REACTION CHART CHART FOR ALKANES GMP (1)
H 2 , Ni →
R–C ≡ CH
200−300°C
or R–CH=CH2
Sabatier senderens reaction Zn −Cu + HCl →
(2)
R–X
(3)
R–Mg–X
(4)
GR
Re d P − Hi, LiAlH 4
2 → RX
(2)
→ R-N
(3)
2 2 7 → Alkyl Sulphonic acid
+ HOH or ROH →
or NH3 or RNH2
Na , dry ether →
RX
Wurtz reaction Zn
→
(5)
RX
(6)
R − C − Cl or ROH || O
Frankland 's reaction
Re d P / Hi →
(4) R–H R–H or R–R R–R or CnH2n+2
(7)
(8)
or
R −C−R || O R −C = O | R
RCHO
Sulphonation H S O
Reed reaction
→ RSO2Cl SO2 + Cl2 h ν
AlCl / HCl
3 → branched alkanes
(6)
→ Alkenes + CH
(7)
→ Aromatic compound
(8) Zn − Hg / Conc. HCl
→
Isomerisation
Pyrolysis
500 −700 °C
4
or C2H6
Cr or Mo or V oxide + Al
2
O3 500°C
CH N
2 2→
step up reaction
Higher alkane
Clemension 's reduction
(9) H N − NH
2 2 →
Wolf / Kishner reduction
or (RCH2CH2)3B
Nitration
(5)
or R −C−R || O
X , h ν or UV light or 400 °C
(1)
+H
O
2 →
(9)
RCOONa
→
(10)
RCOONa
Kolbe's electrolytic synthesis →
NaOH + CaO
S N O B R A C O R D Y H 0 4 f o 2 e g a P
O
2→
∆
Combustion
CO2 + H 2O
S N O B R A C O R D Y H 0 4 f o 3 e g a P
REACTION REACTION CHART FOR ALKENES GR
GMP
(1) R–CH2–CH2–OH (2) R–CH2–CH2–X
(4)
−HX
Zn dust →
(9) R–H (10) CH2=CHCl
HX R–CHX–CH3 → R–CH=CH 2 (3) or HBr , Peroxide (4) → R–CH2–CH2Br C n H 2n HOCl (5) → R–CH(OH)–CH2Cl dil. H 2SO 4
→ R–CH (OH)–CH (6) 2 3 +H O 2
Kolbe's electrolytic synthesis
→
R − C − O − CH 2 − CH 2 − R || O
200 −300 ° C
1 / 2 O 2
→ 200−300°C
(7) (C2H5)4N+OH (8)
X2 (2) → → R–CHX–CH2X
Ni, H 2
(5) R–C ≡ CH (6)
alc. KOH →
− H 2O
for higher alkene −X 2
R − CH − CH 2 | | X X
RCH − COOK | RCH − COOK
→ R–CH2–CH3 (1)
Zn dust →
(3) R–CH2–CH
H 2 , Ni
conc. H 2SO4 →
∆ → → Pyrolysis
→
→ (7) Ag 300°C + CH N 2 2 (8) →
BH 3 (9) → (RCH2CH2)3B
CO + H 2 → (10) HCo ( CO) 4
R − CH − CH 3 R | + CHO
− CH 2 − CH 2
O2
| CHO
→ → CO2 + H2O (11) ∆
Pyrolysis
→
OS O 4 (12) →
CuR 2 →
R − CH − CH 2 | | OH OH
→ (13) 1% alkalineKMnO
R − CH − CH 2 | | OH OH
strong oxidant (14) →
R − C − OH + CO2 + H2O || O
Bayer reagent
4
Per acid
→ (15) Pr iles − chalev's reaction O3 +H 2 O → (16)
Ozonolysis
(17)
+
O
→ 2 → Polyalkene 200°C high P Cl 2
→ Substitution (18) Substitution product 500 °C Al2 (SO 4 )3
→ Isomerisation (19) 200−300 °C acetic anhydride (20) → R–CH2=CH–COCH3 Methyl alkenyl alkenyl ketone keto ne Alkane (21) → Higher alkane
REACTION CHART FOR ALKYNES GR
GMP
(1) CH2Br–CH2Br (2) CH3–CHBr2 (3) CHCl3 (4) CHBr2–CHBr2 CHBr (5) || CHBr (6) CH2=CH–Cl
alc. KOH or NaNH 2 →
(1)
alc. KOH , NaNH 2
(2)
→
Zn dust
→ ∆
Zn →
(8) CaC2 (9) 2C + H2
C2 H 2
(4)
HBr → CH –CHBr
(5)
HOCl → Cl2CHCHO
(6) (7)
2 → CH2=CHCN CH3COOH , Hg 2 → CH3CH(OCOCH3)2
(8)
° 4 2 → CH3CHO
(9)
Conc . H 2SO 4 → CH3CH(HSO4)2
alc. KOH , NaNH 2 →
HC − COONa Kolbe's electrolytic synthesis → (7) || HC − COONa H 2O → electric arc ,1200 °C → Berthelot 's process
(10) CH3–C≡CH
→
(10) CH3–C≡CH
3 →
Ni X2
→ C2H2X4 HBr → CH3BrCH2Br (3) Peroxide
Ag powder →
∆
H2 → C2H4, C2H6
No Peroxide
3
2
HCN , Ba ( CN )
+
Hg +2 , 80 C , dil. H SO
( Kucherov 's reaction )
AsCl3
→ (10) Ca det CHCl=CHAsCl2 & Bunsen reaction C 2 H 5OH / H 2 O
→ CH CHO (11) 3 HgSO 4
(i ) Na ( ii ) R − X
(i ) CH MgI ( ii ) R − X
S N O B R A C O R D Y H 0 4 f o 4 e g a P
CO + HOH
→ CH =CH–COOH (12) 2 Ni ( CO ) 4
CO + EtOH
→ CH =CH–COOEt (13) 2 Ni ,160°C NaNH 2 (14) → Na–C≡C–Na
AgNO3 + NH 4OH
→ Ag–C≡C–Ag (15) ( Tollen 's Re agent ) Cu 2Cl 2 + NH 4 OH (16) → Cu–C≡C–Cu Combustion O2 (17) → CO2 + H2O
CHO
Bayer Re agent (18) → |
→ HCOOH
CHO
(19)
O
3 →
Ozonolysis
+H
O
2 → HCOOH
Trimerisat ion
→ benzene (20) (Re d hot iron tube )
Trimerisation
→ C8H8or 1,3,5,7-cyclo octa tetraene (21) [ Ni ( CN ) 2 ]
Dimerisation
+→ butenyne (22) [ Cu ( NH3 ) 2 ] s
→ (23) ∆
CH3OH
→ (24) ( BF − HgO ) 3
methylal
EXERCISE–I (A) Q.1
In the given reaction HCl C7H12 (A) →
(A) is: (A) Q.2
Q.3
(B)
by the use of which of the following reagents? (A) BD3 followed by HCOOH
(B) BH3 followed HCOOD
(C) BD3 followed by HCOOD
O || (D) BH3 followed by D − C − O − H
Identify (P) in the following reaction:
(A)
Q.6
⊕
H / H 2O → (P)
(B)
(C)
(D)
The reaction of E-2-butene with CH2I2 and Zn–Cu Couple in either medium leads to formation of (A)
Q.5
(D) All of these
1-Methylcyclopentene can be converted into the given compound
+2
Q.4
(C)
(B)
(C)
(E)-3-bromo-3-hexene when treated with CH3O in CH3OH gives (A) 3-hexyne (B) 2-hexyne (C) 2,3-hexadiene
(D)
r
(D) 2,4-hexadiene
The reaction of cyclooctyne with HgSO4 in the presence of aq. H 2SO4 gives (A)
(B)
(C)
(D)
hν + Br2 mixture of product. Among the following which product will formed minimum →
Q.7 amount.
(A)
(B)
(C)
(D)
S N O B R A C O R D Y H 0 4 f o 5 e g a P
H 2SO 4
NBS
→ Q (Major) → P (Major)
Q.8
∆
The structure of Q is
(A)
(B)
(C)
(D)
(C)
(D)
18
↓ (i ) CH3COO OH
→ X ⊕
Q.9
(ii ) H 3O
The probable structure of ‘X’ is
(A)
(B)
Br2 NaI ∆ → P(Alkene) → → Q (Alkene)
Q.10
CCl4
Acetone
Alkene (P) & (Q) respectively are
Q.11
(A) Both
(B)
(C) Both
(D) Both
O
z
o
n
o
l y s i s
o
f
C
H
3–CH=C=CH2 will give
(A) Only CH3CHO (C) Only CO2 Q.12
,
(B) Only HCHO (D) Mixture of CH 3CHO, HCHO & CO2
O-xylene on ozonolysis will give O || CHO (A) | & CH 3 − C − CHO CHO
O CH3 − C = O || | & CH 3 − C − CHO (B) CH3 − C = O
CH 3 − C = O CHO & | (C) | CH 3 − C = O CHO
O CH 3 − C = O || CHO | (D) , CH 3 − C − CHO & | CH 3 − C = O CHO
S N O B R A C O R D Y H 0 4 f o 6 e g a P
S N O B R A C O R D Y H 0 4 f o 7 e g a P
O O (1eq )
4 s → X.
Q.13
H 2O / Acetone
Identify ‘X’.
(A)
(B)
(C)
(D) Reaction will not occur
PdCl 2 ,HOH → Z.
Q.14
CuCl 2 ,O 2
Identify Z.
Q.15
(A)
(B)
(C)
(D) All are correct
CH 3 CH 3 | | O O 4 (1equiv.) s → A ; Identify A CH 3 − C == C − CH 2 − CH = CH 2 ( Acetone / water )
CH 3 CH 3 OH OH | | | | (A) CH 3 − C == C − CH 2 − CH − CH 2
CH 3 CH 3 | | (C) CH 3 − CH — C − CH 2 − CH − CH 3 | | OH OH Q.16
CH 3 | (B) CH 3 − C — | OH
CH 3 | C − CH 2 − CH = CH 2 | OH
(D) Reaction will not occur
1-Penten-4-yne reacts with bromine at – 80°C to produce: (A) 4,4,5,5-Tetrabromopentene (B) 1,2-Dibromo-1,4-pentadiene (C) 1,1,2,2,4,5-hexabromopentane (D) 4,5-dibromopentyne
Q.17
Compound (A) on oxidation with hot KMnO4 / OH gives two compound r
O || CH 3 − CH − COOH & CH 3 − C − CH 2CH 2CH 3 | CH 3
compound A will have structure.
Q.18
(A) CH CH − C == C − CH CH 3 2 2 3 | | CH 3 CH 3
(B) CH 3 − CH − CH = C − CH 2CH 2CH 3 | | CH 3 CH 3
(C) CH CH − C ≡ C − CH 3 3 | CH 3
(D) CH 3 − CH − C ≡ C − CH − CH 3 | | CH 3 CH 3
S N O B R A C O R D Y H 0 4 f o 8 e g a P
Consider the following reaction KMnO / OH − / ∆
4 (A) C6H12 → C5H10O In the above reaction (A) will be
(A) CH3 –CH2 –CH2 –CH2 –CH=CH2
(B) CH 3 − CH − CH 2 − CH = CH 2 | CH 3
(C) CH 3 − CH 2 − CH − CH = CH 2 | CH 3
(D) CH 3CH 2CH 2 − C = CH 2 | CH 3
alcoholic KOH → product
Q.19
Major product is: (A) Q.20 Q.21
(B)
(C)
(D)
Number of required O2 mole for complete combustion of one mole of propane – (A) 7 (B) 5 (C) 16 (D) 10 How much volume of air will be needed for complete combustion of 10 lit. of ethane – (A) 135 lit. (B) 35 lit. (C) 175 lit. (D) 205 lit.
Q.22
When n-butane is heated in the presence of AlCl3 /HCl it will be converted into – (A) Ethane (B) Propane (C) Butene (D) Isobutane
Q.23
The reacting species of alc. KOH is – (A) OH– (B) OR+
Q.24
(C) OK+
(D) RO–
The product of reaction between one mole of acetylene and two mole of HCHO in the presence of Cu2Cl2 – (A) HOCH2 – C ≡ C – CH2OH (B) H2C = CH – C ≡ C – CH2OH (C) HC ≡ C – CH2OH (D) None of these
Q.25
PMA polymer is formed by methyl acrylate, which is prepared as follows – (A) R – C (C) HC
≡
CO + ROH ≡ CH → CO
+
H O
2 → CH
Ni ( CO ) 4
(B) HC
CO + CH OH
3 → ≡ CH
Ni ( CO ) 4
(D) None of these
Q.26
During the preparation of ethane by Kolbe’s electrolytic method using inert electrodes the pH of the electrolyte – (A) Increases progressively as the reaction proceeds (B) Decreases progressively as the reaction proceeds (C) Remains constant throughout the reaction (D) May decrease of the the concentration of the electrolyte is not very high
Q.27
Ethylene forms ethylene chlorohydrin by the action of – (A) Dry HCl gas (B) Dry chlorine gas (C) Solution of chlorine gas in water (D) Dilute hydrochloric acid
Q.28
Anti–Markownikoff’s addition of HBr is not observed in – (A) Propene (B) But–2–ene (C) Butene
(D) Pent–2–ene
Q.29
Which alkene on heating with alkaline KMnO4 solution gives acetone and a gas, which turns lime water milky – (A) 2–Methyl–2–butene (B) Isobutylene (C) 1–Butene (D) 2–Butene
Q.30
Acetylene may be prepared using Kolbe’s electrolytic method employing – (A) Pot. acetate (B) Pot. succinate (C) Pot. fumarate (D) None of these
Q.31
B
Na / NH 3 Lindlar ← R–C≡C–R → A
A and B are geometrical isomers (R–CH=CH–R) – (A) A is trans, B is cis (B) A and B both are cis (C) A and B both are trans (D) A is cis, B is trans Q.32
Q.33 Q.34
Which is expected to react most readily with bromine – (A) CH3CH2CH3 (B) CH2=CH2 (C) CH≡CH
(D) CH3–CH=CH2
By the addition of CO and H2O on ethene, the following is obtained – (A) Propanoic acid (B) Propanal (C) 2–Propenoic acid (D) None of the above An alkyne C7H12 on reaction with alk. KMnO4 and subsequent acidification with HCl yields a mixture CH 3 − CHCOOH | of + CH3CH2COOH. The alkyne is – CH 3 (A) 3–Hexyne (C) 2–Methyl–2–hexyne
Q.35
(B) 2–Methyl–3–hexyne (D) 2–Methyl–2–hexene
A compound (C5H8) reacts with ammonical AgNO3 to give a white precipitate and reacts with excess of KMnO4 solution to give (CH3)2CH–COOH. The compound is – (A) CH2=CH–CH=CH–CH 3 (B) (CH3)2CH–C≡CH (C) CH3(CH2)2C≡CH (D) (CH3)2C=C=CH2
S N O B R A C O R D Y H 0 4 f o 9 e g a P
Q.36
Which of the following reagents cannot be used to locate the position of triple bond in CH3–C≡C–CH3 (A) Br2
Q.37
(C) Cu 22+
(B) O3
CH3–CH2–C≡CH
CH3C≡C–CH3
A and B are – (A) alcoholic KOH and NaNH2 (C) NaNH2 and Lindlar
Q.38
B
(D) KMnO4
BH3 / THF ← −
(B) NaNH2 and alcoholic KOH (D) Lindlar and NaNH2 +
H3O → A
H 2O 2 / OH
A and B are –
Q.39
(A) Both
(B) Both
(C)
(D)
B
BH3 THF HgSO 4 / H 2SO 4 ← CH3–C≡CH → A − H 2O2 , OH
A and B are – O || (A) CH 3CH 2CHO, CH 3 − C − CH 3
O || (B) CH 3 − C − CH 3 CH 3CH 2CHO
(C) CH3CH2CHO (both)
(D) CH 3 − C − CH 3 (both) || O
B2 D6 → product X Q.40 CH3CH=CH2 H O / OH− 2
2
X is –
Q.41
(A) CH 3 − CH − CH 2 D | OH
(B) CH 3 − CH − CH 2OH | D
(C) CH 3 − CH − CH 3 | OD
(D) none is correct
CCl3Br CH2=CH–CH=CH2 → product. The major product is –
(A) Br–CH2–CH=CH–CH2–CCl3
(C) CH 2
=
CH − CH − CH 2 − Br | CCl3
(B) CH 2
=
CH − CH − CH 2 − CCl3 | Br
(D) None is correct
S N O B R A C O R D Y H 0 4 f o 0 1 e g a P
Q.42
Mixture of one mole each of ethene and propyne on reaction with Na will form H2 gas at S.T.P. – (A) 22.4 L (B) 11.2 L (C) 33.6 L (D) 44.8 L
Q.43
Dehydration of 2, 2, 3, 4, 4–pentamethyl–3–pentanol gave two alkenesA and B. The ozonolysis products of A and B are -
(A)
O || A gives (CH 3 )3 C − C − C(CH 3 )3 and HCHO O || B gives CH 3 − C − CH 2 − C(CH 3 )3 and HCHO
(B)
O || A gives (CH 3 )3 C − C − C(CH 3 )3 and HCHO O CH 3 || | B gives CH − C − C − C(CH ) and HCHO 3 3 3 | CH 3
(C)
O || A gives (CH 3 )3 C − C − CH(CH 3 ) 2 and HCHO O || B gives (CH 3 ) − CH 2 − C − C(CH 3 )3 and CH3CH2CHO
(D)
Q.44
None of these NH 4Cl
→ product CH≡CH Cu Cl 2
2
Product is – (A) Cu–C≡C–Cu
Q.45
O3 / H 2O Alkene A →
(B) CH2=CH–C≡CH (C) CH≡C–Cu
(D) Cu–C≡C–NH4
O || CH 3 − C − CH 3 + CH3COOH + CH 3 − C − COOH || O
A can be –
(A)
C(CH 3 ) 2 || (B) CH 3 − C − CH = HC − CH 3
(C) Both correct
(D) None is correct
S N O B R A C O R D Y H 0 4 f o 1 1 e g a P
reagent R 2 ←
Q.46
reagent R1 →
R1 and R2 are – (A) Cold alkaline KMnO4, OsO4 /H2O2 (C) Cold alkaline KMnO4, CH3–O–O–CH3
Q.47
CH 3 | H−C || H−C | CH 3
(B) Cold alkaline KMnO4, HCO3H (D) C6H5CO3H, HCO3H
alkaline KMnO 4 → A, which is true about this reaction?
(A) A is meso 2, 3–butan–di–ol formed by syn addition (B) A is meso 2, 3–butan–di–ol formed by anti addition (C) A is a racemic mixture of d and l, 2, 3–butan–di–ol formed by anti addition (D) A is a racemic mixture of d and l 2,3–butan–di–ol formed by syn addition +
O / H O / Zn Con . H 2SO 4 H3O + CH3MgBr → B 3 2 → C → A
Q.48
A, B and C are – A
B
C
(A)
(B)
(C)
(D)
Q.49
A →
∆
A can be – (A) Conc. H2SO4
(B) alcoholic KOH
(C) Et3N
(D) t-BuOK
S N O B R A C O R D Y H 0 4 f o 2 1 e g a P
Q.50
BrCH2–CH2–CH2Br reacts with Na in the presence of ether at 100 °C to produce – (A) BrCH2–CH=CH2 (B) CH2=C=CH2
Q.51
(C)
(D) All of these
(C)
(D)
(C)
(D)
Which has least heat of hydrogenation – (A)
(B) (1) Hg ( OAc) / H O / THF
2 2 → A. A is –
Q.52
( 2 ) NaBH4 / NaOH / H 2 O
(A)
Q.53
(B)
An organic compound of molecular formula C4H6, (A), forms precipitates with ammoniacal silver nitrate and ammoniacal cuprous chloride. ‘A’ has an isomer ‘B’, one mol of which reacts with one mol of Br2 to form 1, 4-dibromo-2-butene. Another isomer of A is ‘C’, one mole of C reacts with only 1 mol. of Br2 to give vicinal dibromide. A, B & C are (A) CH3–CH2–C≡CH and CH2=CH–CH=CH2 ; (B) CH3–C≡C–CH3 and CH3–CH=C=CH2 ; CH3–C≡C–CH3 CH 2 − CH | | ; CH2 = CH–CH=CH2 C=CH2 and | CH 2 − CH
(C)
(D) CH3–C≡C–CH3 and
; CH2 = CH–CH = CH2
Q.54
x CH3–CH=CH–CH3 product is Y (non–resolvable) then X can be – → cis (A) Br2 water (B) HCO3H (C) Cold alkaline KMnO4 (D) all of the above
Q.55
Electrophilic addition reaction is not shown by (A) CH 2
=
C − CH 3 and Br2 | CH 3
(C) CH3–C≡CH and CH3MgBr
(B) CH≡CH2 and HO–Cl
(D) CH2=CH2and dil. H2SO4 solution
Q.56
A mixture of CH4, C2H4 and C2H2 gaseous are passed through a Wolf bottle containing ammonical cuprous chloride. The gas coming out is (A) Methane (B) Acetylene (C) Mixture of methane and ethylene (D) original mixture
Q.57
In the presence of strong bases, triple bonds will migrate within carbon skeletons by the (A) removal of protons (B) addition of protons (C) removal and readdition of protons (D) addition and removal of protons.
S N O B R A C O R D Y H 0 4 f o 3 1 e g a P
Q.58
CHCOOH CH2=CH–CH=CH 2 + || CHCOOH
→ product X by reaction R. X and R are ∆
(A)
Diels Alder
(B)
Friedel-Crafts
(C)
Diels Alder
(D)
Friedel-Crafts
S N O B R A C O R D Y H 0 4 f o 4 1 e g a P
Q.59
For the ionic reaction of hydrochloric acid with the following alkenes, predict the correct sequence of reactivity as measured by reaction rates: (I) ClCH=CH2 (II) (CH3)2.C=CH2 (III) OHC.CH=CH2 (IV) (NC)2C=C(CN) 2 (A) IV > I > III > II (B) I > IV > II > III (C) III > II > IV > I (D) II > I > III > IV
Q.60
The addition of bromine to 2-cyclohexenyl benzoate in 1,2-dichloroethane produces ____ dibromo derivatives: (A) 2 (B) 3 (C) 4 (D) 6
Q.61
How many products will be formed when methylenecyclohexane reacts with NBS? (A) 3 (B) 1 (C) 2 (D) 4 CH3 − C− NH 2 || O
Mg / dry ether Br → (X) → (Y)
Q.62
The structures of (X) and (Y) respectively are (A) X =
MgBr
(B) X =
(C) X = (D) X = BrMg
MgBr MgBr
;
Y=
;
Y=
;
Y=
;
Y = HO
OH
OH
EXERCISE–I (B)
Q.1
HgSO 4 / H 2SO 4 B → A BH 3 / THF → − H O / OH 2 2
B is identical when A is – (A) (B)
(C)
(D)
Q.2
An alkene on ozonolysis yields only ethanal. There is an isomer of this which on ozonolysis yields: (A) propanone (B) ethanal (C) methanal (D) only propanal
Q.3
CH3– CH = CH–CH3 + CH2N2 ∆ → A A can be
(A) Q.4
(B)
(C)
(D)
Aqueous solution of potassium propanoate is electrolysed. Possible organic products are: (A) n-Butane (B) C2H5COOC2H5 (C) CH3–CH3 (D) CH2=CH2
Re agent R 2 ←
Q.5
Re agent R1 →
R1 and R2 are: (A) cold alkaline KMnO4, OsO4 / H2O2 (B) OsO4 / NaHSO3 ; Ag2O , H3O⊕ (C) cold alkaline KMnO4, C6H5CO3H / H3O⊕ (D) C6H5CO3H ; OsO4 / NaHSO3 Q.6
H / Pt
2 2 → (B) C4H8 3 (A) C4H6 → CH3COOH Hence A and B are (A) CH3C ≡ CCH3, CH3CH = CHCH3 (B) CH2 = CHCH3 = CH2, CH3CH = CHCH3
(C) Q.7
Q.8
O / H O
, CH3CH = CHCH3
Which is / are true statements/ reactions? (A) Al4C3 + H2O → CH4 (C) Mg2C3 + H2O → CH3C ≡CH
(D) None
(B) CaC2 + H2O → C2H2 (D) Me3C–H + KMnO4 →Me3C–OH
A
→ Ph–CH2–CH3 Ph − C − CH 3 || O A could be: (A) NH2NH2, glycol/OH– (C) Red P/HI
(B) Na(Hg)/conc. HCl (D) CH 2 | SH
− CH 2 ; Raney Ni – H2 | SH
S N O B R A C O R D Y H 0 4 f o 5 1 e g a P
S N O B R A C O R D Y H 0 4 f o 6 1 e g a P
t − BuOK
→ Product
Q.9
which is / are correct statements about the product:
Q.10
(A)
CH3
is an endocyclic Saytzeff product
(B)
CH2
is an exocyclic Saytzeff product
(C)
CH2
is an exocyclic Hoffmann product
(D)
CH3
is an endocyclic Hoffmann product
NBS CH2 = CHCH2CH = CH2 → A, A can be
(A) CH 2
=
CH CH CH = CH 2 | Br
(C) CH2 = CH CH2 CH = CHBr
(B) CH2=CHCH=CH–CH2Br
(D) CH 2
=
CH CH 2 C = CH 2 | Br
Q.11
Which are correct statements? (A) meso-2, 3-dibromo-butane on reaction with NaI / acetone gives trans-2-butene (B) d-or l- 2, 3-dibromobutane on reaction with NaI/acetone gives cis-2-butene (C) meso-2, 3-dibromo-butane on reaction with NaI / acetone gives cis-2-butene (D) d-or l- 2, 3-dibromobutane on reaction with NaI/acetone gives trans-2-butene
Q.12
peroxide Ph–CH=CH2 + BrCCl3 →
Product is:
(A)
(B)
(C)
(D)
Q.13
Which of the following elimination reactions will occur to give but-1-ene as the major product? EtOH
(A)
→ CH3.CHCl.CH2.CH3 + KOH
(B)
CH 3 .CH 2 .CH.CH 3 + NaOEt | NMe 3
EtOH →
∆
+
(C)
∆ CH3.CH2.CHCl.CH3 + Me3CoK →
(D)
CH3.CH2.CH(OH).CH3 + conc. H2SO4 → ∆
—
OH
Q.14
The above compound undergoes ready elimination on heating to yield which of the following products?
(A)
(B)
(C)
(D)
Q.15
Which of the following will give same product with HBr in presence or absence of peroxide. (A) Cyclohexene (B) 1-methylcyclohexene (C) 1,2-dimethylcyclohexene (D) 1-butene
Q.16
The ionic addition of HCl to which of the following compounds will produces a compound having Cl on carbon next to terminal. (A) CF3.(CH2)3.CH=CH2 (B) CH3.CH=CH2 (C) CF3.CH=CH2 (D) CH3.CH2CH=CH.CH3
Q.17
Select true statement(s): (A) I2 does not react with ethane at room temperature even though I2 is more easily cleaved homolytically than the other halogens. (B) Stereochemical outcome of a radical substitution and a radical addition reaction is identical. (C) The rate of bromination of methane is decreased if HBr is added to the reaction mixture. (D) Allylic chloride adds halogens faster than the corresponding vinylic chloride.
Q.18
Select true statement(s): (A) Instead of radical substitution, cyclopropane undergoes electrophilic addition reactions in sun light. (B) In general, bromination is more selective than chlorination. (C) The 2,4,6-tri-tert, butylphenoxy radical is resistant to dimerization. (D) The radical-catalysed chlorination, ArCH3 →ArCH2Cl, occurs faster when Ar = phenyl than when Ar = p-nitrophenyl.
Q.19
Nitrene is an intermediate in one of the following reactions: (A) Schmidt rearrangement (B) Beckmann rearrangement (C) Baeyer-Villiger oxidation (D) Curtius reaction
Q.20
Which reagent is the most useful for distinguishing compound I from the rest of the compounds CH3CH2C≡CH CH3C≡CCH3 CH3CH2CH2CH3 CH3CH=CH2 I II III IV (A) alk. KMnO4 (B) Br2 /CCl4 (C) Br2 /CH3COOH (D) Ammonical AgNO3
S N O B R A C O R D Y H 0 4 f o 7 1 e g a P
Q.21
S N O B R A C O R D Y H 0 4 f o 8 1 e g a P
Indicate among the following the reaction not correctly formulated. +SO
Cl
2 2 (A) CH2=CH–CH3 → CH2Cl–CHCl–CH3
UV light
(B) HC≡CH+CH2N2 → photo −
(C) (CH3)3CH + Cl2 → (CH3)3C–Cl as major product halogenation
Na
→ (D) CH3–C≡C–CH2–CH2–CH3 in NH ( liq ) 3
Q.22
List I
List II
Cr2 O3 − Al2O 3 , ∆ (A) n-Hexane →
(B) CH ≡ CH
(1) Substitution reaction
→
(2) Elimination reaction
Re d hot Fe tube
CH 3 | (C) CH 3 − C − X → aq. | CH 3
(3) Aromatisation
(D) CH3–CH2–X → alc. KOH
(4) Cyclization
Q.23
List I
(A) CH 3 − C = CH 2 | CH 3
List II ( i ) BH
3 →
(1) CH3–CH2–CH=CH2
(ii ) H 2O 2 / OH
(i ) Hg ( OAc) / HOH
2 → (B) CH 3 − C = CH 2 (ii ) NaBH4 | CH 3
Cl | (C) CH 3 − CH 2 − CH − CH 3
Cl | (D) CH 3 − CH 2 − CH − CH 3
Codes:
(a) (b) (c) (d)
A 4 4 3 3
(2) CH3–CH=CH–CH3
(3) CH 3 − CH − CH 2OH | CH 3
CH ONa / ∆
3 →
OH | (4) CH 3 − C − CH 3 | CH 3
( CH ) CONa
3 3 →
∆
B 3 3 4 4
C 1 2 2 1
D 2 1 1 2
Q.24
List I (A) Walden Inversion (B) Racemic mixture
List II
(1) Cis addition (2) Trans addition
Baeyer (C) Alkene →
(3) SN1 reaction
Br2 (D) Alkene → Codes: (a) (b) (c) (d)
(4) SN2 reaction
Re agent
Q.25
Q.26
A 3 3 4 4
B 4 4 3 3
C 2 1 1 2
D 1 2 2 1
List I (A) CH3–C≡C– CH3 → cis-2-butene
List II ( 1
)
N
a / N
H
3
(l)
(B) CH 3 − C ≡ C − CH 3 → trans-2-butene
(2) H2 /Pd/BaSO4
(C) CH3C≡C–CH3 → 1-Butyne (D) CH3–CH3–C≡CH → 2-Butyne Codes: A B (a) 2 1 (b) 1 2 (c) 1 2 (d) 2 1
(3) alc. KOH, ∆ (4) NaNH2 , ∆
C 3 4 3 4
D 4 3 4 3
List I
List II
electrolys is
(A) RCOONa → R–R
(1) Corey-Housh reaction
Soda lim e
→ R–CH3 (B) R–CH2–COOH
(2) Kolbe electrolysis
∆
(i ) AgNO
3 → R–Cl (C) RCOOH
(3) Oakwood degradration
(ii ) Cl 2 / ∆
(D) R’–X + R2CuLi → R–R’ A B Codes: (a) 2 3 (b) 1 3 (c) 2 4 (d) 2 4
(4) Hunsdiecker reaction C 4 4 3 1
D 1 2 2 3
S N O B R A C O R D Y H 0 4 f o 9 1 e g a P
Q.27
List I
(A)
→
(1) Birch reduction
(B)
→
(2) Stephen’s reduction
(C)
(D)
Q.28
List II
→
(3) Wolf-Kishner reduction
→
(4) Clemmensen reduction
List I (A) n-Hexane → Benzene (B) CH≡CH → Benzene (CH 2— ) 6 CH 3 (C) CH— 3
→ 2,2,3,3 tetramethyl butane(3) AlCl3 + HCl at 300°C
(D) CH3–CH2–X → n-Butane
Q.29
List II (1) Wurtz reaction (2) Coupling of reactants is taking place
(4) Polymerisation (5) Aromatic procducts is formed (6) Zn + ∆ used as reagent (7) Al2O3 at high temperature
Match List-I with List-II and select the correct answer using the codes given below the lists: List-I (Reaction) List-II (Reagents) (A) CH3–CH=CH2→CH3–CHBr–CH 3 (P) HBr (B) CH3–CH=CH2→CH3–CH2–CH2Br (Q) Br2 (C) CH3–CH=CH2→BrCH2–CH=CH2 (R) HBr / Peroxide (D) CH3–CH=CH2→CH3–CHBr–CH2Br (S) NBS
S N O B R A C O R D Y H 0 4 f o 0 2 e g a P
EXERCISE–II Q.1
Give the product of H O2 BH3 →B → A 2
(a)
1. BD3 .THF → −
(b)
2. D 2 O2 , DO
THF
Q.2
Q.3
CH 3 | CH 3 − C − Cl | CH 3
LiAlH 4 → ?
What are the ozonolysis products of
1. CF3CO3H → A + B. What are A and B?
Q.4
Q.5
2. H 2O / H +
NaOH → A. Write the structure of A. Alc.
NH 2 − NH 2 / H 2O 2 → A. Write the structure of A
Q.6
−
Q.7 Q.8
Q.9
OH → A. Write the structure of major product A.
∆
Give the structure of the alkene that yields on ozonolysis (i) CH3CH2CH2 CHO & HCHO (ii) C2H5COCH3 & CH3CH(CH3) CHO (iii) Only CH3CO.CH3 (iv) CH3.CHO & HCHO & OHC.CH2.CHO (v) Only OHCCH2CH2CH2 –CHO. One of the constituent of turpentine is α-pinene having molecular formula C10H16. The following scheme give reaction of α-pinene. Determine the structure of α–pinene & of the reaction products A through E. E(C10H18O2) A(C10H16Br2) ↑ H2O ↑ Br2 /CCl4 C10H16O(D)
PhCO H Br H O ← α–pinene → B (C10H17OBr) 3
2+
2
S N O B R A C O R D Y H 0 4 f o 1 2 e g a P
Q.10
Propose structures for intermediates & products A to K
Q.11
Identify the following (A to D).
Q.12
What are A to K for the following reactions
(i)
PhC
(ii)
(iii)
(v)
S N O B R A C O R D Y H 0 4 f o 2 2 e g a P
Ar CH C l Li / NH ≡ CH + CH3MgX → A → B → C. alcoholic hγ cold dil. KMnO4 hot KMnO4 → E PhCH2CH2CH3 + Br2 → D → F → G KOH 2
H →
CF3 – CH = CH2
HBr → J
3
(iv)
(vi)
I→
NBS → K
Q.13
What will be the product in the following reaction
(i)
+
→ ∆
(ii)
∆ →
(iv)
→ ∆
+
CO2M e
(iii)
(v)
+
+ Br2 → A
+
∆ →
1. NaNH 2 ( 3equiv.) NH 3 → B 2. CH 3I
Q.14
(i) Compare the reaction of CH2 = CH2 & CF2 = CF2 with NaOEt in EtOH (ii) CCl2 = CCl2 does not decolourise Br2 solution - explain.
Q.15 (i) (ii)
Account for the collowing facts Ozonolysis if carried out in MeOH solvent a hydroxy peroxy ether is formed as unexpected product. When 2, 3 dimethyl 2 butene is treated with O3 in presence of HCHO in CH2Cl2 medium, an ozonide other than that expected of the starting alkene is formed. Identify the unexpected ozonide.
Q.16 (i) (ii) (iii) (iv) (v)
Explain the following: 1, 2 shift does not take place during oxymericuration demercuration. Why? Halogneation of alkene is anti addition but not syn addition. Why? Anti markovnikov addition is not applicable for HCl. Why? 1,4–addition takes place in butadi-ene. Why? C–H bond is stronger than C–C bond but in chlorination C–H bonds get cleaved but not C–C bond. Why?
Q.17
Conversion: (i) C2H2 → racemic 2, 3 dibromobutane (ii) 2 butyne → 2 pentyne (iii) Ethyne → Acetone (iv) Methane → n Butane (v) Ethene → Propionic Acid
Q.18
Conversion: (i) C2H2 → ethylidene diacetate (iii) C2H2 → m nitroaniline
(ii) C2H2 → Butyne diol (iv) cis but 2 ene → Trans but 2 ene
Q.19
Outline a stereospecific synthesis of meso 3, 4 dibromohexane from ethyne.
Q.20
How can you convert (a) Ethane in to meso 2, 3 dimethyl oxiran (b) CaC2 into 1, 3, 5 hexatriene (c) Trimethylsecbutyl amonium hydroxide into 1,4-butan-dial (d) Cyclo hexanol into trans cyclo hexane-1, 2-diol
Q.21
How will you conver (a) Hexane dial in to 1,3,5 hexatriene (b) 1-methyl propyl ethanoate into 1,4-dichloro-2-butene
S N O B R A C O R D Y H 0 4 f o 3 2 e g a P
Q.22
Explain the mechanism of following conversion.
Me2C = CH – CH 2 CH 2 CMe = CHCHO + H3O+ Citral Q.23
→
When citral is allowed to react in presence of dilute acid with olivetol, there is obtained a products, one of which is drug marijuana. Reaction is as follows.
S N O B R A C O R D Y H 0 4 f o mixture of 4 2 e g a P
⊕
Me2C = CH – CH2–CH2–CMe=CH–CHO +
H →
(marijuana) Explain the mechanism. Q.24
The following cyclisation has been observed in the oxymercuration & demercuration of this unsaturated alcohol. Propose a mechanism for this reaction. 1. Hg ( OAC ) → 2
2. NaBH 4
Q.25 (a) (b)
Write the structural formula of limonene from the following observation: Limonene when treated with excess H2 & Pt catalyst, the product formed is 1 isopropyl · 4 methyl cyclohexane When it is treated with O3 & then Zn/H2O the products of the reaction are HCHO & following compound
Q.26
(a)
MeCH2–C≡CBr + CH≡CMe
(b)
Cl | CH 2 − CHCl 2
(c)
OH | CH 2 = CH − CH − CH = CH − CH 3
+
Cu → A
− OH
→ B
MeOH → H2SO4
2+
(d) (e)
(f)
Hg → H 2SO4 / H 2O
D
HOBr Cl3C–CH=CH2 → E
O ZnH 2O H⊕ F 3 → G →
C
Q.27
Acetylene is acidic but it does not react with NaOH or KOH. Why?
Q.28
CH≡C–CH2–CH=CH2, adds up HBr to give CH≡C–CH2–CHBr–CH3 while CH≡C–CH=CH2 adds up HBr to give CH2=C . Br . CH=CH2.
Q.29
Chlorination of ethane to ethyl chloride is more practicable than the chlorination of n-pentane to 1-chloropentane.
Q.30
Why n-pentane has higher boiling point than neopentane?
EXERCISE–III Q.1
0.37 gm of ROH was added to CH3MgI and the gas evolved measured 112 cc at STP. What is the molecular wt. of ROH ? On dehydration ROH gives an alkene which on ozonolysis gives acetone as one of the products. ROH on oxidation easily gives an acid containing the same number of carbon atoms. Gives the structures of ROH and the acid with proper reasoning.
Q.2
An alkane A (C5H12) on chlorination at 3000 gives a mixture four different mono chlorinated derivatives B, C, D and E. Two of these derivatives give the same stable alkene F on dehydrohalogenation, On oxidation with hot alkaline KMnO4 followed by acidification of F gives two products G and H. Give structures of A to H with proper reasoning.
Q.3
There are six different alkene A, B, C, D, E and F. Each on addition of one mole of hydrogen gives G which has the lowest molecular wt hydrocarbon containing only one asymmetric carbon atom. None of the above alkene give acetone as a product on ozonolysis. Give the structures of A to F. Identify the alkenes that is likely to give a ketone containing more than five carbon atoms on treatment with a warm conc. solution of alkaline KMnO4.
Q.4
3, 3−dimethyl−1−butene and HI react to give two products, C6H13I. On reaction with alc. KOH one isomer, (I) gives back 3,3−dimethyl−1−butene the other (J) gives an alkene that is reductively ozonized to Me2C=O. Give the structures of (I) and (J) and explain the formation of the later.
Q.5
Three isomeric alkenes A, B and C, C5H10 are hydrogenated to yield 2−methylbutane A and B gave the same 30 ROH on oxymercuration − demercuration. B and C give different 10 ROH’s on hydroboration -oxidation. Supply the structures of A, B & C.
Q.6
Two isomeric alkyl bromides A and B (C5H11Br) yield the following results in the laboratory. A on treatment with alcoholic KOH gives C and D (C5H10). C on ozonolysis gives formaldehyde and 2 methyl propanal. B on treatment with alcoholic KOH gives only C (C5H10). Deduce the structures of A, B, C and D. Ignore the possibility of geometrical and optical isomerism.
Q.7 (a)
Give the structure of A, B and C. A (C4H8) which adds on HBr in the presence and in the absence of peroxide to give the same product C4H9Br. B (C4H8) which when treated with H2SO4 / H2O give (C4H10O) which cannot be resoslved into optical isomers. C (C6H12 ), an optically active hydrocarbon on catalytic hydrogenation gives an optically inactive compound C6H14.
(b) (c)
Q.8
An alkylhalide, X, of formula C6H13Cl on treatment with potassium tertiary butoxide gives two isomeric alkenes Y and Z (C6H12). Both alkenes on hydrogenation gives 2, 3−dimethylbutane predict the structures of X, Y and Z.
Q.9
Identify a chiral compound C, C10H14, that is oxidized with hot KMnO4 to Ph COOH, and an achiral compound D, C10H14, inert to oxidation under the same conditions.
S N O B R A C O R D Y H 0 4 f o 5 2 e g a P
Q.10
Q.11
S N O B R A C O R D Y H 0 4 f o Three compounds A, B and C are isomers of the formula C5H8. All of them decolorises bromine in CCl4 6 2 e and gives a positive test with Baeyer’s reagent. All the three compounds dissoslve in conc. H2SO4. g a P
C7H14 (A) decolorises Br2 in CCl4 and reacts with Hg(OAC)2 in THF. H2O followed by reduction with NaBH4 to produced a resolvable compound B. (A) undergoes reductive ozonolysis to give the same compound (C) obtained by oxidation of 3−hexanol with KMnO4. Identity A, B and C compound D, and isomers of A reacts with BH3. THF and then H 2O2 /OH to give chiral E. Oxidation of E with KMnO4 or acid dichromate affords a chiral carboxylic acid F. Reductive Ozonolysis of D, gives the same compound G which is obtained by oxidation of 2−methyl−3pentanol with KMnO4 identify D, E, F and G.
Compound A gives a white ppt. with ammonical silver nitrate whereas B and C do not. On hydrogenation in presence of Pt catalyst, A and B both yield n−pentane whereas C gives a product of formula C5H10. On oxidation with hot alkaline KMnO4 (B) gives acetic acid and CH3CH2COOH. IdentifyA, B & C. Q.12
An unsaturated hydrocarbon (A) C6H10 readily gives (B) on treatment with NaNH2 in liquid NH3. When (B) is allowed to react with 1−chloropropane a compound (C) is obtained. On partial hydrogenation in the presence of lindlar’s catalyst, (C) gives (D), C9H18. On ozonolysis, (D) gives 2, 2−dimethylpropanal and 1−butanal with proper reasoning give the structures of (A) (B), (C) and (D).
Q.13
A hydrocarbon A, of the formula C8H10, on ozonolysis gives compound B (C 4H6O2) only. The compound B can also be obtained from the alkylbromide (C3H5Br) upon treatment with magnesium in dry ether, followed by carbondioxide and acidification. IdentifyA, B and C and also give equations for the reactions.
Q.14
An organic compound (A), C6H10 on reduction first gives (B), C6H12 and finally (C), C6H14. (A) on ozonolysis followed by hydrolysis gives two aldehydes (D), C2H4O and (E) C2H2O2. Oxidation of (B) with acidified KMnO4 gives the acid (F), C4H8O2. Determine the structures of the compounds (A) to (F) with proper reasoning.
Q.15
Compound A (C6H12) is treated with Br2 to form compound B (C6H12Br2). On treating B with alcoholic KOH followed by NaNH2 the compound C (C6H10) is formed. C on treatment with H 2 /Pt forms 2-methylpentane. The compound ‘C’ does not rea ct with ammonical Cu2Cl2 or AgNO3. When A is treated with cold KMnO4 solution, a diol D is formed which gives two acids E and F when heated with KMnO4 solution. Compound E is found to be ethanoic acid. Deduce the structures from A to F.
Q.16
An optically active hydrocarbon (A), C8H12 gives an optically inactive compound (B) after hydrogenation. (A) gives no ppt. with Ag(NH3)2+ and gives optically inactive (C), C8H14 with H2 in presence of Pd / BaSO4. Determine the structures, give suitable names for A, B, C & give your reasoning.
Q.17
A organic compound A having carbon and hydrogen, adds one mole of H2 in presence of Pt catalyst to form normal hexane. On vigorous oxidation with KMnO4, it gives a simple carboxylic acid containing 3 carbon atoms. Assign the structure to A.
Q.18
An organic compound A, C6H10 , on catalytic reduction first gives B, C6H12 , and finally C, C6H14. A on ozonolysis followed by hydrolysis gives two aldehydes D, C2H4O and E, C2H2O2. Oxidation of B with acidified KMnO4 gives acid F.
Q.19
A hydrocarbon has 88.89% carbon and 11.11% hydrogen. 0.405 g sample of the hydrocarbon occupies 229.54 ml at 100°C and 1 atm pressure. It decolourises potassium permanganate solution and bromine water without evolving hydrobromic acid. It gave no precipitate with either ammoniacal silver nitrate or cuprous chloride solution. When it reacts with dilute H 2SO4 in presence of mercuric sulphate, under appropriate conditions, methyl ethyl ketone is formed. What is the hydrocarbon. Write the structural formulae of the eight possible isomer of this compound.
Q.20
6g sample of a natural gas consisting of methane (CH4) and ethylene (C2H4) was burned with excess of oxygen and 17.2g of carbon dioxide and some water was obtained as products. What percent by weight of the sample was ethylene.
EXERCISE–IV (A) Q.1
Alcoholic solution of KOH is a specific reagent for – (A) Dehydration (B) Dehydrogenation (C) Dehydro halogenation (D) Dehalogenation
[IIT ‘90]
Q.2
Of the following, unsaturated hydrocarbons are – (A) ethyne (B) cyclohexane (C) n–propane
[IIT ‘90]
Q.3
1-chlorobutane on reaction with alcoholic potash gives – (A) 1–butene (B) 1–butanol (C) 2–butene
(D) ethene [IIT ‘91]
(D) 2–butanol
Q.4
The hybridisation of carbon atoms in C–C single bond of HC≡C–CH=CH2 is – (A) sp3–sp3 (B) sp2–sp3 (C) sp–sp2 (D) sp2–sp2
Q.5
The product(s) obtained via oxymercuation (HgSO4 + H2SO4) of 1–butyne would be – (A) CH 3 − CH 2 − C − CH 3 || O
(B) CH3–CH2–CH2–CHO
(C) CH3–CH2–CHO + HCHO
(D) CH3–CH2–COOH + HCOOH
[IIT ‘91]
Q.6
When cyclohexane is poured on water, it floats, because – [IIT ‘97] (A) Cyclohexane is in ‘boat’ form (B) Cyclohexane is in ‘chair’ form (C) Cyclohexane is in ‘crown’ form (D) Cyclohexane is less dense than water
Q.7
[IIT ‘98] Which of the following compounds will show geometrical isomerism? (A) 2–butene (B) Propene (C) 1–phenylpropene (D) 2–methyl–2–butene
Q.8
In the compound CH2=CH–CH2–CH2–C≡CH, the C2–C3 bond is of the type – (A) sp–sp2 (B) sp3–sp3 (C) sp–sp3 (D) sp2–sp3
Q.9
Which one of the following alkenes will react fastest with H2 under catalytic hydrogenation condition – [IIT ‘2000] (A)
(B)
(C)
[IIT ‘99]
(D)
[IIT ‘2000] (D) AgNO3 in ammonia
Q.10
Propyne and propene can be distinguished by – (A) conc. H2SO4 (B) Br2 in CCl4 (C) dil. KMnO4
Q.11
In the presence of peroxide, hydrogen chloride and hydrogen iodide do not give anti–Markovnikov addition to alkene because – [IIT S‘2001] (A) both are highly ionic (B) one is oxidising and the other is reducing (C) one of the step is endothermic in both the cases (D) All the steps are exothermic in both cases
S N O B R A C O R D Y H 0 4 f o 7 2 e g a P
Q.12
Hydrogenation of the above compound in the presence of poisoned paladium catalyst gives – (A) An optically active compound (B) An optically inactive compound [IIT ‘2001] (C) A racemic mixture (D) A diastereomeric mixture [IIT ‘2001]
Q.13
The reaction of propene with HOCl proceeds via the addition of – (A) H+ in first step (B) Cl+ in first step (C) OH– in first step (D) Cl+ and OH– in single step
Q.14
The nodal plane in the π–bond of ethene is located in – [IIT ‘2002] (A) the molecular plane (B) a plane parallel to the molecular plane (C) a plane perpendicular to the molecular plane which contains the carbon–carbon σ–bond at right angle (D) a plane perpendicular to the molecular plane which contains the carbon–carbon σ–bond
Q.15
Consider the following reactions – H 3C − CH − CH − CH 3 + | | D CH 3
[IIT ‘2002]
→ ‘X’ + HBr
Identify the structure of the major product ‘X’ •
(A) H 3C − CH − CH − C H 2 | | D CH 3 •
(C) H 3C − C− CH − CH 3 | | D CH 3
•
(B) H 3C − CH − C− CH 3 | | D CH 3 •
(D) H 3C − C H − CH − CH 3 | CH 2
Q.16
Identify a reagent from the following list which can easily distinguish between 1–butyne and 2-butyne[IIT ‘2002] (A) bromine, CCl4 (B) H2, Lindlar catalyst (C) dilute H2SO4, HgSO4 (D) ammonical Cu2Cl2 solution
Q.17
C6H5–C≡C–CH3
(A)
HgSO
4 →A
H 2SO 4
(B)
[IIT ‘2003]
(C) C6 H5 − C = CHCH 3 (D) C 6H 5 − CH = C − CH 3 | | OH OH
S N O B R A C O R D Y H 0 4 f o 8 2 e g a P
H+
→
Q.18
− H 2O
x Br2 ( mixture) → 5 compounds of molecular formula C4H8Br2
Number of compounds in X will be: (A) 2 (B) 3 Q.19
[IIT ‘2003]
(C) 4
2–hexyne can be converted into trans–2–hexene by the action of : (A) H2–Pd-BaSO4 (B) Li in liq. NH3 (C) H2–PtO2
(D) 5 [IIT ‘2004]
(D) NaBH4
Q.20
When Phenyl Magnesium Bromide reacts with tert. butanol, which of the following is formed? (A) Tert. butyl methyl ether (B) Benzene (C) Tert. butyl benzene (D) Phenol [IIT ‘2005]
Q.21
1–bromo–3–chlorocyclobutane when treated with two equivalents of Na, in the presence of ether which of the following will be formed? [IIT ‘2005] (A)
Q.22
(B)
(C)
(D)
Cyclohexene is best prepared from cyclohexanol by which of the following: (A) conc. H3PO4 (B) conc. HCl/ZnCl2 (C) conc. HCl (D) conc. HBr [IIT ‘2005]
Q.23
Q.24
CH3–CH=CH2 + NOCl Identify the adduct.
→P
[IIT 2006]
CH 3 − CH − CH 2 | | (A) Cl NO
CH 3 − CH − CH 2 | | (B) NO Cl
NO | CH 3 − CH 2 − CH (C) | Cl
CH 2 − CH 2 − CH 2 | (D) | NO Cl
v
Cl 2 , h fractional distillation → M(isomeric products) → N(isomeric products) C5H11Cl
What are N and M? (A) 6, 6
(B) 6, 4
(C) 4, 4
(D) 3, 3
[IIT 2006]
S N O B R A C O R D Y H 0 4 f o 9 2 e g a P
EXERCISE–IV (B)
Q.1 Q.2
Cl | alc. KOH (CH 3 ) 2 C − CH 2CH 3 → ? alc. KOH
HBr → ? C6 H 5CH 2CHCH 3 → ? heat | Br
[IIT 1992] [IIT 1993]
Q.3
C(C6H12), an optically active hydrocarbon which on catalytic hydrogenation gives an optically inactive compound, C6H14. [IIT 1993]
Q.4
Draw the stereochemical structure of the product in the following reactions. R–C≡C–R
[IIT 1994]
H2 →
Lindlar catalyst
Q.5
Write down the structures of the stereoisomers formed when cis–2–butene is reacted with bromine. [IIT 1995]
Q.6
An organic compound E(C5H8) on hydrogenation gives compound F(C5H12). Compound E on ozonolysis gives formaldehyde and 2–ketopropanal. Deduce the structure of compound E. [IIT 1995]
Q.7
Give the structures of the major organic products from 3–ethyl–2–pentene under each of the following [IIT 1996] reaction conditions. HBr in the presence of peroxide (b) Br2 /H2O Hg(OAc)2 /H2O; NaBH4
(a) (c) Q.8
An alkyl halide, (X) of formula C6H13Cl on treatment with potassium tertiary butoxide gives two isomeric alkenes (Y) and (Z) (C6H12). Both alkenes on hydrogenation give 2, 3–dimethylbutane. Predict the structures of (X), (Y) and (Z) [IIT 1996]
Q.9
3,3–Dimethyl–butan–2–ol loses a molecule of water in the presence of concentrated sulphuric acid to [IIT 1996] give tetramethylethylene as a major product. Suggest a suitable mechanism.
Q.10
One mole of the compound A(molecular formula C8H12), incapable of showing stereoisomerism, reacts with only one mole of H2 on hydrogenation over Pd. A undergoes ozonolysis to give a symmetrical diketone B (C8H12O2). What are the structure of A and B? [IIT 1997]
Q.11
Compound (A) C6H12 gives a positive test with bromine in carbon tetrachloride. Reaction of (A) with alkaline KMnO4 yields only (B) which is the potassium salt of an acid. Write structure formulae and IUPAC name of (A) and (B). [IIT 1997]
Q.12
The central carbon–carbon bond in 1,3–butadiene is shorter than that of n–butane. Why? [IIT 1998] Write the intermediate steps for each of the following reaction [IIT 1998] C6H5CH(OH)C≡CH → C6H5CH=CHCHO
Q.13
Q.14
Write the intermediate steps for each of the following reaction. +
H →
[IIT 1998]
S N O B R A C O R D Y H 0 4 f o 0 3 e g a P
Q.15 Q.16
Discuss the hybridisation of carbon atoms in allene (C3H4) and show the π–orbital overlaps. [IIT 1999] [IIT 1999] Complete the following – 1 2 3 → → →
Q.17
Complete the following –
[IIT 1999]
4 5 6 → → →
Q.18
Explain briefly the formation on the products giving the structures of the intermediates.
HCl →
(i)
+
[IIT 1999]
+ etc.
But
HCl →
(ii)
Explain the non formation of cyclic product in (ii) Q.19
Carry out the following transformation in not more than three steps.
CH3–CH2–C≡C–H
→
[IIT 1999]
O || CH 3 − CH 2 − CH 2 − C − CH 3
Q.20
CH2=CH– is more basic than HC≡C–
[IIT 2000]
Q.21
What would be the major product in each of the following reactions?
[IIT 2000]
(i)
CH 3 | CH 3 − C − CH 2 Br | CH 3
C H OH
5 2 →
∆
(ii)
H2 →
Lindlar 's Catalyst
Q.22
On reaction with 4N alcoholic KOH at 175 °C 1–pentyne is slowly converted into equilibrium mixture of 1.3% 1–pentyne (A), 95.2% 2–pentyne (B) and 3.5% 1,2–pentadiene (C). Give the suitable mechanism of formation of A, B and C with all intermediates. [IIT 2001]
Q.23
Identify X, Y and Z in the following synthetic scheme and write their structures. Is the compound Z optically active? Justify your answer. [IIT 2002] (i ) NaNH
H / Pd − BaSO
alkaline KMnO 4 2 4 → X 2 CH3CH2C≡C–H → Y → Z
(ii ) CH3CH 2 Br
S N O B R A C O R D Y H 0 4 f o 1 3 e g a P
Q.24 (a) (b)
Q.25
A biologically active compound, Bombykol (C16H30O) is obtained from a natural source. The structure S N O of the compound is determined by the following reactions. B On hydrogenation, Bombykol gives a compound A, C16H34O, which reacts with acetic anhydride to R A C give an ester. O Bombykol also reacts with acetic anhydride to give another ester, which on oxidative ozonolysis R D Y (O3 /H2O2) gives a mixture of butanoic acid, oxalic acid and 10-acetoxy decanoic acid. H Determine the number of double bonds in Bombykol. Write the structures of compound Aand Bombykol. 0 4 f o How many geometrical isomers are possible for Bombykol? [IIT 2002] 2 If after complete ozonolysis of one mole of monomer of natural polymer gives two moles of CH2O CH 3 | andone mole of O = C − CH = O . Identifythe monomer and draw the all-cis structure of natural polymer.. [IIT 2005]
Q.26
(i) O
3 H ,∆ → Y. → X (ii ) Zn / Cl COOH +
3
Identify X and Y.
[IIT 2005]
3 e g a P
ANSWER KEY EXERCISE–I (A) Q.1
D
Q.2
B
Q.3
A
Q.4
B
Q.5
A
Q.6
D
Q.7
C
Q.8
C
Q.9
A
Q.10
C
Q.11
D
Q.12
A
Q.13
B
Q.14
A
Q.15
B
Q.16
D
Q.17
B
Q.18
D
Q.19
A
Q.20
B
Q.21
C
Q.22
D
Q.23
A
Q.24
A
Q.25
B
Q.26
A
Q.27
C
Q.28
B
Q.29
B
Q.30
C
Q.31
A
Q.32
D
Q.33
A
Q.34
B
Q.35
B
Q.36
A
Q.37
A
Q.38
D
Q.39
B
Q.40
B
Q.41
A
Q.42
B
Q.43
B
Q.44
B
Q.45
C
Q.46
B
Q.47
A
Q.48
A
Q.49
A
Q.50
C
Q.51
C
Q.52
C
Q.53
A
Q.54
C
Q.55
C
Q.56
C
Q.57
C
Q.58
A
Q.59
D
Q.60
A
Q.61
A
Q.62
C
Q Q
.. 4 4 4
A
,, B B
,, C
,, D D
Q Q
.. 8 8 8
A
,, B B
,, C
,, D D
EXERCISE–I (B) Q.1
A,C
Q
. 2 2
Q
. 5 5
B B
,, C C C
Q Q
.. 6 6 6
Q
. 9 9
A A
, C C
Q.10
A,B
Q.11
A,B
Q.12
A,C
Q.13
B,C
Q.14
B,C,D
Q.15
A,C
Q.16
A,B,D
Q.17
A,C,D
Q.18
B,C,D
Q.19
A,D
Q.20
D
Q.21
A,C,D
Q.22
(A) 3,4 ; (B) 3,4 ; (C) 1,2 ; (D) 2
Q.23
(c)
Q.24
(c)
Q.25
(d)
Q.27
(A) 4; (B) 3; (C) 1; (D) 3,4
Q.29
(A) P; (B) R; (C) S; (D) Q
A A
,, C C C
Q Q
.. 3 3 3
A
,, B B
,, C
A A
,, B B B
Q Q
7 .. 7 7
A
,, B B
,, C C
(a)
(A) =
, (B) =
Q.26
(a)
Q.28
(A)5,7 ; (B) 4,5 ;(C) 3 ; (D) 1,2,6
EXERCISE–II
Q.1
,, D D
(b)
S N O B R A C O R D Y H 0 4 f o 3 3 e g a P
Q.2
O = CH | + O = CH
Q.3
Q.4
S N O B R A C O R D Y H 0 4 f o 4 3 e g a P
CH 3 | CH 3 − C = CH 2
A + B are two enatiomers
Q.5
Q.6
CH − CH 2 − CH 2 − CH 2 − N − CH 3 | CH 3
Q.7
CH 2
Q.8
C | (i) C–C–C–C=C, (ii) C − C − C = C − C − C , (iii) C − C = C − C , (iv) C–C=C–C–C=C, | | | C C C
=
(v)
Q.9
α-pinene →
(D) =
Q.10
, (A) =
, (E) =
, (B) =
, (C) =
, (F) =
ONa OH | | (A) C − C − C − C , (B) C − C − C − C , (C) C–C–C–C–O–C–C, (D) BrMg–C–C–C–C (E) C − C − C − C − C − C − C − C − C , (F) C–C–C–C=C–C–C–C–C , | OH Br | (G) C − C − C − C − C − C − C − C − C , (H) C–C–C–C≡C–C–C–C–C | Br
Q.11
C−C C−C C−C | | | (A) C − C = C − C − C ≡ C − C , (B) cis C − C = C − C − C = C − C , (C) HOOC − C − COOH , | | | | | C C C C C C−C | (D) H − C − COOH | C
Q.12
Ph − CH − Et | , (D) , Br
(A) PhC≡CMgx, (B) Ph–C≡C–CH2Ar, (C)
(E) Ph–CH=CH–Me trans, (F) Ph − CH − CH − Me (threo mix.), (G) Ph–COOH | | OH OH
(H) cold dil. KMnO4, (I) HCO3H, (J) CF3CH2CH2Br, (K)
Q.13
(i)
, (ii)
, (iii)
, (iv) no reaction
Br | (v) (A) , (B) Ph–C≡C–CH3 Ph − CH − CH 2 − Br
Q.14
Q.26
(i) II is faster , (ii) unstable intermediate
(A) MeCH2–C≡C–C≡CMe, (B)
(D)
Q.25 O − CH 3 | , (C) CH 2 == CH − CH == CH − CH − CH 3
Br OH | | , (E) CCl − CH − CH (F) 3 3
, (G)
S N O B R A C O R D Y H 0 4 f o 5 3 e g a P
EXERCISE–III Q.1
CH– CH2OH
Q.2 (A)
(D)
Q.3(A)
CH3 | CH3–CH–CH2CH3
(B)
CH3 | H3C–CH–CH2CH2Cl
(C)
CH3 | H3C–CH–CH–CH3 | Cl CH3 | H3C–C=CH–CH3
CH3 | CH3–C–CH2CH3 | Cl
(E)
CH3 | ClCH2– CH–CH2CH3
(F)
(G)
CH3COCH3
(H)
CH3COOH
CH2CH2CH3 | CH2=C–CH2CH3
(B)
CHCH2CH3 || CH3– CCH2CH3
(C)
CH=CH–CH3 | CH3–CH–CH2CH3
(D)
CH2–CH=CH2 | CH3–CH–CH2CH3
(E)
CH2CH2CH3 | CH3–C=CH–CH3
(F)
CH2CH2CH3 | CH3–CH–CH=CH2
(G)
CH2CH2CH3 | H3C–C–CH2CH3 | H CH3 CH3 | | CH3–C – CH–CH3 | I
Q.4
(I)
CH3 | CH3CH–C–CH3 | | I CH3
(J)
Q.5
(A) CH3–C=CH–CH3 | CH3
(B)
CH2=CCH2CH3 | CH3
(B)
CH2Br–CH2–CH–CH3 | CH3
Q.6 (A)
(D)
Br | CH3–CH–CH
CH3 CH3
CH3CH=C–CH3 | CH3
(C) CH2=CHCH(CH3)2
(C)
Q.7 (A)
Q.8
CH3–CH=CH–CH3
(X)
(B)
(C)
CH3–CCl–CH–CH3 (Y) | | CH3 CH3
CH2=C –– CH–CH3 | | CH3 CH3
CH2=CH–CH–CH3 | CH3
CH=CH2 | CH3–C* –H | CH2CH3 ] (Z)
CH3–C = C–CH3 | | CH3 CH3
S N O B R A C O R D Y H 0 4 f o 6 3 e g a P
Q.9
(A) PhCH(CH3)CH2CH3
(B) PhC(CH3)3
Q.10
OH | (A) CH3–CH2–C–CH2–CH2CH3 (B) CH3CH2–C–CH2CH2CH3 (C) CH3CH2–C–CH2CH3 || | || CH2 CH3 O (D) CH3 (E) CH3 (F) CH3 | | | CH3–CH–C–CH2CH3 CH3CH–CH–CH2CH3 CH3–CH–CH–CH2–CH3 || | | CH2 CH2OH CO2H (G) CH3–CH – C–CH2–CH3 | || CH3 O
Q.11
(A) CH3CH2CH2–C≡CH3
(B) CH3CH2C≡C–CH3
Q.12
(C) Cyclopentene
CH3 | (A) CH3–C–C≡CH | CH3
CH3 | (B) H3C–C–C≡C– Na+ | CH3
CH3 | (C) H3C–C–C≡C–CH2CH2CH3 | CH3
CH3 | (D) H3C–C–C=CH–CH2CH2CH3 | CH3
Q.13
(A)
Q.14
(A) CH3– CH= CH–CH=CH–CH3 (B) CH3CH2CH2CH=CHCH3 (C) CH3CH2CH2CH2–CH2CH3 (D) CH3CHO (E) CHO (F) CH3CH2CH2COOH | CHO
Q.15
(A) CH3–CH=CH–CH–CH3 | CH3
(B)
(D) CH3–CH – CH – CH–CH3 | | | OH OH CH3
(C)
(B) CH3–CH–CH–CH–CH3 | | | Br Br CH3 (E) CH3COOH
(C)
(F)
H3C–CH–COOH | CH3
Q.16
H | = − − CH 3 C C C − C ≡ C − CH 3 | | | H H CH 3
Q.17
CH3CH2CH = CHCH2CH3
Q.18
(A) CH3–CH=CH–CH=CH–CH3, (B) CH3–CH2–CH=CH–CH2–CH3,
CHO
(C) CH3–CH2–CH2–CH2–CH2–CH3, (D) CH3CHO, (E) |
CHO
CH3–C≡C–CH–CH3 | CH3
, (F) CH3CH2COOH
S N O B R A C O R D Y H 0 4 f o 7 3 e g a P
Q.19
Isomer are : C≡C–C–C, C=C–C=C, C=C=C–C,
Q.20
23.7
,
,
,
,
EXERCISE–IV (A) Q.1
C
Q.2
A,D
Q.3
A
Q.4
C
Q.5
A
Q.6
D
Q.7
A,C
Q.8
D
Q.9
A
Q.10
D
Q.11
C
Q.12
B
Q.13
B
Q.14
A
Q.15
B
Q.16
D
Q.17
A
Q.18
B
Q.19
B
Q.20
B
Q.21
D
Q.22
A
Q.23
A
Q.24
B
EXERCISE–IV (B)
Q.1
Q.3
CH 3 − C = CH − CH 3 | CH 3
Q.2
H | CH 3 − CH 2 − C − CH = CH 2 | CH 3 (C6 H12 )
Q.5
alc. KOH → heat
HBr →
Q.4
Q.6
CH 2 CH 2 || || (E) CH − C − CH 3
Q.7
OH Br | | (a) (CH3–CH2) CH − CH − CH 3 ; (b) (CH3–CH2)2 C − CH − CH 3 ; (c) (C2H5)3C–OH | Br
Q.8
CH 3 CH 3 | | CH 3 − C − CH − CH 3 ; (Y) CH 2 | Cl
CH 3 CH 3 CH 3 CH 3 | | | | = C − C − CH 3 ; (Z) CH 3 − C = C − CH 3
S N O B R A C O R D Y H 0 4 f o 8 3 e g a P
