Hydrocarbons

Step 1: Understand the reaction. The given reaction shows the conversion of a carboxylic acid to an alcohol with the elimination of carbon dioxide. This type of reaction is typical for decarboxylation, which occurs in the presence of NaOH.
Step 2: Conclusion. Thus, the reaction takes place in the presence of NaOH.

Step 1: Identifying the reaction type.
The Diels-Alder reaction is a cycloaddition reaction between a diene and a dienophile, often involving -unsaturated carbonyl compounds.

Step 2: Conclusion.
Thus, the correct answer is option (B).

Final Answer:

Step 1: Structure of epoxide.
An epoxide is a three-membered ring with two carbon atoms and one oxygen atom. It is a cyclic ether with highly strained bonds.

Step 2: Conclusion.
Thus, the correct answer is option (A).

Final Answer:

Step 1: Grignard reagents.
Grignard reagents are organomagnesium compounds, where magnesium forms an organometallic bond with carbon.

Step 2: Conclusion.
Thus, the correct answer is option (C).

Final Answer:

Step 1: Identify the reaction type.
Ethane reacts with nitric acid at high temperature (around 675 K) to give nitroalkanes via free radical substitution.
Step 2: Explain product formation.
In ethane, both carbon atoms are equivalent, but radical substitution can lead to formation of two nitro products due to different radical fragmentation pathways:
- Nitroethane: CH CH NO
- Nitromethane: CH NO
Step 3: Conclude the correct option.
Thus, the mixture of CH CH NO and CH NO is obtained.
Final Answer:

Step 1: Recall reaction of secondary nitroalkanes.
Secondary nitroalkanes ( ) upon hydrolysis in acidic medium form ketones.
This is known as Nef reaction.
Step 2: Nef reaction condition.
The nitroalkane is first converted into nitronate salt in base, then acid hydrolysis gives carbonyl compound.
But the final conversion to ketone needs aqueous acid (HCl).
Step 3: Conclusion.
Thus reagent Y is aqueous HCl.
Final Answer:

Step 1: Identify reaction shown.
Benzene is converted into benzaldehyde in presence of and HCl.
This is the Gattermann–Koch reaction.
Step 2: Reactants of Gattermann–Koch reaction.
The formyl group is introduced using:

in presence of (and often CuCl).
Step 3: Conclusion.
So the missing reactant is carbon monoxide .
Final Answer:

Step 1: Recall hydroboration-oxidation reaction.
Hydroboration-oxidation converts an alkene into an alcohol by anti-Markovnikov addition.
Reagents:

Step 2: Nature of addition.
OH group attaches to the less substituted carbon of the double bond (anti-Markovnikov rule).
Step 3: Apply to 4-methyl-octene.
Since the alkene has 8 carbon chain with a methyl group at position 4, the product remains an octanol derivative.
So the product formed is 4-methyl octanol.
Final Answer:

Step 1: Understand Lindlar reduction.
Lindlar catalyst reduces an alkyne to a cis-alkene.
So starting alkyne becomes an alkene .
Step 2: Use ozonolysis clue.
Ozonolysis breaks the double bond to give carbonyl compounds.
If ozonolysis gives only one type of product, the alkene must be symmetric.
Step 3: Check which alkyne gives symmetric alkene.
(A) is symmetric, gives cis-2-butene.
cis-2-butene on ozonolysis gives only one kind of product: acetaldehyde (2 moles).
Other options are unsymmetrical, giving two different products.
Step 4: Hence correct alkyne is option (A).
Final Answer:

Step 1: Determine molecular formula of .
Assume :


Ratio:

Empirical formula:

Step 2: Use vapour density to find molar mass.

Empirical mass of .
So molecular formula:

Thus (ethene).
Step 3: Reaction with .
Ethene gives chlorohydrin:

So .
Step 4: Reaction with followed by hydrolysis.


So:

Final Answer:

Step 1: Use ozonolysis rule.
Ozonolysis breaks the double bond and converts both carbon atoms of double bond into carbonyl groups.
Step 2: Identify products.
Products are:
- Acetaldehyde:
- Acetone:
Step 3: Reconstruct the alkene.
Acetaldehyde comes from a carbon having and .
Acetone comes from a carbon having two groups.
Thus alkene must be:

Step 4: Name the alkene.
This is -methyl- -butene.
Final Answer:

Step 1: Identify .
with alcoholic KOH undergoes dehydrohalogenation (elimination) to form ethene:
Step 2: Identify .
Ethene adds bromine in presence of to form 1,2-dibromoethane:
Step 3: Identify .
1,2-dibromoethane reacts with KCN and both bromines are replaced by CN groups, giving dicyanoethane:
Step 4: Final hydrolysis to acid .
On acidic hydrolysis, both groups convert to :
This is succinic acid.
Final Answer:

Step 1: Understand the Wurtz reaction.
In the Wurtz reaction, two alkyl halides react in the presence of sodium to form a higher alkane. In this case, the reaction leads to the formation of , which is the expected product.
Step 2: Conclusion.
Thus, the product of the Wurtz reaction for is .
Final Answer:

Step 1: Understand the reaction mechanism.
The ozonolysis of alkenes typically results in cleavage of the double bond to produce aldehydes or ketones. The given products glyoxal and 2, 2-dimethyl butane-1, 4-dial point towards the structure of the alkene.
Step 2: Conclusion.
The alkene must be one that results in these specific products upon ozonolysis, which corresponds to option (B).
Final Answer:

The compound is a conjugated system with both a double bond and a triple bond. Its IUPAC name is derived by identifying the positions of the functional groups. The correct name is pent-3-en-1-yne.

Step 2: Conclusion.
The correct IUPAC name of the compound is pent-3-en-1-yne, corresponding to option (b).

The Wurtz reaction can be used to prepare alkanes by the coupling of two alkyl halides in the presence of sodium. All of the given compounds can be synthesized through this reaction.

The reaction is an example of a nucleophilic addition reaction, where the mercury ion adds to the double bond, leading to the formation of the product.

The Wurtz reaction of leads to the formation of isobutane by coupling two ethyl groups in the presence of sodium.

The false statements are: - Statement I: A primary carbocation is less stable than a tertiary carbocation. This is false because a tertiary carbocation is more stable due to alkyl groups providing inductive and hyperconjugative stabilization. - Statement IV: Isopropyl carbocation is more stable than ethyl carbocation. This is false because ethyl carbocation is more stable than isopropyl carbocation, as the inductive effects of alkyl groups play a role.

Step 1: Carbocation Formation.
When an alcohol (OH) is treated with an acid, it undergoes dehydration to form carbocations. Depending on the structure, 1°, 2°, or 3° carbocations can be formed. All types can form under the right conditions.
Step 2: Conclusion.
The correct answer is (D), All the three.

Step 1: When HBr reacts with alkenes in the presence of anhydrous conditions, it adds across the double bond.
Step 2: The product formed is a bromoether, where the Br atom adds to the carbon adjacent to the ether group.

Final Answer:

The given structure corresponds to a nitrated aromatic compound. The possible starting materials could include various sources of nitration such as conc. nitric acid, conc. sulfuric acid, and a chlorination source like with aluminum chloride.
Final Answer:

The ozonolysis of the alkene breaks it into aldehydes and ketones. Based on the product distribution, the alkene with the given formula is -trimethyl-3-hexene.
Final Answer:

The reaction between acetylene and methanol under these conditions forms methyl vinyl ether through an addition reaction.
Final Answer:

To calculate the value of , we use the equation: Using the relation , we can substitute the values and simplify the expression, resulting in .

Step 1: Hydrolysis of Xenon Hexafluoride.
On partial hydrolysis, xenon hexafluoride reacts with water to form various compounds. The products formed are Xenon oxyfluoride and xenon dioxide fluoride.

Step 2: Oxidation states.
The oxidation state of Xenon in XeOF is +6, and in XeO F , it is also +6.

Ethers have an oxygen atom bonded to two alkyl/aryl groups.