ORGANIC CHEMISTRY
ALKYL HALIDE
1 1.
Which alkyl halide will react fastest with aqueous ethanol ethanol ? (A) (CH3)3C – Br (B) (CH3)3C – Cl (C) (CH3)2CH – Br (D) CH3 – CH2 – CH2 – Br
2.
When the concentration of alkyl halide is tripled and the concentration of OH¯ ion is reduced to half, the rate of SN2 reaction increases by: (A) 3 times (B) 2 times
(C) 1.5 times
3.
An S N2 reaction at an asymmetric asymmetri c carbon of a compound always always gives: (A) an enantiomer of the substrate (B) a product with opposite optical rotation (C) a mixture of diastereomers (D) a single stereoisomer
4.
In the given reaction, CH3CH2 – X + CH3SNa The fastest reaction occurs when ‘X’ is (A) – OH (B) – F (C) – OCOCF3
5.
(D) OCOCH3
2
Which of the following involv es an “ SN ” route of reaction: Na
6.
(D) 6 times
(A) CH3Cl ether
(B) CH3MgCl + CH3Cl
(C) CH3Cl + LiAlH4
(D) All
The compound which undergoes fastest reaction with aq. KOH soluti on is : (SN2) (A) C 6H5 CH OCH3 (B) C6H5 CH CH3 (C) C6H5 – CH2 – CH2 – Cl (D) C6H5 CH CH2 CH3 | Cl
7.
8.
| Cl
| Cl
The rate of S N1 reaction is fastest with: (A)
(B)
(C)
(D)
In the following foll owing reaction the most probable product will be:
(A)
(B)
(C)
Alkyl Halide
(D)
1 1.
Which alkyl halide will react fastest with aqueous ethanol ethanol ? (A) (CH3)3C – Br (B) (CH3)3C – Cl (C) (CH3)2CH – Br (D) CH3 – CH2 – CH2 – Br
2.
When the concentration of alkyl halide is tripled and the concentration of OH¯ ion is reduced to half, the rate of SN2 reaction increases by: (A) 3 times (B) 2 times
(C) 1.5 times
3.
An S N2 reaction at an asymmetric asymmetri c carbon of a compound always always gives: (A) an enantiomer of the substrate (B) a product with opposite optical rotation (C) a mixture of diastereomers (D) a single stereoisomer
4.
In the given reaction, CH3CH2 – X + CH3SNa The fastest reaction occurs when ‘X’ is (A) – OH (B) – F (C) – OCOCF3
5.
(D) OCOCH3
2
Which of the following involv es an “ SN ” route of reaction: Na
6.
(D) 6 times
(A) CH3Cl ether
(B) CH3MgCl + CH3Cl
(C) CH3Cl + LiAlH4
(D) All
The compound which undergoes fastest reaction with aq. KOH soluti on is : (SN2) (A) C 6H5 CH OCH3 (B) C6H5 CH CH3 (C) C6H5 – CH2 – CH2 – Cl (D) C6H5 CH CH2 CH3 | Cl
7.
8.
| Cl
| Cl
The rate of S N1 reaction is fastest with: (A)
(B)
(C)
(D)
In the following foll owing reaction the most probable product will be:
(A)
(B)
(C)
Alkyl Halide
(D)
9.
In the following reaction :
NaI Acetone
‘X’ [The structure of ‘X’ can be]
(A)
10. 10.
(B)
(C)
(D)
The following foll owing reaction is described as: as: NaOH
(A) S N1 with racemisation (B) intramolecular SN2 with walden inversion (C) intramolecular SN2 with retention of configuration (D) intermolecular S N1 with recemisation 11.
12. 12.
The insecticide chlordane is warmed with dilute NaOH solution for some time. The expected product would be :
(A)
(B)
(C)
(D)
Which Whic h configuation will be adopted by the poduct at cabon atoms marked (1) and and (2) respectively in the given reaction. Ether + CH3 CH2 CH CH3 | (D) Br
(A) D, D
(B) D, L
(C) L, L
Alkyl Halide
(D) L, D
13.
Which of the following will undergo fastest elimination reaction with alcoholic KOH. (A)
14.
(B)
(D)
When all-cis isomers of C6H6Cl6 (1, 2, 3, 4, 5, 6 - Hexachlorocyclohexane) is heated with alc. KOH, the most probable product is :
(A)
15.
(C)
(B)
(C)
(D)
The most probable product in the following reaction is :
Trans (A)
16.
(B)
(C)
(D)
Which product(s) will be formed in the following reaction :
alc. KOH
(A) No reaction 17.
(B) Trans alkene
(C) cis alkene
(D) cis + trans alkene
What will be the major product of the following reaction CH3 CH3 | | CH3 C C CH CH3 | | | H CH3 Br CH3 CH3 | | (A) CH3 C C | | OCH3 H CH3 | (C) CH3 C | H
CH3 | C CH3 | H
CH3 CH3 | | C C CH3 | | OCH3 H
CH3OH, 30C
CH3 CH3 CH3 | | | (B) CH3 C C CH CH3
CH3 | (D) CH3 C | H
Alkyl Halide
CH3 | C CH CH2 | CH3
18.
Which of the following reaction will undergo an elimination reaction and an alkene will be formed in the products.
Zn dust,
(A)
(B)
alc. KOH,
(C)
conc H SO ,
2 4
(D)
19.
When (1R, 2R)-1, 2-dibromo-1, 2-diphenyl ethane is treated with alcoholic solution of KOH, the most probable product would be : (A) trans-1, 2-diphenyl ethene (B) A mixture of cis, trans alkenyl bromide (C) cis-alkenyl bromide (D) trans-alkenyl bromide
20.
Which of the following products is formed when m-bromochlorobenzene is treated with magnesium in the presence of ether ?
(A)
21.
(B)
(C)
(D)
(C) C2H5SH
(D) none of these
The end product Y of the reaction is : H O
(A) C2H6
Y C2H5MgBr + S X 3 (B) C2H5SMgBr
swers EXERCISE-1 1.
A
2.
C
3.
D
4.
C
5.
D
6.
A
7.
A
8.
B
9.
A
10.
B
11.
D
12.
B
13.
A
14.
B
15.
C
16.
C
17.
C
18.
B
19.
D
20.
A
21.
C
Alkyl Halide
2 1.
Predict the compound in each pair that will undergo the SN2 reaction faster. (a) (i)
(ii)
(c) (i)
(e) (i) 2.
(b) (i)
(ii)
(ii)
(ii)
(d) (i)
(ii)
(f) (i)
(ii)
Predict the compound in each pair that will undergo solvolysis (in aqueous ethanol) more rapidly. (a) (i) (CH3CH2)2CH – Cl (ii) (CH3)3C – Cl (b) (i)
3.
(ii)
(c) (i)
(ii)
(d) (i)
(ii)
(e) (i)
(ii)
(f) (i)
(ii)
The correct decreasing order of relativ e reactivity of the following bromides towards hot aqueous KOH solution is: (I) CH3 – CH2 – Br (II) CH3 – CH2 – CH2 – Br (III) OH – CH2 – CH2 – Br
(IV) CH3 –
(A) I > III > II > IV
(C) I > II > IV > III
(B) I > II > III > IV
– CH2 – Br (D) III > I > II > IV
4.
For a given alkyl halide, the ease of formation of Grignard reagent is: (A) RI > RBr > RCl (B) RI > RBr < RCl (C) RI < RBr < RCl (D) RI > RCl > RBr
5.
The correct order of reactivity of the following halides for bimolecular elimination is : (I) (CH3)2 CH – CHI – CH3 (II) (CH3)2CH – CH2 – CH2 – I (III) CH3 – CH2 – CHI – CH3
(IV) (CH3 )2 C CH2 CH3 | I
(A) II > III > I > IV
(C) I > II > III > IV
(B) III > II > I > IV
Alkyl Halide
(D) IV > I > III > I
6.
The relative reactivity of following halides towards ethanol follows the order: (I) CH3 –
– Cl
(A) II > IV > III > I 7.
(II) CH3CH2Cl
(III) CH3 – O – CH2 – Cl (IV) C6H5 – CH2 – Cl
(B) I > IV > III > II
(C) IV > III > II > I
The increasing order of reactivity of the following isomeric halides with alcoholic AgNO3 is : (I) C6H5 – CH = CH – CH2 – CH2 – Cl
8.
9.
10.
(D) I > III > IV > II
(II) C 6H5 CH CH CH CH3 | Cl
(III) C 6H5 C CH CH2 CH3 | Cl
CH3 | (IV) C 6H5 C CH CH2 | Cl
(A) III < IV < II < I
(C) III < I < II < IV
(B) I < III < IV < II
(D) I < II < IV < III
Rate of S N1 reaction is :
(I)
(II)
(III)
(A) IV > I > III > II
(B) II > III > I > IV
(C) I > III > II > IV
(IV) (D) IV > II > III > I
Rate of reaction with aqueous AgOH solution follows the order.
(I)
(II)
(III)
(IV)
(A) II > IV > I > III
(B) IV > II > I > III
(C) III > II > IV > I
(D) I > IV > II > III
In the following compound, arrange the reactivity of different chlorine atoms towards NaSH in decreasing order in non aqueous solvent.
(A) II > IV > I > III (C) II > IV > I > III 11.
(B) I > II > III > IV (D) II > III > I > IV
The reactivity order of CH3MgBr with the following reagents is (I) PhOH
(A) I > III > II > IV (C) IV > I > III > II
(II) CH3CHO
(III) R–C CH
(B) I > II > III > IV (D) I > III > IV > II
Alkyl Halide
(IV) CH3 C Cl || O
Correct decreasing order of reactivi ty to wards SN2 reaction : Q. 12 to Q. 18
I
II
III
IV
12.
CH3Cl (A) I > II > III > IV
CH3CH2Cl (B) III > II > I > IV
CH3CH2 –CH2 –Cl (C) IV > III > I > II
CH3CHClCH3 (D) None
13.
CH3CH2 CH2Cl (A) IV > I > II > III
CH3 CH2CH2Cl (B) III > II > I > IV
CH3CH2CH2CH2Cl (C) IV > I > III > II
CH3CH2CH2 (D) II > I > IV > III
CH3 | CH3 C CH2 Cl | CH3
CH3 | CH3 – C – Cl | CH3
CH3 | CH3 C Cl | H
CH3 – CH2 – Cl
(A) II > III > I > IV
(B) II > I > IV > III
(C) IV > III > I > II
(D) I > III > II > I
Br – CH2 – CH2 – Br
N C – CH2 – Br
CH3 CH CH3 | Br
CH3 – CH2 – Br
(A) II > I > IV > III
(B) II > III > I > IV
(C) II > I > III > IV
(D) IV > I > II > III
16.
CH3CH2Cl (A) II > III > I > IV
CH2 = CH – CH2Cl (B) II > I > IV > III
CH3Cl (C) IV > III > II > I
CH2 = CHCl (D) none
17.
CH3 –O–CH2 –Cl
H C CH2 Cl || O
CH3 –CH –Cl 2
CH3 –Cl
(A) IV > III > II > I
(B) II > III > I > IV
(C) II > I > IV > III
(D) none
HO–H2C–CH2 –Cl
CH3 –CH2 –Cl
CH3 –CH2 –CH2 –Cl
CH3 C Cl || O
(A) IV > I > II > III
(B) IV > I > III > II
(C) II > I > IV > III
(D) II > III > I > IV
14.
15.
18.
Arr ang e the f ol lo wi ng in decreasi ng order of react iv it y tow ards SN 1 reaction : (19 to 29)
I 19.
20.
21.
II
III
IV
CH3Cl
CH3CH2Cl
CH3CH2 –CH2 –Cl
CH3CHClCH3
(A) IV > I > II > III
(B) I > II > III > IV
(C) IV > III > II > I
(D) III > IV > I > II
CH3 –CH2 –CH2 –Cl
CH3 | CH3 CH CH2 Cl
CH3 | CH3 C CH2 Cl | CH3
CH3 | CH3 C Cl | CH3
(A) IV > III > II > I
(B) I > II > III > IV
(C) III > II > I > IV
(D) IV > I > II > III
CH3CH2 CH2Cl (A) IV > III > II > I
CH3 CH2CH2Cl (B) IV > I > II > III
CH3CH2CH2CH2Cl (C) I > II > III > IV
CH3CH2CH2 (D) IV > II > I > III
(A) I > II > III > IV
(B) IV > III > II > I
(C) IV > II > I > III
(D) IV > I > II > III
22.
Alkyl Halide
I
II
III
IV
23.
(A) IV > III > I > II
(B) IV > I > II > III
(C) I > II > III > IV
(D) II > I > III > IV
(A) III > II > I > IV
(B) IV > I > III > II
(C) I > II > III > IV
(D) IV > III > II > I
CH2=CH–CH2 –Cl
CH3 | CH2 CH CH Cl
(A) IV > III > I > II
(B) II > III > I > IV
24.
25.
CH3 –CH=CH–CH2 –Cl CH3 C CH CH3 | Cl (C) I > II > IV > III (D) III > II > I > IV
CH3 | C6H5 C Br | CH3
O || CH3 C CH Br | CH3
(CH3)2CHBr
(B) I > II > IV > III
(C) II > IV > I > III
(D) III > IV > I > II
(C6H5)2CHCl
CH2 = CH CH2Cl
C6H5 CH2Cl
(A) I > III > II > IV
(B) IV > III > II > I
(C) II > IV > III > I
(D) I > II > III > IV
(A) I > II > III
(B) II > III > I
(C) II > I > III
(D) III > II > I
(A) I > II > III > IV
(B) IV > III > II > I
(C) II > III > I > IV
(D) IV > II > III > I
26.
(A) II > I > IV > III
27.
28.
29.
swers EXERCISE-2 1.
(a) (i) > (ii) (d) (ii) > (i)
(b) (ii) > (i) (e) (ii) > (i)
(c) (ii) > (i) (f) (i) > (ii)
2.
(a) (ii) > (i) (d) (ii) > (i)
(b) (ii) > (i) (e) (ii) > (i)
(c) (i) > (ii) (f) (ii) > (i)
3. D
4. A
5. D
6. D
7. C
8. D
9. B
10. C
11. D
12. A
13. B
14. C
15. A
16. A
17. C
18. A
19. C
20. A
21. B
22. D
23. C
24. A
25. B
26. A
27. C
28. C
29. D
Alkyl Halide
3 The major p roduct of the foll owin g reactio ns (Q.1 - Q.4)
(A)
2.
+
1.
CH3 I
(B)
(C)
(D)
(C)
(D)
+
iodomethane (methyl iodide)
(A)
3.
4.
(B)
CH3CH2CH2CH2CH2Br + 1 bromopen tan e
(A) CH3 – CH2 – CH2 – CH2 – CH3
(B) CH3 – CH2 – CH2 – CH2 – CH2 – SH
(C) CH3 CH2 CH CH2 CH3 | SH
(D) CH3 CH CH2 CH2 CH3 | SH
CH3CH2CH2CH2Cl + : NH3 1 chlorobu tane ammonia (n butyl chloride ) ( excess )
(A) CH3 – CH2 – CH2 – CH3
(B) CH3 CH2 CH CH3 | NH2
(C) CH3 CH2 CH CH3 | Cl
(D) CH3 – CH2 – CH2 – CH2 – NH2
Predict the products of t he following S N2 reacti on s (Q. 5 - Q. 12) 5.
CH CH OH
3 2 CH3CH2ONa + CH3CH2Cl (A) CH3 – CH2 – CH2 – O – CH3
(C) CH3 – CH2 – CH2 – CH2 – OH
(B) CH3 – CH2 – O – CH2 – CH3 (D) CH3 CH2 CH CH3 | OH
Alkyl Halide
acetone
+ NaCN
6.
(A)
(B)
(C)
(D)
7.
(A)
8.
(B)
(C)
(D)
acetone
CH3(CH2)8CH2Cl + NaI (A) CH3 (CH2 )7 CH CH3 |
(B) CH3 (CH2 ) 8 CH2 I
(C) CH3 – (CH2)8 – CH3
(D) CH3 – (CH2)7 – CH = CH2
I
+ CH3I
9.
(A)
10.
(B)
(C)
(D)
(CH3 )3 C CH2CH2Br + NH3 excess (A) (C) (CH3)3C – CH2 – CH2 – CH3
(B) (CH3)3C – CH2 – CH2 – NH2 (D) [(CH3)3C – CH2 – CH2]4N
+ NaOH
11.
(A)
(B)
(C)
NaOH
12.
CH3OH
(A)
(B)
(C)
(D)
Alkyl Halide
(D)
ether + Mg
13.
(A)
14.
(B)
(C)
CH I
CuI 3 (CH3)2 CHCH2I Li
(A) (CH3)2CH – CH3 (C) (CH3)2CH – CH2 – CH2I 15.
(D)
(B) (CH3)2CH – CH2 – CH3 (D) None of these ( C2H5 )2 CuLi
CH3 CH2 CH C CH3 | Br (A) CH3 CH2 CH C CH3 | C 2H5
(B) CH3 CH2 CH C CH3 + C2H5Cl + LiBr | C 2H5
(C) CH3 – CH2 – CH = CH – CH 3 + 2C2H5Cu
(D) CH3 CH2 CH C CH3 LiBr | C2H5
( i) Et 2O
+ (CH3)2CH Mg Br ( ii) H O
16.
2
17.
18.
(A) CH3 – (CH2)4 – CH2 – OH
(B) CH3 CH CH CH CH3 | C 2H5
(C)
(D) CH3 CH CH2 CH3 | OH
C6H5COOH + CH3MgI (A) C6H5COOMgI (B) CH4
(C) Both A & B
(D) none
H2O
2-Butanone + Ethyl Magnesium bromide H | CH2 CH3 CH C (A) 3 | OH
CH3 | (B) CH3 C CH2 CH3 | OH
C 2H5 | (C) CH3 C CH2 CH3 | OH
(D) All of these
Alkyl Halide
The major product of the following reactions (Q.19 - Q.23) 19.
CH3CH2CH2 – Br + CH3O¯ (A) CH3 – CH2 – CH2 – OH & SN2 (C) CH3 – CH2 – CH2 – O – SN1
20.
(B) CH3 – CH2 – CH2 – O – CH3 & SN2 (D) CH3 – CH2 – O – CH2 – CH3 & E2 50C
CH3 – CH2 – CH2 – Br + (CH3)3CO¯
(CH3 )3 COH
(A) CH3 – CH2 – CH3 & SN2 (C) CH3 – CH = CH2 & E2
(B) CH3 – CH = CH2 & SN1 (D) None
(A) CH3 – CH2 – CH2 – CH2 – SH & SN1
CH3 | (B) CH3 C SH & SN2 | H
(C)
(D)
21.
22.
& E2
(CH3CH2)3 C – Br + OH¯
(A)
CH2 CH3 | (B) CH3 C CH3 & SN2 | CH2 CH3
& E2
(C) CH3 – CH = CH – C2H5 23.
& SN2
(D) None
(CH3CH2)3 C – Br (A) CH3OC (CH2CH3)3 & SN2 (C) CH3OC (CH2CH3)3 & SN1
(B) CH3OCH (CH2CH3)2 & SN1 (D) None
swers EXERCISE-3 1.
A
2.
C
3.
B
4.
D
5.
B
6.
A
7.
D
8.
B
9.
C
10.
A
11.
C
12.
B
13.
D
14.
B
15.
B
16.
C
17.
C
18.
C
19.
B
20.
C
21.
D
22.
A
23.
C
Alkyl Halide
4 Which of the following is the correct option of reagents for the given conversion : (Q.1 - Q.6) 1.
2.
3.
Propane 2-Iodopropane (A) I2 / h (C) F2 / h, NaI / acetone
(B) NaI / acetone (D) Br2 / h, NaI / acetone
CH3 – C C – H CH3 – CH2 – CH3 (A) Br 2 / h, alc. / KOH, Br 2 / h, NaNH2 (C) Br 2 / h, NaNH2
(B) Br 2 / CCl4, alc. / KOH, Br 2 / h, NaNH2 (D) Br 2 / h, aq. KOH, Br 2 / h, NaNH2
CH3 | CH3 C D CH3 – CH2 – CH3 | H
(A) Br 2 / h, NaNH2, D2 / Ni (B) Br 2 / h, Mg / Et2O, D2O (C) Br 2 / h, Mg / Et2O, CH3 – Br, Br 2 / h, D2O (D) Br 2 / h, Mg / Et2O, C2H5Br, Br 2 / h, D2O
4.
CH3 | CH3 CH COOH
CH4
(A) Br 2 / h, CH3CH2MgCl, Br 2 / h, HCOOH (B) Cl2 / h, CH3CH2MgCl, Br 2 / h, KCN, H3O+ (C) Br 2 / h, CH3CH2MgCl, Br 2 / h, NaNH2, HCN, H3O+ (D) Cl 2 / h, CH3MgBr, Br 2 / h, CH3 – COOH
5.
O || OHC (CH2 )3 C CH3
(A) Mg / Et2O, CH3Cl, Br 2 / h, alc. KOH, hot KMnO4 (B) OH¯ , H2SO4 / , O3 / Zn, H2O (C) CH3MgBr, Br 2 / h, alc. KOH, hot KMnO4 (D) Mg / Et2O, CH3Cl, Br 2 / h, alc. KOH, O3
6.
(A) alc. KOH, Br 2 / CCl4, NaNH2, Na / liq NH3 , Br 2 / CCl4 (B) alc. KOH, Br 2 / CCl4, NaNH2, H2 / Pd-CaCO3, Br 2 / CCl4 (C) alc. KOH, Br 2 / CCl4 (D) NaNH2, cold KMnO4, excess HBr
swers EXERCISE-4 1. D
2. A
3. B
4. B
Alkyl Halide
5. D
6. A
5 Read the followin g questions and cho ose :
(A) If both Assertion and Reason are true and the Reason is correct explanation of the Assertion. (B) If both Assertion and Reason are true, but Reason is not correct explanation of the Assertion. (C) If Assertion is true, but Reason is f alse. (D) If Assertion is false, but the Reason is true. 1.
Assert ion : The SN 2 reaction of neopentyl bromide, (CH3)3CCH2Br, with sodium ethoxide,
Na+(OCH2CH3)¯ , proceeds about 0.00001 times as fast as the reaction of bromoethane. Reason : The bulky groups present at -position in neopentyl bromide exert a strong steric hinderance to the approaching nucleophile. (A) A (B) B (C) C (D) D 2.
Assertion : The major product of the reaction of methylenecyclohexane with N-bromosuccinimide
is 1-(bromomethyl)cyclohexene. Reason : In the reaction of methylenecyclohexane with N-bromosuccinimide a more stable secondary as well as allylic radical intermediate is formed which gives 2-Bromo-1methylenecyclohexne (A) A (B) B (C) C (D) D 3.
Assert ion : Aryl halides are less reactive than alkyl halides towards nucleophilic reagents. Reason : In aryl halides halogens create negative charge centres to the ortho & para position of
benzene ring which favours electrophilic substitution reactio but not the nucleophilic substitution reaction. (A) A (B) B (C) C (D) D 4.
Assertion : The trans isomer of 2-chlorocyclohexanol is hydrolysed with many times faster rate
than the cis isomer on heating with NaOH solution. NaOH
Reason : The trans isomer causes less steric crowding in the SN2 reaction to the nucleophile.
(A) A 5.
(B) B
(C) C
(D) D
Assert ion : CH3CH2CH2CH2Cl reacts with 0.01 M NaCN in ethanol to yield primarily CH 3
CH2CH2CH2CN whereas under the same conditions (CH 3)3CCl reacts to give primarily (CH3)3 COCH2CH3. Reason : The first reaction is a SN2 reaction in which CN¯ is a strong nucleophile, in the second reaction the reactant forms a carbocation intermediate which is solvent stabilised by C2H5OH & SN1 reaction takes place. (A) A (B) B (C) C (D) D 6.
Assert ion : Rate of hydrolysis of (CH3)3C – F increases under strongly acidic conditions. Reason : It is an example of electrophilic catalysis. F¯ is a poor leaving group H-bonding with
strong acid encourages its departure. (A) A (B) B
(C) C
Alkyl Halide
(D) D
7.
Assertion : Hydrolysis of n-Butylchloride in aqueous ethanol is accelerated in presence of NaI. Reason : In all SN1 reactions the reactivity increases in a polar medium in which the carbocation
is solvent stabilised. (A) A 8.
(B) B
(C) C
(D) D
Assert ion : When the alkyl bromides were subjected to hydrolysis in a mixture of ethanol &
water (80% C2H5OH, 20% H2O) at 55ºC, the rates of reaction showed the following order : (CH3)3C – Br > CH3 – Br > CH3CH2 – Br > (CH3)2 CH – Br Reason : In SN1 reaction the reactivity order of different alkanes is R3CH > R2CH2 > RCH3 > CH4 since the carbocation produced in the rate determining step has the stability order as R3C+ > R2CH+ > RCH2+ > CH3+ . (A) A (B) B (C) C (D) D 9.
Assertion: In contrast to SN2 reactions,SN1 reactions show relatively little nucleophile selectivity. Reason : The reactivity of nucleophile is less in SN1 as compared to SN2.
(A) A 10.
(B) B
(C) C
(D) D
Assert ion: CH3CH2 – CH2 – CH2 – Cl reacts with 0.01 M NaCN in ethanol to yield primarily
CH3 – CH2 – CH2 – CH2 – CN whereas under the same conditions (CH 3)3 C – Cl reacts to give primarily (CH3)3 COCH2 –CH3. Reason : With different alkyl halide the nucleophilicity of nucleophiles changes. (A) A (B) B (C) C (D) D
Alkyl Halide
swers EXERCISE - 5 1.
A
2.
B
Explanation : The intermediate allylic radical reacts at the more accesible site and gives the more highly substituted double bond. NBS
3.
B
Explanation : due to : Halogen atoms has double bond character and p-orbital of benzene ring repels the nucleophile therefore SN2 reactivity is also very low. 4.
C
Explanation :
NaOH
NGP
does not hav e NGP..
(NGP = Neighboring group participation or anchimeric assistance) 5.
A
Explanation : NaCN C2H5OH CH3CH2CH2CH2Cl CH3CH2CH2CH2CN SN2 NaCN C H OH
5 2
SN1
ionises in C2H5OH and solvolysis (ethanolysis) takes place. while in CH3CH2CH2CH2Cl, CN¯ (being strong anionic nucleophile) attacks and bimolecular substitution takes place.
Alkyl Halide
6.
A
Explanation : CH3 CH3 CH3 | | 1| H3C C.....F .....H O H H3C C OH + HF + H+ H3 C C F + H O H | | | | | CH H CH H CH3 3 3 7.
B
Explanation : NaI
CH3 – CH2 – CH2 – CH2 – I + NaCl CH3 – CH2 – CH2 – CH2 – Cl CH3 – CH2 – CH2 – CH2 – OH is a good nucleophile as well as a good leaving group. 8.
B
Explanation : (CH3)3 C – Br hydrolyses by SN1 route, so due to ionisation reaction it is fastest. CH3 – Br will hydrolyse by SN2, and rate of SN2 will be faster in CH3 Br than CH3 – CH2 – Br. (CH3)2 CH – Br will be hydrolysed by mixed mechanism (SN1 + SN2) and both routes will have moderate speed. 9.
B
Explanation : In SN1 reactions the first step is slow ionisation of alkyl halide. This is the rate determining step so the rate of overall reaction depends upon nature and concentration of alkyl halide only. There is no direct dependence of rate on nature or concentration of nucleophile. 10.
C
Explanation : NaCN C2H5OH CH3 – CH2 – CH2 – CH2 – Cl CH3 – CH2 – CH2 – CH2 – CN SN2 NaCN C H OH
5 2
SN1
ionises in C2H5OH and solvolysis (ethanolysis) takes place. while in CH3 – CH2 – CH2 – CH2 – Cl, CN¯ (being strong anionic nucleophile) attacks and bimolecular substitution takes place.
Alkyl Halide
Identify the unk nown compo und/s or reagent/s in the fol lowi ng reactions : (Q.1 - Q.16)
1.
CH3 | CH3 C CH CH2 | Br
NaI, Acetone (X) (Allylic SN2)
CH3 | (a) CH2 C CH CH2
CH3 | (b) CH3 C CH CH2 |
I CH3 | (c) CH3 C CH CH2 I
CH3 | (d) CH2 C CH CH3 |
I NaOH
CH CHCH Br
2 2 (X)
2.
(a)
(b)
(c)
(d) NaOH, (X)
3.
(a)
(b)
(c)
(d)
X (major)
4.
(a)
(b)
(c)
(d)
Alkyl Halide
H O, Ag 2 W (major product)
5.
(a)
(b)
(c)
‘X’
6.
7.
8.
(a)
(b)
(c)
(d)
CH22 + 2Na+ CH¯
(a) HOOC – CH2 – CH2 – CH2 – COOH
(b)
(c)
(d)
V (C6H14)
Cl2 W + X + Y Sunlight
(Monochlorides)
( CH3 )3 CO ¯ K W or X Z (C6H12) ( CH3 )3 COH
C 2H5OH KOH Y No reaction.
Alkyl Halide
(d)
(a) V : CH3 – CH2 – CH2 – CH2 – CH2 – CH3
CH3 | (b) V : CH3 C CH2 CH3 | CH3
CH3 | (c) Y : Cl CH2 C CH2 CH3 | CH3
(d) Y : Cl – CH2 – CH2 – CH2 – CH2 – CH3
alc. KOH Br / h (X) 2
9.
(a)
.
(c)
(d)
C H O ¯ K C H OH 2 5 2 5
X
10.
11
(b)
(a)
(b)
(c)
(d)
CH MgBr
CH3MgBr
CH3MgBr
3 Y Z CH2 COOC 2H5 X | CH2 C CH2 CH2 OH || O
CH2 COCH3 | (a) Z : CH2 C(CH3 ) CH2 CH2 OMgBr | OMgBr
CH2 COOC2H5 | (b) Y : CH2 C(CH3 ) CH2 CH2 OMgBr | OMgBr
CH2 COCH3 | (c) Y : CH2 C CH2 CH2 OMgBr || O
CH2 COCH3 | (d) X : CH2 C CH2 CH2 OH || O
Alkyl Halide
12.
( i) ( C H ) O
2 5 2 n – C7H15OH CH3 – CH2 – CH2 – CH2 – MgBr + (X)
( ii) H3O
13.
(a)
Br – CH2 – CH2 – CH2 – OH
(b)
(c)
H2O+ – CH2 – CH2 – CH2 – OH
(d)
(1) Li
Y CH3 CH CH2 Br 2, 7-Dimethyloctane ( 2) CuI X | CH3
(a) X :
(c) Y :
14.
CH3 CH CH2 CuLi | CH3
CH CH CH2 CuLi (b) X : 3 | CH 3
Br CH2 CH2 CH2 CH CH3 | CH3
(d) Y :
CH3 CH2 CH2 CH CH2 Br | CH3
(b)
(c)
(d)
(b)
(c)
(d)
(c)
(d)
W + (CH3)2CuLi
(a)
16.
2
W + CH3 –CH2 –CH2 –Br
(a)
15.
TsO – CH2 – CH2 – CH2 – OH
(CH3CH2)2CuLi + W
(a)
(b)
17.
A monoprotic acid ‘A’ on reduction with red P + HI gives an alkane ‘B’ which on reaction with Cl2 / h gives a mixture of two isomeric monochloro alkanes. ‘B’ can be prepared by catalytic hydrogenation of a hydrocarbon ‘C’ whose one mole on reaction with excess C2H5MgI gives 4 moles of ethane. Identify the incorrect option / options. (a) A is an acid (b) B is an alkane (c) C is an alkene (d) C is an alkyne
18.
Optically active compound (A) C4H8Br 2 is treated with CH3ONa / CH3OH to obtain compound (B). C4H7Br. ‘B’ reacts with Gilman reagent of n-butyl iodide and gives ‘C’ (C 8H16). ‘C’ on reductive ozonolysis gives D and E. D gives red precipitate with fehling solution but E does not. Both respond positively towards I2 / NaOH solution. Identify A. CH3 | (a) CH2 CH2 C H | | Br Br
CH3 | (b) CH3 CH2 C Br | Br
Alkyl Halide
CH3 | (d) CH3 CH C H | | Br Br
Br CH3 | | (c) CH3 C C H | | Br H 19.
One mole of compound A (C3H8O3) is treated with excess of CH3MgBr in ether. The liberated gas measured 67.2 litres at S.T.P. Identify the structure of A.
CH2 OH | (a) CH OH | CH2 OH
20.
21.
CH3 | O | (c) HC O H | O | CH3
CHO | (d) CHOH | CHO
A primary alkyl halide (A), C4H9Br, reacted with alcoholic KOH to give compound (B).Compound (B) reacted with hydrogen bromide to give (C), an isomer of (A). When (A) was treated with sodium, it gave a compound (D), C8H18, which was different from the compound produced when n-butyl bromide was reacted with sodium. The structural formula of A is. CH3 | (a) CH3 C CH3 | Br
CH3 | (b) CH3 C CH2 Br | H
(c) CH3 – CH2 – CH2 – CH2 – Br
(d) CH3 CH2 CH CH3 | Br
When 1-chloromethyl-4-methyl-1, 3-cyclopentadiene (A) is treated with H 2O / Ag+ , three isomeric alcohols are obtained. Then which is not among these isomeric alcohols.
(a)
22.
CH3 | O (b) | CH2 | O | CH3
(b)
(c)
(d)
An organic compound (A), C7H15Cl on treatment with alcoholic caustic potash gives a hydrocarbon (B) C7H14.(B) on treatment with ozone and subsequent hydrolysis gives acetone and butyraldehyde. What (B) ? H | (a) H3C C C C CH3 | | | CH3H H
(b) H3 C C CHCH2CH2CH3 | CH3
H | (c) H3C C CH2 CH CH2 | CH3
(d) H3 C C CH2 CH2 CH3 | CH3
Alkyl Halide
23.
Treatment of 2-bromobutane with hot alcoholic KOH gives a mixture of three isomeric butenes (A), (B) and (C). Ozonolysis of the minor product (A), gives formaldehyde and another aldehyde in equimolar amounts. Both (B) and (C) gave the same single product (D) on ozonolysis. What is the structural formula of (D) ?
(a) CH3 CH2C O | H 24.
(b)
HC O | H
The reaction of 2-chloro-2-methyl-1-phenylpropane with methanol yields a mixture composed of a substitution product A (54%) and two elimination products, B (27%) and C (19%). which structure is written with incorrect percentage. CH3 | (b) Ph CH2 C CH2 (27%)
(a)
CH3 | (d) PhCH 2 C CH2 (19%)
(c) Ph CH C CH3 (27%) | CH3 25.
(c) CH3 C O | H
O || CH C CH3 3 (d)
Compound X is optically inactiv e and has the formula C16H16Br 2. On treatment with strong base, X gives hydrocarbon Y, C16H14. Compound Y absorbs 2 equivalents of hydrogen when reduced over a palladium catalyst and reacts with ozone to give two fragments. One fragment, Z, is an aldehyde with formula C7H6O. The other fragment is glyoxal, (CHO) 2. suggest structure for X. (a) Ph CH2 CH CH CH2 Ph | | Br Br
(b) Br CH2 CH CH CH2 Br | | Ph Ph
(c) Ph CH CH2 CH2 CH Ph | | Br Br
Ph | (d) Ph C CH CH CH Br 2 2 | | H Br
26.
Organic compound (A) C5H11Cl on treatment with sodamide form compound (B). Compound (B) reacts with dilute acid to form (C), C 5H12O. When compound (C) is heated with Al2O3 ,it forms (D) which is isomeric to (B). Combined ozonolysis of (B) and (D) forms four compounds two of which reduces Tollen’s reagent and three of which show haloform reaction. What will be the nature of (C). (a) A primary alcohol (b) A secondary alcohol (c) A tertiary alcohol (d) An ether
27.
The alkyl halide C4H9Cl (A) reacts with alcoholic KOH and gives alkene (B) which reacts with bromine to form a dibromide (C). (C) is transformed with sodamide to a gas (D) which forms a precipitate with ammonical silv er nitrate solution. Give the structures (C). (a) CH3 – CH2 – CH(Br)CH – Br
(b) CH3 CH CH2 CH2 | | Br Br
(c) CH2 CH2 CH2 CH2 | | Br Br
(d) CH3 – CH2 – CH2 – CHBr 2
Alkyl Halide
28.
Five isomeric alkyl halides (C6H13Cl) produce the same alkane (A) (C 6H14) on reduction with Zn/Cu couple. Identify (A). CH3 | CH CH C CH3 (a) 3 2 | H
CH3 | (b) H C CH CH CH CH 3 2 2 3
CH3 | (d) CH3 CH2 C CH3 | CH3
CH3 | (c) H3C CH2 CH CH2 CH3
29.
A chloroderivative (X) on treatment with Zn-Cu couple in ethanol gives a hydrocarbon (Y). When (X) is dissolved in ether and treated with sodium, 2, 2, 5, 5 - tetramethylhexane is obtained. Identify how many different types of hydrogen atoms are present in Y. (a) 1 (b) 2 (c) 3 (d) 4
30.
An alkane C6H14 (A) on chlorination at 300° gives a mixture of 4 monochlorinated derivative B, C, D and E. Two of these (C, D) derivatives giv e the same stable alkene (F) on dehydrohalogenation. The other two alkyl halides (B, E) give alkenes (G) and (H) respectively. Which on ozonolysis give methanal as the common product. (H) gives a ketone while (G) gives an aldehyde. Give struct ure of (H). CH3 | (a) CH2 CH C CH2 CH3 | H
CH3 | (b) CH3 CH C CH2 CH3
CH2 || (c) CH CH C CH CH 3 2 2 3 31.
CH2 || (d) CH C CH CH CH 3 2 2 3
A compound A has molecular formula C6H13Br. When it is treated with magnesium metal in ether, dilute acid produces the compound n–hexane as a product.W hen compound A is reacted with sodium metal in ether, the product is 4, 5 –diethyloctane .From the above information deduce the structural formula of (A). (a) CH3 – CH2 – CH2 – CH2 – CH2 – CH2 – Br (b) CH3 – CH2 – CH2 – CH2 – CHBr – CH3
(c) CH3 – CH2 – CH2 – CH(Br)CH2 – CH3
32.
CH3 | (d) CH3 CH2 C CH2 CH3 | Br
The D-isomer of an optically active organic compound A (C 8H17Br) on reaction with Na I in acetone produced another optically pure compound B (C8H17I). A on hydrolysis with NaOH solution in H2O (20%) + Acetone (80%) mainly produced optically active compound C (C8H18O). B also on hydrolysis under similar conditions produced another compound (D). Both A and B on reaction with CH3MgBr in ether produced 2-methyloctane. Indicate how C and D are related ? (a) Identical (b) Enantionmers (c) Diastereomers (d) Structural isomers
Alkyl Halide
33.
34.
35.
A mixture of isobutane and iso butene is catalytically hydrogenated and the entire alkane is monobrominated in the presence of light at 1270C, which exclusive product would be formed? H | (a) CH C CH 2 3 | | Br CH3
Br | (b) CH3 C CH3 | CH3
CH3 CH2 (c) | Br CH2 CH2
(d) CH3 – CH2 – CHBr – CH3
(W) and (X) are optically active isomers of C5H9Cl.(W) on treatment with one mol of H2 is converted to an optically inactive compounds (Y), but (X) gives an optically active compounds (Z) under the same conditions. Give configurations of (W), in fischer projections.
(a)
(b)
(c)
(d)
An alkyl halide (A) C4H9Cl on reaction with CH 3MgCl gives on alkene (B) (C 4H8) as major product with CH4. Only one dichloro derivative can be obtained by C4H9Cl. Give structure of B. (a) CH2 = CH – CH2 – CH3
(b)
(c)
(d) CH3 C CH2 | CH3
36.
An alkyl halide (A) on reaction with Mg in ether gives a Grignard reagnet (B). B on reaction with C2H5Cl gives C5H12 (C). Monochlorination of (C) gives 4 structures. Out of which how many structures contain one chiral carbon ? (a) 1 (b) 2 (c) 3 (d) 4
37.
One mole of hydrocarbon C 9H12 (A) on reaction with CH3MgX gives 2 moles of CH4. On complete hydrogenation A gives C9H20 (B). B is optically inactive & on monohalogenation it gives only 2 structurally different monohalo derivatives. Give structures of these derivatives. CH2 CH3 | (a) CH3 CH C CH2 CH3 | | X CH2 CH3
CH2 CH3 | CH2 CH3 CH CH C (b) 2 | | X CH2 CH3
CH2 CH3 | CH2 CH2 CH2 X CH C (c) 3 | CH2 CH3
CH2 CH3 X | (d) CH3 C CH2 CH2 CH3 | CH2 CH3
Alkyl Halide
38.
An organic compound P (C5H8O) decolourises Br 2 water & Baeyer’s reagent. P does not responds to 2, 4-DNP but on reaction with PhMgBr, (P) gives 2 moles of benzene. On complete hydrogenation ‘P’ gives (Q) which on reaction with Luca’s reagent gives immediately white turbidity. Give possible structure of P. OH | (a) H3C CH2 C CH3 | CH3
OH | (b) H2 C CH C CH3 | CH3
OH | (c) H C C C CH3 | CH3
OH | (d) H C C C CH2 CH3 | H
swers EXERCISE - 6 1.
c
2.
b
3.
c
4.
c
5.
b
6.
d
7.
a
8.
b, c
9.
a
10.
c
11
a, b
12.
b
13.
b, c
14.
d
15.
a
16.
d
17.
b
18.
c
19.
a
20.
b
21.
a,b
22.
b
23.
c
24.
b
25.
c
26.
c
27.
a
28.
b
29.
a
30.
c
31.
c
32.
b
33.
b
34.
c
35.
d
36.
b
37.
a, b
38.
c
.
Alkyl Halide
CH2CH2Br CHO
? The final product of the reaction is
1. CH3
(A)
2.
(B)
(B) Z > Y (D) Y > X
The most polar as well as stable conformation of 1-chloropropane is
(A)
4.
(D)
Which one is correct for the products of the given reaction
(A) only X and Z are formed (C) X > Y 3.
(C)
(B)
(C)
(D)
Dehydrohalogenation by a strong base is slowest in.
(A)
(B)
(C)
Alkyl Halide
(D)
5.
In which case an alkene is formed on reaction with a strong anionic base.
(A)
6.
(B)
(C)
Which statement is correct for the following reaction, regarding rate, and stereorientation at C1 and C2 respectively (for the major product)
(A) Very slow, inversion and retention (C) Very fast, retention and retention 7.
(D)
(B) Very fast, inversion and retention (D) Very fast inversion and inverstion
The correct reactivity order for the following reaction is NaI / Acetone R – I + NaBr RBr ..
R:
CH3
I (A) I > II > III > IV 8.
10.
CH3 O .. CH2
CH2 CH
II (B) III > I > II > IV
III (C) IV > III > I > II
IV (A) III > IV > I > II
The major product of the foll owing reaction is
(A)
9.
CH3 CH2
(B)
(C)
(D)
The rate of reaction of CH3CH2Br is fastest with (A) (CH3)3CLi (B) NaNH2
(C) CH3ONa
(D) NaOH
In which of the following reactions an optically active single product is formed
(A) CH3ONa +
(C)
(B)
+
(D)
Alkyl Halide
11.
Which one is correct for the products of the given reaction
(A) only X and Z are formed (C) X > Y 12.
(B) Z > Y (D) Y > X
Read the foll owing question and choose the correct answer : (A) If both Assertion and Reason are true and the Reason is correct explanation of the Assertion. (B) If both Assertion and Reason are true, but Reason is not correct explanation of the Assertion. (C) If Assertion is true, but Reason is f alse. (D) If Assertion is false, but the Reason is true.
Ass erti on : In the above reaction a mixture of two stereoisomeric products can be formed Reason : The O O group of the molecule provides anchimeric assistance (neighbouring group
participation) in substitution of carbon atoms. 13.
so there is retention of configuration at both the asymmetric
Which one of the following hexachlorocyclohexane is least reactive and which one is most reactive in E2 reactions with a strong base for dehydrohalogenation.
(A) I least & II most (C) III least & I most
(B) II least & I most (D) III least & II most
Alkyl Halide
14.
Which of the following reaction is not feasible. CH 3 CH3OH
(A)
(B)
(C)
CH 3
+
OCH 3
+
C H OH
5 2
+
CH3 | (D) CH3 C Cl | CH3
+
+ HC + O
Section - B (Reading comp rehensio n Q. 15 - 19)
Observe the fol lowi ng reactions and give answers of questions. (Q.13 – Q.17)
+
CH3 | CH3 CH2 C CH2
+
Alkyl Halide
15.
In Ist reaction the structure of reaction interm ediate should be
(A) 16.
(B)
(C)
(D)
Which statement seems to be most appropriate for reaction II. (A) It is an E1 reaction (B) The product is a mixture of stereoisomers (C)
is a good nucleophile but a poor base which can cause dehalogenation in vicinal dihalides
(D) It is SN1 reaction at 3° C – Cl bond while SN2 reaction at 1° C – Cl bond. 17.
Which of the following structure represents the transition state of slow step of reaction III.
(A)
(B)
(C)
(D)
18.
Which of the reaction (I – VI) involv es rearrangement of reaction intermediate. (A) III only (B) III, V (C) II, III, V (D) I, III, V
19.
Planar reactive intermediates are formed in reactions (A) III, V (B) I, III, V (C) V, VI
(D) I, III, V, VI
swers EXERCISE - 7 1. 4.
D
2.
D
C
3.
(C)
All ‘Cl’ are having trans orientation with respect to each other. So the required anti orientation of – H and ‘Cl’ is not available. 5. 6.
(B) H and Cl are in anti position to each other only in case B. 7. C B
8.
(B)
9.
(A)
10.
D
15.
B
NaI / Acetone 2 (First SN )
The nucleophilicity order is 11. D 16.
C
>
>
>
12.
D
13.
B
14.
D
17.
D
18.
B
19.
D
Alkyl Halide
1.
1. CH MgBr ( excess )
(i)
3 C6H5 – CH2CO2CH3 ?
(ii)
2 C6H5C2H5 ?
2. H
1. Br , Heat , Light
[JEE - 94]
2. NaCN
NaOH ?
2.
[JEE - 94]
S
N2
Which statement is incorrect about the following reaction:
3.
[JEE - 1995]
(A) only one product is formed (B) the product will have walden inversion (C) the reaction mixture will have optical rotation zero (D) the product is a mixture of D + L isomers Optically active 2-iodobutane on treatment with NaI in acetone gives a product which does not [JEE - 95] show optical activity. Explain briefly.
4.
NaI C Acetone
5.
[JEE - 96]
6.
An alkyl halide (X) of formula C 6H13Cl on treatment with potassium tertiary butoxide giv es two isomeric alkenes (Y) and (Z) of formula C 6H 12. Both alkenes on hydrogenation give [JEE - 96] 2,3-dimethylbutane. Predict (X), (Y) and (Z).
7. (i)
A hydrocarbon A, of the formula C8H10, on ozonolysis gives compound B(C4H6O2) only. The compound B can also be obtained from the alkyl bromide, C (C 3H5Br) upon treatement with magnesium in dry ether, followed by carbon dioxide and acidific ation. Identify A, B and C and also give equations for the reactions.
(ii) A compound D(C8H10O) upon treatment with alkaline solution of iodine giv es a yellow precipitate.
The filtrate on acidification gives a white solid E (C7H6O2). Write the structures of D, E and explain [JEE - 1996] the formation of E.
8.
? Cl – CH2 – CH2 – CH2 – COPh + KOH + MeOH + KNH2 A
9. 10.
11. 12.
[JEE - 97]
Which of the following is an organometallic compound (A) Lithium methoxide (B) Lithium acetate (C) Lithium dimethylamide (D) Methyl lithium (CH3)3CMgCl on reaction with D2O produces : (A) (CH3)3CD (B) (CH3)3OD
[JEE-1997] [JEE-1997]
(C) (CD3)CD
(D) (CD3)3OD
Which of the following is the correct method for synthesizing methyl–t–butylether and why ? (i) (CH3)3C Br +
13.
[J EE - 97]
NaOMe
(ii) CH3Br + NaO–t–Bu
[JEE - 97]
Complete the following giving the structure of the principal organic products. [JEE - 1997]
Alkyl Halide
14.
15.
The reaction
with HBr gives :
(A)
(B)
(C)
(D)
Which of the following will react with water : (A) CHCl3 (B) Cl3CCHO
(C) CCl4
[JEE-1998]
[JEE-98]
(D) ClCH2CH2Cl
alcoholic KOH, heat
16.
C6H5CH2CHClC6H5
17.
Benzyl chloride (C6H5CH2Cl) can be prepared from toluene by chlorination with : (A) SO2Cl2 (B) SOCl2 (C) Cl 2 / h (D) NaOCl
[JEE - 98]
[JEE - 1998]
18.
19.
(b) Show the steps to carry out the following transformations. (ii) Ethylbenzene 2-phenylpropionic acid CH3 | CH3 C CH2Br | CH3
[JEE - 1998]
C2H5OH
[J EE - 2000]
20.
An SN2 reaction at an asymmetric carbon of a compound always gives (A) an enantiomer of the substrate (B) a product with opposite optical rotation [JEE-2001] (C) a mixture of diastereomers (D) a single stereoisomer
21.
An alkene (A) C16H16 on ozonolysis gives only one product (B) C8H8O. Compound (B) reacts with NaOH / I2 yields sodium benzoate. Compound (B) reacts with KOH / NH 2NH2 forming [JEE - 2001] (C) C8H10. Write the structures of compound (A), (B) & (C).
22.
Identify the set of reagent / reaction conditions ‘X’ and ‘Y’ in the following set of transformations X
Y Product CH3 – CH2 – CH2Br 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 (C) X = dilute aqueous NaOH, 20°C Y = Br 2 / CHCl 3, 0°C (D) X = Conc. alcoholic NaOH, 80°C Y = Br 2 / CHCl 3, 0°C
23.
[JEE-2002]
Consider the following reaction :
CH3 CH CH CH3 Br X HBr | | D CH3
Identify the structure of the major product ‘X’ :
(A) CH3 CH CH C H2 | | D CH3
(C) CH3 C CH CH3 | | D CH3
(B) CH3 CH C CH3 | | D CH3
(D) CH3 C H CH CH3 | CH3
Alkyl Halide
[JEE-2002]
24.
CH3MgBr P. The product P will be Ethylester excess
(A)
25.
(B)
[JEE-2003]
(C)
(D)
The order of reactivity of compound I - III towards phenyl magnesium bromide is
PhCOPh
CH3CHO
I
II
(A) I > II > III 26.
[JEE-(scr.)2004]
CH3COCH3
III
(B) II > I > III
(C) II > III > I
(D) III > II > I
The number of chiral compounds produced upon monochlorination of 2-methylbutane is [JEE-(scr.)2004]
(A) 2 27.
(D) 8
(B)
(D)
(D)
Phenyl magnesium bromide reacting with t-Butyl alcohol gives (A) Ph – OH
29.
(C) 6
1-Bromo-3-chlorocyclobutane will react with two moles of Na in ether producing [JEE-(scr.)2005]
(A)
28.
(B) 4
(B) Ph – H
(C)
[JEE-(scr.)2005]
(D)
Compound (X)
is reacted with aqueous acetone it gives
following products.
(A) K, L 30.
(B) K, M
(C) 'L' only
Explain the following observations
[JEE-(scr.)2005]
(D) 'M' only [JEE(M)-05]
Acidic solution
Neutral solution
Alkyl Halide
swers LEVEL - JEE
1.
(i)
(ii)
2.
3.
D
5.
6.
7.
(X)
(2-chloro -2, 3-dimethyl butane)
(Y)
(2, 3, 4-trimethyl but-1-ene)
(Z)
(2, 3-dimethyl but-2-ene)
(i)
(A)
(B)
(ii)
(D)
(E)
(C)
8.
9.
(A) Ph – C C – Ph
( = Elimination)
10.
D
A
12.
equation (ii) is better method. In equation (1) Isobutene will be the major product due to elimination. It is Ullmann’s reaction
13.
11.
Heat ( CuI2 )
14.
B
17.
(A, C)
15.
B
16.
Alkyl Halide
+
