ORGANIC CHEMISTRY Expand each of the following condensed formulas into their complete structural formulas. i. CH3CH2COCH2CH3 H H O H H H C C C C C H H H H H CH3CH=CH(CH2)3CH3 H H H H H H H C C C C C C C H H H H H H H

How many ∏ & sigma are present in each of the molecules? HC CCH CHCH3 CH2 C CHCH3 Solution SigmaC C :4; sigmaC H:6; ∏C C:1; ∏C C:2 Sigma C C :3; sigma C H:6; ∏ C C:2 What is the type of hybridization of each carbon atom in the following compound? CH3Cl : sp3 (CH3)2CO : sp3,sp2

CH3CN : sp3,sp HCONH2 : sp2 CH3CH CHCN : sp3,sp2,sp2,sp H2C O : sp2 , Trigonal Planar CH3F : sp3 ,Tetrahedral HCN : sp , Linear 4) For each of the following compounds ,write the condensed formula and also their bond line formula. i. HOCH2CH2CH2CH(CH3)CH(CH3)CH3 ii. N C CH C N OH

Solution: Condensed formula: i. HO(CH2)3CH(CH3)CH(CH3)2 HOCH(CN)2 Bond -line formula: i. HO OH ii. NC CN

5) Expand each of the following bond-line formulas to show all the atoms including carbon and hydrogen. i. H2C CH2 CH3 H2C HC CH H2C CH2 CH3 ii. H H H H H H H H H C C C C C C C C H H H H H H H H H

iii. H H H H C C C C C H OH H H H iv iii. H H H H C C C C C H OH H H H iv. H3C CH3 H C C H 6)Explain why the names given in parentheses are incorrect i. CH3CHCH2CH2CHCHCH2CH3 CH3 H3C CH3

[and not 3,4,7-Trimethyloctane] Sol n :Lowest locant number,2,5,6 is lower than 3,4,7. ii. CH3CH2CHCH2CHCH2CH3 C2H5 CH3 3-Ethyl-5-methylheptane [and not 5-Ethyl-3-methylheptane] Sol n : substituents are at equivalent position ; lower number is given to the one that comes first in the alphabetical order. 8)Write the IUPAC name of the following org. compound.

i. i. 1CH3 2CH2 3CH 4CH2 5CH2 6CH 7CH2 8CH3 OH CH3 Sol n : 6-Methyloctan-3-ol 6CH3 5CH2 4C 3CH2 2C 1CH3 O O Sol n : Hex-2,4-dione. 6CH3 5C 4CH2 3CH2 2CH2 1COOH O Sol n : 5-Oxohexanoic acid.

8)Derive the structure of: i. 2-Chlorohexane CH3 CHC H2CH2CH2CH3 Cl iv. 6CH 5C 4CH 3CH 2CH 1CH2 Sol n : Hex-1,3-dien-5-yne. 8)Derive the structure of: i. 2-Chlorohexane CH3 CHC H2CH2CH2CH3 Cl ii.Pent-4-en-2-ol CH3CHCH2CH CH2 OH iii. 3-Nitrocyclohexene. 1 2 6 3 NO2 5 4

v. 6-Hydroxy-heptanal iv. Cyclohex-2-en-1-ol 5 6 4 1 OH 3 2 3 2 v. 6-Hydroxy-heptanal Soln : CHOCH2CH2CH2CH2CHCH3 OH

i. o- Ethylanisole OMe C2H5 ii. p- Nitroaniline Soln : O2N NH2 9) Write the structural formula for : i. o- Ethylanisole OMe C2H5 ii. p- Nitroaniline Soln : O2N NH2

2,3 –Dibromo-1-phenylpentane Br Soln : Br 4-Ethyl-1-fluoro-2-nitrobenzene. F C2H5 O2N

10)Using curved-arrow notation , show the formation of reactive intermediates when the following covalent bonds undergo heterolytic cleavage: i . CH3 SCH3 ; ii.CH3 CN ; iii. CH3 Cu Soln : i. CH3 SCH3 +CH3 + -SCH3 ii.CH3 CN +CH3 + -CN iii. Cu CH3 +Cu + -CH3

11) Categorise as nucleophilic or electrophilic: HS-,BF3,C2H5O-,(CH3)3N:,Cl+,CH3C+ O,H2N:,N+O2 Soln : Nucleophiles: HS-, C2H5O-,(CH3)3N:,H2N:- . Electrophilic: Cl+,CH3C+ O,N+O2, ,BF3 13) Identify the electrophilic centre in the following. i. CH3CH O ; ii. CH3CN ; iii. CH3I Soln: AmongCH3HC∆ O, CH3C∆ N, H3C∆ I , the carbon atoms marked with delta(∆) are electrophilic centers' as they will have partial positive charge due to the polarity of the bond.

13) Which bond is more polar in the following pairs of molecules. i 13) Which bond is more polar in the following pairs of molecules? i . H3C H , H3C Br ; ii. H3C NH2 , H3C OH iii. H3C OH, H3C SH Soln : i. C Br , because Br is more electronegative than H. ii. C O , because O is more electronegative than N. iii. C O , because O is more electronegative than N. 14) In which C C bond of 3CH32CH21CH2 Br , the inductive effect is expected to be the least? Soln : Magnitude of inductive effect diminishes as the number of intervening bond increases. Hence , the effect is least in the bond between carbon-3 and H.

15) Write resonance structures of CH3COO- and show the movement of electrons by curved arrows. Soln : O•••• ••O••••- CH3 C CH3 C ••••O••- O•• •• 16)Write resonance structures of CH2 CH CHO. Soln : i . ii. CH2 CH C H +CH2 CH C H •• O•• •• O•• ••-

iii. ••CH2 - CH C H •• O •• + Stability : i>ii>iii. 17)Explain why the following structures , i and ii cannot be the major contributors for the real structure of CH3COOCH3.. CH3 C+ ••O•• CH3 CH3 C ••+O•• CH3 •• -O•• •• i. •• -O•• •• ii. Soln : The two structures are less important contributors as they involve charge separation. Additionally , structure i contains a carbon atom with an incomplete octet.

18)Explain why (CH3)3 C+ is more stable than CH3 C+H2 and C+H3 is the least stable cation. Soln : Hyperconjugation interaction in (CH3)3 C+ is greater than in CH3 C+H2 as the (CH3)3 C+ has nine C H bonds. In C+H3 ,vacant p orbital is perpendicular to the plane in which C H bond lie; hence cannot overlap with it. Thus, C+H3 lacks hyper conjugative stability. 19)On complete conbustion,0.246 g of an organic compound gave 0.198 g of CO2 and 0.1014 g of water. Determine the percentage composition of carbon and hydrogen in the compound. Soln : % of carbon=12 0.198 100 21.95% 44 0.246

% of H = 2 0.1014 100 4.58% 18 0.246 20)In Dumas’ method for estimation of nitrogen, 0.3 g of an organic compound gave 50 mL of nitrogen collected at 300K temperature and 715mm pressure. Calculate the percentage composition of nitrogen in the compound. (Aqueous tension at 300K=15mm) Soln : Volume of nitrogen collected at 300K & 715 mm pressure is 50ml Actual pressure =715-15=700mm Volume of nitrogen at STP= 273 700 50 41.9 mL 300 760

22,400 mL of nitrogen at STP weighs = 28 g 41 22,400 mL of nitrogen at STP weighs = 28 g 41.9 mL of nitrogen weighs=28 41.9 g 22400 % of nitrogen = 28 41.9 100 17.46% 22400 0.3 21)During estimation of N2 present in an organic compd. ,by Kjeldahl’s method , the NH3 evolved from 0.5 g of the compd in Kjeldahl’s estimation of N2 , neutralized 10 mL of 1 M H2SO4. Find out the % of N2 in the compd. Soln : 1 M of 10 mL H2SO4= 1 M of 20 ml NH3 1000 mL of 1 M ammonia contains 14 g nitrogen

188 g AgBr contains 80 g Br 0.12 g AgBr contains 80 0.12 g Br 188 20 mL of 1 M ammonia contains 14 20 g nitrogen 1000 % ofN2 =14 20 100 56% 1000 0.5 22) In Carius method of estimation of halogen, 0.15 g of an organic compound gave 0.12 g of AgBr . Find out the % of bromine in the compound. Soln : Molar mass of AgBr = 108 +80 = 188 g mol-1 188 g AgBr contains 80 g Br 0.12 g AgBr contains 80 0.12 g Br 188

% of Br = 80 0.12 100 34.04% 188 0.15 23) If 0.157 g of an organic compound gave 0.4813 g of BaSO4. What is the % of S in the compd.? Soln : Molecular mass of BaSO4 = 233 g 0.4813 g BaSO4 contains 32 0.4813 g of S 233 % of S= 32 0.4813 100 233 0.157