All questions of Hydrocarbons for NEET Exam

The correct answer is option c

Thus, a racemic mixture is obtained. A racemic mixture is one that has an equal amount of left and right handed enantiomers of a chiral molecule.

Isocyanide is a compound and it is not a functional group.

When propene on ozonolysis it yields a new structure called ozonide 
and there cleavage takes place and it yields two products namely 
1.acetaldehyde
2.formaldehyde

In option c, there are only two different position which can be chlorinated, 
So, option c is correct.

Ethylene reacts with HBr to form Ethyl bromide. The reaction propagates as follow:-
H₂C=CH₂  +  HBr → H₂C+-CH₃ →H₂BrC-CH₃ 
Since π cloud is electron rich, so HBr dissociates into H⁺ and Br^-. H⁺ attacks on alkene to give a carbocation and then Br^- attacks to get ethyl bromide.

This is the case of allylic substitution. The radical will delocalise itself to two locations. And so, we will get two different positions for radical substitution.

There is also a possibility at C₆, but it is not in our option.

For SO₂Cl₂: The reactivity patterns of SO₂Cl₂ and SOCl₂ are quite different. SOCl₂ is a good electrophile, and can be thought of as a source of Cl− ions. These ions can go on to react in their typical nucleophilic fashion. SO₂Cl₂ however is often a Cl₂ source, as it readily decomposes giving off sulfur dioxide. Usually, much easier/safer to use this than measuring out (and getting into solution) chlorine gas. The chlorination of simple alkanes by Cl₂ gas (or something that makes it in solution) happens by a radical mechanism i.e. Cl⋅ not Cl
For Cl₂ and heat/light:

For Cl with AlCl₃: It is used for chlorination of compounds like benzene
For HCl: It is used for halogenations of a double bond.

The correct answer is Option B. 

There are 3 monobromo derivatives exists for anthracene:
1-Chloroanthracene
2-Chloroanthracene
and 9-Chloroanthracene

Termination is the last step. So there shouldn't be any free radical atom remaining. In first option there is Cl• remaining it can't be termination step.The steps in free radical halogenation are as

 To determine the number of different isomers of alkenes (including stereoisomers) that upon catalytic hydrogenation yield the same 2,2,3,5-tetramethylhexane, 

1. Identify the Structure of the Product: The product, 2,2,3,5-tetramethylhexane, has a hexane backbone with methyl groups attached at the 2nd, 2nd, 3rd, and 5th carbon atoms.

2. Determine Possible Positions for the Double Bond: The double bond in the precursor alkene can be located between different pairs of carbon atoms in the hexane chain. Possible positions are: - Between C1-C2 - Between C2-C3 - Between C3-C4 - Between C4-C5

3. Consider Stereoisomers: For each position where the double bond is placed, determine if cis/trans (geometric) isomers are possible: - Between C1-C2: Only one structural isomer is possible since substituents do not allow for different geometrical arrangements. - Between C2-C3: Two stereoisomers are possible (cis and trans) due to the presence of different substituents on the double-bonded carbons. - Between C3-C4: Two stereoisomers are possible (cis and trans) for the same reason as above. - Between C4-C5: Only one structural isomer is possible.

4. Calculate the Total Number of Isomers: - C1-C2: 1 isomer - C2-C3: 2 isomers (cis and trans) - C3-C4: 2 isomers (cis and trans) - C4-C5: 1 isomer Adding these up: 1 + 2 + 2 + 1 = 6 isomers.

5. Consider Additional Structural Isomers: There is an additional structural isomer due to the possibility of a different branching pattern that still leads to the same saturated product upon hydrogenation. This increases the total count to 7 isomers. Therefore, there are 7 different isomers of alkenes that satisfy the given condition.

When chlorine and Ethane are taken with chlorine in excess only then we have more than one product like chloroethane, dichloroethane, trichloroethane etc. To avoid this we should take Ethane in excess because when we will take it then in excess then we will have only single time chlorination and we will get monochloroethane.

Benzene do not show addition reaction like other unsaturated hydrocarbons. However it show substitution reactions. Due to resonance all the C – C bonds have the same nature, which is possible because of the cycli c del oca lisati on of π-electr on s in benzene. Monosubstitution will give only a single product.

Compound I is having the highest acidic strength due to the -I effect of five CF₃ substituents.

Compound II is having less acidic strength than I but more than the rest due to the extremely stable conjugate anion formed after deprrotonation.

So, Option B is correct.

The reductive ozaonalysis of 2,3 -Dimethyl-2-butene yields acetone.The reaction is as follows: 2,3 -Dimethyl-2-butene acetone

Hydrogen abstraction by halogen radical in the propagation step is exothermic in chlorination but endothermic in bromination. Hence, the option (A) is an incorrect statement.

CH₃ – CH = CH₂ → CH₃ – CH⁺ – CH₂
The reason for this is only that carbocation is formed which has maximum stability. In this case, we have 6 α-H while for option a, b and d; we have 0, 2 and 2 α-H respectively. So only carbocation in option c forms.

For boiling point, we have to consider both branching and Molecular mass. In 4  bromopentane molecular mass is nearly the same as compared to 3 chloro pentane but we have 3,3-dichloropentane extended into 2 directions so the boiling point of 3,3-dichloropentane will be more and the other order will be followed by option C.

The correct answers are Options C and D.
Aromatic compounds are those which follow Huckel's rule i.e, they have (4n+2) π electrons, n must be an integer.
In option C, there are 5 π bonds which means 10 π electrons; so 4n+2 = 10 i.e, n= 2 which is an integer.
In option D, Nitrogen has a lone pair which contains 2 electrons therefore this compound also have 10 π electrons; so n= 2.

The correct answers are option B.
As only 4 hydrogen atoms are increased after hydrogenation, there should be only 1 3× bond. So option B is correct.
 

The correct answers are Options A, B and C.

After free radical halogenation of methyl cyclobutane, we have its 8different isomers. They are as follow:-


From i) and ii), we get only positional isomers. From iii) we will have 2 isomers, cis and Trans. They won't show a chiral centre.
In iv) we have 2 chiral centres which will give us 4 isomers.So, in total there would be 4+2+1+1 = 8 isomers.

Hence, Option A, B and D are correct.

The correct answer is option  A,B.
(a)  Tollen's reagent produces a white precipitate of acetylide with terminal alkyne.
(b) Fehling solution produces a red precipitate with terminal alkyne.
So $$1-butyne$$ is terminal alkyne.

Free radical bromination reaction is highly selective, occurs mainly at the carbon where most stable free radical is formed.
We know that the stability of free radical is in the order,
Tertiary radical > Secondary radical > Primary radical
In (a), (b) and (c), the bromination occurs at secondary carbon whereas in (d) the bromination occurs at tertiary carbon. Since, tertiary radicals are more stable than secondary radical the major product of monobromination of ethyl cyclohexane is (d).
The stability of tertiary radical is due to the higher number of α−Hygrogens which give more hyperconjugation effect than secondary.

Peroxide effect is observed only in case of HBr. Therefore, addition of HCl to propene even in the presence of benzyoyl peroxide occurs according to Markovnikov’s rule :

The given compound will turn itself to
To gain aromaticity, it will transfer the electrons as follow:-

It is clear that the compound is heterocyclic(the ring constitutes other than C and H). Due to -ve charge on outer O atom, it has high affinity for BF₃. However, NaBH₄  has no reaction with this. As the compound will turn itself to latter, there is no aldehyde or ketone group present in the compound.

Temperature

While at 220º – 250ºC it forms ether

We know that

1-3-5-7-cyclooctatetraene it has 8 pi electrons, and like stated above, fits the criteria of 4n, to be antiaromatic. to avoid this state of anti-aromaticity (less stable then expected), it becomes non-planar, so it can be more stable then it would be in the antiaromatic state. cyclooctatetraene can do this because it can fold, however other 6 carbon compounds that have 4n electrons and are planar can not and result in an antiaromatic compound.
Potassium cyclooctatetraene is formed by the reaction of cyclooctatetraene with potassium metal:
2 K + C8H8 → K2C8H8
The reaction entails 2-electron reduction of the polyene and is accompanied by a color change from colorless to brown.

Of all the options listed CH₃CH₂CH₂CH₃ has the least number of C-atoms and hence has the lowest b.p.

The correct answer is Option C.
The C-Br bond in the case of 7-bromo-1, 3, 5-cycloheptatriene is broken easily because the intermediate carbocation formed is very stable (aromatic as it contains (4n + 2)π e- ie, follows Huckle rule) while it does not break easily in the case of S-bromo-1, 3-cyclopentadiene because carbocation formed here is highly unstable as it is anti aromatic ie, does not follow Huckel rule. (It contains 4π electrons).
 

Correct answer is 4 because heat of hydrogenation is directly proportional to number of carbon atoms present in it

When both double and triple bonds are present,  then triple bond is considered as the principal group.

This is the electrophilic addition reaction in which addition takes place via more stable carbocation according to the Markovnikov's rule.