All questions of Hydrocarbons for NEET Exam

The cis isomer in this case has a boiling point of 60.3 degC, while the trans isomer has a boiling point of 47.5 degC. In the cis isomer the two polar C-Cl bond dipole moments combine to give an overall molecular dipole, so that there are intermolecular dipole–dipole forces (or Keesom forces), which add to the London.

The answer is c. But-3-ene. But-3-ene is not a structural isomer of C4H8 because it is the same molecule as But-1-ene, just numbered differently. The structural isomers of C4H8 are But-1-ene, But-2-ene, and 2-methylpropene. But-1-ene and But-2-ene are position isomers, differing in the position of the double bond. 2-methylpropene is a branched isomer. But-3-ene is not a distinct isomer because the numbering of the carbon chain starts from the end closest to the double bond, making it identical to But-1-ene.

7 is correct.

Halogens can act as electrophiles to attack a double bond in alkene. Double bond represents a region of electron density and therefore functions as a nucleophile. 

The alkene produced as a byproduct during the free radical bromination of isobutane is 2-methyl propene (option c). Let's understand the process in detail:

1. Free Radical Bromination:
During free radical bromination, isobutane (C4H10) reacts with a bromine radical (Br•) to form a primary alkyl radical (C4H9•). This reaction is initiated by the presence of light or heat.

C4H10 + Br• → C4H9• + HBr

2. Disproportionation of the Intermediate Alkyl Free Radical:
The intermediate alkyl free radical (C4H9•) can undergo different reactions, including abstraction of a hydrogen atom from another isobutane molecule or rearrangement. However, in this case, it undergoes disproportionation to form a new alkene.

C4H9• → Alkene + Alkane

3. Formation of 2-Methyl Propene:
The disproportionation of the C4H9• radical leads to the formation of 2-methyl propene. This alkene is obtained when the alkyl radical undergoes a β-scission, resulting in the formation of a double bond.

C4H9• → 2-methyl propene + Alkane

Hence, during the free radical bromination of isobutane, 2-methyl propene is produced as a byproduct through the disproportionation of the intermediate alkyl free radical.

To summarize:
- Isobutane reacts with a bromine radical to form a primary alkyl radical.
- The primary alkyl radical can undergo various reactions, but in this case, it undergoes disproportionation.
- The disproportionation of the alkyl radical leads to the formation of 2-methyl propene as a byproduct.

The compound that reacts most readily with Br(g) is C3H6. Here's why:

Explanation:
When a compound reacts with Br(g), it undergoes a substitution reaction called bromination. In this reaction, a Br atom replaces a hydrogen atom in the compound. The reactivity of a compound towards bromination depends on its structure and the stability of the resulting product.

Comparing the compounds:
Let's compare the given compounds and analyze their structures to determine which one is most reactive towards bromination.

a) C2H2:
C2H2 is an alkyne with a triple bond between two carbon atoms. This triple bond is very strong and stable, making it difficult for Br(g) to break it and substitute a hydrogen atom. Therefore, C2H2 is less reactive towards bromination.

b) C3H6:
C3H6 is an alkene with a double bond between two carbon atoms. The double bond is weaker and less stable than a triple bond. Therefore, it is easier for Br(g) to break the double bond and substitute a hydrogen atom. This makes C3H6 more reactive towards bromination compared to C2H2.

c) C2H4:
C2H4 is also an alkene with a double bond between two carbon atoms, similar to C3H6. It has the same structure as C3H6, but it has fewer carbon atoms. Since the number of carbon atoms does not significantly affect the reactivity towards bromination, C2H4 is also reactive towards bromination, but less reactive compared to C3H6.

d) C4H10:
C4H10 is an alkane with only single bonds between carbon atoms. Alkanes are generally less reactive towards bromination because the single bonds are strong and stable. Breaking a single bond to substitute a hydrogen atom is more difficult for Br(g) compared to breaking a double or triple bond. Therefore, C4H10 is the least reactive towards bromination among the given compounds.

Conclusion:
Based on the structural analysis and the stability of the bonds, C3H6 is the most reactive compound towards bromination among the given options.

Photochemical chlorination of alkane take place by free radical mechanism which are possible by Homolysis of C - C bond

Propene, also known as propylene, is an unsaturated hydrocarbon with the chemical formula C3H6. When propene reacts with HBr (hydrogen bromide) in the presence of peroxide, it undergoes a radical addition reaction to form n-propyl bromide (n-C3H7Br).

Here is a detailed explanation of the reaction:

1. Radical Initiation:
The presence of peroxide (often represented as ROOR, where R is an alkyl group) initiates the reaction by undergoing homolytic cleavage to produce two alkyl radicals, represented as R•. In this case, the peroxide could be tert-butyl peroxide (C(CH3)3OO•).

ROOR → 2R•

2. Radical Propagation:
The alkyl radical (R•) reacts with propene (C3H6) to form a more stable secondary radical intermediate.

R• + C3H6 → RCH2CH2•

The secondary radical intermediate then reacts with HBr to produce the alkyl bromide.

RCH2CH2• + HBr → RCH2CH2Br

This reaction step can occur at any position on the propene molecule, resulting in different possible alkyl bromide products.

3. Radical Termination:
The reaction can also undergo radical termination steps where two alkyl radicals combine to form a non-radical species. These termination steps play a role in consuming any excess alkyl radicals and stopping the chain reaction.

R• + R• → R-R

In the case of propene reacting with HBr in the presence of peroxide, the major product formed is n-propyl bromide (n-C3H7Br). This is because the secondary radical intermediate reacts preferentially with HBr at the terminal carbon of propene, resulting in the formation of n-propyl bromide. The other possible products, such as allyl bromide (C3H5Br), isopropyl bromide (C3H7Br), and 3-bromopropane (C3H7Br), are less favored due to the stability of the radical intermediate and the reactivity of HBr at the terminal carbon.

In summary, when propene reacts with HBr in the presence of peroxide, the major product obtained is n-propyl bromide (n-C3H7Br). This reaction follows a radical mechanism, initiated by the homolytic cleavage of peroxide and proceeds through radical propagation steps to form the alkyl bromide product.

B and C is correct

Understanding Alkenes and Their Reactivity
Alkenes are hydrocarbons characterized by the presence of at least one carbon-carbon double bond (C=C). This unique structure significantly influences their chemical properties, especially their reactivity compared to alkanes.
Incorrect Statement: Alkenes are less reactive than alkanes
- Alkenes are generally more reactive than alkanes. This increased reactivity is due to the presence of the double bond, which is more susceptible to reactions than the single bonds found in alkanes.
- The double bond consists of one sigma bond and one pi bond. The pi bond is a region of high electron density that can engage in electrophilic addition reactions, making alkenes quite reactive.
Comparison of Alkenes and Alkanes
- Physical Properties: Alkenes have similar physical properties to alkanes, such as boiling and melting points, primarily due to their similar structures and molecular weights.
- Reactivity: Alkenes react readily with electrophiles, undergoing reactions like hydrogenation, halogenation, and polymerization, enhancing their chemical activity.
Key Reactions of Alkenes
- Polymerization: Alkenes can undergo polymerization, a process where small alkene molecules (monomers) combine to form large polymer chains. This reaction is a fundamental aspect of synthetic materials.
- Electrophilic Addition: Alkenes are also involved in electrophilic addition reactions, where an electrophile reacts with the double bond, leading to the formation of more complex molecules.
In summary, the assertion that alkenes are less reactive than alkanes is incorrect. Alkenes are indeed more reactive due to their double bonds, allowing them to participate in a variety of chemical reactions.

Ans.is 6

It is a type of HYDROBORATION OXIDATION REACTION (alcohol and phenol and ether chapter ) 1. 2-Methylcyclohexene reacts with (BD3)2 to form a intermediate compound 2. In which the deturium deficient site the deterium will go and attach 3. And the peroxide will give a OH- and it will attach to the next to the deturium attach (it a kind of anti markowiffs rule)