Alcohols and Phenols both are acidic but phenols are more acidic. Why?

Alcohols and phenols both are acidic in nature because they Liberate H2 with active metals. But alcohols are weaker acids than phenols because,  

(i) Alcohols fail to turn blue litmus red:

(ii) Alcohols fail to neutralize alkalies whereas phenols do so:

(1) Both phenol and phenoxide ions exhibit resonance, whereas alcohols and alkoxide ion do not.
(2) Due to resonance, positive charge 's developed over "O" of phenol this weakens O-H bond and facilitates release of a proton. On the other hand due to electron donating nature of alkyl group -O-H bond is strengthened. Therefore it hinders the release of proton.      

(3) The phenoxide ion is more stable than phenol, due to greater dispersal of negative charge, therefore. Equilibrium get shifted to forward direction. whereas alkoxide ion is less stable than alcohols due to intensification of negative charge, so equilibrium get shifted towards backward direction. Hence phenols are more acidic than alcohols. 

Related Questions:

Benzamide is less easily hydrolysed than methyl benzoate. Why ?

Highly branched carboxylic acids are less acidic than unbranched acids. Explain why?

Phenoxide ion is more stable than an alkaoxide. why?

CCl4 can not be hydrolysed but SiCl4 can be. Why?

Carbon tetrachloride (SiCl4) easily hydrolysed due to availability of vacant d – orbitals in valence shell of their central atom ( Silicon), hence it can easily extend their coordination number beyond four but this is not possible in case of carbon tetrachloride (CCl4) due to absence of vacant d – orbitals in carbon atom.

Related Questions:

(1) What are the SESQUI OXIDES ? give the examples.

(2) Arrange in increasing order of extent of hydrolysis [ CCI4, MgCI2, AICI3, PCl5, SiCI4].

(3) Although Sulphur contain vacant d-orbital but SF6 does not under go hydrolysis. Why ?

(4) CCl4 can not be hydrolysed but SiCl4 can be. Why?

(5) Silianol (SiH3OH) is more acidic than methanol (CH3OH) why?

(6) What happened when Graphite treated concentrated acids like HNO3/HF or KMnO4 ?.

(7) Why Boric acid become strong acid in the presence of cis 1,2-diol or 1,3-diol ?

(6) What is the molecular formula of Borax ?

(9) What is oil of vitriol ?

(11) Is all the C-C bond length in fullerene is equal ?

(12) Why CO2 is a gas, but SiO2 is a high melting solid?.

(13) CCl4 can not be hydrolysed but SiCl4 can be. Why?

(14) Why the C-C bond length in graphite is shorter than C-C bond length of diamond?

(15) How many six membered ( hexagons) and five membered (Pentagons) carbon ring present in (C60) fullerene ?.

What is Laporte Selection rule?

Laporte Selection Rule is given by Otto Laporte a German American Physicist

According to Laporte selection rule only allowed transitions are those occurring with a change in parity (flip in the sign of one spatial coordinate.) OR During an electronic transition the azimuthal quantum number can change only by ± 1 (Δ l = ±1) The Laporte selection rule reflects the fact that for light to interact with a molecule and be absorbed, there should be a change in dipole moment.

Practical meaning of the Laporte rule:

Laporte allowed transitions: are those which occur between gerade to ungerade or ungerade to gerade orbitals.

Laporte forbidden transitions: are those which occur between gerade to gerade or ungerade to ungerade orbitals. 

Gerade = symmetric with respect to centre of inversion i.e. atomic or molecular orbital with center of symmetry or number of nodal plane = 0, 2, 4 (even number)

Ungerade = anti symmetric with respect to centre of inversion i.e. atomic or molecular orbital without center of symmetry or number of nodal plane = 1, 3, 5, (odd numbers)

Important Note: 

This rule affects Octahedral and Square planar complexes as they have center of symmetry. Tetrahedral complexes do not have center of symmetry therefore this rule does not apply

Related Question:

What is the Selection rule for colour in complexes?

Although both [Mn(H2O)6]2+ and [FeF6]3- have a d5 configuration and high-spin complexes. But the dilute solutions of Mn2+ and Fe +3 complexes are therefore colorless. Why?

Which of the Complex of the following pairs has the highest value of CFSE?
Colour of Complexes due to charge transfer:
Why violet colour of [Ti(H2O)6]Cl3 disapear (colourless) on heating heating ? 
Why [Ni(CN)4]-2 is colourless while [Ni(H2O)4]-2 although both have +2 oxidation state and 3d*8 configuration ?
Why [FeF6]3– is colourless whereas [CoF6]3– is coloured ? 
Why Fe(CO)5 is colourless while Fe(bipy)(CO)3 is intensely purple in colour ?
Why all the tetrahedral Complexes are high spin Complexes ?

Although both [Mn(H2O)6]2+ and [FeF6]3- have a d5 configuration and high-spin complexes. But the dilute solutions of Mn2+ and Fe +3 complexes are therefore colorless. Why?

Both [Mn(H₂O)₆]²⁺ and [FeF₆]^3- have a d⁵ configuration and high-spin complexes, but electronic transitions are not only Laporte-forbidden, but also spin-forbidden. thus the dilute solutions of Mn²⁺ and Fe ⁺³ complexes are colorless.

Important Note:

For first transition series d⁵ system, weak ligand field , and coordination number six (6) Complexes are found to be colourless due to violation of selection rule.

What is the Selection rule for colour in complexes?

Spin selection rule states that transitions that involve a change in spin multiplicity as compare to ground state are forbidden. 

(1) According to this rule, any transition for which Δ S = 0 (it means no change in spin multiplicity after d-d transition) is allowed. 

(2) If Δ S ≠ 0 (change in spin multiplicity after transition) then it is forbidden (transition not allowed)

Intensity of colour due to d-d transition:

(1) Intensity of colour due to d-d transition will found to be high if transition follow laporte selection rule.

(2) Intensity of colour due to d-d transition will found to be poor due violation of laporte selection rule.

(3) Intensity of colour in tetrahedral Complexes for (non centre of symmetry) is found to be higher than octahedral (centre of symmetry).

Important Note:

For first transition series d⁵ system, weak ligand field, and coordination number six (6) Complexes are found to be colourless due to violation of selection rule.

Related Question:

Although both [Mn(H2O)6]2+ and [FeF6]3- have a d5 configuration and high-spin complexes. But the dilute solutions of Mn2+ and Fe +3 complexes are therefore colorless. Why?

Which of the Complex of the following pairs has the highest value of CFSE?
Colour of Complexes due to charge transfer:
Why violet colour of [Ti(H2O)6]Cl3 disapear (colourless) on heating heating ? 
Why [Ni(CN)4]-2 is colourless while [Ni(H2O)4]-2 although both have +2 oxidation state and 3d*8 configuration ?
Why [FeF6]3– is colourless whereas [CoF6]3– is coloured ? 
Why Fe(CO)5 is colourless while Fe(bipy)(CO)3 is intensely purple in colour ?
Why all the tetrahedral Complexes are high spin Complexes ?