C2-Allens Made Chemistry Exercise {PART-1}

EXERCISE-01 CHECK YOUR GRASP SELECT THE CORRECT ALTERNATIVE (ONLY ONE CORRECT ANSWER) 1 . Which of the following reacts with water to form very stable product ? (A) CH3Cl (B) CCl4 (C) CCl3CHO (D) CH2ClCH2Cl (D) pentanal 2 . The compound that gives a positive iodoform test is :- (D) CH3CH2CHO (D) Formaldehyde (A) 1-pentanol (B) 2-pentanone (C) 3-pentanone 3 . Which of the following does not undergo aldol condensation ? (A) HCHO (B) CH3CHO (C) CH3COCH3 4 . Cannizzaro reaction is not given by : (A) Trimethylacetaldehyde (B) Acetaldehyde (C) Benzaldehyde 5 . In the reaction : O+ O OH /  [X] [X] will be :- O OH OH O O OH (B) (C) OH (A) (D) 6 . In a Cannizaro reaction, the intermediate that will be the best hydride donor is : H H (C) H (D) H O– O– H3CO O– O2N O– O– O– (A) OH (B) O– 7 . If 3-hexanone is reacted with NaBH4 followed by hydrolysis with D2O, the product will be : (A) CH3CH2CH(OH)CH2CH2CH3 (B) CH3CH2CD(OH)CH2CH2CH3 (C) CH3CH2CH(OD)CH2CH2CH3 (D) CH3CH2CD(OD)CH2CH2CH3 8 . Predict the product 'B' in the sequence of reaction CH  CH 30H%gHS2OS4O4 A dil.1N0aOCH  B (A) CH3COONa (B) CH3COOH (C) CH3CHO (D) CH3–CH–CH2–CHO OH 9 . In the reaction CH2Br + Mg dry ether  A+ CH 3CHO  adduct H3O  B . The product (B) is : (A) CH3CH CH2Br (B) CH2CHCH3 OH OH CH2CHCH3 (C) CH3–CH2 OH (D) CH2=CH CH2Br

10 . Which one of the following reagents is suitable for the conversion of 2-cyclohexenone into 3-methylcyclohexanone ? (A) CH3MgI (B) (CH3)2CuLi (C) CH3I2 and Zn (D) CH3AlCl2 1 1 . Acetaldehyde reacts with NaOH to form :- OH H OH H (A) CH3 CH2 CH C O (B) CH3 CH CH2 C O (C) CH3 CH C CH3 (D) CH2 CH2 CH2 C O OH O OH H 1 2 . Benzaldehyde reacts with formaldehyde in the presence of alkali to form :- (A) Methyl alcohol and sodium benzoate (B) Benzyl alcohol and sodium formate (C) Benzoic acid and ethanol (D) Formic acid and benzyl alcohol 1 3 . The compounds A, B and C in the reaction sequence CH3 O AlIk2ali  ( A ) Ag  B dil H2SO4 C C H g 2 ppt CH3 are given by the set :- (B) Iodoform, acetylene, acetaldehyde (A) Iodoform, ethylene, ethyl alcohol (D) Iodoform, 2–propanol, propanone (C) Iodoform, propyne, acetone 1 4 . In the reaction sequence RCOCl + H2 PdBaSO4  A HCN B H3O  C A,B and C are given by the set :- (A) RCHO, RCH(OH)CN, RCH(OH)CH2NH2 (B) RCHO, RCH(OH)CN, RCH(OH)COOH (C) RCHO, R C CH2, R C CH2 C OH (D) RCHO, R–CH2–CN, R–CH2–COOH O CN O O 1 5 . Oxidation of 2-methyl propane–1,2–diol with periodic acid gives :- (A) Propionic acid and formaldehyde (B) Acetone and formaldehyde (C) Acetone and acetic acid (D) Acetone and propionic acid 1 6 . A carbonyl compound gives a positive iodoform test but does not reduce Tollen's reagent or Fehling's solution. It forms a cyanohydrin with HCN, which on hydrolysis gives a hydroxy acid with a methyl side chain. The compound is :- (A) Acetaldehyde (B) Propionaldehyde (C) Acetone (D) Crotonaldehyde 1 7 . A carbonyl compound 'A' reacts with hydrogen cyanide to form a cyanohydrin 'B' which on hydrolysis gives an optically active alpha hydroxy acid 'C'. 'A' gives a positive iodoform test 'A', 'B' and 'C' are given by the set :- H OH H OH (A) HCHO; C ;C H CN H COOH CH3 OH CH3 OH (B) CH3CHO; C; C H CN H COOH

C2H5 OH C2H5 OH (C) CH3CH2CHO; C; C H COOH H CN CH3 O; CH3 OH CH3 OH (D) CH3 C C ; CH3 C COOH CH3 CN 1 8 . In which of the following reactions aldehydes and ketones are distinguished : (A) Reaction with phenyl hydrazine (B) Reaction with hydroxylamine (C) Reaction with semicarbazide (D) Reaction with silver nirate mixed with ammonia 1 9 . Cyanohydrin of the following compound on hydrolysis gives optically active product : (A) HCHO (B) CH3CHO (C) CH3COCH3 (D) All the above 2 0 . The major organic product formed from the following reaction is :– O (i) CH3NH2  (ii) LiAlH4 (iii) H2O NHCH3 (B) NHCH3 NHCH3 (D) ONHCH3 (A) (C) OH OH 2 1 . Which one of the following on treatment with 50% aq. NaOH yields the corresponding alcohol and acid (A) C6H5CHO (B) CH3CH2CH2CHO (C) CH3COCH3 (D) C6H5CH2CHO 2 2 . Which compound gives reaction with 2,4-dinitro phenyl hydrazine? (A) CH3 C O (B) CH3 C O H (D) All of these CH3 H (C) C O H 2 3 . Which of the following reaction leads to the formation of secondary alcohol ? O O CH3 (i()iLi)iHA2lHO 4  (A) C6H5 C CH3 (i) C(iHi)3HMgBr  (B) C6H5 C (C) CH 3— CHO (i)(Liii)AHlH 4  O CH3 ((iii)) HBr2  (D) CH3 C 2 4 . Which compound is unable to react with NaHSO3 ? (A) CH3CHO (B) CH3 CH OCH3 (C) CH3 C O (D) HCHO CH3 CH3 2 5 . Which of the following compounds gives a ketone with grignard reagent (A) formaldehyde (B) Ethyl alcohol (C) Methyl cyanide (D) Methyl iodide CHECK YOUR GRASP ANSWER KEY EXERCISE -1 Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Ans. C B A B D C C D B B B B B B B C B D B B Que. 21 22 23 24 25 Ans. A D B B C

EXERCISE–02 BRAIN TEASERS SELECT THE CORRECT ALTERNATIVES (ONE OR MORE THEN ONE CORRECT ANSWERS) 1 . End products of the following sequence of reaction is O CH3 ((iii))NHaOI,,  A  B O A & B are : I I and CO2H I O (A) I CH3 (B) I and O O I I O I O CO2H (C) I and (D) I and I I 2 . Which of the following compound on treatment with LiAlH4 will give a product that will give positive iodoform test? (A) CH3CH2CHO (B) CH3CH2CO2CH3 (C) CH3CH2OCH2CH3 (D) CH3COCH3 3 . Among the given compounds, the most susceptible to nucleophilic attack at the carbonyl group is (A) CH3COCl (B) CH3CHO (C) CH3COOCH3 (D) CH3COOCOCH3 4 . Which of the following does not give iodoform reaction ? (A) CH3CH2CH2OH (B) CH3OH (C) CH3CHO (D) PhCOCH3 5 . Which of the following compounds undergo periodic oxidation O OCH3 O OCH3 O (A) (B) (C) CH–CH–CH3 (D) OH HO OH OH OH OH OCH3 O 6 . Ph Ph OH  A. A is :- O Ph OH Ph OH Ph (D) None (A) (B) CO2 (C) Ph OH O Ph Ph O

7 . In the reaction sequence, [X] is ketone : CH3 [X] KMnO4/ OH/  HOOC(CH2)3 – CH – COOH [X] will be :- O O O O (A) CH3 (C) CH3 (D) H3C CH3 (B) 8 . In the given reaction : H2C O NaBH4 (A) (i) BH3/THF (ii) H2O2/OH (B) (A) and (B) are : (A) H2C OH and HOCH2 O (B) H3C OH and HOH2C O (C) H2C and HOH2C O (D) H3C OH and CH3 O HO 9 . In the Cannizzaro reaction given below, — 2Ph – CHO OH Ph – CH2OH + PhCO  2 the slowest step is : (A) the attack of OH at the carbonyl group (B) the transfer of hydride to the carbonyl group (C) the abstraction of proton from the carboxylic acid (D) the deprotonation of Ph – CH2OH 1 0 . Which of the possible compound will be formed in the following sequence of reaction. CH2 = CH2 HBr X Hydrolysis Y Nl2ae2xCcOes3s  Z : (A) C2H5Br (B) C2H5OH (C) CHI3 (D) CH3CHO 1 1 . Which of the following statements are correct ? (A) Benzaldehyde reduces Fehling's solution (B) C6H5CHO + C6H5CHO NaOH C6H5CH = CHC6H5 + O2 is a Claisen-Schmidt reaction. (C) pKa (formic acid) is less than pKa (acetic acid) O OH (D) CH3CCH3 + CH3CHO NaOH CH3– C – CH2 CHO is an example of aldol condensation. CH3

1 2 . CH3 – CHO OH CH3CH(OH)CH2CHO In the aldol condensation of acetaldehyde represented above, which of the following intermediate species are obtained ? (A) H2C O (B) H2C– O O O O CH2 H H (C) H3C (D) H3C O H H 1 3 . When m-Chlorobenzaldehyde is treated with 50% KOH solution, the products obtained is (are) COO– CH2OH (A) (B) Cl Cl OH OH OH OH CH – CH CH – CH (C) (D) Cl Cl OH OH O O 1 4 . The conversion : O can be effected by using the reagent O (A) H2O, H2SO4 (B) O2 (C) C6H5 COOH (D) CrO3, H2SO4 1 5 . Which of the following hydrogens will be the most acidic? (d) O H C CH CH2 CH HH (b) (a) H (c) (A) a (B) b (C) c (D) d 1 6 . An organic compound with the molecular formula C3H6O does not respond positively to the silver mirror test with Tollens reagent but produces an oxime. The compound is (A) CH2= CHCH2OH (B) CH3CH2CHO (C) CH2= CHOCH3 (D) CH3COCH3

1 7 . Cyclohexanone on being heated with NaOH solution forms OH (B) (A) OH OH O O (D) (C) O O 1 8 . In the sequence of reactions O NBS aq.KOH CCl4 A B the final product (B) is O COOK O O (B) O OH (C) (D) (A) 1 9 . 2- Methylcyclohexanone is allowed to react with metachloroperoxobenzoic acid. The major product formed in the reaction is O O O H3C O CHO (D) O O O (A) (B) (C) CH3 CH3 O 2 0 . In the reaction 1. (CH3)2CuLi A. 2.H3O The product (A) is : O O H3C OH CH3 OH (B) (C) (D) (A) CH3 CH3 CH3 BRAIN TEASERS ANSWER KEY EXERCISE -2 Qu e. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 An s . D D Qu e. 16 17 A A,B A,B,C B B A B B,C,D C,D A,B,C A,B C C An s . D C 18 19 20 C CC

EXERCISE–03 MISCELLANEOUS TYPE QUESTIONS TRUE OR FALSE : 1. The reaction of methyl magnesium bromide with acetone followed by hydrolysis gives secondary alcohol. 2. Aldehydes are more reactive than ketones. 3 . The yield of ketone, when a secondary alcohol is oxidised, is more than the yield of aldehyde when primary alcohol is oxidised by K2Cr2O7/H 4 . Both aldehydes and ketones reduce Tollen's reagent. 5. Aldol condensation is given by all carbonyl compounds. 6. Acetaldehyde and acetone can be distinguished by iodoform test. 7. LiAlH4 converts ketones into secondary alcohols. 8. Methanol can be distinguished from ethanol by haloform reaction. 9. Propanone does not show tautomerism. 10. Ketones restore pink colour of Schiff's reagent. FILL IN THE BLANKS : 1 . To prepare ethanol, CH3MgI is treated with the other reagent .................. 2. Urotropine is formed by the action of .................. with ................... 3 . The conversion of acid chlorides into aldehydes by reduction is termed .................. 4. Aldehyde show reducing properties due to their ready conversion into .................. 5 . Hydrazone of an aldehyde when heated with sodium ethoxide forms .................. This is known as .................. reaction. 6. Cannizzaro's reaction is followed by those aldehydes which .................. -hydrogen atom. 7 . Two separate solutions, Fehling's solution A(.........) and Fehling's solution B(NaOH +..........) are at first mixed up together and is then heated with the aldehyde. A .................. precipitate is formed. 8 . Tollen's reagent gives .................. with acetaldehyde. 9. Aldehydes have boiling points lower than those of .................. and higher than those of .................. of comparable molecular masses. 10. When calcium acetate is distilled alone .................. is formed. MATCH THE COLUMN 1 . Match the compounds/ions in Column I with their properties/reactions is Column. II Column -I Column -II ( A ) C6H5CHO ( p ) gives precipitate with ( B ) CH3C  CH 2,4dinitrophenylhydrazine (C ) CN– ( q ) give preciptitate with AgNO3 ( D ) I– ( r ) is a nucleophile ( s ) is involved in cyanohydrin formation

2 . Match the column I with column II. Column-II Column-I ( p ) Ketone ( A ) RMgX + HCHO  ( q ) 1ºAlcohol (n + 1) carbon n carbo n ( r ) Acid (n + 1) carbon Adduct H3O  ( s ) 1ºAlcohol (n + 2) carbon (B) RMgX + (CH2)2O  n carbo n Adduct H3O  (C) RMgX + CO2  n carbo n Adduct H3O  (D) RMgX + Ph–CN  n carbo n Adduct H3O  3 . Aldol dondensation proceeds by carbon-carbon bond fromation between an enolate donor and a carbonyl acceptor. For each of the following aldol products (1 through 4) OH O O CO2C2H5 OH CH CO2C2H5 CHO (4) (1) (2) (3) match the donor and acceptor compound. Column -II [Acceptor] Column -I [Donor] (A) CH2– CHO ( p ) CH2 = O (B) O ( q ) CHO (C) O (r) O CO2C2H5 ( s ) CH2 – CHO ( D ) CH2 CO2C2H5 ASSERTION & REASON QUESTION : These questions contains, Statement-I (assertion) and Statement-II (reason). (A) Statement-I is True, Statement-II is True ; Statement-II is a correct explanation for Statement-I (B) Statement-I is True, Statement-II is True ; Statement-II is NOT a correct explanation for Statement-I (C) Statement-I is True, Statement-II is False. (D) Statement-I is False, Statement-II is True. 1 . Statement-I : 2, 2,-dimethyl propanol undergoes Cannizaro reaction with concentrated NaOH. Because Statement-II : Cannizarro reaction is a disproportionation reaction.

2 . Statement-I : Benzaldehyde does not undergoes aldol condensation. Because Statement-II : Benzaldehyde does not contains acidic –Hydrogen. 3 . Statement-I : Acetaldehyde is less reactive than trichloro acetaldehyde. Because Statement-II : Chlorine atom exhibit –I effect in trichloro acetaldehyde. 4 . Statement-I : Benzaldehyde gives a positive test with Benedict's and Fehling's solution. Because Statement-II : Benzaldehyde gives silver mirror with Tollen's reagent. 5 . Statement-I : R–CO+ is more stable than R–C+=O. Because Statement-II : Resonance in carbonyl compound provides C— and O–. 6 . Statement-I : Rate of addition of HCN on carbonyl compounds increases in presence of NaCN. Because Statement-II : Reaction involves the addition of CN– in rate determining step. 7 . Statement-I : Fehling's solution can be used to distinguish benzaldehyde from acetaldehyde. Because Statement-II : Both benzaldehyde and acetaldehyde reduces tollen's reagent. 8 . Statement-I : Ketones are less reactive than aldehydes. Because Statement-II : Ketones do not give Schiff's test. 9 . Statement-I : Benzaldehyde is more reactive than ethanol towards nucleophilic attack. Because Statement-II : The overall effect of –I and +R effect of phenyl group decreases the electron density on the carbon atom of >C = O group in benzaldehydes. 1 0 . Statement-I : In formaldehyde all the four atoms are in same plane. Because Statement-II : The carbon atom in formaldehyde is sp2 hybridised. COMPREHENSION BASED QUESTIONS : Comprehension # 1 Aldehyde, ketone, acid and acid derivatives contain >C = O group. Aldehyde and ketones gives nucleophilic addition reactions where as acid and acid derivatives gives nucleophilic addition followed by elimination reactions. Nucleophilic addition reactions followed by elimination of acid derivatives is known as acyl substitution reaction. This substitution reaction takes places by formation of tetrahedral intermediate. 1 . For the given reaction O O L + Nu Nu+L R R which of these is correct ? (A) L must be better leaving group than Nu (B) Nu– must be strong enough nucleophile to attack carbonyl carbon (C) Carbonyl carbon must be enough electrophilic to react with Nu– (D) All of these

2 . Which of the following compounds has very poor leaving group ? O O O O (A) R (B) R (C) R (D) R H Cl OH OR' 3 . Which one of the following is least reactive compound for nucleophilic acyl substitution. O H3C O NH NH H3C O H3C (A) H3C O (B) NH (C) (D) Cl H3C O2N Comprehension # 2 Aldehydes and ketones are amphoteric. Thus they can act both as acids and bases. Under acidic conditions, the carbon of the protonated carbonyl group is much more electrophilic, reacting even with weak nucleophilie. Carbonyl compound gives nucleophilic addition reaction. In this reaction the nucleophilic attack proceedes the electrophilic attack. 1 . Which of these statements are correct ? (A) Carbonyl compound is amphoteric in character (B) Acid catalyst makes the carbonyl carbon more electrophilic (C) basic catalyst makes the nucleophilic attack more faster. (D) All of these 2 . Which of the statements are/is correct ? (A) The rate determining step of addition reaction is the addition of nucleophile (B) The rate-determining step is addtion of electrophile (C) The reaction intermeidate of the reaction is alkoxide ion (D) both (A) and (C) 3 . Which one of the carbonyl compounds is more reactive towards NaCN/H+ ? (A) H5C6 O H3C CHO H (B) O (C) H3C O (D) NC O H H 4 . Carbonyl compounds gives nucleophilic addtion with (A) carbon nucleophile (B) oxygen nucleophile (C) Nitrogen nucleophile (D) All of these Comprehension # 3 Aldehyde and ketones are specially susceptible to nucleophile addition because carbonyl group C=O is polar (due to electronegativity different between carbon and oxygen). + – C=O

Positive charge on carbon makes it reactive towards the nucleophile. This addition is catalysed by acid. Reactivity of carbonyl compound towards nucleophilic addition increases with increase in the electron deficiency at carbonyl carbon. Thus, (–I.E.) group increase while (+I.E.) groups decrease the reactivity of carbonyl compound. 1 . Which of the following is most reactive to give nucleophilic addition? (A) FCH2CHO (B) ClCH2CHO (C) BrCH2CHO (D) ICH2CHO 2 . Carbonyl compounds show nucleophilic addition with : (A) HCN (B) NaHSO3 (C) (CH3OH + HCl) (D) all of these 3 . Which among the following carbonyl compounds is most polar? C2H5 C=O CH3 C=O CH3 C=O H (A) (B) (C) (D) C=O CH3 CH3 H H 4 . Select the least reactive carbonyl compound for nucleophilic addition: O O O O (A) C6H5––C––C6H5 (B) C6H5––C––CH3 (C) C6H5––C––H (D) CH3––C––H 5 . Which among the following isomeric compound is most reactive? O O (A) CH3––CH2––CH2––CH2––C––H (B) CH3––CH2––CH2––C––CH3 O (D) All are equally reactive (C) CH3––CH2––C––CH2––CH3 MISCELLANEOUS TYPE QUESTION ANSWER KEY EXERCISE -3  True / False 3. T 4. F 5. F 1. F 2. T 6. F 7. T 8. T 9. F 10. F  Fill in the Blanks 1. HCHO 2. HCHO, NH 3. Rosenmund reduction 3 6. do not have 4. acids 9. alcohol, alkane 5. alkane, wolf-kishner 7. CuSO , roschell salt, Cu O 42 8. silver mirror 10. acetone  Match the Column 1. (A) p,s ; (B) q,r ; (C)  q,r,s ; (D)  q,r 2. (A) q ; B s ; (C)  r ; (D) p 3. (A) s ; (B) p ; (C)  q ; (D) r  Assertion - Reason Questions 1. D 2. A 3. A 4. D 5. B 6. A 7. B 8. B 9. A 10. A  Comprehension Based Questions 3. (B) 4. (D) 3. (D) 4. (A) Comprehension #1 : 1. (D) 2. (A) 3. (A) 5. (A) Comprehension #2 : 1. (D) 2. (D) Comprehension #3 : 1. (A) 2. (D)

EXERCISE–04 [A] CONCEPTUAL SUBJECTIVE EXERCISE 1 . Which of the carbonyl groups in p-MeOC6H4COMe and p-NO2C6H4COMe protonates more readily in acid solution and why ? 2 . Identify (A) and (B) in the given sequence of reaction PhCH2CHO SeO2  (A) (i)c(oiin)H.OH– (B) 3. (a) Distinguish between PhCOEt and p-MeC6H4COMe by a chemical method. (b) Arrage the following in the order of increasing reactivity towards nucleophilic addition reactions. 4. (i) CH3CHO, C6H5COC6H5, CH3COC6H5,CH3COCH3 5. (a) (ii) CH3CHO, CF3CHO, CH2 = CHCHO Compound X with molecular formula C9H10O form a semicarbazone and give negative Tollen's and Iodoform (b) tests. Upon reduction it gives n-propyl benzene. Deduce the structure of X. 6. The double bond in aldehydes and ketones is reactive towards nucleophilic reagents like CN— whereas that in an alkene is not. Alkenes undergo electrophilic addition whereas aldehydes and ketones undergo nucleophilic addition. Complete the following equations giving the structures of the major organic product. O (ii) Cl OEt base ? (i) SeO2  O+ O 7 . Complete the following equation : (a) H3C O CH (i ) NaNH2 ? + HC (ii ) H3O CH3 O Cl (b) H3C + O NaNH2  C2H5 CH3 O 8 . Two organic compounds (A) and (B) have same empirical formula CH2O. Vapour density of (B) is twice the vapour density of (A). (A) reduces Fehling solution but does not react with NaHCO3. Compound (B) neither reacts with NaHCO3 nor reduces Fehling solution. What are (A) and (B) ? Also report an isomer of (B) if it reacts with NaHCO3. 9 . Complete the following equation : OH I2 HC  CH NaNH2  ? CH3Br ? Hg2/ H ? H H3C NaOH H2O CH3 O H3C 1 0 . A comopound has two isomers (A) and (B) of formula C5H10O. Isomer (A) on treating with NaOH (aq.) give 2, 2-demethylpropan-1-ol and 2, 2-dimethylpropanoic acid salt. The isomer (B) on treating with NaOH (aq.) gives 3-hydroxy-2-propylheptanal. What are A and B ?

CONCEPTUAL SUBJECTIVE EXERCISE ANSWER KEY EXERCISE -4(A) 1 . MeO C Me + O– O MeO C Me (1) (–OMe having +R) O– (1) C Me O– + C Me + O + N N O O O– (–NO2 having –R) I will be more readily protonated than (II). Alternative protoned (I) is more stabilised by resonance than protonated (II). MeO + + OH OH MeO C Me C Me (X) O– + O– + + OH + OH N N + C Me O O– C Me (Y) In (X) there is extended conjugation and only one charge is involved. In (Y) there is not this extended conjugation and the relative close proximity of two positive charges is a destabilising factor. Hence (X) is more stable than (Y). 2 . (A)  C8H6O2 = C6H5 – C–CHO ; (B) C8H8O3 = C6H5–CHOHCO2H O 3 . (a) I does not undergo haloform reaction whereas II does. (b) A = C6H5COC6H5 < CH3COC6H5 < CH3COCH3 < CH3CHO B = CH2=CHCHO < CH3CHO < CF3CHO 4 . (i) X froms semicarbazone and thus possesses carbonyl group. >C=O + H2N.NHCONH2  >C=N.NHCONH2 (ii) It does not given Tollen's reagent test and thus it is ketone. (iii) It does not given iodiform test and thus it is not methyl ketone. (iv) Keepting in view of the above facts and molecular formula ; X is H5C6 C.CH2.CH3 O (v) This on reduction will given n-propyl benzene.

5 . (a) The carbonyl group in aldehydes and ketones add on CN— resulting in the fromation of an anion where the negative charge resides on oxygen. However if a nucleophile adds on to an alkene the negative charge resides on carbon. Since carbon is much less strongly electron attracting than oxygen, this species is less stable and hence not readily formed. O– C CN (b) In alkenes the double bond joins two carbon atoms and there is no resultant polarity. In carbonyl compounds, the carbonyl group is highly polar and the high partial positive charge on the C atom makes it subsceptible to nucleophilic attack. 6 . (i) O (ii) O O OEt ; O 7 . (a) H3C OH ; (b) O O H3C H3C HC H3C C2H5 O 8 . A = HCHO (Formaldehyde) B = HCOOCH3 (Methyl methanoate) O OH O 9 . HC  C—Na+ ; H3C—C  CH ; H3C ; H3C + CHI3 CH3 CH3 ONa CH3 1 0 . A = H3C CH3 ; B = H3C CHO CHO

EXERCISE–04 [B] BRAIN STORMING SUBJECTIVE EXERCISE 1 . Identify (A) to (D) as reactant, reagent, product as name of the reaction in following : (i) 3(CH3)2C=O HCl (A) Aldol condensation (ii) CH3COCl + H2 (B) (C) Rosenmund's reaction (iii) (D) (i) NH2NH2 CH3CH2CH3 ( ii ) C2H5ONa (iv) (D) OBr CHBr3 + CH3COO 2 . Two different Grignard reagents, (X) and (Y) produce C6H5CH2C(CH3)2OH on reaction with (P) and (Q) respectively. Give structures of (X), (Y) and (Q). 3. (a) 2-methyl-1,3-cyclohexanedione is more acidic than cyclohexanone - explain with reason (b) Explain why HCN will add to the double bond in CH2=CHCOOH but not in RCH=CHR 4 . A compound C5H10O does not reduce Fehling's solution, form a phenyl hydrazone, shows the haloform reaction, and can be converted into n-pentane by Zn—Hg and conc. HCl. What is this comopound ? 5. (a) Convert PhCHO into PhCH = CHCOPh (b) Identify A, B, C and D in the following reaction. C7H6O Alc. C14H12O2 [A] KCN [B] CH3COONa C9H8O2 Br2 C9H8O2Br2 [C] [D] Acetic anhydride 6 . A ketone (A) which undergoes haloform reaction gives compound B on reduction. B on heating with sulphuric acid gives compound C, which form mono ozonide D. D on hydrolysis in presence of Zn dust gives only acetaldehyde. Identify A, B and C. Write down the reactions involved. 7 . Complete the following reactions : O (i) +  NH3 H 3O  (A) dil H2SO4 (B) (–H 2 O) (C) (i )OsO4 /H2O,OH (D) Low Temp. Hg2 ( ii ) HIO4 H3C—CCNa Write the structures of A to D and give the IUPAC name of (D). O CH3 H2ONa.1O0H0C (A) (ii) H3C O 8 . The sodium salt of a carboxylic acid, A, was produced by passing a gas, B, into an aqueous solution of caustic alkali at an elevated temperature and pressure. A on heating in presence of sodium hydroxide followed by treatment with sulphuric acid gave a dibasic acid C. A sameple of 0.4 g of C on combustion gave 0.08 g of water and 0.39 g of carbondioxide. The silver salt of acid weighin 1g on ignition yielded 0.71 g of silver as residue. Identify A, B and C.

9 . Give reasons for the following : (a) Ketones are less electrophilic than aldehydes. (b) Aldehydes are reducing agents and ketones are not. (c) HBr fails to give addition products with carbonyl compounds. 1 0 . Identify A, B and C ? O CH3 Br2 (1eq.) (A ) NaBH4 (B ) dil.OH(C ) BRAIN STORMING SUBJECTIVE EXERCISE ANSWER KEY EXERCISE -4(B) 1 . (i) H3C H3C CHCOCH CH3 phorone. 3 O dryHCl CH3 H3C H3C (ii) CH3COCl + H2 Pd / BaSO4  CH3CHO + HCl quinoline H3C  NH2NH2 C2H5ONa (iii) CO CH3CH2CH3 H3C OO 2 . If X is Ph—CH2MgX, then P is CH3 – C – CH3and if Y is CH3–MgX, then Q is Ph CH3 3. (a) 2-methyl-1, 3-cyclohexanedione is more acidic because its enolate ion is stabilized by an additional resanance structure. OO O O CH3 CH3 CH3 CH3 O O H –H OO Resonance stabilized conjugate base. (b) In the case of the acid the double bond is activated due to the presence of a –COOH group O OO H2C = CH – C – OH + – CH = C – OH HCN NC – CH2 – HC = C – OH H2 C OO  NC – CH2 – CH – C – OH H NC – CH2 – CH2 – C – OH 4 . (i) Since C5H10O forms a phenyl hydrazone, hence it is an aldehyde or ketone. (ii) As it does not reduce Fehling's solution, hence it is not aldehyde but a ketone. O (iii) Since it undergoes haloform reaction, therefore it is CH3 – C – R type of ketone. The value of R – can be derived as follows : R= —C3H7 O O CH3 Hence this ketone is H3C CH3 CH3 or H3C

(iv) Ketone on Clemmensen reduction yields n-pentane; hence R is n-propyl and not isopropyl. 5. (a) CH3MgI, H3O+, H+/K2Cr2O7,  PhCHO/ OH, (b) A = C6H5CHO ; B = PhCH(OH)COPh ; C = C6H5CH=CH—COOH ; D = C6H5CHBr—CHBr—COOH O H3C 4[H] H3C CH3 CH3 Zn–Hg + HCl pentan-2-one pentane C6H5 2CuO No reaction NH C6H5NHNH2 N CH3 – H2O H3C O 3I2+4NaOH/ CHI3 + H3C O Na + 3NaI + 3H2O H3C CH3 ; B = H3C CH3 ; C = H3C CH3 ; H3C O CH3 6. A = D= O O OH O 7 . (i) A = OH OH CH3 O O CH O ; C= CH3 ; D = OHC CH3 ;B= 6,7-dioxooctanal O O (ii) CH3 8 . A = HCOOH ; B = CO ; C = COOH COOH 9 . (a) The positive inductive effect of the second alkyl radical reinforces that of the first one decreasing still further the partial positive charge on the carbonyl carbon atom.This reduces the attraction of the atom for nucleophilic reagents. Hence ketones are less electrophilic. (b) The >C=O group in aldehydes activates the H atom attached to the carbonyl group. This is due to the relaying of the –I effect of the oxygen atom to the C—H bond so that partial positive charge is created on the H atom. The result of this activation is that the H atom of the –CHO group can be oxidised readily to a (OH) group. Thus aldehydes are reducers. (c) HBr is strongly polar and is hence readily added to the polarized >C=O group. The addition product H3C OH H3C Br is however unstable and decomposes to give the original carbonyl compound and HBr. 1 0 . (A) : O OH O CH2Br ; (B) : CH2Br ; (C) : CH2

EXERCISE–05 PREVIOUS YEARS QUESTIONS 1 . The formation of cyanohydrin from a ketone is an example of [IIT-90] (A) Electrophilic addition (B) Nucleophilic addition (C) Nucleophilic substitution (D) Electrophilic substitution 2 . The enolic form of acetone contains - [IIT-90] (A) 9 sigma bonds, 1 pi bond and 2 lone pairs (B) 8 sigma bonds, 2 pi bond and 2 lone pairs (C) 10 sigma bonds, 1 pi bond and 1 lone pairs (D) 9 sigma bonds, 2 pi bond and 1 lone pairs 3 . m–chlorobenzaldehyde on reaction with conc. KOH at room temperature gives [IIT -91] (A) Potassium m-chlorobenzoate and m-hydroxybenzaldehyde (B) m-hydroxy benzaldehyde and m-chlorobenzyl alcohol (C) m-chlorobenzyl alcohol and m-hydroxybenzyl alcohol (D) Potassium m-chlorobenzoate and m-chlorobenzyl alcohol 4 . Hydrogenation of benzoyl chloride in the presence of Pd on BaSO4 gives [IIT-92] (A) Benzyl alcohol (B) Benzaldehyde (C) Benzoic acid (D) Phenol 5 . An organic compound C3H6O does not give a precipitate with 2,4-Dinitrophenyl hydrazine reagent- [IIT-93] (A) CH3CH2CHO (B) CH3COCH3 (C) CH2=CH–CH2OH (D) —OH 6 . Under Wolff Kishner reduction conditions, the conversion which may be through about is - [IIT-95] (A) Benzaldehyde into Benzyl alcohol (B) Cyclohexanol into Cyclohexane (C) Cyclohexanol into Cyclohexanol (D) Benzophenone into Diphenylmethane 7. In the reaction, P is, [IIT-95] CH3 CO SeO2 P + Se + H2O CH3 (A) CH3COCHO (B) CH3COOCH3 (C) CH3COCH2OH (D) None 8. In the Cannizzaro reaction given below 2 Ph –CHO OH Ph–CH2OH + PhCO  the slowest step is- 2 (A) the attack of OH at the carbonyl group [IIT-96] (B) the transfer of hydride to the carbonyl group (C) the abstraction of proton from the carboxylic acid (D) the deprotonation of Ph–CH2OH 9 . Which of the following will give yellow precipitate with I2/NaOH- [IIT-97] (A) ICH2COCH2CH2 (B) CH3COOCOCH3 (C) CH3CONH2 (D) CH3CH(OH)CH2CH3 1 0 . Among the given compounds, the most succeptible to nucleophilic attack at the carbonyl group is - (A) MeCOCl (B) MeCHO [IIT-97] (C) MeCOOMe (D) MeCOOCOMe

1 1 . In a Cannizzaro reaction the intermediate which is the best hydride donor is - [IIT-97] H H (C) H (D) H (A) C6H5– C – O (B) C6H5– C – O O2N O CH3O O OH O O O 1 2 . CH3CHO + H2NOH  CH3 –CH = N – OH. The above reaction occurs at - [IIT-97] (A) pH = 1 (B) pH ~ 4.5 (C) Any value of pH (D) pH = 12 13. Among the following compounds, which will react with acetone to give a product containing >C = N– [IIT-98] (A) C6H5NH2 (B) (CH3)3N (C) C6H5NHC6H5 (D) C6H5NHNH2 14. The product obtained via oxymercuration (HgSO4+H2SO4) of 1-butyne would be - [IIT-98] O (B) CH3CH2CH2CHO (A) CH3CH2– C–CH3 (C) CH3CH2CHO + HCHO (D) CH3CH2COOH+HCOOH 1 5 . Which of the following will not undergo aldol condensation - [IIT-98] (A) Acetaldehyde (B) Propanaldehyde (C) Benzaldehyde (D) Trideutero acetaldehyde 1 6 . Which of the following will react with water to form a stable hydrate - [IIT-98] (A) CHCl3 (B) Cl3CCHO (C) CCl4 (D) ClCH2CH2Cl 1 7 . The enol form of acetone, after treatment with D2O gives - [IIT-99] OH OH OH OD (A) CH2–C–CH2D (B) CD2=C–CD3 (C) CH2=C–CH2D (D) CD2=C–CD3 1 8 . Which of the following has the most acidic hydrogen- [IIT-2000] (A) 3-hexanone (B) 2, 4-hexanedione (C) 2, 5-hexanedione (D) 2, 3 hexandione 1 9 . The appropriate reagent for the following transformation - [IIT-2000] O CH2CH3 CH3  OH OH (A) Zn (Hg), HCl (B) NH2NH2,OH– (C) H2/Ni (D) NaBH4 20. Read the following statement and explanation and answer as per the option given below : [IIT-01] Statment-I : Dimethylsulphide is commonly used for the reduction of an ozonide of an alkene to get the carbonyl compounds. Statment-II : It reduces the ozonide giving water soluble dimethyl sulphoxide and excess of it evaporates. (A) If both assertion and reason are correct, and reason is the correct explanation of the assertion (B) If both assertion and reason are correct, but reason is not correct explanation of the assertion (C) If assertion is correct but reason is incorrect (D) If assertion is incorrect but reason is correct

2 1 . A mixture of benzaldehyde and formaldehyde on heating with aqueous NaOH solution gives - [IIT -01] (A) benzyl alcohol and sodium formate (B) sodium benzoate and methyl alcohol (C) sodium benzoate and sodium formate (D) benzyl alcohol and methyl alcohol 2 2 . Compound A (molecular formula C3H8O) is treated with acidified potassium dichromate to from a product B (molecular formula C3H6O). B forms a shining silver mirror on warming with ammonical silver nitrate. B when treated with an aqueous solution of H2NCONHNH2. HCl and sodium acetate gives a product C. Identify the structure of C- [IIT-02] (A) CH3CH2CH=NNHCONH2 (B) CH3–C=NNHCONH2 CH3 (C) CH3–C=NCONHNH2 (D) CH3CH2CH=NCONHNH2 CH3 2 3 . Any one of the product formed is : [IIT-02] CHO OHC (i) NaOH(excess) 100ºC CHO OHC (ii) H/H2O COOH HOOC CH2OH CH2OH (A) (B) COOH HOOC CH2OH CH2OH CH2OH COOH O O (C) (D) COOH CH2OH O O OCOCH3 [IIT-03] Acidic 2 4 . P OCOCH3 Hydrolysis Q Product formed by P & Q can be differentietod by : (A) 2, 4-DNP (B) Lucas reagent (ZnCl2+ conc. HCl) (C) NaHSO3 (D) Fehlings solution 2 5 . Which of the reagent is used to convert 2–Butanone into propanoic acid - [IIT-05] (A) NaOH, I2/H (B) Tollen's reagent (C) Fehling solution (D) NaOH, NaI/H

O [IIT-06] 2 6 . CH3–C–CH3+ next homologue NH2OH mixture Following statements is/are correct about mixture : (A) mixture is 3-types of oximes (B) mixture is 2-types of oximes (C) all are optically active (D) one is optically active 2 7 . CH3–CH2–CH2–Cl P (i) O2/ Q + Phenol, P & Q are - [IIT-06] AlCl3 (ii) H3O O O (A) and CH3–CH2–C–H (B) and CH3–C–CH3 O O (C) and CH3–C–CH3 (D) and CH3–CH2–C–H O C 2 8 . O The above reagent can be synthesise from the following compounds in alkaline medium following by acidification - [IIT-06] CHO COOH COOCH3 COOCH3 (A) CHO (B) CHO (C) CH3 (D) COOH 2 9 . An unknown compound of carbon, hydrogen and oxygen contains 69.77%C and 11.63% H, and has a molecular weight of 86. It does not reduce Fehling solution but forms a bisulphate addition compound and gives a positive iodiform test. What are the possible structures ? [IIT-87] 3 0 . Iodoform is obtained by the reaction of acetone with hypoiodite but not with iodide. Why ? [ I I T - 9 1 ] 3 1 . An organic compound 'A' on treatment with ethyl alcohol gives a carboxylic acid 'B' and compound 'C'. Hydrolysis of 'C' under acidic conditions gives 'B' and 'D'. Oxidation of 'D' with KMnO4 also gives 'B'. 'B' on heating with Ca(OH)2 gives 'E' (C3H6O). E does not give Tollen's test and does not reduce Fehling's solution but form a 2,4-dinitrophenyl hydrazone. Identify (A), (B), (C), (D) and (E). [IIT-92] 3 2 . Complete the following : (i) H3CO— —CHO+ HCHO KOH (A )  (B ) [IIT-92] 33. Complete the following : C6H5 – CHO + CH3 – COOC2H5 NaOC2H5 in absolute (A) [IIT-95] C2H5OH and heat

3 4 . Complete the following reaction with appropriate structure : [IIT-96] CH3 O (i ) KCN / H2SO4 (X) (ii ) LiAlH4 H 3 5 . Suggest appropriate structures for the missing compounds. (The number of carbon atoms remains the same throughout the reactions). [IIT-96] CH3 dil KMnO4  (A) HIO4  (B) OH– (C)  CH3 3 6 . Complete the following : [IIT-98] O O [A ] CHC6H5 (i(ii))HLiA,IhHe4at [B] 3 7 . Compound 'A' (C8H8O) on treatment with NH2OH.HCl given 'B' and 'C'. 'B' and 'C' rearrange to give 'D' and 'E', respectively, on treatment with acid. B, C, D and E are all isomers of molecular formula (C8H9NO). When 'D' is boiled with alcoholic KOH an oil 'F' (C6H7N) separates out. 'F' reacts rapidly with CH3COCl to give back 'D'. On the other hand, È' on boiling with alkali followed by acidification gives a white solid 'G' (C7H6O2). Identify 'A' to 'G'. [IIT-99] 3 8 . Carry out the following transformation in not more than three steps. [IIT-99] H3C H3C O H CH3 3 9 . An organic compound (A), C6H10O, on reaction with CH3MgBr followed by acid treatment gives compound (B). The compound (B) on ozonolysis gives compound (C), which in presence of a base give 1-acetyl cyclopentene (D). The compound (B) on reaction with HBr gives compound (E). Write the strucures of (A), (B), (C) and (E). Show how (D) is formed from (C). [IIT-2000] 4 0 . An organic compound (A) reacts with H2 to give (B) and (C) successively. On ozonolysis of (A), two aldehydes (D) C2H4O and (E) C2H2O2 and formed. On ozonolysis of (B) only propanal is formed. What are (A) to (E) ? [IIT-01] 4 1 . + X CH3COONa [IIT 2005] What is X? (B) BrCH2, COOH (C) (CH3CO)2O (D) CHO–COOH (A) CH3COOH 4 2 . The smallest ketone and its next homologue are reacted with NH2OH to form oxime. (A) Two different oximes are formed (B) Three different oximes are formed (C) Two oximes are optically active (D) All oximes are optically active [JEE 2006] 4 3 . Cyclohexene on ozonolysis followed by reaction with zinc dust and water gives compound E. Compound E on further treatment with aqueous KOH yields compound F. Compound F is [JEE 2007] (A) (B) (C) (D)

4 4 . Statement-1 : Glucose gives a reddish-brown precipitate with Fehling’s solution. because Statement-2 : Reaction of glucose with Fehling’s solution gives CuO and gluconic acid. (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1. (C) Statement-1 is True, Statement-2 is False. (D) Statement-1 is False, Statement-2 is True. [JEE 2007] 4 5 . Match the compounds/ion in column I with their properties/ reaction in Column II. Indicate your answer by darkening the appropriate bubbles of the 4 × 4 matrix given in the ORS. [JEE 2007] Column I Column II (A) C6H5CHO (P) gives precipitate with 2,4-dinitrophenylhydrazine (B) CH3CCH (Q) gives precipitate with AgNO3 (R) is a nucleophile (C) CN– (S) is involved in cyanohydrin formation (D) I– Paragraph for Question No. 46 to 48 In the following reaction sequence, products I, J and L are formed. K represents a reagent. 1. Mg / ether Hex-3-ynal 1. NaBH4  I 2.3C.OH23O J K  Me Cl H2  L || Pd/BaSO4 2. PBr3 O quinoline 4 6 . The structure of the product I is [JEE 2008] (A) Me Me Br Br (B) (C) Me Me Br Br (D) 4 7 . The structures of compounds J and K, respectively, are [JEE 2008] (A) Me COOH and SOCl2 (B) Me OH || and SO2Cl2 O (C) Me and SOCl2 (D) Me COOH and CH3SO2Cl COOH 4 8 . The structure of product L is [JEE 2008] Me CHO (B) Me CHO (D) Me CHO (A) (C) Me CHO

Paragraph for Question No. 49 to 51 A tertiary alcohol H upon acid catalysed dehydration gives a product I. Ozonolysis of I leads to compounds J and K. Compound J upon reaction with KOH gives benzyl alcohol and a compound L, whereas K on reaction with KOH gives only M. H3C O Ph M= Ph H [JEE 2008] 4 9 . Compound H is formed by the reaction of O O || || (A) Ph + PhMgBr (B) Ph + PhCH2MgBr CH3 CH3 O O Me (C) || (D) || + PhCH2MgBr + MgBr Ph H Ph H Ph 5 0 . The structure of compound I is [JEE 2008] Ph CH3 H3C Ph Ph CH3 H3C CH3 (A) (B) (C) (D) H H Ph H Ph H CH2Ph Ph 5 1 . The structures of compounds J, K and L, respectively, are [JEE 2008] (A) PhCOCH3, PhCH2COCH3 and PhCH2COO¯K+ (B) PhCHO, PhCH2CHO and PhCOO¯K+ (C) PhCOCH3, PhCH2CHO and CH3COO¯K+ (D) PhCHO, PhCOCH3 and PhCOO¯K+ Paragraph for Question Nos. 52 to 54 A carbonyl compound P, which gives positive iodoform test, undergoes reaction with MeMgBr followed by dehydration to give an olefin Q. Ozonolysis of Q leads to a dicarbonyl compound R, which undergoes intramolecular aldol reaction to give predominantly S. P 1.MeMgBr Q 1.O3  R 1.OH¯ S 2.H ,H2O 2. Zn ,H 2O 2.  3.H2SO4 , 5 2 . The structure of the carbonyl compound P is [JEE 2009] (A) Me O Me (B) O Me O (C) (D) O Et Me

5 3 . The structure of the products Q and R, respectively, are [JEE 2009] O O (A) Me , H (B) , H COMe CHO Me Me Me Me Me Me Me Me (C) , O (D) Me O Me Et , CH3 H CHO CHO Me Me Et Me Et 5 4 . The structure of the product S is [JEE 2009] OO O Me (A) (B) (C) (D) Me MeO Me Me Me Paragraph for Questions Nos. 55 to 56 An acyclic hydrocarbon P, having molecular formula C H , gave acetone as the only organic product 6 10 through the following sequence of reactions, in the which Q is an intermediate organic compound. P (i) dil H2SO4/HgSO4 (i) conc.H2SO4 O [JEE 2011] (ii) NaBH4/ethanol (Catalytic amount) (C6H10) (iii) dil.acid 2C Q (–H2O) H3C CH3 (ii) O3 (iii) Zn/H2O 5 5 . The structure of compound P is - (B) H3CH2CC–CH2CH3 (A) CH3CH2CH2CH2–CC–H H3C H3C (D) H3C–C–CC–H (C) H–C–CC–CH3 H3C H3C 5 6 . The structure of the compound Q is - H3C OH H3C OH (A) H–C–C–CH2CH3 (B) H3C–C–C–CH3 H3C H H3C H H3C OH OH (C) H–C–CH2CHCH3 (D) CH3CH2CH2CHCH2CH3 H3C

5 7 . The number of aldol reaction(s) that occurs in the given transformation is [JEE 2012] [JEE 2013] CH3CHO + 4HCHO conc. aq. NaOH OH OH HO OH (A) 1 (B) 2 (C) 3 (D) 4 5 8 . Among P, Q, R and S, the aromatic compound(s) is / are : Cl P NaH Q AlCl3 OO (NH4 )2 CO3 R O 100115ºC HCl S (A) P (B) Q (C) R (D) S 5 9 . After completion of the reactions (I and II), the organic compound(s) in the reaction mixtures is(are) O Reaction I : H3C CH3   Br2 (1.0 mol) [JEE 2013] aqueous /  NaOH (1.0 mol) O Reaction II : H3C CH3 BCr2H(31C.0OmOoHl) (1.0 mol) O O OO O CHBr3 U H3C CH2Br H3C CBr3 Br3C CBr3 BrH2C CH2Br H3C ONa P Q RS T (A) Reaction I : P and Reaction II : P (B) Reaction I : U, acetone and Reaction II : Q acetone (C) Reaction I : T, U, acetone and Reaction II : P (D) Reaction I : R, acetone and Reaction II : S acetone

PREVIOUS YEARS QUESTIONS ANSWER KEY EXERCISE -5 1. (B) 2. (A) 3. (D) 4.(B) 5. (C, D) 6. (D) 7.(A) 8. (B) 9. (A) 10. (A) 11. (D) 12. (B) 13. (A) 14. (A) 15. (C) 16. (B) 17. (C) 18. (B) 19. (B) 20. (A) 21. (A) 22. (A) 23. (C) 24. (D) 25. (A) 26. (A) 27. (C) 28. (A) For empirical formula 29. (i) Element % Relative no. of atoms Simplest ratio C 69.77 5.76 5 H 11.63 11.63 10 O 19.20 1.2 1  Empirical formula of compound is C5H10O. and Empirical formula wt. = 86 Also Molecular wt. = 86  Molecular formula of compound is C5H10O. (ii) Compound form bisulphate addition compound and thus, has carbonyl group. i.e., aldehyde or ketone. (iii) It does not reduce Fehling solution and thus, it is not aldehyde but it is ketone. (iv) It gives positive iodoform test and thus, it has O H3C (v) Above facts reveals that compound is CH3.CH2.CH2.C.CH3 H3C CH3 or O H3C O Pentan-2-one 3-methylbutan-2-one 3 0 . The formation of CHI3 takes place as follows : CH3COCH3 OI CH3COCI3 OH  CH3COO– + CHI3 I– cannot bring about this reaction. The active species is OI–, OI– is an oxidant and an iodinating agent. OO H3C O H3C O 31. A = O ; B = H3C ; C= O ; D = H5C2 – OH ; E = H3C H3C OH CH3 H5C2 O 3 2 . A = HOH2C— —OCH3 B = HCOOK 3 3 . C6H5CH = CHCOOC2H5 3 4 . KCN + H2SO4  KHSO4 + HCN CH3 H5C2 OH H5C2 OH O + HCN  H CN LiAIH4  H NH2 H Cyanohydrine 1-aminobutan-2-ol

35. A = CH3 O O OH OH ; B= ; CH3 OH H3C CH3 C= CH3 O (Aldol) 3 6 . A = C6H5CHO, base B = C6H5 O H5C6 – C – CH3 H5C6 – C – CH3 CH3 C= 37. A = ; B= N – OH ; OH – N H3C– C – NH E = H5C6– C – NH ; F = H5C6 – NH2 D = O C6H5 ; O CH3 G = H5C6 O OH 38. CH3CH2C  CH Na ln  CH3CH2C  CNa C– NHa3XX  CH3CH2C  C—CH3 liquid NH3 H3C O HgSO4  H2SO4 pent-2-one CH3 O CH3 O H ; B= ; C = H3C O 39. A = COCH3 CH3 D= ;E= Br 4 0 . A = H3C CH3 B = H3C CH3 C = H3C CH3 D = H3C – CHO E = CHO CHO 41. C 42. B 43. A 44. C 4 5 . (A) P, S; (B) Q; (C) Q, R, S; (D) Q,R 46. D 47. A 48. C 49. B 53. A 54. B 55. D 56. B 50. A 51. D 52. B 57. C 5 8 . A,B,C,D 5 9 . C



EXERCISE-01 CHECK YOUR GRASP SELECT THE CORRECT ALTERNATIVE (ONLY ONE CORRECT ANSWER) H OH 1 . The product formed in the reaction + SOCl  is- 2 Cl H H O Cl H SO3H (A) (C) (D) (B) H Cl 2. The reaction + SOCl N Cl + SO + N+ Cl– 2 HO H 2 H H proceeds by the mechanism (A) SN1 (B) SN2 (C) S i (D) SE2 N (D) hexane 3 . 1, 3- Dibromopropane reacts with metallic zinc to form (A) propene (B) cyclopropane (C) propane 4 . Consider the following reaction sequence, CH C  CH aq.H2SO4  A PCl5  B. 3 H gSO 4 Heat The products (A) and (B) are, respectively, (A) CH COCH and CH CCl CH (B) CH CH CHO and CH CH CHCl 33 3 23 32 32 2 (C) CH3CHOHCH3 and CH3CHClCH3 (D) CH3CH2CH2OH and CH3CH2CH2Cl 5 . Which of the following has highest dipole moment: (A) CH Cl (B) CH F (C) CH Br (D) CH I 3 3 3 3 6 . In SN1 the first step involves the formation of (A) free radical (B) carbanion (C) cabocation (D) final product 7 . To form alkane isonitrile, alkyl halide is reacted with: (A) KCN (B) AgCN (C) HCN (D) NH CN 4 8 . Which one of the following compounds most readily undergoes substitution by SN2 mechanism ? CH3 Cl H3C CH3 C3H7 (A) H3C (B) H3C (C) H3C (D) Cl H3C Cl Cl 9 . Sec. Butyl chloride undergo alkaline hydrolysis in the polar solvent by (A) SN2 (B) SN1 (C) SN1 and SN2 (D) none of these

1 0 . Grignard reagent can be prepared by (A) CH —CH —Cl + Mg edthryer  (B) CH3 CH CH2 + Mg edthryer  32 Cl OH CH3 (D) All of them (C) CH3 C OH + Mg edthryer  CH3 1 1 . Most stable carbocation formed from (CH ) C–Br, (C H ) CBr,(C H ) CHBr and C H CH Br would be 33 6 53 6 52 65 2     (A) CH 5 C H 2 (B) (CH ) C (C) (C H ) 3 C (D) (C H) CH 6 33 6 5 6 52 1 2 . For the reaction CH CH(X)CH CH alc. KOH CH3 CH CH CH3 3 23 CH2 CH CH2 CH3 (A) CH —CH CH—CH predominates (B) CH CH—CH —CH predominates 33 2 23 (C) Both are formed in equal amounts (D) The product ratio depends on the halogen 1 3 . The products of reaction of alcoholic silver nitrite with ethyl bromide are (A) Ethane (B) Ethene (C) Ethyl alcohol (D) Nitro ethane 14. The reaction ,CH Br + OH–  CH OH + Br– obeys the mechanism 3 3 (A) SN1 (B) SN2 (C) E (D) E 1 2 1 5 . Ethylidene chloride can be prepared by the reaction of HCl and (A) Ethane (B) Ethylene (C) Acetylene (D) Ethylene glycol 1 6 . 1–phenyl–2–chloropropane on treating with alc. KOH gives mainly (A) 1–phenylpropene (B) 2–phenylpropene (C) 1–phenylpropane–2–ol (D) 1–phenylpropan–1–ol 1 7 . Grignard reagent is obtained when magnesium is treated with (A) Alkyl halide in presence of alcohol (B) Alkyl halide in presence of phenol (C) Alkyl halide in presence of dry ether (D) Alkyl halide in presence of alcoholated ether 1 8 . Ethylene reacts with bromine to form - (A) Chloroethane (B) Ethylene dibromide (C) Cyclohexane (D) 1-bromo propane 1 9 . C H Br2 X KCN  Y ; Y is 24 (A) CH CH CN (B) NC—CH —CH —CN 32 22 (C) Br—CH —CH CN (D) Br—CH CHCN 22 2 0 . Reactivity order of halides for dehydrohalogenation is (A) R – F > R – Cl > R – Br > R – I (B) R – I > R – Br > R – Cl > R – F (C) R – I > R – Cl > R – Br > R – F (D) R – F > R – I > R – Br > R – Cl 2 1 . Which of the following is least reactive in a nucleophilic substitution reaction (A) CH CHCl (B) CH CH Cl (C) CH CHCH Cl (D) (CH ) C–Cl 2 32 22 33

2 2 . The correct reactivity order of alcohols towards H–X will be (I) CH CH—OH (II) H3C OH (III) CH —CH —OH (IV) CH3 CH CH3 2 CH2 32 OH (A) II > I > III > IV (B) IV > III > II > I (C) II > IV > I > III (D) II > IV > III > I 23. Identify 'Z' in the following reaction series, CH .CH CH Br aq.NaOH (X) AHl2eOa3t  (Y) HOCl (Z) : 3 22 (A) Mixture of CH3CH CH2 and CH3CH CH2 (B) CH3 CH CH2 OH Cl Cl Cl OH OH (C) CH3 CH CH2 (D) CH3 CH CH2 Cl OH Cl Cl 2 4 . For the reaction, C H OH + HX ZnX2  C H X, the order of reactivity is 25 25 (A) HI > HCl > HBr (B) HI > HBr > HCl (C) HCl > HBr > HI (D) HBr > HI > HCl 2 5 . Ethyl alcohol reacts at a faster rate with HI than with HCl in forming the corresponding ethyl halides under identical conditions mainly because - (A) HI, being a stronger acid, protonates ethyl alcohol at oxygen much better and helps substitution (B) the bond length in HI is much shorter than that in HCl (C) I– is a much better leaving group (D) I– is a much better nucleophile than Cl– CHECK YOUR GRASP ANSWER KEY EXERCISE -1 Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Ans. B B B A A C B B C A C D D B C Que. 16 17 18 19 20 21 22 23 24 25 Ans. A C B B B A D B B D

EXERCISE–02 BRAIN TEASERS SELECT THE CORRECT ALTERNATIVES (ONE OR MORE THEN ONE CORRECT ANSWERS) 1 . Which of the following does/do produce a white precipitate of AgCl on warming with alcoholic silver nitrate? (A) Allyl chloride (B) t-Butyl chloride (C) Benzyl chloride (D) Vinyl chloride 2 . What is the order of reactivity of the following compounds towards nucleophilic substitution? Cl Cl Cl Cl (I) (II) (III) (IV) (A) I < II < III < IV (B) IV < III < II < I (C) IV < I < II < III (D) IV < II < I < III 3 . The order of decreasing nucleophilicity of the following is (A) H2O > OH– > CH3COO– > CH3O– (B) CH3O– > OH– > CH3COO– > H2O (C) CH COO– > CH O–> OH– > H O (D) HO– > CH O–> CH COO–> H O 33 2 33 2 4 . The order of decreasing S 1 reactivities of the halides N CH3CH2CH2Cl CH2= CHCHClCH3 CH3CH2CHClCH3 I II III (A) I > II > III (B) II > I > III (C) II > III > I (D) III > II > I 5 . Consider the following anions. O O O– O CF3–S–O– C6H5–S–O– (III) CH3–C–O– O O (IV) (I) (II) When attached to sp3- hybridized carbon, their leaving group ability in nucleophilic substitution reactions decreases in the order (A) I > II > III > IV (B) I > II > IV > III (C) IV > I > II > III (D) IV > III > II > I 6 . The basicity of RO–,HO–, RCOO–, ROH, and H2O are of the order - (A) HO– > RO– > H O > ROH > RCOO– 2 (B) RO– > HO– > RCOO– > ROH > H O 2 (C) H O > ROH > RCOO– > HO– > RO– 2 (D) ROH > H O > HO– > RCOO– > RO– 2 7 . Which of the following are aprotic solvents : (A) DMSO (B) DMF (C) H2O (D) CH3COOH 8 . Which is/are true statements (s) : (A) Protonation increases electrophilic nature of carbonyl group (B) C F3 SO – is better leaving group t han C H 3 S O – 3 3 (C) Benzyl carbonium ion is stabilised by resonance (D) CCl3CH OH is stable,due to H-Bonding OH

Me 9. Ph C—OH SOCl2  inC5H5N H Which statement is true for the above reaction ? (A) Retention of configuration (B) Inversion of configuration (C) Inversion and Retention both (D) None 1 0 . Which of the following undergoes hydrolysis most easily Cl Cl (A) (C) (C) Cl NO2 NO2 NO2 NO2 Cl (D) NO2 NO2 1 1 . A compound 'A' formula of C H Cl on reaction with alkali can give 'B' of formula C H O or 'C' of formula 36 2 36 C H . 'B' on oxidation gave a compound of the formula C H O . 'C' with dilute H SO containing Hg2+ ion 34 362 24 gave 'D' of formula C H O, which with bromine and alkali gave the sodium salt of C H O . Then 'A' is 36 242 (A) CH CH CHCl 2 (B) CH CCl CH (C) CH ClCH CH Cl (D) CH CHClCH Cl 32 3 23 2 22 32 1 2 . Isobutyl magnesium bromide with dry ether and absolute alcohol gives (A) CH3 CH CH2OH and CH CH MgBr (B) CH3 CH CH2 CH2 CH3and Mg (OH) Br 32 CH3 CH3 (C) CH3 CH CH3 , CH CH and Mg(OH)Br (D) CH3 CH CH3 and CH CH OMgBr 2 2 32 CH3 CH3 1 3 . Following reaction is H  H CH3(CH2)5 Br OH HO (CH2)5CH3 C C H3C CH3 (A) E (B) SN1 (C) E (D) SN2 1 2 1 4 . On treatment with chlorine in presence of sunlight, toluene gives the product - (A) o-chloro toluene (B) 2, 5-dichloro toluene (C) p-chloro toluene (D) Benzyl chloride 1 5 . In SN1 reaction an optically active substrates mainly gives : (A) Retention in configuration (B) Inversion in configuration (C) Racemic product (D) No product 1 6 . Alkyl iodides can be prepared by :- (A) RCH2COOAg + I2 CCl4  RCH2I (B) RCH2Cl +NaI acetone  RCH2I + NaCl   (C) R—OH + HI  RI + H2O (D) CH4 + I2  CH3I

1 7 . Which of the following reagents can be used to prepare an alkyl halide :- (A) NaCl (B) HCl + ZnCl2 (C) SOCl2 (D) PCl5 1 8 . Which of the following reactions depict the nucleophilic substitution of C2H5Br :- (A) C2H5Br + C2H5SNa  C2H5SC2H5 + NaBr (B) C2H5Br NaC2H5OH C2H6 +HBr (C) C2H5Br + AgCN  C2H5NC + AgBr (D) C2H5Br + KOH  C2H5OH+ KBr 1 9 . For an SN2 reaction, which of the following statements are true :- (A) The rate of reaction is independent of the concentration of the nucleophile (B) The nucleophile attacks the C-atom on the side of the molecule opposite to the group being displaced (C) The reaction proceeds with simultaneous bond formation and rupture (D) None of these 2 0 . Which of the following is an SN2 reaction :- (A) CH3CH2Br + KOH  CH3CH2OH + KBr (B) CH3CH2Br + CH3CH2ONa  CH3CH2OCH2CH3 + NaBr (C) (CH3)3CBr + KOH  (CH3)3COH+ KBr CH3 CH3 (D) CH3—CH2—C—Br + KOH CH3CH2—C—OH + KBr CH3 CH3 BRAIN TEASERS ANSWER KEY EXERCISE -2 Q u e. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 A n s . A ,B,C C B C B B A ,B A ,B,C ,D B D A D D D C Que. 16 17 18 19 20 A n s . B .C B,C ,D A ,C ,D B,C A ,B

EXERCISE–03 MISCELLANEOUS TYPE QUESTIONS TRUE OR FALSE : 1 . Alkyl halides follow the reactivity sequence, R—I > R—Br > R—Cl > R—F 2 . Vinyl chloride reacts with dilute NaOH to form vinyl alcohol. 3 . Allyl chloride is more reactive than vinyl chloride. 4 . Tertiary butyl bromide undergoes SN1 reactions. 5 . Both vic- and gem-dihalides on heating with zinc dust in presence of alcohol form same alkene. FILL IN THE BLANKS : 1 . The interaction of elemental sulphur with Grignard reagent gives ...................... . 2 . An alkyl halide may be converted into alcohol by ......................reaction 3 . Diethyl ether is obtained from ethyl bromide by treating it with ...................... and the name of the reaction is .................. . 4 . Allyl chloride is a ...................... compound while vinyl chloride is inert towards nucleophilic substitution. 5 . The dihalides in which halogen atoms are attached to adjacent carbon atoms are termed as............... . 6 . Alkyl halides are formed when thionyl chloride and ............... are refluxed in presence of pyridine. 7 . Ethylene chloride on hydrolysis with aq. KOH forms............... . MATCH THE COLUMN 1 . Match the column I with column II. Column-II (Mechanism) Column-I (reaction) (A) CH3 C OH CH3 C Cl ( p ) SN1 Ph (+) H + SOCl2 Ph (+) H (B) CH3 C OH Pyridine CH3 C Cl ( q ) SN2 Ph (+) H + SOCl2 H5C6 (–) H – ( r ) SNi (C) CH3 (D) CH3 C OH  CH3 C Cl (s ) E2 H5C6 (+) +HCl C6H5 (±) H H 2 . Match the column I with column II. Column-II Column-I (Relative rate of solvolysis in 50% aqueous ethanol at 45°C) (Substrate) (p) 7700 ( A ) Cl (q) 1 ( B ) Cl (r) 91 H3C Cl (C) ( s ) 1,30,000 H3C ( D ) Ph Cl

ASSERTION & REASON QUESTION : These questions contains, Statement-I (assertion) and Statement-II (reason). (A) Statement-I is True, Statement-II is True ; Statement-II is a correct explanation for Statement-I (B) Statement-I is True, Statement-II is True ; Statement-II is NOT a correct explanation for Statement-I (C) Statement-I is True, Statement-II is False. (D) Statement-I is False, Statement-II is True. 1 . Statement-I : Iodination of akanes is carried out by heat in presence of readucing agent. Because Statement-II : Iodination of alkanes takepalce explosively. 2 . Statement-I : Chloropropane has higher boiling point than chloroethane. Because Statement-II : Haloalkanes are polar molecules. 3 . Statement-I : Polar solvent slows down SN2 reaction. Because Statement-II : CH3–Br is less reactive than CH3Cl. 4 . Statement-I : Primary benzylic halides are more reactive than primary alkyl halides towards SN1 reaction. Because Statement-II : Reactivity depends upon the nature of the nucleophile and the solvent. 5 . Statement-I : Vinylic halides are reactive towards nucleophilic substitution reaction. Because Statement-II : Reactivity is due to the polarity of carbon-halogen bond. 6 . Statement-I : Aryl halides undergo electrophilic substitution less readily than benzene. Because Statement-II : Ar yl halide gives only meta product w.r.t. electrophilic substitution. 7 . Statement-I : Optically active 2-iodibutane on treatment with NaI in acetone undergoes racemisation. Because Statement-II : Repeated Walden inversions on the reactant and its product eventually gives a racemic mixture. 8 . Statement-I : Free radical chlorination of n-butane gives 72% of 2-chlorobutane and 28% of 1-chlorobutane though it has six primary and four secondary hydrogens. Because Statement-II : A secondary hydrogen is abstracted more easily than the primary hydrogen. 9 . Statement-I : Nucleophilic substitution reaction on an optically active alkyl halide gives a mixture of enantiomers. Because Statement-II : The reaction occurs by SN1 mechanism. 1 0 . Statement-I : Boiling point of alkyl halide increases with increasse in molecular weight. Because Statement-II : Boiling point of alkylhalides are in the order RI > RBr > RCI > RF.

COMPREHENSION BASED QUESTIONS : Comprehension # 1 An organic compound A has molecular formula C10H17Br and it is non-resolvable. A does not decolourize brown colour of bromine water solution. A on treatement with (CH3)2COK/(CH3)3COH yields B as major product. B on treatment with H2/Pt yields (C10H16) which on treatment with Cl2/hv yields three monochloro derivative. Also B on boiling with acidic permanganate solution yields C(C10H16O3). C on heating with sodalime yields D (C9H16O). D on reducing with LiAlH4 followed by heating the product with concentrated H2SO4 yields E (C9H16) as major product. E on treatment with ozone followed by work-up with Zn–H2O yields 6-Ketononanal. 1 . Compound A is : Br Br Br (D) none of these (A) (B) (C) 2 . Compound B is : (A) (B) (C) (D) 3 . Compound C is : COOH O COOH OO H CH2 (D) CH2 (A) (B) (C) O C=C CH2 CH2 O O H 4 . Compound D is : O O (C) O (D) CH3 CH3 (A) (B) COOH Comprehension # 2 Nucleophilic substitution reactions generally expressed as Nu– + R — L  R— Nu + L– Where Nu–  Nucleophile ; R—L  substrate ; L  leaving group The best leaving groups are those that become the most stable ions after they depart. Since most leaving groups leave as a negative ion, the best leaving groups are those ions that stabilize a negative charge most effectively. A good leaving group should be (a) electron-withdrawing to polarize the carbon (b) stable once it has left (not a strong base) (c) polarisable- to maintain partial bonding with the carbon in the transition state (both SN1 and SN2). This bonding helps to stabilise the transition state and reduces the activation energy.

1 . Among the following which is feasible ? (A) X– + CH3—CH2—H  CH3—CH2—X + H– (B) X– + CH3—OH  CH3—X + OH (C) X–+ H3C   CH3—X + H2O —OH H (D) X– + CH3—CH3  CH3—X + CH3 2 . Among the following which is false statement ? (A) The weaker the base after the group departs, the better the leaving group (B) A reactive leaving group would raise the energy of the product, driving the equilibrium towards the reactants (C) Relative leaving group ability may vary with change of solvent (D) Better leaving group only increases SN2 rate, not SN1. 3. CH3Br CH3F CH3OH CH3OSO2CF3 (I) ( II ) (III) (IV ) The correct order of decreasing reactivity of the above compounds towards CH3O– in an SN2 reaction is : (A) I > IV > II > III (B) IV > I > II > III (C) IV > I > III > II (D) IV > II > I > III 4. Cl– CH3O– CH3S – I– (I) (II) ( II ) (IV ) The correct order of increasing leaving group capability of above anoins (A) III < IV < II < I (B) II < III < I < IV (C) II < IV < III < I (D) I < III < II < IV Comprehension # 3 Nucleophilic aliphatic substitution reaction is mainly of two types : SN1 and SN2. The SN1 mechanism is a two step process. Reaction velocity of SN1 reaction depends only on the concentration of the substrate. Since product formation takes place by the formation of carbocation, optically active substrate gives (+) and (–) forms of the product. In most of the cases the product usually consits of 5-20% inverted product and 80-95% racemised species. The more stable the carbocation, the greater is the proportion of racemisation. In solvolysis reaction, the more nucleophilic the solvent, the greater is the proportion of inversion. 1 . Which one of the following compound will give SN1 reaction predominantly ? CH3 (B) H3C (C) H3C—Br (D) All of these (A) H5C6——–Br Br CH3 2 . Which of the following compounds will give SN1 and SN2 reactions with considerable rate ? I. C6H5—CH2—Br II. CH2=CH—CH2—Br III. CH3—CH(Br)CH3 IV. H3C CH3 CH3 Br Select the correct answer from the codes given below (A) I, II and III (B) I, II and IV (C) II, III and IV (D) I, III and IV

3 . For the given reaction R1 R1 R——–X HOH R——–OH R2 R2 Which substrate will give maximum racemisation ? CH3 H2C CH3 (A) H5C6——–Br (B) Br C2H5 C2H5 Br —OCH3 Br —NO2 H5C6——– H5C6——– (C) (D) CH3 + NH3 MISCELLANEOUS TYPE QUESTION ANSWER KEY EXERCISE -3  True / False 2. F 3. T 4. T 5. F 1. T  Fill in the Blanks 1. thioalcohol 2. nucleophilic substitution 3. sodium ethoxide, williamson's synthesis 4. reactive 5. vic-dihalides or alkylene halides 6. alcohols 7. glycol  Match the Column 1. (A) r ; B q ; (C)  s ; (D) p 2. (A) q ; (B) r ; (C)  s ; (D) p  Assertion - Reason Questions 1. D 2. B 3. C 4. B 5. D 6. C 7. A 8. A 9. A 10. B 4. (C)  Comprehension Based Questions 4. (B) Comprehension #1 : 1. (A) 2. (C) 3. (A) Comprehension #2 : 1. (C) 2. (D) 3. (B) Comprehension #3 : 1. (A) 2. (A) 3. (C)

EXERCISE–04 [A] CONCEPTUAL SUBJECTIVE EXERCISE 1 . Arrange the following compounds in order of : Ph Decreasing S 1 reaction rate : N Ph Ph Cl Cl Cl Cl (I) (II) (III) (IV) 2 . Select the member of each pair that shows faster rate of S 2 reaction with KI in acetone. N (a) CH –CH –CH –CH –Cl and CH3–CH–CH2–Cl 322 2 CH3 (I) (II) (b) CH –CH –CH –Cl and CH –CH –CH –Br 322 322 (I) (II) (c) CH3–CH–CH2–CH2–Cl CH3 CH3 and CH3–C–CH2Cl (I) CH3 (II) Br CH3 Br (d) CH3—CH2—CH2—CH—CH3 and CH3—CH—CH2—CH—CH3 (I) (II) 3 . Of the following statements which are true for S 1 reaction. N (a) Tertiary alkyl halides react faster than secondary. (b) The absolute confuguration of the product is opposite to that of the reactant when an optical active substrate is used. (c) The reaction shows first order kinetics. (d) The rate of reaction depends markedly on the nucleophilicity of the nucleophile. (e) The mechanism is two step. (f) Carbocations are intermediate. (g) Rate  [Alkyl halides] (h) The rate of the raction depends on the nature of the leaving group. 4 . Of the following statements, which are true for S 2 reaction. N (a) Tertiary alkyl halides reacts faster than secondary. (b) The absolute configuration of product is opposite to that of the reactant when an optically active substrate is used. (c) The reaction shows first order kinetics. (d) The rate of the reaction depends markedly on the nucleophilicity of the attacking reagent. (e) The mechanism is one step.

(f) Carbocations are intermediate. (g) Rate  [Alkyl halides] (h) The rate of the raction depends on the nature of the leaving group. 5 . Arrange the isomers of molecular formula C H Cl in order of decreasing rate of reaction with sodium iodide 49 in acetone. 6 . There is an overall 29-fold difference in reactivity of 1-chlorohexane, 2-chlorohexane towards potassium iodide in acetone. Which one is the most reactive ? why ? CH2CH = CH2 7 . Identify the product when A reacts with (A) (a) Br /Fe (b) Br /CCl (c) NBS (d) HBr 2 24 8. Identify major product in the following : ClCH CHCl O H – CH =CCl 22 22  EtO– (b) CH CH CH CH Br EtO– (a) CH3CH2CHCH3 3222 Br EtO– CH3 CH2CH2CHCH3 (c) Br (d) H3CCCH2CH2Br (CH3)3COK CH3 (CH3)3COH CH2Cl EtO– (e) EtOH 9 . Which is faster in the following pairs of halogen compounds undergoing S 2 reactions? N (a) Cl and Cl (b) I and Cl Cl Br (c) and (d) Br and Cl 1 0 . R – Mg – Br (A) on reaction with H O forms a gas (B), which occupied 1.4 L/g at NTP. What is product 2 when R – Br reacts with benzene in presence of AlCl ? 3

CONCEPTUAL SUBJECTIVE EXERCISE ANSWER KEY EXERCISE -4(A) 1 . IV > I > II > III 2 . (a) I (b) II (c) I (d) I 3 . (a) T (b) F (c) T (d) F (e) T (f) T (g) T (h) T 4 . (a) F (b) T (c) F (d) T (e) T (f) F (g) F (h) T 5 . 1° > 2° > 3° Anion of acetic acid is more stabilised by resonance than phenoxide ion. 6 . 1-chlorohexane Because it follows Sn2 path. CH2CH = CH2 7 . (a) and para Br Ortho Br CHCH= CH2 (c) Br CH2CHCH2 (b) Br CHCH2CH3 CH=CHCH2CH3 (d) Br (1, 2-H– shift gives more stable benzylic carbocation) 8 . (a) CH CH = CHCH (b) CH CH = CHCH (c) 3 33 3 CH3 CH3 (d) H3CCCH= CH2 (e) CH3 9 . Ease of backside attack (less steric hindrance) decides which undergoes S 2 faster (except in (b) in which N iodide is better leaving group). In all cases first one is fater than the other for S 2 reaction. N 1 0 . Gas B is CH , hence A is CH MgBr. CH Br forms CH3 on reaction with benzene. 4 33

EXERCISE–04 [B] BRAIN STORMING SUBJECTIVE EXERCISE 1 . Explain the following observations: (a) Azide ion (N –) react with 2- bromopentane thousand times faster than with neopentyl bromide in a 3 S 2 reaction though former is a secondary halide while latter is primary. N (b) What will happen to the stereochemistry of product of the following reaction:\\ Br H CH3 + N– SN2 3 D (c) What will happen to the rate if the concentration of alkyl bromide in (b) is doubled? (d) What will happen to the rate if the concentration of azide ion in (b) is doubled? (e) How the sign of optical rotation of reactant and product are related in (b) (f) When allowed to stand in dilute H SO , laevo-rotatory 2-butanol slowly loses optical activity. 24 2 . Provide structure of major product in the following reaction indicating stereochemistry where appropriate: D D (a) C OH TsCl NaCN (b) C OH HBr NaCN CH3 H CH3 H Heat 3 . Propose mechanism of the following reactions: Cl O (a) O + CH3O– CH3OH O O Br CH3O– O (b) O O O OH (c) + HO– HO O 4 . Which of the following alkyl halide could be successfully used to synthesize Grignard reagent and why other fail? Br N Br HO (II) (I) O Br OH H2N Br (III) (IV)

5 . An alkyl bromide A has molecular formula C H Br and four different structures can be drawn for it, all 8 17 optically active. A on refluxing with ethanolic KOH solution yields only one elimination product B(C H ) 8 16 which is still enantiomeric. B on treatment with H /Pt yields C(C H ) which does not rotate the plane 2 8 18 polarized light, B on ozonolysis followed by work-up with H O yields D(C H O) as one product which 22 7 14 is still resolvable. Deduce structures of A to D. 6 . Identify A to G in the following. (a) Br2CCl4 A KCN B H3O+ C (b) O Br2CCl4 D KCN E H3O+ G 7 . Br Mg/ether D2O Na/ether Cl B CD (A) Na/ether HCHO/H3O+ (with two mol of A) F Identify B to F 8 . Vinyl chloride does not give S reaction but allyl chloride gives. Explain. N 9 . Arrange the following in the increasing order of their ability as a leaving group: (a) CH S–, CH O–, CF – and F– 333 (b) CF SO –, CH SO – and CH COO– 33 33 3 1 0 . RBr when treated with AgCN in a highly polar solvent gives RNC whereas when it is treated with NaCN it gives RCN. Explain.

BRAIN STORMING SUBJECTIVE EXERCISE ANSWER KEY EXERCISE -4(B) 1 . (a) Though neopentylbromide is primary, bulky tertiary butyl group possess very large steric hindrance to the attack of bulky nucleophile N –. 3 Br N3 H (b) H CH3+ N– SN2 CH3 3 DD (c) Rate will double (d) Rate will double (e) not related (f) Recemization occur through carbocation intermediate D DD 2. C CH3 (b) C CN + C (a) H3C H NC CH3 NC H H Cl O– O O O + Cl– 3 . (a) O – + OCH3 Cl (b) Br – Br OCH3 + OCH3 O– OCH3 O –Br– O O O– O O– O– (c) + O–H O– O HO HO Product H2O HO O O 4 . Only II can be used for successful synthesis of Grignard reagent, rest all contain acidic proton and will react with R– (from Grignard reagent) forming alkane. Br BC O 5. D CN A (B) 6 . (A) Br CN Br

COOH O O Br CN (C) (D) (E) COOH Br CN O COOH O (F) (G) COOH COOH (decarboxylation takes place on heating when there is a keto group at - position) 7 . (B) Cl Mg– Br (C) Cl D (D) D D (E) Cl Cl (F) Cl CH2OH 8 . In Vinyl chloride, C – Cl bond is stable due to resonance (as in chlorobenzene) CH2 •••• + CH Cl CH2 CH Cl •• Hence S reaction in which Cl is replaced by nucleophile is not possible. In addition to this, sp2- hybridised N carbon is more acidic than sp3- carbon, hence removal of proton (H+) is easier than removal of halide (Cl–) In allyl chloride, S reaction is easier since allyl carbocation formed after removal of Cl– is stabilised by N resonance. CH2 CHCH2Cl CH2 + CHCH2+Cl– Allyl carbocation + + CH2 CH CH2 CH2 CH CH2 9 . (a) CF – < CH O– < CH S– ; (b) CH COO– < CH SO – < CF SO – 33 3 3 33 33 1 0 . As [CN]– is an ambident nuicleophile which ahve two nucleophile which have two nucleophilic sites and can attack from either side. In a highly polar solvent, AgCN promotes the formation of carbocation R+, precipitation of AgBr. •• C •• •• C N R—BR + Ag+ [CN–] N R+ + CN– + Ag Br  fast R–N+ C – •• •• slow In the absence of such promotion by Ag+, with Na+[CN]–, the resulting SN2 reaction is found to proceed with preferential attack on the atom in the nucleophile which is more polarisable i.e. C. NC–+R – Br [NC –....R....Br  ]  N  C – R + Br– Transition State

EXERCISE–05 PREVIOUS YEARS QUESTIONS 1 . Chlorination of toluene in the presence of light and heat followed by treatment with aqueous NaOH gives- (A) o-cresol (B) p-cresol [IIT-90] (C) 2, 4 dihydroxytoluene (D) Benzoic acid 2 . Aryl halides are less reactive towards nucleophilic substitution reaction as compared to alkyl halides due to- (A) The formation of less stable carbonium ion (B) Resonance stabilization [IIT-90] (C) The inductive effect (D) sp2 hybridised carbon attached to the halogen 3 . 1-Chlorobutane on reaction with alcoholic potash gives - [IIT-91] (A) 1–butene (B) 1–butanol (C) 2–butene (D) 2–butanol 4 . In the addition of HBr to propene in the absence of peroxides, the first step involves the addition of - (A) H+ (B) Br– (C) H (D) Br [IIT-93] 5 . Arrange the following compounds in order of increasing dipole moment : [IIT-96] (I) Toluene ; (II) m-dichlorobenzene ; (III) o-dichlorobenzene ; (IV) p-dichlorobenzene (A) I < IV < II < III (B) IV < I < II < III (C) I < IV < II < III (D) IV < II < I < III 6 . In the reaction of p-chloro toluene with KNH2 in liq. NH3, the major product is - [IIT-97] (A) o-toluidine (B) m-toluidine (C) m-chloroaniline (D) p-chloroaniline 7 . (CH3)3CMgCl reaction with D2O produces : IIT-97] (A) (CH3)3CD (B) (CH3)3OD (C) (CD3)3CD (D) (CH3)3OD 8 . The intermediate during the addition of HCl to propene in presence of peroxide is - [IIT-97] (A) CH3CHCH2Cl + (C) CH3CHCH2 (D) CH3CH2CH2 (B) CH3CHCH3 9 . The number of possible enantiomeric pairs that can be produced during monochlorination of isopentane- (A) 2 (B) 3 (C) 4 (D) 1 [IIT-97] 1 0 . During debromination of meso-2,3-dibromobutane, with Zn dust/CH3COOH the major compound formed is - [IIT-97] (A) n-butane (B) 1-butene (C) cis-2-butene (D) trans-2-butene 1 1 . Benzyl chloride (C6H5CH2Cl) can be prepared from toluene by chlorination with [IIT-98] (A) SO2Cl2 (B) SOCl2 (C) PCl5 (D) NaOCl 1 2 . Toluene, when treated with Br2/Fe, gives o and p-bromotoluene, because the CH3 group - (A) is ortho and para directing (B) is meta directing [IIT-99] (C) deactivates the ring by hyperconjugation (D) deactivates the ring 1 3 . A solution of (+) 2–chloro–2phenylethane in toluene racemises slowly in the presence of small amount of SbCl5 due to the formation of - [IIT-99] (A) Carbanion (B) Carbene (C) Free-radical (D) Carbocation 1 4 . The order of reactivity of the following alkyl halides for a SN2 reaction is - [IIT-2000] (A) RF > RCl > R –Br > R–I (B) R–F > R –Br > R–Cl > R–I (C) R–Cl > R –Br > R–F > R–I (D) R–I > R –Br > R–Cl > R–F 1 5 . Which of the following has the highest nucleophilicity : [IIT-2000] (A) F– (B) OH– (C) CH3– (D) NH2– 1 6 . An SN2 reaction at an asymmetric carbon of a compound always gives : [IIT-2001] (A) an enantiomer of the substrate (B) a product with opposite optical rotation (C) a mixture of diastereomers (D) a single stereoisomer

1 7 . The number of isomer for the compound with molecular formula C2BrCl FI is - [IIT-01] (A) 3 (B) 4 (C) 5 (D) 6 1 8 . In the presence of peroxide, hydrogen chloride and hydrogen iodide do not give anti-Markovnikov's addition to alkenes because : [IIT-01] (A) both are highly ionic (B) one is oxidising and the other is reducing (C) one of the steps in endothermic in both the cases (D) all the steps are exothermic in both the reactions 1 9 . Identify the set of reagents/reaction conditions 'X' and 'Y' in the following set of transformations -[IIT-02] CH3–CH2–CH2Br X  Product Y  CH3–CH–CH3 Br (A) X = dilute aqueous NaOH, 20°C, Y= HBr/acetic acid, 20°C (B) X = concentrated alcoholic NaOH, 80°C; Y = HBr/acetic acid, 20° (C) X = dilute aqueous NaOH 20°, Y = Br2/CHCl3, 0°C (D) X = concentrated alcoholic NaOH, 80°C; Y = Br2/CHCl3, 0°C 20. F NO2 (CH3)2NH (A) (i) NaNO2+ HCl 0°–5°C (B) [IIT-03] DMF (ii) H2/Catalytic Reduction (A) O2N NH2 H3C N NH2 (B) H3C H3C N NO2 H3C (C) H2C (D) N NO2 H3C NH2 2 1 . MeO CH3H CH3 NO2 on hydrolysis in presence of acetone : [IIT-05] H Cl CH3 (K) MeO CH3H CH3 NO2 (L) MeO H H CH3 NO2 H OH CH3 OH CH3 CH3 (M) MeO CH3H CH3 NO2 (A) K & L H CH3 OH (C) M only (D) K & M (B) only L