Chapter 9: Hydrocarbons

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A compound is only said to be aromatic if it completely satisfies the following conditions;

• It should have a planar structure and should be cyclic.

• The $\pi $-electrons of the compound must be completely delocalized in the ring.

• The total number of $\pi $-electrons present in the ring should be equal to (4n+ 2), where n = 0, 1, 2 … etc. {Huckel’s rule}.

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In the given compound, one carbon atom is $s{{p}^{3}}$  hybridized which signifies that it is tetrahedral (not planar). As for the compound to be aromatic, it should be planar. Thus, the given compound is not aromatic in nature.

ii)

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Ans:

In the given compound, one carbon atom is $s{{p}^{3}}$ hybridized which signifies that it is tetrahedral (not planar).  As for the compound to be aromatic, it should be planar. 

Also, for the given compound, the number of $\pi -$ electrons is 4 so, by Huckel’s rule;

\[4n+2=4\]

\[n=\frac{1}{2}\]

For a compound to be aromatic, the value of n must be an integer i.e. 0, 1, 2… etc. which is not  satisfied for the given compound. Therefore, it is not aromatic in nature.

iii) 

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For the given compound, the number of $\pi -$electrons is 8 so, by Huckel’s rule;

\[4n+2=8\] 

\[n=\frac{2}{3}\] 

For a compound to be aromatic, the value of n must be an integer i.e. 0, 1, 2… etc. which is not  satisfied for the given compound. Therefore, it is not aromatic in nature.

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The reactions are given as;

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1. m-nitrochlorobenzene

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The reactions are given as;

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1. p -nitrotoluene

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The reactions are given as;

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1. acetophenone

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The reactions are given as;

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The given alkane can be represented as;

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• Primary carbon atoms are those which are bonded to only one carbon atom or none. i.e., they have only 1 carbon atom as their neighbor or none (in case of methane). The given structure has 5 primary carbon atoms and 15 hydrogen atoms attached to it.

• Secondary carbon atoms are those which are bonded to 2 carbon atoms i.e., they have 2 carbon atoms as their neighbors. The given structure has 2 secondary carbon atoms and 4 hydrogen atoms attached to it.

• Tertiary carbon atoms are those which are bonded to 3 carbon atoms i.e., they have 3 carbon atoms as their neighbors. The given structure has 1 tertiary carbon atom and only 1 hydrogen atom is attached to it.

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Alkanes mainly experience intermolecular Van der Waals forces. The stronger the force, the greater the boiling point of the alkane.

As branching increases further, the surface area of the molecule decreases which eventually results in a small area of contact. Thus, the Van der Waals force also decreases which can be overcome at a relatively lower temperature. Therefore, the boiling point of an alkane chain decreases with an increase in branching.

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Addition of HBr to propene is an example of an electrophilic addition reaction. Hydrogen bromide provides an electrophile, ${{H}^{+}}$ . This electrophile attacks the double bond to form primary and secondary carbocations as shown:

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Secondary carbocations are comparatively more stable than primary carbocations. Thus, the former predominates since it will form at a faster rate. Thus, now $B{{r}^{-}}$ attacks the carbocation to form 2 – bromopropane as the major product.

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This reaction follows Markovnikov’s rule.

Now,

In the presence of benzoyl peroxide, an additional reaction takes place by anti-Markovnikov’s  rule. The reaction follows a free radical chain mechanism as;

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Here, 1 – bromopropane is obtained as the major product.

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In the presence of peroxide, Br free radical acts as an electrophile. Therefore, two different products are obtained in addition of HBr to propene according to the absence and presence of peroxide.

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o-xylene has two resonating structures showing different reactions as follows;

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The three products are formed i.e., methyl glyoxal, 1,2-demethylglyoxal, and glyoxal from two Kekule structures. Since all three products cannot be obtained from any one of the two structures, this proves that o-xylene is a resonance hybrid of two Kekule structures.

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Acidic character of any species is defined on the basis of its ease with which it can lose the H– atoms. The hybridization state of carbon in the given compound is given as;

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According to the hybridization criterion, as the s–character increases the electronegativity of carbon increases and the electrons of C–H  bond pair lie closer to the C atom.

The s–character increases in the order:

\[s{{p}^{3}}<s{{p}^{2}}<sp\] 

Thus, the decreasing order of acidic behavior is Ethyne > Benzene > Hexane.

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Benzene is a planar molecule having delocalized electrons above and below the plane of the ring.  Thus, this makes it electron-rich. As a result, it is highly attractive to electron deficient species i.e., electrophiles. This is the reason; benzene undergoes electrophilic substitution reactions very easily.

Whereas, nucleophiles are electron-rich. Hence, they are repelled by benzene. Therefore, benzene undergoes nucleophilic substitutions with much difficulty.

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The reactions are given as;

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The reactions are given as;

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The reactions are given as;

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