Alcohols: Preparation & Properties

Hint: Ether mixed with alcohol then no reaction occurs

Acetyl chloride and acetic anhydride react with alcohol and form an ester as a product. The reaction is an example of a nucleophilic acyl substitution reaction. In a nucleophilic substitution reaction, strong nucleophiles substitute weak nucleophiles.

The reaction is as follows:

But when Alcohol reacts with ether then no reaction occurs because ether and alcohol both act as a nucleophile and two nucleophiles do not react with each other.  Hence, ether is the correct answer.

HINT: Temperature is very effective factor in dehydration of alcohols.

Explanation: 

Temperature is the effective factor for dehydration of alcohol by Al₂O₃. At higher temperautre (350^∘C−380^∘C) 1^∘ alcohol undergoes intramolecular dehydration to yield an alkene:

$RCH_{2}CH_{2}OH \: \: \xrightarrow[Al_{2}O_{3}]{350-380^{0}C}\: \: RCH=CH_{2} \, \, +\, \, H_{2} O \, \, \, (Intramolecular \: \:\, \, \, \, dehydration)$

while at lower temperature (220^∘C − 250^∘C) it undergoes intermolecular dehydration to yield to ether.

$RCH_{2}CH_{2}OH \, \, + \, \, \,OH CH_{2}CH_{2}R \xrightarrow \: \: RCH_{2}CH_{2}O CH_{2}CH_{2}R \, \, +\, \, H_{2} O \, \, \, (Intermolecular \, \, \, dehydration)$

HINT: Iodoform reaction.

Explantion: 

Iodoform test is used to check the presence of carbonyl compounds with the structure R-CO-CH₃ or alcohols with the structure R-CH(OH)-CH₃ in a given unknown compound. 

Among the given compound , CH₃OH doesnt’t have -CH(OH)-CH₃ group so it won’t give iodoform reaction.

HINT: On oxidation both gives different products.

Explanation: 

On treating n-propyl alcohol with acidified potassium dichromate, we get an aldehyde.  

On treating secondary alcohol with potassium dichromate in an acidic medium, it acts as an oxidation agent and gives propanone (ketone) as the product.

Ozonolysis is the addition of Oxygen to alkenes or alkynes and making it saturated. Therefore, it cannot be used to distinguish between them.

Hint: In this reaction, I^– will attack the less substituted carbon.

Explanation:

Since mixed ether is given in this reaction, note that iodide will bond with the smaller alkyl group to minimise the steric repulsion as iodide ion is bulky itself.

The reaction is as follows:

${\mathrm{CH}}_3-\stackrel{}{}\stackrel{|}{\mathrm{CH}}{\mathrm{CH}}_3-{\mathrm{CH}}_2-\mathrm{O}-{\mathrm{CH}}_2-{\mathrm{CH}}_3+\mathrm{HI}\stackrel{∆}{\to }{\mathrm{CH}}_3-\stackrel{}{}\stackrel{|}{\mathrm{CH}}{\mathrm{CH}}_3-{\mathrm{CH}}_2\mathrm{OH}+{\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{I}$

Hint: The compound containing methyl ketone and secondary alcohol gives the iodoform test.

The compound containing  and gives the iodoform test.

Step 1:

Compounds having ${\mathrm{H}}_3\mathrm{C}-\mathrm{CO}-$ group and ${\mathrm{CH}}_3\mathrm{CH}\left(\mathrm{OH}\right)-$ gives iodoform when heated with Iodine and NaOH.

Step 2:

Among the following compounds CH₃CH₂OH, CH₃COCH₃  and propan-2-ol  will give iodoform test.
Ethanol is the only primary alcohol to give the triiodomethane (iodoform) reaction.

Hence, the second option is the correct answer.

Hint: The first reaction is an oxidation reaction and the second reaction is an electrophilic aromatic substitution reaction.

Alkaline KMnO₄is a strong oxidizing agent and therefore it oxidizes the whole carbon chain directly attached to the benzene nucleus into the –COOH group.

Furthermore, the Br₂ group in the presence of a halogen carrier causes bromination whereas ethyl alcohol in an acidic medium results in esterification. 

HINT : Glycol is an antifreeze agent.

Explanation:

Antifreeze is a liquid that prevents the radiator in cars from freezing or overheating. It’s also known as engine coolant. It’s key ingredient is ethylene glycol, which lowers water’s freezing point and raises its boiling point. 

So, option (1) is the correct answer.