Organic Chemistry Some Basic Principle And Techniques Solved

Organic Chemistry- Some Basic Principle And Techniques. Solved examples

Q. Expand each of the following condensed formulas into their complete structural formulas. a. CH 3 CH 2 COCH 2 CH 3 b. CH 3 CH=CH(CH 2)3 CH 3 Sol. a. H H O H H H C C C H H H b. H H H H C C C C H H H H

Q. Structures and IUPAC names of some hydrocarbons are given below. Explain why the names given in the parentheses are incorrect. a. CH 3 CH CH 2 CH 3 CH 3 2, 5, 6 -Trimethyloctane [and not 3, 4, 7 Trimethyloctane] b. CH 3 CH 2 CH 3 CH CH 2 CH 3 3 -Ethyl -5 –methylheptane [and not 5 -Ethyl-3 methylheptane] Sol: - a. Lowest locant number , 2, 5, 6 is lower than , 3, 5, 7. b. Substituent are in equivalent position; lower number is given to the one that comes first in the name according to alphabetical order

Q. Write the IUPAC names of the compound from their given structures. CH 3 CH 2 CH 3 OH CH 3 Sol: - 1. the functional group present is an alcohol. hence the suffix is ‘-ol’ 2. The longest chain containing –OH has 8 carbon atoms hence the corresponding saturated hydrocarbon is octane. 3. The –OH is on carbon atom 3. In addition a methyl group is attached at 6 th carbon. Hence the systematic name of this compound is 6 Methyloctan-3 -ol.

O O b. CH 3 CH 2 C CH 3 Sol. The functional group present is ketone (>C=O), hence suffix ‘-one’. Presence of 2 keto group is indicated by ’di’, hence suffix becomes ‘dione’. The 2 keto groups are at carbon 2, 4. The longest chain contains 6 carbon atoms , hence, parent hydrocarbon is hexane. thus the systematic name is Hexane -2, 4 -dione. O

O c. CH 3 C CH 2 COOH Sol: - Here, 2 functional group namely ketone , carboxylic acid are present The principal functional group is the carboxlic acid group ; hence the parent chain will be suffixed with ’oic’ acid. Numbering of the chain starts from carbon of – COOH functional group. The keto group in the chain at the carbon 5 is indicated by ‘oxo’. The longest chain including the principal functional group has 6 carbon atoms ; hence their parent hydrocarbon is hexane. The compound is, therefore , named as 5 -Oxohexanoic acid.

d. CH CH 2 Sol: - The two C=C functional groups are present at carbon atoms 1, 3 , while the Carbon , carbon triple bond functional group is a present at carbon 5. These group are indicated by suffixes ‘diene’ and ‘yne’ respectively The longest chain containing the functional group has 6 carbon atoms ; hence the parent hydrocarbon is hexane. The name of compound , therefore , is Hexa 1, 3 dien-5 -yne. Q. Derive the structure of (a) 2 -Chlorohexane, (b)Pent-4 en-2 -ol, (c)3 -Nitrocyclohexene, (d) Cyclohex-2 -en-1 -ol, (e) 6 -Hydroxy-heptanal.

Sol: - (a) ‘hexane’ indicates the presence of 6 carbon atoms in the chain. The functional group chloro is present at carbon 2. Hence , the structure of the compound is CH 3 CH 2 CH 2 CH(Cl)CH 3. (b) ‘pent’ indicates that parent hydrocarbon contain 5 carbon atoms in the chain. ’en’ and ‘ol’ corresponds to the functional groups C=C, -OH at carbon atoms 4, 2 respectively. Thus , the structure is CH 2=CHCH 2 CH(OH)CH 3 (c) Six membered ring containing a carbon – carbon double bond is implied by cyclohexane , which is numbered as shown in (i). The prefix 3 -nitro means that a nitro group is present on Computer-3. Thus, complete structural formula of the compound (ii). Double bond is suffixed functional group whereas NO 2 is preffixed functional group therefore double bonds gets preference over –NO 2 group:

NO 2 (i) (ii) (d) ‘ 1 -ol’ means that a –OH group is present at Computer 1. OH is suffixed functional group and gets preference over C=C, bond , Thus the structure is as shown in (ii) OH (i) (ii)

(e) ‘heptanal’ indicates the compound to be an aldehyde containing 7 carbon atoms in the parent chain. The ‘ 6 hydroxy’ indicates that –OH group is present at carbon 6. Thus the structural formulas of the compound is : CH 3 CH(OH)CH 2 CH 2 CHO. Carbon atom of –CHO group is included while numbering the carbon chain. Q. How many sigma and pi bonds are present in each of the following molecules. (a) HC=CCH=CHCH 3 (b) CH 2=C=CHCH 3 Sol: - (a) 4 sigma: C-C, 6 sigma: C-H , 1 pi: C=C. 2 pi: C triple bond C. (b)3 sigma: C-C, 6 sigma: C-H, 2 pi C=C,

Q. What is the type of hybridisation of each carbon in the following compounds (a)CH 3 Cl , (b) (CH 3)2 CO , (c)CH 3 CN, (d)HCONH 2, (e)CH 3 CH=CHCN. Sol: - (a) sp 3; (b) sp 3, sp 2; (c) sp 3, sp; (d) sp 2; (e) sp 3, sp 2, sp Q. Write the state of hybridisation of carbon in the following compounds and shapes of each of the molecule. (a)H 2 C=O, (b) CH 3 F, (c) CH=N Sol: - (a) sp 2 hybridised carbon , trigonal planar; (b) sp 3 hybridised carbon , tetrahedral ; (c) Sp hybridised carbon, linear.

Q. Write the structural formula of: (a) o-Ethylanisole, (b) p-Nitroaniline, (c)2, 3 -Dibromo-1 phenylpentane, (d) 4 -Ethyl-1 -fluoro-2 -nitrobenzene. Sol: NH 2 OMe C 2 H 5 Br NO 2 F Br C 2 H 5

Q. Using curved –arrow notation , show the formation of reactive intermediates when the following covalent bonds undergo heterolytic cleavage (a). CH 3 -SCH 3, (b) CH 3 -CN, (c) CH 3 -Cu Sol: - (a) CH 3 SCH 3 (b) CH 3 CN (c) CH 3 Cu CH 3 +SCH 3 + CN CH 3 + Cu

Q. Giving justification , categorise the following molecules /ions as nucleopile or electropile HS , BF 3 , C 2 H 5 O (CH 3)3 N , Cl , CH 3 -C=O, H 2 N , NO 2 Sol: - Nucleophiles: HS , C 2 H 5 O , (CH 3)3 N , H 2 N , These species have unshared pair of electrons , which can be donated and shared with an electrophile. Electrophiles: - BF 3 , CH 3 -C=O, NO 2 These reactive sites have only six valence electron , can accept electron pair from a nucleophile. Q. Identify the electrophilic center in the following : CH 3 CH=O, CH 3 CN , CH 3 I.

Sol: -Among CH 3 H C=O, H 3 CC=N , and H 3 C-I, the starred carbon atoms are electrophilic centers as they will have partial charge due to polarity of the bond. Q. Which bond is more polar in the following pairs of molecules : (a) CH 3 -H, CH 3 -Br , (b) CH 3 -NH 2, CH 3 -OH, (c) CH 3 -OH , CH 3 -SH. Sol: - (a) C-Br, since Br is more electronegative than H, (b) C-O , (c) C-O. Q. In which C-C bond of CH 3 CH 2 Br, the inductive effect is expected to be the least. Sol: - Magnitude of inductive effect diminishes as the number of intervening bond increases. Hence , the effect is least in the bond between carbon -3 and hydrogen.

Q. Write resonance structures of CH 3 COO and shows the movement of electron by curved arrows. Sol: - First, write the structures and put unshaped pairs of valence electrons on appropriate atoms. Then draw the arrows one at a time moving the electrons to get te other structure. O CH 3 C O CH 3 O C O

Q. Write the resonance structures of CH 2=CH-CHO. Indicates relative stability of the contributing structures. Sol: O O CH 2=CH-C-H CH 2 -CH=C-H (1) (2) O CH 2 -CH=C-H (3) Stability : 1>2>3 [1: Most stable , more number of covalent bonds each carbon and oxygen atom has an octet and no separation of opposite charge 2: negative charge on more electronegative atom and positive charge on more electropositive atom. 3: dose not contribute as oxygen has positive charge and carbon has negative charge hence least stable.

Q. Explain why the following two structures , 1 and 2 cannot be the major contributors to real structures of CH 3 COOCH 3. O O CH 3 -C-O-CH 3 -C=O-CH 3 (1) (2) Sol: - The structures are less important contributors as they involve charge separation. Additionally structures 1 contains a carbon atom with an incomplete octet.