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Ncert Solutions Chemistry Class 11th

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6 months ago

Ncert Solutions Chemistry Class 11th Hydrocarbons

13.17. Write down the products of ozonolysis ofl, 2-dimethylbenzene (o-xylene). How does the result support Kekule structure of benzene?
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V
Vishal Baghel

Contributor-Level 10

o-Xylene may be regarded as a resonance hybrid of the following two Kekule structures. Ozonolysis of each one of these gives two products as shown below:

Thus, in all, three products are formed. Since all the three products cannot be obtained from any one of the two Kekule structures, this shows that o-xylene is a resonance hybrid of the two Kekule structures (I and II).

New answer posted

6 months ago

Ncert Solutions Chemistry Class 11th Hydrocarbons

13.16. Addition of HBr to propene yields 2-bromopropane, while in the presence of benzoyl peroxide, the same reaction yields 1-bromopropane. Explain and give mechanism.
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V
Vishal Baghel

Contributor-Level 10

Addition of HBr to propene is an ionic electrophilic addition reaction in which the electrophile, i.e., H+ first adds to give a more stable 2° carbocation. In the 2nd step, the carbocation is rapidly attacked by the nucleophile Br~ ion to give 2-bromopropane.

In presence of benzoyl peroxide, the reaction is still electrophilic but the electrophile here is a Br free radical which is obtained by the action of benzoyl peroxide on HBr.
In the first step, Br radical adds to propene to generate the more stable 2° free radical. In the second step, the free radical thus obtained rapidly abstracts a hydrogen atom from HBr to give 1-bromopr

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New answer posted

6 months ago

Ncert Solutions Chemistry Class 11th Hydrocarbons

13.15. What effect does branching of an alkane chain has on its boiling point?
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Vishal Baghel

Contributor-Level 10

Branching of carbon atom chain decreases the boiling point of alkane.

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6 months ago

Ncert Solutions Chemistry Class 11th Classification of Elements and Periodicity in Prop

3.43. Assertion: Their synthesis and characterisation of new elements with very high atomic numbers require highly sophisticated costly equipment and laboratory.

Reason: The new elements with very high atomic numbers are so unstable that only minute quantities, sometimes only a few atoms of them are obtained.
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New answer posted

6 months ago

Ncert Solutions Chemistry Class 11th Hydrocarbons

13.14. In the alkane, H3C—CH2—C(CH3)2—CH2—CH(CH3)2, identify 1°, 2°, 3° carbon atoms and give the number of H atoms bonded to each one of these.
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Vishal Baghel

Contributor-Level 10

The expanded formula of the given compound is

New answer posted

6 months ago

Ncert Solutions Chemistry Class 11th Classification of Elements and Periodicity in Prop

3.42. Assertion: Elements of the same group exhibit similar chemical properties.

Reason: Elements of the same group have similar valence shell electronic configuration.
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Payal Gupta

Contributor-Level 10

3.42. (a) Elements of the same group have similar valence shell electronic configuration and, therefore, exhibit similar chemical properties. However, the elements of the same period have an incrementally increasing number of electrons from left to right, and, therefore, have different valences.

New answer posted

6 months ago

Ncert Solutions Chemistry Class 11th Hydrocarbons

13.13. How will you convert benzene into (i)p-nitrobromobenzene (ii) m-nitrochlorobenzene (iii) p-nitrotoluene (iv) acetophenone?
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Vishal Baghel

Contributor-Level 10

(i) The two substituents in the benzene ring are present at p-positions. Therefore, the sequence of reactions should be such that first an o, p-directing group, i.e., Br atom should be introduced in the benzene ring and this should be followed by nitration. Thus,

 
(ii) Here since the two substituents are at p-position w.r.t. each other, therefore, the first substituent in the benzene ring should be a o, p-directing group (i.e., CH3) and then the other group (i.e., NO2) should be introduced. Therefore, the sequence of reactions is:
 
(iii) Here since the two substituents are at m-position w.r.t. each other, therefore

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New question posted

6 months ago

Ncert Solutions Chemistry Class 11th Classification of Elements and Periodicity in Prop

3.41. Assertion: The size of an anion will be larger than that of the parent atom.

Reason: The addition of one or more electrons would result in increased repulsion among the electrons and a decrease in effective nuclear charge.
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New answer posted

6 months ago

Ncert Solutions Chemistry Class 11th Hydrocarbons

13.12. Explain why the following systems are not aromatic?
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V
Vishal Baghel

Contributor-Level 10

 

Due to the presence of a sp3-hybridized carbon, the system is not
planar. It does contain six n-electrons but the system is not fully conjugated since all the six n-electrons do not form a single cyclic electron cloud which surrounds all the atoms of the ring. Therefore, it is not an aromatic compound.
 
(ii)
Due to the presence of a sp3 -carbon, the system is not planar. Further, it contains only four n-electrons, therefore, the system is not aromatic because it does not contain planar cyclic cloud having (4n + 2) n-electrons.
 
(iii)

Cyclo-octatetraene is not planar but is tub shaped. It is, therefore, a non-planar system

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New question posted

6 months ago

Ncert Solutions Chemistry Class 11th Classification of Elements and Periodicity in Prop

3.41. Assertion: The size of an anion will be larger than that of the parent atom.

Reason: The addition of one or more electrons would result in increased repulsion among the electrons and a decrease in effective nuclear charge
0 Follower 2 Views

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