Organic Chemistry Some Basic Principles and Techniques

Chemistry Class 11

NCERT

1   What are hybridisation states of each carbon atom in the following compounds?$\\$ $CH _2=C=O, CH _3 CH=CH_ 2 , (CH _3 )_ 2 CO, CH _2=CHCN, C_ 6 H _6$

Solution :

(i)$\underset{1}CH_2=\underset{2}C=0$$\\$ $C-1$ is $sp^2$ hybridised$\\$ $C-2$ is $sp$ hybridised$\\$$\\$ (ii)$\underset{1}CH_3-\underset{2}CH=\underset{3}H_2$$\\$ $C-1$ is $sp^3 $ hybridised.$\\$ $C-2$ is $sp^2$ hybridised.$\\$ $C-3$ is $sp^2$ hybridised.$\\$ $\\$ (iii)$\underset{1}CH_3-{\overset{{\overset{O}{||}}}{\underset{2}{C}}}-\underset{2}{C}H_3$$\\$ $C-1$ and $C-3$ are $sp^ 3$ hybridised.$\\$ $C-2$ is $sp^ 2$ hybridised.$\\$ (iv)$\underset{1}CH_2=\overset{2}CH-\overset{3}C \equiv N$$\\$ $C-1$ is $ sp^2$ hybridised.$\\$ $C-2$ is $sp^ 2$ hybridised.$\\$ $C-3$ is $sp$ hybridised.$\\$$\\$ (v) $C_ 6 H_ 6$$\\$ All the $6$ carbon atoms in benzene are $sp^ 2$ hybridised.

2   Indicate the $\sigma$ and $\pi $ bonds in the following molecules:$\\$ $C_ 6 H_ 6 , C_ 6 H_{ 12} , CH _2 Cl_ 2 , CH _2 = C = CH _2 , CH_ 3 NO_ 2 , HCONHCH_ 3$$\\$

Solution :

(i) $C _6 H _6$$\\$ $\\$ There are six $C-C $ sigma ( $\sigma _{C - C}$ ) bonds, six $C-H$ sigma ( $\sigma_{ C - H} $) bonds, and three $C = C$ pi ( $\pi_{ C - C}$ ) resonating bonds in the given compound.$\\$ (ii) $C _6 H _{12}$$\\$ There are six $C-C$ sigma ( $\sigma _{ C -C}$ ) bonds and twelve $C-H$ sigma ( $\sigma_{C - H}$ ) bonds in the given compound.$\\$ (iii)$CH_2Cl_2$$\\$ There two $C-H$ sigma ( $\sigma _{ C - H}$ ) bonds and two $C-Cl $ sigma ( $\sigma_{ C -Cl }$) bonds in the given compound.$\\$ (iv) $CH_ 2 = C = CH_ 2$$\\$ There are two $C-C$ sigma ($ \sigma _{C - C}$ ) bonds, four $C-H$ sigma ( $\sigma _{C - H}$ ) bonds, and two $C=C$ pi ( $\pi_{C -C }$) bonds in the given compound.$\\$ (v) $CH _3 NO_ 2$$\\$ There are three ${C-H}$ sigma ( $\sigma_{ C - H}$ ) bonds, one $C-N $sigma ( $\sigma _{C -N}$ ) bond, one N-O sigma ( $\sigma_{N - O}$ ) bond, and one N=O pi ( $\pi_{ N - O}$ ) bond in the given compound.$\\$ (vi) $HCONHCH _3$$\\$ There are two $C-N$ sigma ( $\sigma _{C - N}$ ) bonds, four $C-H$ sigma ( $\sigma_{ C - H}$ ) bonds, one $N-H $ sigma bond, and one $C=O$ pi ( $\pi_{C - C}$ ) bond in the given compound.

3   Write bond line formulas for: Isopropyl alcohol, $2,3-$Dimethyl butanal, Heptan$-4-$one.

Solution :

The bond line formulae of the given compounds are: (a) Isopropyl alcohol$\\$ (b) $2, 3-$dimethyl butanal$\\$ (c) Heptan$-4-$one$\\$

4   Give the IUPAC names of the following compounds:$\\$ (a)$\\$ (b)$\\$ (c)$\\$ (d)$\\$ (e)$\\$ (f)$Cl_2CHCH_2OH$

Solution :

(a)3-phenyl propane$\\$ (b)2-methyl-1-cyanobutane$\\$ (c)2, 5-dimethyl heptane$\\$ (d)3-bromo-3-chloroheptane$\\$ (e)3-chloropropanal$\\$ (f) $Cl_ 2 CHCH _2 OH$$\\$ 1, 1-dichloro-2-ethanol

5   Which of the following represents the correct $IUPAC$ name for the compounds concerned?$\\$ (a) 2,2-Dimethylpentane or 2-Dimethylpentane $\\$(b) 2,4,7-Trimethyloctane or 2,5,7- Trimethyloctane $\\$(c) 2-Chloro-4-methylpentane or 4-Chloro-2-methylpentane $\\$(d) But-3-yn-1-ol or But-4-ol-1-yne

Solution :

(a) The prefix di in the $IUPAC$ name indicates that two identical substituent groups are present in the parent chain. Since two methyl groups are present in the $C-2$ of theparent chain of the given compound, the correct $IPUAC$ name of the given compound is 2, 2- dimethylpentane.$\\$ (b) Locant number 2, 4, 7 is lower than 2, 5, 7. Hence, the $IUPAC$ name of the givencompound is 2, 4, 7-trimethyloctane.$\\$ (c) If the substituents are present in the equivalent position of the parent chain, thenthe lower number is given to the one that comes first in the name according to thealphabetical order. Hence, the correct $IUPAC$ name of the given compound is 2-chloro-4-methylpentane.$\\$ (d) Two functional groups -alcoholic and alkyne - are present in the given compound.The principal functional group is the alcoholic group. Hence, the parent chain will besuffixed with ol. The alkyne group is present in the $C-3$ of the parent chain. Hence, thecorrect $IUPAC$ name of the given compound is But-3-0-yn-1-ol.

6   Draw formulas for the first five members of each homologous series beginning with the following compounds. $\\$(a)$ H-COOH $ $\\$(b)$ CH_ 3 COCH _3 $ $\\$(c)$ H-CH=CH_ 2$

Solution :

The first five members of each homologous series beginning with the given compounds are shown as follows:$\\$ (a) $H-COOH :$ Methanoic acid $\\$ $CH 3 -COOH : $Ethanoic acid $\\$ $CH 3 -CH _2 -COOH : $Propanoic acid $\\$ $CH 3 -CH _2 -CH _2 -COOH :$ Butanoic acid $\\$ $CH 3 -CH _2 -CH _2 -CH 2 -COOH :$ Pentanoic acid $\\$ (b) $CH _3 COCH _3 :$ Propanone$\\$ $CH _3 COCH _2 CH _3 :$ Butanone$\\$ $CH _3 COCH _2 CH _2 CH _3 :$ Pentan-2-one$\\$ $CH_ 3 COCH _2 CH_ 2 CH _2 CH_ 3 :$ Hexan-2-one$\\$ $CH _3 COCH _2 CH _2 CH_ 2 CH _2 CH _3 :$ Heptan-2-one $\\$ (c) $H-CH=CH_ 2 :$ Ethene$\\$ $CH _3 -CH=CH _2 :$ Propene$\\$ $CH _3 -CH _2 -CH=CH _2 : $1-Butene$\\$ $CH _3 -CH- 2 -CH _2 -CH=CH _2 :$ 1-Pentene$\\$ $CH _3 -CH _2 -CH _2 -CH _2 -CH=CH _2 :$ 1-Hexene

7   Give condensed and bond line structural formulas and identify the functional group(s) present, if any, for :$\\$ (a) 2,2,4-Trimethylpentane$\\$ (b) 2-Hydroxy-1,2,3-propanetricarboxylic acid$\\$ (c) Hexanedial

Solution :

The functional group present in the given compound is aldehyde $(-CHO).$

8   Identify the functional groups in the following compounds

Solution :

The functional groups present in the given compounds are:$\\$ (a) Aldehyde $(-CHO),$ $\\$ Hydroxyl $(-OH),$ $\\$ Methoxy $(-OMe),$ $\\$ (b) Amino $(-NH _2 ),$ $\\$ Ketone $(C = O),$ $\\$ Diethylamine $(N(C_ 2 H _5 ) _2 )$ $\\$ $-OCH _2 CH _{2 -}$ $\\$ (c) nitro group $(-NO_ 2 )$ group$\\$ $C=C $ double bond $(- C = C -)$, ,

9   Which of the two: $O _2 NCH _2 CH _2 O ^-$ or $CH_ 3 CH _2 O^ -$ is expected to be more stable and why?

Solution :

$NO _2$ group is an electron-withdrawing group. Hence, it shows $-I $ effect. By withdrawing the electrons toward it, the $NO_ 2$ group decreases the negative charge on the compound, thereby stabilising it. On the other hand, ethyl group is an electron-releasing group. Hence, the ethyl group shows $+I$ effect. This increases the negative charge on the compound, thereby destabilising it. Hence, $O _2 NCH _2 CH _2 O ^-$ is expected to be more stable than $CH _3 CH_ 2 O ^-$ .

10   Explain why alkyl groups act as electron donors when attached to a $\pi $ system.

Solution :

When an alkyl group is attached to a $\pi $ system, it acts as an electron-donor group by the process of hyperconjugation. To understand this concept better, let us take the example of propene.

In hyperconjugation, the sigma electrons of the $C-H $ bond of an alkyl group are delocalised. This group is directly attached to an atom of an unsaturated system. The delocalisation occurs because of a partial overlap of a $sp ^3 -s$ sigma bond orbital with an empty p orbital of the $\pi $ bond of an adjacent carbon atom.$\\$ The process of hyperconjugation in propene is shown as follows:

This type of overlap leads to a delocalisation (also known as no-bond resonance) of the $\pi$ electrons, making the molecule more stable.

11   What is smog? How is classical smog different from photochemical smog?

Solution :

The word smog is a combination of smoke and fog. It is a type of air pollution that occurs in many cities throughout world. Classical smog occurs in cool humid climate. It is also called as reducing smog form by combination of smoke, dust and fog containing sulphur oxides. Whereas photochemical smog occurs in warm and dry sunny climate. It has high concentration of oxidizing agents and therefore, it is also called as oxidizing smog.