(i) Add NaHCO₃ to each. Phenol will not react, benzoic acid will give brisk effervescence due to CO₂.
(ii) Add I₂ and NaOH. 2-Propanol will give yellow ppt. of iodoform, whereas 1-propanol will not give yellow ppt.

methyl iodide and phenol

The mechanism of acid dehydration of ethanol to yield ethene is shown above. First step is the protonation of oxygen atom of −OH group. Second step is the loss of a molecule of water to form carbonium ion. Last step is deprotonation to form carbon carbon double bond.
Step 1:
Protonation of ethanol to form ethyl oxonium ion:

Step 2:
Formation of carbocation (rate determining step):
 
Step 3:
Elimination of a proton to form ethene:

(i) 


(ii) 

(i) Propene to propan-1-ol

(ii) ethanal to propan-2-ol

In phenols, the lone pairs of electrons on the oxygen atom are delocalised over the benzene ring due to resonance and hence, are not easily available for protonation to destabilise O-H bond thus, making protonation difficult.

Benzene react with chlorine in the presenc of ferric chloride which form chlorobezene, which on further reaction with sodium hydroxide gives sodium phenoxide. Then sodium phenoxide acidified and form phenol.

3-pentanol immediately gives turbidity with Lucas reagent but 2-pentanol gives turbidity in around five minutes.

The reaction of alcohols with conc.HCI and ZnCI₂ (Lucas reagent) occurs through intermediate formation of carbocations. Obviously, more stable the carbocation, more reactive is the alcohol. Since the stability of carbocations. Obviusly, more stable the carbocation, more reactive is the alcohol. Since the stability of carbocations follows the order : \({ 3 }^{ \circ }>{ 2 }^{ \circ }>{ 1 }^{ \circ }\), therefore, reactivity of alcohols towards Lucas reagent follows the same order, i.e., \({ 3 }^{ \circ }>{ 2 }^{ \circ }>{ 1 }^{ \circ }\).

\({ C }H_{ 3 }-\underset { \overset { \parallel }{ O } }{ C } -{ C }H_{ 3 }+{ C }H_{ 3 }MgBr\overset { Dry }{ \underset { ether }{ \longrightarrow } } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \\ { C }H_{ 3 }-\overset { \underset { | }{ { C }H_{ 3 } } }{ \underset { \overset { | }{ { C }H_{ 3 } } }{ C } } \quad -OMgBr\overset { { H }^{ + }{ H }_{ 2 }O }{ \longrightarrow } \underset { tert-butyl\quad alcohol }{ \quad { C }H_{ 3 }-\overset { \underset { | }{ { C }H_{ 3 } } }{ \underset { \overset { | }{ { C }H_{ 3 } } }{ C } } \quad -OH } \)

Boiling points of ethers are much lower than the boiling points of alcohols. The large difference in boiling points of alcohols and ethers is due to the presence of H-bonding in alcohols.

The reactions showing acidic nature of phenol are:
(i) Reaction with sodium. Phenol reacts with active metals like sodium to liberate H₂ gas.

(ii) Reaction with NaOH. Phenol dissolves in NaOH to form sod. phenoxide and water.

Comparison of acidic character of phenol and ethanol. Phenol is more acidic than ethanol. This is due to the reason that phenoxide ion left after the loss of a proton from phenol is stabilized by resonance (Refer to structures,VI to X on page 11/25) while ethoxide ion (left after loss of a proton from ethanol) is not.

o-Nitrophenol is steam-volatile due to chelation (intramolecular H-bonding) and hence can be separated by steam distillation from p-nitrophenol which is not steam volatile because of intermolecular H-bonding. 

The addition of diborane to alkenes to form trialkylboranes followed by their oxidation with alkaline hydrogen peroxide to form alcohols is called hydroboration-oxidation reaction.

For example,
\(\underset { Propene }{ { CH }_{ 3 }CH={ CH }_{ 2 } } \overset { { BH }_{ 3 } }{ \underset { THF }{ \longrightarrow } } ({ CH }_{ 3 }{ CH }_{ 2 }{ CH }_{ 2 })\overset { { OH }^{ - },{ H }_{ 2 }{ O }_{ 2 } }{ \longrightarrow } \underset { Propene-1-ol }{ { 3CH }_{ 3 }CH{ CH }_{ 2 } } OH+B({ OH })_{ 3 }\)

It is because alcohol can form H-bonds with H₂O, whereas hydrocarbons cannot, that is why alcohols are soluble in water.

Cresol is methylphenol. It has three positional isomers: o-cresol, m-cresol and p-cresol. Due to +l-effect of CH₃ group, electron density in the O-H bond increases. Therefore, it becomes difficult to break the O-H bond as compared to that in phenol and hence all cresols are less acidic than phenol or phenol is a stronger acid than cresols.

(Due to + l-effect of CH₃ group, electron density in the O- H bond increases thereby making the cleavage of O - H bond difficult)

It is due to C -O bond is less polar due to electron-withdrawing effect of benzene ring, whereas in methanol C--O bond is more polar due to electron releasing effect of methyl group.

(i) The major product of the given reaction is 2-methylprop-1-ene. It is because sodium ethoxide is a strong nucleophile as well as a strong base. Thus elimination reaction predominates over substitution.
(ii) 

Propan-1-ol, 4-methylphenol, phenol,  3-nitro-phenol, 3,5-dinitrophenol,  2,4,6-trinitrophenol is increasing order of acid strength.

(a) Methanol < ethanol < propan-1-ol < butan-2-ol < butan-1-ol < pentan-1-ol
(b) n-Butane < ethoxyethane < pentanal < pentan-1-ol.

1-Propanol will not react with I₂/NaOH to give iodoform.

(i)

(ii)
 
(iii)
 
(iv)

To prepare phenol, cumene is first oxidized in the presence of air of cumene hydro-peroxide.
 
Then, cumene hydroxide is treated with dilute acid to prepare phenol and acetone as by-products.

(i) If propene is allowed to react with water in the presence of an acid as a catalyst, then propan-2-ol is obtained.

(ii) If benzyl chloride is treated with NaOH (followed by acidification) then benzyl alcohol is produced.
 
(iii) When ethyl magnesium chloride is treated with methanal, an adduct is the produced which gives propan-1-ol on hydrolysis.
 
(iv) When methyl magnesium bromide is treated with propane, an adduct is the product which gives 2-methylpropane-2-ol on hydrolysis.