > d > a

Example: Compare the basicity of the numbered nitrogen atoms:Solution: The planarity of ring will be destroyed if L.P. will take part in Resonance.

Basicity order of Nitrogen follows the order:
N(sp³) > N(sp²) > N(sp)

(In this sp², l.p. is in Resonance with ring hence will be less available for H+ therefore it will be least basic)

Example: Compare the basic strength of the following:
(a)
(b)
(c)
Solution: In part (a) NO₂ is at p-position Hence will attract e- density by both -M and -I.
In part (b) NO₂ is at m-position hence will attract e- density by -I only There is no such effect in part (c)
→ Availability of L.P. on nitrogen in part (a) is minimum followed by b and then c.
c > b > a

Ortho effect: 
The ortho substituted aniline are less basic than aniline and ortho substituted benzoic acids are more acidic than benzoic acid.

Ortho effect is valid only for benzoic acid and aniline.

Example: Compare the basic strength of the following:
(a)

(b)

(c)

(d)

Solution: a > b > d > c
* Due to ortho effect d > c
if c is less basic than d then it will be certainly less basic than b as b is more basic than d.

Example: Compare the basic strength of the following:
(a)
(b)
(c)
(d)
Solution: Do yourselves.

S.I.P → Steric inhibition of Protonation (ortho effect)
After protonation, repulsion increases therefore ortho substituted aniline is less basic than aniline.

S.I.R → Steric inhibition of resonance
(a)

(b)