Test: Alkanes & Cycloalkanes - NEET MCQ

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.

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

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.

Correct Answer :- b

Explanation : The 2o hydrogens are 20 times more reactive than the 1o ones.

Bromination in the presence of light is done by free radical mechanism. The radical on the carbon chain is formed where the radical is most stable. We can’t put radical on the ring as the aromaticity of the ring will be lost. So, it should be put either on the CH₃. However, the CH₃ which is attached to NO₂ ring will not bear stable radical as NO will decrease the stability of radical.(NO₂ is an electron withdrawing group). So the only position is the other CH₃ in option d. There also we have stability due to resonance.

Hammond's postulate states that the transition state of a reaction resembles either the reactants or the products, to whichever it is closer in energy. In an exothermic reaction, the transition state is closer to the reactants than to the products in energy

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.

As shown in the above mechanism 1-butene and 2-butene cannot be formed by this free radical. release of 1 hydrogen radical gives 2-methyl propene as disproportionated product.

Hence, the correct option is C.

Product 1-Does not have any chiral carbon so only one product
Product 2-Two chiral centres so RR, RS and SR possible
Product 3-Two chiral centres so RR, RS and SR possible
Product 4-Two chiral centres but plane of symmetry exists, so only two isomers.

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.

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.

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.

The statements (B) and (C) are true regarding free radical iodination of the alkane.
(B) Direct iodination of alkane with iodine in presence of light is impractical.
During this reaction, Hl, a strong reducing agent is obtained as a byproduct, which catalyzes the reverse reaction thereby preventing direct iodination of alkane.
(C) Iodination of alkane can be achieved successfully using an oxidizing agent catalyst. 
Presence of an oxidizing agent oxidizes unwanted byproduct Hl and enables iodination of alkane to proceed.

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.

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

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.

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