Alcohols

Step 1: Understand the reaction with LiAlH . Lithium aluminum hydride (LiAlH ) is a strong reducing agent. When alcohols react with LiAlH , they are reduced to the corresponding aldehydes or alcohols depending on the condition.
Step 2: Conclusion. Thus, R–OH reacts with LiAlH to form a primary alcohol, RCH OH.

Step 1: Understand the reactions. Alkaline KMnO does not oxidize aldehydes like CH CHO under normal conditions. It only oxidizes aldehydes in acidic conditions.
Step 2: Conclusion. Thus, CH CHO does not react with alkaline KMnO .

Step 1: Reaction of diethyl ether with water. When diethyl ether is heated with water in the presence of an acid catalyst, it undergoes hydrolysis to form ethyl alcohol.
Step 2: Conclusion. Thus, the product is ethyl alcohol.

Step 1: Hybridization of oxygen in alcohol. The oxygen atom in alcohol (R-OH) is bonded to two atoms, and it has two lone pairs. This corresponds to an sp hybridization.
Step 2: Conclusion. Thus, the hybridization of oxygen in an alcohol molecule is sp .

Correct answer is (b)

Step 1: Effect of substituents on acidity.
Electron-withdrawing groups like or increase the acidity of phenols by stabilizing the negative charge on the oxygen after deprotonation. Conversely, electron-donating groups like reduce the acidity of phenol.

Step 2: Conclusion.
Thus, phenol is less acidic than p-methoxyphenol, but more acidic than p-chlorophenol. Therefore, options (A) and (C) are correct.

Final Answer:

Step 1: Reaction of ethanol with conc. H SO .
Concentrated sulphuric acid acts as a dehydrating agent.
Step 2: Dehydration of ethanol.
On heating ethanol at about , it loses water and forms ethene.

Step 3: Identify product.
The product is ethene .
Final Answer:

Step 1: Identify anisole.
Anisole is methoxybenzene:

Step 2: Formation from phenol.
Phenol reacts with methyl halide in presence of base to form anisole via Williamson ether synthesis.

Step 3: Name of reaction type.
Formation of ether from phenol is called etherification.
Final Answer:

Step 1: Identify ethylene glycol.
Ethylene glycol is a dihydric alcohol:

Step 2: Strong oxidation converts glycol to dicarboxylic acid.
Under strong oxidizing conditions, both terminal groups are oxidized to .
Step 3: Oxidant required for complete oxidation.
Alkaline KMnO is a strong oxidizing agent that oxidizes ethylene glycol to oxalic acid:

Step 4: Conclusion.
Thus alkaline KMnO gives oxalic acid.
Final Answer:

Step 1: First chlorination of ethanol.
Chlorine acts as an oxidizing agent in presence of light and gives acetaldehyde:

So:

Step 2: Further chlorination of acetaldehyde.
Acetaldehyde undergoes successive substitution of -hydrogens to form chloral:

So:

Final Answer:

Step 1: Reaction of glycerol with oxalic acid.
When glycerol is heated with oxalic acid at , oxalic acid decomposes and produces formic acid.
This is a known preparation method of formic acid in lab.
Step 2: Note temperature dependence.
At higher temperature ( ), glycerol gives acrolein by dehydration.
But at , the major product is formic acid.
Final Answer:

Step 1: Identify the reagent.
represents diazomethane ( ).
Diazomethane is a methylating agent.
Step 2: Role of .
Strong acid protonates diazomethane to generate -like species.
This methyl group then reacts with alcohol oxygen.
Step 3: Product formed.
Alcohol is converted into methyl ether:
Final Answer:

Step 1: Meaning of carbinol.
Carbinol is old name for methanol.
Substituted carbinols are named depending on groups attached to carbon having .
Step 2: Ethyl carbinol structure.
Ethyl carbinol means one ethyl group attached to the carbon bearing .
So structure becomes:
which is isopropyl alcohol (2-propanol).
Final Answer:

Step 1: Recall Victor Meyer test.
Victor Meyer test distinguishes between primary, secondary and tertiary alcohols.
Step 2: Colour obtained.
- Primary alcohol gives red colour.
- Secondary alcohol gives blue colour.
- Tertiary alcohol gives no colour.
Step 3: Identify type of given alcohols.
- n-propyl alcohol is primary.
- isopropyl alcohol is secondary.
- tert-butyl alcohol is tertiary.
- sec-butyl alcohol is secondary.
Step 4: Therefore red colour is given by primary alcohol.
So correct is n-propyl alcohol.
Final Answer:

Step 1: Understand the formation of peroxides.
Peroxides are highly reactive compounds that can form during the storage of ethers. The accumulation of peroxides in ether can lead to explosive reactions when the ether is distilled.
Step 2: Conclusion.
Thus, the presence of peroxides is responsible for the explosion that can occur during the distillation of ethers.
Final Answer:

Step 1: Role of glycerine in preservation.
Glycerine acts as a preservative by preventing microbial growth due to its hygroscopic nature, which reduces the water activity in food.
Step 2: Conclusion.
Thus, glycerine is used to prevent microbial growth and extend the shelf life of fruits and vegetables.
Final Answer:

Secondary alcohol, on reaction with anhydrous and conc. (Lucas reagent) gives an oily layer product after five minuts.
oily layer
product (turbidity) after 5 min.

When glycerol reacts with excess iodine and heat, it undergoes substitution to form allyl iodide. The reaction involves the substitution of the hydroxyl group with an iodine atom.

Step 2: Conclusion.
The product of the reaction is allyl iodide, corresponding to option (b).

The compound that is most readily dehydrated is the one that can form the most stable carbocation or intermediate during dehydration. can form a stable secondary carbocation, making it the most easily dehydrated compared to the other compounds.

Step 2: Conclusion.
The compound is most readily dehydrated, corresponding to option (c).

Step 1: Grignard reagent reaction.
Grignard reagents react with carbonyl compounds to form an alcohol. In this case, the reaction with the ketone forms a secondary alcohol.

Step 2: Conclusion.
The product is a secondary alcohol, , corresponding to option (3).

Step 1: Analyzing the reaction.
Glycerol reacts with HCl, which leads to the formation of a product where an additional chlorine atom replaces one of the hydroxyl groups. The molecular formula of the product is .

Step 2: Conclusion.
The correct product structure is , corresponding to option (2).

oxygen atom can donate lone pair of electron more easily, therefore, it is more basic than -oxygen.

(a)
No reaction

No reaction

No reaction

Fructose has multiple chiral centers, leading to the formation of 16 optical isomers. This can be calculated by the formula , where is the number of chiral centers. Fructose has 4 chiral centers, so the number of optical isomers is .

Step 2: Conclusion.
Thus, the correct answer is 16.

The parent chain has three carbons with OH on second carbon.

In (CH ) CHOH, the carbon bearing –OH is attached to two other carbons.