Isomerism Solved Problems

Page # 14 ISOMERISM SOLVED PROBLEMS Ex.1 Number of isomeric forms of C 7 H 9 N having benzene ring will be (A) 7 (B) 6 (C) 5 CH3 CH. CH. CH3 | | CH3 CH3 (D) 4 (Ans. C) Sol. (i) (ii), (iii), (iv) Ex.2 Which of the following is an isomer of (iii) (v) (B) (CH3)2CHOH (D) (C2H5)2CHOH (Ans. A) Sol. Diethyl ether has 4 carbon atoms, among (iv) Ex.6 different alternative alochols only (CH3)3COH has 4 carbon atoms. Ex.3 (C) 4 (D) 5 C4H8 is (A) 2 Ex.7 CH3CH2CH2 CCH Ex.4 How many isomers of C5H11OH will be primary alcohols (A) 2 (B) 3 (C) 4 CH3CH2CH2CH2CH2OH (i) (C) 4 CH3CH2CH = CH2 cis- butene- 2 CH3 CHCH2 CH2 OH | CH3 (D) 5 butene - 1 trans- butene-2 CH3CHCl2 and CH2Cl. CH2Cl show which type of isomerism – (A) Functional (B) Chain (C) Position (D) Metamerism (D) 5 (Ans. C) Sol. (B) 3 (Ans. B) Sol. (Ans.B) Sol. The maximum number of unbranched alkene isomers with molecular formula The possible number of alkynes with the formula C5H8 is (A) 2 (B) 3 (iv) CH 3 | CH 3 — C — CH 2 CH 3 | CH 3 diethyl ether (A) (CH3)3COH (C) C3H7OH CH3 CH2 CHCH2 CH3 | CH3 (Ans. C) Sol. The two isomers have same functional group, but at different positions. Ex.8 Which of the following is an example of position isomerism – (A) Isopentane and neopentane (ii) (B) Glucose and fructose CH3 | HOCH2 — CHCH2 CH3 CH3 — C — CH2 OH | | CH3 CH3 (iii) Ex.5 (C) Ethanol and dimethyl ether (D)  - Naphthol and  - naphthol(Ans.D) Sol. -naphthol differ in the position of the alcoholic group. (iv)  - naphthol How many chain isomers can be obtained from the alkane C6H 14 – (A) 4 (B) 5 (C) 6 Ex.9 (D) 7 (Ans. B) Sol. CH3CH2CH2CH2CH2CH3 CH3 CH. CH2 . CH2 CH3 | CH3 (i) (ii) As the name indicates  - naphthol and  - naphthol Keto - enol tautomerism is observed in – O || (A) C 6H5 — C — H O || (C) C 6H5 — C — C 6H5 O || (B) C 6H5 — C — CH3 Page # 15 ISOMERISM O CH3 || | (D) C 6H5 — C — C — C 6H5 | CH3 Sol. CH3 | ClCH2 . CH. CH2 CH3 (Ans.B) The compound must contain  - hydrogen (iv) atom for showing keto enol tautomerism. Ex.10 The enolic form of acetone contains – CH3 | CH3 C . CH2 CH3 | Cl (v) CH3 CH3 | | CH3 — CH — CH — CH3 CH3 CHCH2 CH2 Cl | Cl (A) 9  bonds, 1 bond and 2 lone pairs (B) 8  bonds, 2  bonds and 2 lone pairs (C) 10  bonds, 1  bond and 1 lone pair (D) 9  bonds, 2  bonds and 1 lone pair (vi) (Ans. A) Sol.  :O || CH3 — C  CH2 O || CH3 — C — CH3 keto form (vii) CH3 | H3 C — C — CH2 Cl | CH3 enolic form No. of  bonds in enolic form : 3 + 1 + 1 + 1 + 1 + 2 = 9 (viii) No. of  bonds in enolic form : 1 No. of lone pairs of electrons in enolic form = 2 Ex.11 An alkane can show structural isomerism Ex.13 The compound C4H10O can show (A) Metamerism (B) Functional isomerism if it has..........number of minimum carbon (C) Positional isomerism atoms – (A) 1 (B) 2 (C) 3 (D) 4 (Ans. D) Sol. (D) All types Sol. CH4, CH3.CH3, CH3.CH2.CH3 exist only in one structural form, while CH3CH2CH2CH3 can possible for C5H11Cl – (A) 6 (B) 8 (C) 10 show metamerism while alcohols can show position isomerism. Ex.14 Which of the following can exhibit cistrans isomerism – (D) 12 ( A) HC  CH (Ans. B) Sol. (B) ClCH = CHCl (C) CH3.CHCl. COOH(D) ClCH2 – CH2Cl (Ans. B) CH3CH2CH2CH2CH2Cl, Sol. (i) The molecular formula C4H10O reminds us that the compound can be an ether which is isomeric with alcohol. Further ethers can exist in more than one structural form. Ex.12 How many structural formulae are (Ans. D) Remember that ci s-tans geometri cal isomerism is possible only in alkenes and Cl Cl | | CH3 CH2 CH2 CHCH3 , CH3 CH2 CHCH2 CH3 (ii) (iii) further only in those alkenes in which the doubly bonded carbon atoms individually have different atoms/groups. Page # 16 ISOMERISM Ex.15 Give the E-Z designation of the following Ex.20 A Fischer projection of (2R, 3S)-2,3- compound - C6H5 C=C CH3NH butanediol is : - I HO (A) HO Br (A) E (B) Z (C) E-Z (D) E-E (Ans. A) Sol. Precedence order = (C) , H HO CH3 H H CH3 HO (B) H CH3 OH H CH3 (D) H H CH3 H OH CH3 CH3 OH H CH3 (Ans. A) Ex.16 Keto-enol tautomerism is observed in : (A) C6H 5 – CHO 1 (B) C6H5COCH3 Sol. (C) C6H5COC6H 5 (D) C6H5COCH2COCH3 Sol. HO 2 H HO 3 H C6H5COCH3, C6H5COCH2COCH3 Ex.21 The statement about COOH COOCH (B) A compound with two aldehyde (C) An alkenoic acid C 4H 6O 2 correct (Ans. D) COOH H OH H OH H H OH H OH HO COOH (I) represents A diketone (II) (III) (CI and III are enantiomers Alkenoic acid Ex.18 Esters are functional isomers of : (A) Hydroxy aldehyde (B) Ketone Sol. (D) Diols(Ans. A) Esters for eg. CH3 – COOCH3 is an isomer of hydroxy aldehyde. (D) I and II are enantiomers Sol. A H COOCH3 (B) I and II are diastereomers Compound having two aldehydes OH COOCH3 (A) I and II are identical (C) Diketone the compounds I, II and III (A) A diketone (D) An alkanoic acid 4 S CH3 (Ans. B, D) Ex.17 C4H 6O2 does not represent : – Sol. R CH3