H
2
O
2
R
2
O
2
Hydrocabon
AIPMT/NEET:
1. Which one of the following compounds will react with NaHCO
3
solution to give sodium salt and carbon
dioxide ? [AIPMT 1999 ]
(*1) acetic acid (2) n-hexanol (3) phenol (4) both (2) and (3)
Sol. Only acetic acid reacts with NaHCO
3
to give sodium salt and CO
2
, because NaHCO
3
is a base and it reacts
with acid only.
This is one of the distinuishing test of carboxylic acids.
R – COOH + NaHCO
3
R – COONa + CO + H
2
O
2. Acetylene and dil. H
2
SO
4
reacts to produce : [AIPMT 1999, AIIMS 2002, RPMT 2003,04]
(1) CH
3
COOH (*2) CH
3
CHO (3) CH
3
COCH
3
(4) None of these
Sol. Acetylene with dil. H
2
SO
4
in the presence of Hg ion forms aldehyde.
2+
Note : Hydration of ethyne yields aldehyde but hydration of higher homologous of alkyne yields ketone.
3. Among the following alkenes, [AIPMT 2000]
1-butene cis-2-butene trans-2-butene
I II III
the decreasing order of stability is :
(1) II > I > III (*2) III > II > I (3) III > I > II (4) I > II > III
Sol. (I)
C C C C
C C
(II)
C C
(III)
C
C C
H
H=2
Stability : III > II > I
H H
H=6
H
C
H=6
4. Propan–1–ol may be prepared by reaction of propene with [AIPMT 2000]
O
||
(1) CH
3
C O O H
(2) H
3
BO
3
(3*) B
2
H
6
/NaOH-H
2
O
2
(4) H
2
SO
4
/H
2
O
Sol. C
C
C

B
2
H
6
/N
aOH
C
C
C
(Anti markonikoff rule)
OH
5. An organic compound A(C
4
H
9
Cl) on reaction with Na/diethyl ether gives a hydrocarbon which on
monochlorination gives only one chloro derivative, then A is [AIPMT 2001]
(1*) t-butyl chloride (2) s-butyl chloride (3) iso-butyl chloride (4) n-butyl chloride
C C C
Sol.
C C Cl
Na/ether
C C C C
C C C
6. When CH
3
CH
2
CHCl
2
is treated with NaNH
2
, the product formed is [AIPMT 2002]
(1) CH
3
–CH=CH
2
(2*) CH
3
–CCH (3) CH
3
CH
2
CH (4) CH
3
CH
2
CH
Sol. CH
3
CH
2
CHCl
2
NaNH
2
CH
3
CCH
7. Reaction of HBr with propene in the presence of peroxide gives : [AIPMT 2004]
(1) Isopropyl bromide (2) 3-bromo propane (3) Allyl bromide (4*) n-propyl bromide
Sol. C
C
C
HB
r
C
C
C
(Anti markonikoff rule)
Br
2
3 2 2 3
8. Camphor is often used in molecular mass determination because : [AIPMT 2004 ]
(1) it is readily available (2) it has a very high cryoscopic constant
(3*) it is volatile (4) it is solvent for organic substances
Sol. Comphor is used in molecular mass determination due to volatile nature. The method is called Rast's comphor
method. Comphor acts as a solid solvent which is volatile, hence can be removed easily.
9. The best method for the separation of naphthalene and benzoic acid from their mixture is : [AIPMT 2005]
(1) Chromatography (2) Crystallisation (3) Distillation (4*) Sublimation
Sol. The best method for the separation of naphthalene and benzoic acid from their mixture is sublimation because
it is applicable for those organic compounds which pass directly from solid to vapour state on heating and
vice - versa on cooling. In these compounds naphthalene is volatile and benzoic acid is non-volatile due to the
formation of dimer Via hydrogen bonding (intermolecular).
10. In the following reaction : [AIPMT 2006; RPMT 2007]
C
2
H
2
(X) CH
3
CHO
what is (X) ?
(1) CH
3
CH
2
OH (2*) CH
2
=CHOH (3) CH
3
CH
2
CHO (4) CH
3
–O–CH
3
OH
Sol. C H
HB
r
CH
CH
Tauto
CH CHO
2 2
R
2
O
2
2
merism
3
11. The order of decreasing reactivity towards electrophilic reagent for the following : [AIPMT 2007]
(a) Benzene (b) Toluene (c) Chloro benzene (d) Phenol
(1) b > d > a > c (2) d > c > b > a (*3) d > b > a > c (4) a > b > c > d
12. For the following : (a) I
–
(b) Cl
–
(c) Br
–
[AIPMT 2007]
the increasing order of nucleophilicity would be :
(*1) Cl
–
< Br
–
< I
–
(2) I
–
< Cl
–
< Br
–
(3) Br
–
< Cl
–
< I
–
(4) I
–
< Br
–
< Cl
–
CH
3
13.
In the reaction :
CH
3
— CH — CH
2
— O — CH
2
— CH
3
+ HI  ............
Which of the following compounds will be formed :
[AIPMT 2007]
(1) CH
3
— CH — CH
3
CH
3
+ CH
3
— CH
2
— OH
(2)
CH
3
— CH — CH
2
— OH + CH
3
— CH
3
CH
3
(*3) CH
3
— CH — CH
2
— OH + CH
3
— CH
2
— I (4) CH
3
— CH — CH
2
—I + CH
3
— CH
2
—OH
CH
3
Sol.
CH — CH — CH — O + CH — CH
HI
CH
3
CH
3
— CH — CH
2
— OH + CH
3
— CH
2
— I
CH
3
CH
3
14. Predict the product ‘C’ obtained in the following reaction of 1-butyne : [AIPMT 2007]
CH
3
CH
2
CCH + HCl  (B)

H
(C)
(1) (*2)
(3) (4)
Sol. CH
3
CH
2
–C CH + HCl

CH
3
– CH
2
– C CH
2
|
Cl
H
I
I
|
CH
3
CH
2
– C – CH
3
|
Cl
15.
CH
3
H
3
C — CH — CH CH
2
+ HBr  A
A (Predominantly) is :
(Markonikoff rule)
[AIPMT 2008]
CH
3
Br CH
3
(*1) CH
3
— C — CH
2
CH
3
Br
(2)
CH
3
— CH — CH — CH
3
CH
3
Br
(3)
CH
3
— CH — CH — CH
3
CH
3
(4)
CH
3
— CH — CH
2
— CH
2
Br
Sol.
CH
3
H
3
C — CH — CH CH
2
+ HBr
CH
3
CH
3
— C — CH
2
CH
3
CH
3
— C — CH
2
— CH
3
Br Br
Br¯
Br
CH
3
— C — CH
2
— CH
3
CH
3
16. Which one of the following is most reactive towards electrophilic attack ? [AIPMT 2008]
(1)
NO
2
(*2)
OH
(3)
Cl
(4)
CH
2
OH
Sol.
OH
is most reactive towards electrophilic attack due to +M effect of OH group.
17. The homologue of ethyne is : [AIPMT 2008]
(1) C
2
H
2
(2) C
2
H
6
(3) C
3
H
8
(*4) C
3
H
4
Sol. CH
CH

Hom
olo
gue
s
C H
( CH
2
) 3 4
18. H
2
COH.CH
2
OH on heating with periodic acid gives [AIPMT 2009]
|
(1) 2CO
2
(2) 2HCOOH (3)
CHO
(4*) 2 C = O
CHO
Sol.

HIO
4
2
19. Benzene reacts with CH
3
Cl in the presence of anhydrous AlCl
3
to form : [AIPMT 2009]
(1) Xylene (*2) Toluene (3) Chlorobenzene (4) Benzylchloride
CH
3
Sol.
CH
3
Cl
AlCl
3
20. Nitrobenzene can be prepared from benzene by using a mixture of conc. HNO
3
and conc. H
2
SO
4
. In the
mixture, nitric acid acts as a/an : [AIPMT 2009]
(1) Catalyst (2) Reducing agent (3) Acid (*4) Base
Sol. HNO
3
acts as Base and H
2
SO
4
acts as Acid.
21. The reaction of toluene with Cl
2
in presence of FeCl
3
gives 'X' and reaction in presence of light gives 'Y'. Thus,
'X' and 'Y' are : [AIPMT 2009]
(1) X = Benzyl chloride, Y = m-chlorotoluene
(2) X = Benzal chloride, Y = o-chlorotoluene
(3) X = m-chlorotoluene, Y = p-chlorotoluene
(*4) X = o- and p-chlorotoluene, Y = Trichloromethyl benzene
22. The correct order of increasing reactivity of C–X bond towards nucleophile in the following compounds is :
X X
NO
2
(CH
3
)
3
C—X
(CH
3
)
2
CH—X
[AIPMT 2010]
(I)
NO
2
(II)
(III) (IV)
(1) III < II < I < IV (*2) I < II < IV < III (3) II < III < I < IV (4) IV < III < I < II
23. Which of the following species is not electrophilic in nature? [AIPMT 2010]
(1)
Cl
(2) BH
3
(*3)
H
3
O
(4) NO
2
Sol. Electrophiles are electron deficient species. Among the given, H
3
O
has lone pair of electrons for donation,
thus it is not electron deficient and hence, does not behave like an electrophile.
24. Which one is a nucleophilic substitution reaction among the following ? [AIPMT Pre-2011]
(1) CH
3
— CH CH
2
+ H
2
O
H
+
CH
3
— CH — CH
3
OH
(2)RCHO + R'MgX R — CH — R'
OH
2
3
3
CH
3
CH
3
(*3)
CH
3
— CH
2
— CH — CH
2
Br + NH
3
CH
3
— CH
2
— CH — CH
2
NH
2
(4) CH
3
CHO + HCN  CH
3
CH(OH)CN
Sol. (1) It is addition Reaction
(2) It is addision reaction
(3) It is substitution reaction
(4) It is addition reaction
25. In Duma's method of estimation of nitrogen 0.35 g of an organic compound gave 55 mL of nitrogen collected
at 300 K temperature and 715 mm pressure. The percentage composition of nitrogen in the compound would
be : (Aqueous tension at 300 K = 15 mm) [AIPMT 2011]
(1) 15.45 (2*) 16.45 (3) 17.45 (4) 14.45
Sol. In Duma’s method of estimation of nitrogen :
Calculation :- volume of N
2
at NTP (By gas equation)
–
1
v
273
V ml.
t 273
760

% of nitrogen in given compound
28
V
× 100
22400 W
Here, W = 0.35 gm.
= 715 mm (Pressure at which N collected)
= aqueous tension of water = 15 mm.
(t + 273) K = 300 K
v ml = volume of moist nitrogen in nitrometer = 55 ml.
so volume of N at NTP = (V) =
(715 – 15) 55
273
= 46.098 ml.
2
300 760
% of nitrogen =
28
22400
46.098
100
= 16.45 %
0.35
26. The Lassaigne’s extract is boiled with conc. HNO
3
while testing for halogens. By doing so it :
[AIPMT (Scre) 2011]
(1*) decomposes Na
2
S and NaCN, if formed. (2) helps in the precipitation of AgCl.
(3) increases the solubility product of AgCl. (4) increases the concentration of NO
–
ions.
Sol. NaCN + HNO
3
— NaNO + HCN
27. Which of the following compunds undergoes nucleophilic substitution reaction most easily ?
[AIPMT Mains-2011]
(1)
Cl
Cl
(*2)
NO
2
(3)
Cl
CH
3
(4)
Cl
OCH
3
Cl
Sol.
NO
2
undergoes nucleophilic substitution reaction most easily due to –M effect of NO
2
group.
28. Considering the state of hybridization of carbon atoms, find out the molecule among the following which is
linear ?
[AIPMT 2011]
(1) CH –CH=CH–CH (2*) CH –CC–CH
3 3 3 3
(3) CH =CH–CH –CCH (4) CH –CH –CH –CH
2 2 3 2 2 3
Sol. CH –C C – CH (linear)
3 3
29. In the following reaction :
H
2
O/H
A
+
B

Major Product
Minor Product
The major product is : [AIPMT 2012]
(1*) (2)
(3) (4)
Sol.
30. Which of the following reagents will be able to distinguish between 1-butyne and 2-butyne? [AIPMT 2012]
(1*) NaNH
2
(2) HCI (3) O
2
(4) Br
2
Sol. 1-Butyne and 2-butyne are distinguish by NaNH
2
because 1-Butyne react with NaNH
2
due to active hydrogen.
31. Acetone is treated with excess of ethanol in the presence of hydrochloric acid. The product obtained is :
(1)
(CH
3
)
2
C
OH
OC
2
H
5
(*2)
(CH
3
)
2
C
OC
2
H
5
[AIPMT Pre-2012]
OC
2
H
5
Sol.
O
(3)
CH
3
CH
2
CH
2
– C – CH
3
C O + CH
3
CH
2
OH
H
3
C
HCl
H
3
C
H
3
C
O
(4)
CH
3
CH
2
CH
2
– C – CH
2
CH
2
CH
3
OC
2
H
5
C
OC
2
H
5
32. Among the following compunds the one that is most reactive towards electrophilic nitration is :
[AIPMT Pre-2012]
(*1) Toluene (2) Benzene (3) Benzoic acid (4) Nitrobenzene
Sol. More electron rich benzen ring is most reactive towards electrophilic nitration.
33. When trans-2-Butene is reacted with Br
2
then product is formed : [AIPMT-2013]
(1) Racemic-2, 3-dibromobutane (*2) Meso-2, 3-dibromobutane
(3) d-2, 3-dibromobutane (4) -2, 3-dibrobutane
H
3
C
3 3
H
3
C
Sol.
H
3
C
C C
H
CH
3
Br
2
Anti add.
H
H
3
C
H
3
C
H
Br
(meso)
Br
34. Which of the following compounds will not undergo Friedal-Craft's reaction easily : [NEET UG-2013]
(1) Toluene (2) Cumene (3) Xylene (*4) Nitrobenzene
Sol. Nitrobenzene will not undergo Friedal-Craft's reaction easily due to electron deficient benzene ring.
35. Among the following ethers, which one will produce methyl alcohol on treatment with hot concentrated HI ?
[NEET UG-2013]
(1) CH
3
– CH — CH
2
— O — CH
3
(2) CH
3
— CH
2
— CH
2
— CH
2
— O — CH
3
CH
3
CH
3
(3)
CH
3
– CH
2
— CH — O — CH
3
(*4)
CH — C —O — CH
CH
3
CH
3
CH
3
CH
3
Sol.
CH —C—O—CH
HI
CH —C—I + CH
OH
3 3
SN'
3 3
CH
3
CH
3
36. Which one of the following is not a common component of Photochemical Smog? [AIPMT-2014]
(1) Ozone (2) Acrolein (3) Peroxyacetyl nitrate (*4) Chloroflurocarbons
Sol. CFC (Chloroflurocarbons)
37. Which of the following organic compounds has same hybridization as its combustion product –(CO
2
) ?
[AIPMT 2014]
(1) Ethane (*2) Ethyne (3) Ethene (4) Ethanol
Sol. In Ethyne (CHCH) both carbon atoms are sp hybrid as the hybridisation of combustion product, carbon
atom of O=C=O(CO
2
).
38. In the Kjedahl’s method for estimation of nitrogen present in soil sample, ammonia evolved from 0.75g of
sample neutralized 10ml. of 1M H
2
SO
4
The percentage of nitrogen in the soil is: [AIPMT 2014]
(*1) 37.33 (2) 45.33 (3) 35.33 (4) 43.33
Sol. Volume of 1M H
2
SO
4
= 10 m mol
Volume of NH
3
consumed = 20 m mol
Weight of N =
14 20
g = 0.280 g
1000
% N =
0.280
100 37.33%
0.75
39. Identify Z in the sequence of reactions : [AIPMT 2014]
(*1) CH
3
–(CH
2
)
3
–O–CH
2
CH
3
(2) (CH
3
)
2
CH
2
–O–CH
2
CH
3
(3) CH
3
(CH
2
)
4
–O–CH
3
(4) CH
3
CH
2
–CH(CH
3
)–O–CH
2
CH
3
C
Sol. CH
3
CH
2
CH=CH
2
HBr
H
2
O
2
Br
CH
3
CH
2
CH–CH
2
H
C
2
H
5
ONa
CH
3
–(CH
2
)
3
–OCH
2
CH
3
HBr in presence of peroxide gives anti Markovnikoff addition product.
1ºalkyl halide on reaction with C
2
H
5
ONa gives S
N
2 reaction.
40. A single compound of the structure :- [AIPMT-2015]
CH
3
CH
3
OHC
C
H
2
C
C
O
H
2
is obtainable from ozonolysis of which of the following cyclic compounds ?
(1)
H
3
C
H
3
C
(2)
H
3
C
CH
3
(3)
CH
3
CH
3
(*4)
H
3
C
CH
Sol.
H
3
C
CH
3
O
3
Zn/H
2
O
H
3
C
CH
3
CH
2
CH O
41. Given :- [AIPMT-2015]
H
3
C
CH
3
H
3
C
CH
2
H
2
C
CH
2
CH
3
CH
3
CH
2
(I) (II) (III)
The enthalpy of the hydrogenation of these compounds will be in the order as :-
(*1) III > II > I (2) II > III > I (3) II > I > III (4) I > II > III
1
Sol. Heat of Hydrogenation
Stability
42. The total number of -bond electrons in the following structure is :- [AIPMT-2015]
H
3
C
H
3
C
H H
H
CH
3
H
2
C
H
CH
3
(1) 8 (2) 12 (3) 16 (*4) 4
Sol. Two -bonds contains electrons.
43. The reaction of C
6
H
5
CH = CHCH
3
with HBr produces: [AIPMT-2015]
O
3
H
( 2) CH CH Br ( 2) CH CH Br
CH=CHCH
3
(1)
C
6
H
5
CH
2
CHCH
3
|
Br
(2) C
6
H
5
CH
2
CH
2
CH
2
Br (3)
Br
(*4) C
6
H
5
CHCH
2
CH
3
|
Br
Sol. C H CHCHCH

HBr
C
H
CH
CH
CH
(Markonikoff rule)
6 5 3
6 5 2 3
Br
44. In the reaction with HCl, an alkene reacts in accordance with the Markovnikov's rule, to give a product
1-chloro-1-methylcyclohexane. The possible alkene is : [Re-AIPMT-2015]
CH
2
CH
3
CH
3
(1) (2) (*3) (1) and (2) both (4)
Sol.
CH
2
HCl
CH
3
HCl
45. 2,3-Dimethyl-2-butene can be prepared by heating which of the following compounds with a strong acid ?
[Re-AIPMT-2015]
(1)
(CH
3
)
2
CH – CH – CH = CH
2
CH
3
(*2) (CH
3
)
3
C – CH = CH
2
(3) (CH
3
)
2
C = CH – CH
2
– CH
3
(4) (CH
3
)
2
CH – CH
2
– CH = CH
2
H
3
C
CH
3
Sol.
(CH
3
)C–CH=CH
2
S.A.
(CH
3
)
3
C–CH–CH
3
(CH
3
)
2
–C–CH–CH
3
CH
3
H
3
C
C=C
CH
3
46. In the reaction [NEET 2016- Phase -I]
H—C CH
(1)N
aNH
2
/liq
.NH
3
x
(1)N
aNH
2
/liq
.NH
3
y
3 2 3 2
X and Y are :
(1) X = 1-Butyne ; y = 2-Hexyne (*2) X = 1-Butyne ; y = 3-Hexyne
(3) X = 2-Butyne ; y = 3-Hexyne (4) X = 2-Butyne ; 2 = 2-Hexyne
CH
3
Cl
CH
3
Cl
3 2
0
C
0C
NaNH
2
/liq.NH
3
CH 
NaNH
Sol. H—C CH
C
H
CH
Br
HC
C¯

3
C
H
2
HC
C–CH
2
–CH
3

2
¯C
C–CH
2
–CH
3

CH
3
C
H
2
B
r
CH –CH –C
C–CH –CH
3 2 2
3
47. Which of the following compounds shall not produce propene by reaction with HBr followed by elimination or
direct only elimination reaction ? [NEET-2016 Phase -II ]
H
2
(1)
H
3
C—
C
—CH
2
Br (2) CH
2
CH
2
C
H
2
H
2
(3) H
3
C— C —CH
2
OH (*4) H
2
C=C=O
48. In the given reaction : +

HF
P. The product P is [NEET -2016 Phase -II ]
(1) (2) (3) (*4)
Sol. 
HF

49. Predict the correct intermediate and product in the following reaction [NEET -2017 ]
H C — C CH
H
2
O, H
2
SO
4
intermediate product
3 HgSO
4
(A)
(B)
(1)
A : H
3
C–C=CH
2
SO
4
(3) A : H
3
C–C=CH
3
O
B : H
3
C–C–CH
3
O
B : H
3
C–CCH
(2)
A : H
3
C–C=CH
2
OH
(*4) A : H
3
C–C=CH
2
OH
B : H
3
C–C=CH
2
SO
4
B : H
3
C–C–CH
3
O
Sol. H C — C CH
H
2
O, H
2
SO
4
3
HgSO
H
3
C–C=CH
2
OH
O
H
3
C–C–CH
3
50. Which one is the correct order of acidity ? [NEET -2017 ]
(1) CH
2
= CH
2
> CH
3
– CH = CH
2
> CH
3
– C CH > CH CH
(*2) CH CH > CH
3
– C CH > CH
2
= CH
2
> CH
3
– CH
3
(3) CH CH > CH
2
= CH
2
> CH
3
– C CH > CH
3
– CH
3
(4) CH
3
– CH
3
> CH
2
= CH
2
> CH
3
– C CH > CH CH
F
F
4
3 2 3
Sol. Stability negative is depends on hybridisation EN of sp > sp
2
> sp
3
51. The most suitable method of separation of 1 : 1 mixture of ortho and para-nitrophenols is [NEET -2017 ]
(1) Sublimation (2) Chromatography (3) Crystallisation (*4) Steam distillation
Sol. The most suitable method of separation of 1 : 1 mixture of ortho and para-nitrophenols is steam distillation.
52. Hydrocarbon (A) reacts with bromine by substitution to form an alkyl bromide which by Wurtz reaction is
converted to gaseous hydrocarbon containing less than four carbon atoms. (A) is : [NEET - 2018 ]
(1) CH
3
–CH
3
(2) CH
2
= CH
2
(3) CH CH (*4) CH
4
Sol. By wurtz reaction only even alkane formation is possible.
53. Which of the following molecules represents the order of hybridisation sp
2
, sp
2
, sp, sp from left to right
atoms ? [ NEET - 2018 ]
(1) CH
2
= CH – CH = CH
2
(*2) CH
2
= CH – C
CH
(3) HC
C – C
CH (4) CH
3
– CH = CH – CH
3
sp
2
sp
2
sp sp
Sol.
CH
2
CH – C CH
54. The most suitable reagent for the following conversion is:- [ NEET -2019 ]
H
3
C
C
C
CH
3
H
3
C
H
CH
3
H
cis-2-butene
(1) Na/liquid NH
3
(*2) H
2
, Pd/C, quinoline
(3) Zn/HCl (4) Hg
2+
/H
+
, H
2
O
Sol. By H
2
, Pd/C, quinoline syn addition takes place.
55. An alkene 'A' on reaction with O
3
and ZnH
2
O gives propanone and ethanal in equimolar ratio. Addition of HCl
to alkene 'A' gives 'B' as the major product. The structure of product 'B' is:- [ NEET -2019 ]
CH
3
(1)
Cl CH
2
CH
2
CH
CH
2
Cl
(2) H C CH CH CH
CH
3
3 2 3
CH
3
(*3)
H
3
C
CH
2
C
CH
3
Cl
CH
3
(4)
H
3
CCH CH
Cl
CH
3
O
Sol.
CH
3
—C—CH
3
propanone
+ OHC–CH
3
H
3
C
H
3
C
CCH–CH
3
CH
3

HCl
H
C
CH
C
CH
Cl
AIIMS:
1. Which of the following compound is formed when CH
2
=CH(CH
2
)
2
COOH react with HBr ? [AIIMS 2000]
(1) CH
3
CHCH
2
CH
2
BrCOOH (2*) CH
3
CHBrCH
2
CH
2
COOH
H
2
O
2
/ OH
(3) CH
2
BrCH
2
(CH
2
)
2
COOH (4) CH
3
CH
2
CH
2
BrCH
2
COOH
Sol. CH =CH(CH ) COOH + HBr  CH CHBrCH CH COOH
2 2 2 3 2 2
2. With ammonical cuprous chloride solution a reddish brown precipitate is obtained on treating with :
[AIIMS 2001]
(1) CH
4
(2) C
2
H
4
(3*) C
2
H
2
(4) C
3
H
6
Sol. HCCH CuCCCu
red ppt.
3. Acetone can easily be converted into propane by the action of : [AIIMS 2002]
(1*) HI (2) H PO
(3) HNO
(4) HIO
3 3 3 3
4. The number of and bonds present in pent-1-ene-4-yne is : [AIIMS 2002]
(1) 3, 10 (2) 9, 4 (3) 4, 9 (4*) 10, 3
Sol. Pent-1-en-4-yne contain 10 and 3 bonds.
5. Propan -1-ol can be prepared from propene by : [AIIMS 2003]
(1) H
2
O / H
2
SO
4
(2) Hg (OAc)
2
/ H
2
O followed by NaBH
4
(3*) B
2
H
6
followed by H
2
O
2
(4) CH
3
CO
2
H / H
2
SO
4
Sol. Propanol-1 can be prepared from propene as follows :
CH
3
CH = CH
2

B
2
H
6

(CH
3
CH
2
CH
2
)
3
B

CH CH = CH
H
O
H

(This addition follow Markownikoff's rule)
3 2
dil.H
2
SO
4
6. 3-phenylpropene on reaction with HBr gives (as a major product) : [AIIMS 2005]
(1*) C
6
H
5
CH(Br)CH
2
CH
3
(2) C
6
H
5
CH
2
CH (Br) CH
3
(3) C
6
H
5
CH
2
CH
2
CH
2
Br (4) C
6
H
5
CH(Br) CH = CH
2
Sol. The addition of HBr on unsymmetrical alkene take palce on the basis of Markownikoff's addition.
Br
–
C H CH(Br)CH CH
6 5 2
3
7. Which of the following gives propyne on hydrolysis? [AIIMS 2005]
(1) Al
4
C
3
(2*) Mg
2
C
3
(3) B
4
C (4) La
4
C
3
Sol. Mg C + H O  CH CH
2 3 2 3
8. Which of the following sequence of reactions (reagents) can be used for the conversion of C
6
H
5
CH
2
CH
2
OH
into C
6
H
5
CH = CH
2
? [AIIMS 2006]
(1) SOCl ; H O (2*) SO Cl ; alc. KOH (3) Cl / h ; H O (4) SOCl ; alc. KOH
2 2 2 2 2 2 2
Sol. C H CH CH OH 
SO
2
Cl
2
C H CH CH Cl 
alc.
KO
H
C H CH = CH
6 5 2 2
6 5 2 2
6 5 2
9. The hydrocarbon which does not decolourise alkaline KMnO
4
solution and also does not give any precipitate
with ammonical silver nitrate is : [AIIMS 2007]
(1*) benzene (2) acetylene (3) propyne (4) 1-butyne
Sol. Terminal alkyne decolourise alkaline KMnO
4
solution.
2
10. The compound which is obtained by treating chloropropane with alcoholic KOH, when reacts BH
3
/THF followed
by acetic acid gives : [AIIMS 2009]
(1) CH
3
CH
2
CH
2
OH (2*) CH
3
CH
2
CH
3
(3) CH
3
CH(OH)CH
3
(4) CH
3
CH
2
CHOHCH
3
Sol. CH CH CH Cl 
alc.
KO
H
CH –CH = CH

BH
3
/T
HF
CH CH CH
3 2 2
3 2 2
CH
3
COOH
3 2 3
11. Formation of alcohol by oxymercuration demercuration of alkenes : [AIIMS 2009]
(1) Involves carbocations and rearrangement (2) Involves carbanions and rearrangement
(3) Is stereospecific (4*) Does not involve rearrangement and carbocation
Sol. Formation of alcohol by oxymercuration demercuration of alkenes does not involve rearrangement and
carbocation.
12.
Sol.
2
Butene
HBr
[AIIMS 2010]
Major products of above reaction will be :
(1) 1 (*2) 2 (3) 4 (4) 3
2
Butene
HBr
CH CHBrCH CH
3 2 3
13. Which of the following reaction has zero activation energy ? [AIIMS 2010]

(1) CH
4
+
Cl
CH
3
+ HCl (2) Cl
2
Cl
(3*)
CH
3
+
CH
3
CH – CH
(4)
CH
3
+ Cl – Cl

CH – Cl +
Cl
3 3 3
Sol. In chain terminating step, activation energy does not require.
14. Structural formula of Lewisite is : [AIIMS 2010]
CHCl
CHCl
CHCl
CH
2
(1)
||
CHAsCl
3
(2)
||
CHAsCl
(3*)
||
(4)
CHAsCl
2
||
CHAsCl
2
Sol. Lewisite is obtained when acetylene reacts with arsenic chloride.
CH CH

AsC
l
3
CHCl = CHAsCl
Lewisite
15. The number of -and -bonds present in pent-1-en-4-yne is : [AIIMS 2011]
(1*) 10, 3 (2) 4, 9 (3) 3, 10 (4) 9, 4
Sol. (1) Pent-4-ene-1-yne
CH
CH–CH –C
CH
2 2
No. of –bonds: 4 (C – C) + 6 (C – H) = 10
No. of –bonds: 1 (C == C) + 2 (C

C) = 3
17. Which one of the following alkenes will react faster with H
2
under catalytic hydrogenation conditions ?
[AIIMS, 2012]
R
H
R R
R
R R R
(*1)
R
(2)
H
R
(3)
R
H
(4)
H
R
H
18. In a hydrocarbon, mass ratio of hydrogen and carbon is 1:3, the empirical formula of hydrocarbon is
[AIIMS 2012]
(1*) CH
4
(2) CH
2
(3) C
2
H (4) CH
3
Sol. Mass ratio of H : C = 1 : 12
However, given mass ratio of H : C = 1 : 3 Therefore, for every C atom, there are 4 H atoms, hence empirical
formula = CH
4
.
19. Consider the following statements : A hydrocarbon of molecular formula C
5
H
10
is a [AIIMS 2012]
I. monosubstituted alkene
II. disubstituted alkene
2
(i) B
2
H
6
/THF
(ii) H O /OH
–
2 2
3
3
III. trisubstituted alkene
Which of the following statement(s) is (are) correct?
(1*) I, II and III (2) I and II (3) II and III (4) I and III
20. Which one of the following cannot be prepared by Wurtz reaction ? [AIIMS 2012]
(1*) CH
4
(2) C
2
H
6
(3) C
3
H
8
(4) C
4
H
10
Sol. CH
4
has only one carbon atom, hence it can’t be prepared by Wurtz reaction, which involves two molecules
of alkyl halide.
21. What would be the expected product of the reaction of propyne with Br
2
/H
2
O if the mechanism of this reaction
is analogous to that of propene ? [AIIMS, 2013]
(1) 2-brompropanol (*2) Bromoacetone
(3) 2-bromo-2-propanol (4) Bromopropanol
22. Point out incorrect Sawhorse projection for the following compund. [AIIMS, 2015]
CH
3
H Br
H Br
(*1)
CH
3
H
H
Br
Br
CH
3
(2)
CH
3
Br
H
Br
H
CH
3
CH
3
(3)
H
CH
3
(4)
H
CH
3
Br
23. In the following reaction, [AIIMS, 2017]
(B)
CH
2
H O
+
(A)
(A) and (B) respectively, are
(1)
CH
2
OH
(3)
CH
2
OH
and
Sol.
CH
2
OH
CH
3
OH
CH
2
H O
+
(2)
(*4)
CH
3
OH
CH
3
and
OH
CH
3
OH
CH
2
OH
24. The basic strength of [AIIMS, 2017]
CH C, CH
2
CH, CH
3
CH
2
I II III
Will be in order
(*1) I < II < III (2) II < III < I (3) III < II < I (4) III < I < II
25. For geometrical isomers of 3-hexene, which of the following statement is correct ?
[AIIMS, 2018]
H
C
H
5
C
2
C
2
H
5
C
H
H
C
H
5
C
2
H
C
C
2
H
5
trans-form cis-form
(1) M.P. is high and dipole moment is also high for trans form
(2) M.P. is low and dipole moment is also low for trans form
(*3) M.P. is high but dipole moment is low for trans form
(4) M.P. is low but dipole moment is high for trans form
CH
3
H
Br
Br
CH
3
Br
H
(i) B
2
H
6
/THF
(ii) H O /OH
–
2 2
2 2 3
26. Ph —CH —CH —C C —CH 

Cl
2
/hv
X, where X is [AIIMS, 2018]
(1) Ph —CH(Cl) —
CH
2
H
(*2) Ph —CH(Cl) —
CH
2
CH
3
C C
H
H
C
C
H CH
3
(3) Ph —CH —CH(Cl) —CH CH CH —CH
(4) None of these
Other Exam:
1. Baeyer’s reagent is used : [AFMC 1999]
(1) In Benedict solution (2) For oxidation
(3*) For detection of unsaturation (4) For reduction
Sol. The Baeyer’s reagent is alkaline KMnO
4
solution which is used for the detection of unsaturation
2. Halogenation of alkanes is a example of : [RPMT 2000]
(1) Nucleophilic substitution (2) Oxidation
(3) Electrophilic substitution (4*) Free radical substitution
Sol. Halogenation of alkanes is a example of free radical substitution.
3. Which of the following will have least hindered rotation about carbon-carbon? [RPMT 2000]
(1*) C H (2) CH == CH
(3) CH CH (4) Hexa chloro ethane
2 6 2 2
4. Hydrogen bromide is added to CH
3
CH = CH
2
in the presence of peroxide. The resultant compound formed
is : [RPMT 2000]
(1*) CH
3
CH
2
CH
2
Br (2) CH
3
CHBr.CH
3
(3) C
2
H
5
CH
2
Br (4) None of these
Sol. Antimarkownikoff addition of HBr takes places in presence of peroxide
5. In the following sequence of reaction the end product is :
Hg
2
HC CH

A
H
2
SO
4

CH
3
M
g
X
H
2
O
B 
O
C [RPMT 2000]
(1*) CH
3
COCH
3
(2) CH
3
CHO (3) C
2
H
5
OH (4) CH
3
CH
2
CH
2
OH
Hg
2
Sol. HC CH

CH CHO

CH
3
M
g
X
CH CH(OH)CH

O
CH COCH .
H
2
SO
4
3
H
2
O
3
3 3 3
6. Alkynes shows mainly : [RPMT 2000]
(1) substitutions reaction (2) polymerisation
(3*) electrophilic addition reaction (4) all of the above
Sol. Alkynes shows mainly electrophilic addition reaction .
7. Electrolysis of sodium propionate solution produces ? [RPMT 2001]
(1*) Ethane (2) Acetylene (3) Ethyl alcohol (4) None of these
8. The heat of hydrozination of CH
2
= CH
2
is 30 kcal/mole, what is the heat of hydrozination of 1,3-butadiene?
[RPMT 2001]
(1) 63 kcal/mole (2*) 57 kcal/mole (3) 60 kcal/mole (4) 30 kcal/mole
3
9. According to Markownikoff rule addition of water on 1-butene gives : [RPMT 2001]
(1) Primary alcohol (2) tertiary alcohol (3*) Secondary alcohol (4) none of these
10. Addition of Br
2
on CH
2
== CH–CH
2
–CH
3
gives : [RPMT 2001]
(1) (2*)
(3) (4)
Sol. CH
2
== CH–CH
2
–CH
3
+ Br
2

11. General formula of paraffin is : [RPMT 2002]
(1) C
n
H
2n
(2) C
n
H
2n – 2
(3*) C
n
H
2n + 2
(4) C
2n
H
2n
Sol. General formula of paraffin is C
n
H
2n + 2
.
12. Which compound increase antiknocking property of gasoline? [RPMT 2002]
(1*) (C
2
H
5
)
4
Pb (2) (C
2
H
5
)
2
Pb (3) (CH
3
)
4
Pb (4) (CH
3
)
2
Pb
Sol. (C
2
H
5
)
4
Pb increase antiknocking property of gasoline
13. Normal butane convert into isobutane by : [RPMT 2002]
(1) LiAlH
4
(2*) AlCl
3
(3) NaBH
4
(4) Zn/HCl
Sol. Isomerisation of alkane takes place in presence of AlCl
3
14. Which of the following is not a member of homologus series? [RPMT 2002]
(1) Ethene (2) 1-butene (3) 2-butene (4*) 2-butyne
Sol. In homologus series functional group remains same.
15. Which of following gives H
2
gas with Na? [RPMT 2002]
(1) CH
4
(2) C
2
H
6
(3) C
2
H
4
(4*) C
2
H
2
Sol. HC CH

2N
a
NaC CNa + H
2

16. Calcium carbide on hydrolysis from : [RPMT 2002, AFMC 2006]
(1) ethane (2) ethylene (3) methane (4*) acetylene
17. Both CH
4
and C
2
H
6
can be prepared in one step by the reaction of [RPMT 2003]
(1*) CH
3
Br (2) CH
3
CH
2
OH (3) CH
3
OH (4) CH
3
COCH
3
18. When C
2
H
4
is reacted with H
2
is presence of Ni, then the product formed is : [RPMT 2003]
(1) CH
4
(2*) C
2
H
6
(3) HCOOH (4) HCHO
19. CH CH CH OH

PC
l
5
A 
alc
KO
H
B [AFMC 2003]
3 2 2
B is identified as :
(1) propanal (2) propane (3) propyne (4*) propene
Sol. ROH

PC
l
5
o
r S
OC
l
2
o
r
RCl
PX
3
(XCl, Br,I) or
ZnCl
2
/ HCl
CH CH CH OH
PC
l
5
CH
3
CH CH
2
3 2 2
HCl
Alkene
20. Ethylidenebromide on heating with metallic sodium in ether solution produce : [RPMT 2003]
(1) 1-Butene (2) Ethylene (3) Ethane (4*) 2-Butene
H
HSO
21. Anti Markownikoff rule addition reaction is not observed in : [RPMT 2003]
(1) pent-2-ene (2*) but-2-ene (3) 1-butene (4) propene
22. The hybridization of carbon in CH
3
– CH = C = CH
2
is : [RPMT 2003]
(1) sp
3
sp
3
sp
2
sp
2
(2) sp
3
sp
2
sp sp
3
(3) sp
3
sp
3
sp
2
sp
3
(4*) sp
3
sp
2
sp sp
2
23. When CH
4
, C
2
H
2
, C
2
H
4
and C
2
H
6
gasses are passed through the Wolf bottle of AgNO
3
, then which one the
following gas does not come out from the bottle? [RPMT 2003]
(1) CH
4
(2*) C
2
H
2
(3) C
2
H
4
(4) C
2
H
6
24. Which hydrocarbon is mainly present in gobar gas ? [AFMC 2003]
(1) Butane (2) Propane (3*) Methane (4) Ethane
25. Which of the following product is obtained by treating 1-butyne with HgSO
4
and H
2
SO
4
?
[AFMC 2003]
(1*) CH
3
CH
2
COCH
3
(2) CH
3
CH
2
CH
2
CHO
(3) CH
3
CH
2
CH
2
COOH (4) CH
3
CH
2
CH=CH
2
26. Which of the following yields both alkane and alkene ? [AFMC 2004]
(1*) Kolbe’s reaction (2) Williamson’s synthesis
(3) Wurtz reaction (4) Sandmeyer’s reaction
27. “The addition of unsymmetrical reagents to unsymmetrical alkenes occurs in such a way that the negative
part of the addendum goes to that carbon atom of the double bond which carries lesser number of hydrogen
atoms”. is called by : [AFMC 2004]
(1) Saytzeff rule (2*) Markownikoff’s rule (3) Kharasch effect (4) Anti-Saytzeff rule
28. Octane number can be changed by : [AFMC 2004]
(1) Isomerisation (2) Alkylation (3) Cyclisation (4*) All of these
29. Propene on adding with H, gives : [RPMT 2004]
(1*) CH
– CH – CH
(2) CH
– CH – CH 
3 3 3 2 2
(3) CH – CH – CH (4) None of these
3 2
30. The compound, which decolourise bromine water but does not react with Tollen’s reagent is :
[RPMT 2004]
(1*) ethene (2) ethyne (3) propane (4) ethane
31. Marsh gas is : [RPMT 2004]
(1*) CH
4
(2) C
2
H
6
(3) C
3
H
8
(4) None of these
32. Natural gas is : [RPMT 2004]
(1) CH
4
(2) CO + H
2
+ CH
4
(3*) CH
4
+ C
2
H
6
+ C
3
H
8
(4) All of these
33. Acetylene on treating with Cu
2
Cl
2
gives : [RPMT 2004]
(1*) Cu
2
C
2
(2) CuCHCl (3) CH
4
(4) None of these
34. The only alcohol that cannot be prepared by the indirect hydration of alkene is : [AFMC 2005]
(1) ethyl alcohol (2) propyl alcohol (3) isobutyl alcohol (4*) methyl alcohol
Sol. Methyl alcohol cannot be prepared by hydration of alkene as simplest alkene has two carbons so alcohol of
at least two carbon atoms can be formed.
H C = CH
H
3
C
|


4

2
H
3
C
2
(H
2
SO
4
)
H
2
C
HOH
|
CH
2
OSO
3
H
H
2
C = CH – CH
3

H
2
SO
4
CHCH
2
OH
Ethanol
2 4 3
4 3 3
4
3

HO
H

H C CH CH
HO
CH
CH CH
H
2
SO
4
(Markowniko ff's
rule)
3 3
|
OH
Major product
2 2 3
Minor product
35. X is heated with soda lime and gives ethane. X is : [AFMC 2005]
(1) Ethanoic acid (2) Methanoic acid (3*) Propanoic acid (4) Either (1) or (3)
36. Which gas is liberated when Al
4
Cl
3
is hydrolysed ? [AFMC 2005]
(1*) CH
4
(2) C
2
H
2
(3) C
2
H
6
(4) CO
2
37. The number of double bonds in gammexane is : [RPMT 2005]
(1*) 0 (2) 1 (3) 2 (4) 3
38. Which kind of fission is favoured by sunlight? [RPMT 2006]
(1) Heterolytics fission (2*) Homolytic fission (3) Both (1) and (2) (4) None of these
39. Propionic acid and KOH reacts to produce which one of the following ? [RPMT–2006]
(1*) Potassium propionate(2) Propyl alcohol (3) Propionaldehyde (4) Does not react
40. Which of the following has highest knocking property ? [AFMC 2007]
(1) Aromatic hydrocarbons (2) Olefins
(3) Branched chain paraffins (4*) Straight chain paraffins
41. Which of the following is the most stable alkene ? [AFMC 2007; Ludhiana CMC (Med.) 2008]
(1*) R
2
C = CR
2
(2) RCH=CHR (3) RCH=CH
2
(4) H
2
C=CH
2
42. The compounds formed at anode in the electrolysis of an aqueous solution of potassium acetate, are
[RPMT 2008]
(1*) C
2
H
6
and CO
2
(2) C
2
H
4
and CO
2
(3) CH
4
and H
2
(4) CH
4
and CO
2
43. Baeyer’s reagent decolourises which of the following : [RPMT 2009]
(1) Alkane (2) Alkene
(3*) Alkene and alkyne both (4) None of the above
44. In the free-radical halogenation of alkanes, chain propogating step is : [RPMT 2011]
(1) Cl

h
2Cl
(2) CH + Cl
CH Cl + H
(3*) CH + Cl

CH
+ HCl (4) CH
+ Cl

CH
3
Cl
Sol. CH + Cl

RD
S
CH
+ HCl
It is chain propogation step of halogenation of alkane.
AIEEE:
45. Which of these will not react with acetylene? [AIEEE 2002]
(1*) NaOH (2) ammonical AgNO
3
(3) Na (4) HCl.
Sol. Acetylene reacts with the other three as :
46. On mixing a certain alkane with chlorine and irradiating it with ultraviolet light, it forms only one
monochloroalkane this alkane could be : [AIEEE 2003]
(1) propane (2) pentane (3) isopentane (4*) neopentane.
Sol. The number of monohalogenation products obtained from any alkane depends upon the number of different
2
types of hydrogen it contains. Compound containing only one type of hydrogen gives only one
monohalogenation product.
CH
3
CH
2
CH
3
— two types of hydrogen
propane (two monohalogenation structural product)
CH
3
CH
2
CH
2
CH
2
CH
3
— three types of hydrogen
pentane (two monohalogenation structural product)
— four types of hydrogen (four monohalogenation structural product)
— one types of hydrogen (one monohalogenation structural product)
Thus the given alkane should be neopentane.
47. The prussian blue colour obtained during the test of nitrogen by Lassaigne’s test is due to the formation of :
[AIEEE 2004]
(A*) Fe
4
[Fe(CN)
6
]
3
(B) Na
3
[Fe(CN)
6
] (C) Fe(CN)
3
(D) Na
4
(Fe(CN)
5
NOS]
Sol. 6NaCN + FeSO
4
Na
4
[Fe(CN)
6
]
Na
4
[Fe(CN)
6
] + 4Fe
3+

xH
2
O
Fe
4
[Fe(CN)
6
]
3
. xH
2
O
Ferric ferrocyanide (Prussian blue)
48. Of the five isomeric hexanes,the isomer which can give two monochlorinated compounds is ?
[AIEEE 2005]
(1) n–Hexane (2*) 2,3-Dimethylbutane (3) 2,2-Dimethylbutane (4) 2-Methylpentane
Sol. 2,3-Dimethylbutane has two chemically different hydrogen atoms so it can give two monochlorinated
structural compounds.
49. Alkyl halides react with dialkyl copper reagents to give [AIEEE 2005]
(1) alkenes (2) alkyl copper halides (3*) alkanes (4) alkenyl halides
Sol. In Corey house synthesis of alkane, alkyl halide reacts with lithium dialkyl cuprate.
R CULi + R’X RR’ + RCu + LiX
50. Reaction of one molecule of HBr with one molecule of 1, 3-butadiene at 40°C gives predominantly
(1) 3-bromobutene under kinetically controlled conditions [AIEEE 2005]
(2*) 1-bromo-2-butene under thermodynamically controlled conditions
(3) 3-bromobutene under thermodynamically controlled conditions.
(4) 1-bromo-2-butene under kinetically controlled conditions.
Sol. 1, 2-addition product is kinetically controlled product while 1, 4-addition product is thermodynamically controlled
product and formed at comparitively higher temperature.
CH
2
= CH – CH = CH
2
CH
2
(Br) CH = CHCH
3
+ CH
3
CH(Br) – CH = CH
2
1, 4–addition 1, 2–addition
Therefore, 1 - bromo-2-butene will be the main product under themodynamically dcontrolled conditions.
51. Elimination of HBr from 2-bromobutane result in the formation of : [AIEEE-2005]
(1) Predominantly 2-butyne (2) Predominantly 1-butene
(3*) Predominantly 2-butene (4) Equimolar mixture of 1 and 2-butene
Sol.
52. Reaction of trans 2–phenyl –1 – bromocyclopentane on reaction with alcoholic KOH produces:
[AIEEE-2006]
3
3
(1) 2–phenylcyclopentene (2) 1–phenylcyclopentene
(3*) 3-phenylcyclopentene (4) 4-phenylcyclopentene
Sol.

alc.
KO
H
3–phenylcyclopentene
It is anti elimination reaction so hydrogen atom from second carbon will not eliminated as it is in syn-position
rather hydrogen atom from 5
th
carbon will be eliminated.
53.

Bas
e
/

/
The alkene formed as a major product in the above elimination reaction is : [AIEEE-2006]
(1) (2*) CH
2
= CH
2
(3)
(4)
Sol.
54. Which of the following reactions will yield 2, 2- dibromopropane? [AIEEE-2007, 3/120]
(1) CH CH + 2HBr (2) CH – CH = CH + HBr 
2
(3*) CH
3
– C CH + 2HBr (4) CH CH = CHBr + 2HBr 
Sol. HC CH

H
Br
= CH
2

H
Br

CH
3
– CH = CH
2

H
Br
CH
3
–
CH
3
– C CH

H
Br
CH
3
– = CH
2

H
Br
55. In the following sequence of reactions, the alkene affords the compound 'B'
CH CH = CHCH

O
3
A

H
2
O
B compound B is [AIEEE 2008, 3/105]
3 3
Zn
(1) CH
3
CH
3
CHO (2) CH
3
COCH
3
(3) CH
3
CH
2
COCH
3
(4*) CH
3
CHO
Sol. CH CH = CHCH

O
3
/
Z
n
2CH CHO.
3 3
H
2
O
3
56. The hydrocarbon which can react with sodium in liquid ammonia is [AIEEE-2008, 3/105]
(1*) CH CH C CH (2) CH CH = CHCH
3 2 3 3
(3) CH CH C CCH CH (4) CH CH CH C CCH CH CH
3 2 2 3
3 2 2
2 2 3
H
Ph

57. The main product of the following reaction is : [AIEEE-2010, 4/144]
C
6
H
5
CH
2
CH(OH)CH(CH
3
)
2

con
c.
H
2
S
O
4
(1*) (2)
(3) (4)
Sol.

Co
nc
. H
2
SO
4

H
shift

+
H / 
58. One mole of a symmetrical alkene on ozonolysis gives two moles of an aldehyde having a molecular mass
of 44 u. The alkene is : [AIEEE-2010, 4/144]
(1) propane (2) 1-butene (3*) 2-butene (4) ethene
Sol. C
n
H
2n
O = 44
C
n
H
2n
= 44 – 16
C
n
H
2n
= 28
n = 2
CH –CH=CH–CH 
O
3
/ Z
n
CH –CH=O
3 3 3
59. Ozonolysis of an organic compound 'A' produces acetone and propionaldehyde in equimolar mixture.
Identify 'A' from the following compounds : [AIEEE 2011, 4/120]
(1) 1-Pentene (2) 2-Pentene
(3*) 2-Methyl-2-pentene (4) 2-Methyl-1-pentene
Sol.
60. How many chiral compounds are possible on monochlorination of 2- methyl butane ? [AIEEE-2012]
(1) 8 (2) 2 (3*) 4 (4) 6
Sol.
Four monochloro derivatives are chiral.
61. Which branched chain isomer of the hydrocarbon with molecular mass 72u gives only one isomer of mono
substituted alkyl halide ? [AIEEE-2012]
(1) Tertiary butyl chloride (2*) Neopentane
(3) Isohexane (4) Neohexane
B C
D E
Sol. single product.
62. 2– Hexyne gives trans –2–Hexene on treatment with : [AIEEE-2012]
(1) Pt/H
2
(2*) Li / NH
3
(3) Pd/BaSO
4
(4) Li AlH
4
Sol.
63. Which of the following would not give 2-phenylbutane as the major product in a Friedel-Craft alkylation
reaction? [JEE Mains Online-2013]
(1) Butyl chloride + AlCl
3
(2) 1-butene + HF
(*3) Butanocylchloride + AlCl
3
then Zn, HCl (4) 2-butanol + H
2
SO
4
Sol.
64. For which of the following compounds Kjeldahl method can be used to determine the percentage of Nitrogen?
[JEE Mains Online-2013]
(1) Nitrobenzene (*2) Alanine (3) Pyridine (4) Diazomethane
65. The major organic compound formed by the reaction of 1, 1, 1– trichloroethane with silver powder is
[JEE Mains 2014]
(1) Ethene (*2) 2-Butyne (3) 2-Butene (4) Acetylene
66. In the presence of peroxide, HCl and HI do not give anti-Markownikoff's addition to alkenes because :
(*1) One of the steps is endothermic in HCl and HI [JEE Mains Online-2014]
(2) Both HCl and HI are strong acids
(3) HCl is oxidizing and the HI is reducing
(4) All the steps are exothermic in HCl and HI.
67. Phthalic acid reacts with resorcinol in the presence of concentrated H
2
SO
4
to give :
[JEE Mains Online-2014]
(1) Phenolphthalein (2) Alizarin (3) Coumarin (*4) Fluorescein
O
68. Which one of the following substituents at para-position is most effective in stabilizing the phenoxide
ion ?
[JEE Mains Online-2014]
(1) –CH
3
(2) –OCH
3
(*3) –COCH
3
(4) –CH
2
OH
69. In a set of reactions p-nitrotoluene yielded a product E : [JEE Mains Online-2014]
CH
3
NO
2
Br
2
FeBr
3
Sn/HCl
NaNO
2
HCl
CuBr
HBr
2
3
The product E would be :
(1)
Br
(*2)
CH
3
Br
Br
(3)
CH
3
Br
Br
(4)
CH Br
Br
70. The synthesis of alkyl fluorides is best accomplished by [JEE Mains 2015]
(*1) Swarts reaction (2) Free radical fluorination
(3) Sandmeyer's reaction (4) Finkelstein reaction
71. Which of the following compounds will exhibit geometrical isomerism? [JEE Mains 2015]
(1) 1, 1-Diphenyl-1-propane (*2) 1-Phenyl-2-butene
(3) 3-Phenyl-1-butene (4) 2-Phenyl-1-butene
72. Which compound would give 5-keto-2-methyl hexanal upon ozonolysis? [JEE Mains 2015]
(1)
H
3
C
CH
3
(2) (2)
CH
3
(*3)
CH
CH
3
(4)
CH
3
CH
3
COOK
73.
Electrolysis
A
; A is [JEE Mains Online-2015]
COOK
(1) (*2)
(3) (4)
n
74. Match the organic compounds in Column-I with the Lassaigne's test results in Column-II appropriately :
[JEE Mains Online-2015]
Column-I
Column-II
(A)
Aniline
(i)
Red color with FeCl
3
(B)
Benzene sulfonic acid
(ii)
Violet color with sodium nitroprusside
(C)
Thiourea
(iii)
Blue color with hot and acidic solution of FeSO
4
(1) (A)-(ii); (B)-(i); (C)-(iii) (*2) (A)-(iii); (B)-(ii); (C)-(i)
(3) (A)-(iii); (B)-(i); (C)-(ii) (4) (A)-(ii); (B)-(iii); (C)-(i)
CH
3
Br
Br
CH
3
Br
H
75. 2-chloro-2methylpentane on reaction with sodium methoxide in methanol yields : [JEE Mains 2016]
CH
3
(a)
C
2
H
5
CH
2
C – OCH
3
CH
3
(b)
C
2
H
5
CH
2
C = CH
2
CH
3
(c)
C
2
H
5
CH = C – CH
3
CH
3
(1) All of these (2) (a) and (c) (*3) (c) only (4) (a) and (b)
76. The reaction of propene with HOCl (Cl
2
+ H
2
O) proceeds through the intermediate : [JEE Mains 2016]
(1) CH
3
– CH
+
– CH
2
– OH (*2) CH
3
– CH
+
– CH
2
– Cl
(3) CH
3
– CH(OH) – CH
+
(4) CH
– CHCl – CH
+
77. The gas evolved on heating CH
3
MgBr in methanol is [JEE Mains Online-2016]
(*1) Methane (2) Ethane (3) Propene (4) HBr
78. The hydrocarbon with seven carbon atoms containing a neopentyl and a vinyl group is :
[JEE Mains Online-2016]
(1) 2, 2-dimethyl-4-pentane (*2) 4, 4-dimethylpentene
(3) Isopropyl-2-butene (4) 2, 2-dimethyl-3-pentene
79. The major product obtained in the following reaction is [JEE Mains 2017]
C
6
H
5
(+)
C
6
H
5
t
BuOK
(1) (–)C
6
H
5
CH(O
t
Bu)CH
2
C
6
H
5
(2) (±)C
6
H
5
CH(O
t
Bu)CH
2
C
6
H
5
(*3) C
6
H
5
CH = CHC
6
H
5
(4) (+)C
6
H
5
CH(O
t
Bu)CH
2
C
6
H
5
80. 3-Methyl-pent-2-ene on reaction with HBr in presence of peroxide forms an addition product. The number of
possible stereoisomers for the product is [JEE Mains 2017]
(*1) Four (2) Six (3) Zero (4) Two
81. Which of the following statements is not true about partition chromatography?
(*1) Mobile phase can be a gas separation depends upon.
(2) Equilibration of solute between a mobile and a stationary phase.
(3) Stationary phase is a finely divided solid adsorbent.
(4) Paper chromatography is an example of partition chromatography. [JEE Mains Online-2017]
82. The trans-alkenes are formed by the reduction of alkynes with : [JEE Mains 2018]
(1) Sn–HCl (2) H
2
–Pd/C, BaSO
4
(3) NaBH
4
(*4) Na/Liq. NH
3
83. The total number of optically active compounds formed in the following reaction is :
O
HBr
[JEE Mains Online-2018]
(1) Two (*2) Four (3) Six (4) Zero
84. The major product of the following reaction is [JEE Mains Online-2018]
.
.
.
.
.

CH=CHCH
3
HBr
Br CHCH
2
CH
3
CH CH CH Br
CH CH CH
CH=CHCH
3
(*1)
(2)
2 2
(3)
3
Br
(4)
Br
85. 1-methyl ethylene oxide when treated with an excess of HBr produces : [JEE Mains online 2020]
(1)
Ans. (4)
O
Sol.
Br
CH
3
HBr
excess

(2)
Br
Br
Br
Br
CH
3
(3)
Br
CH
3
(4)
Br
Br
CH
3
86. The major products A and B in the following reactions are : [JEE Mains online 2020]
CN
Peroxide
Heat
[A]
[A] + B
(1) A =
CN
and B = CN (2) A =
CN
CN
and B =
(3) A =
CN
and B =
CN
(4) A =
CN
and B =
CN
Ans. (1)
Sol. A =
CN
and B = CN
87. The major product [B] in the following sequence of reactions is : [JEE Mains online 2020]
CH C CH CH CH
(i) B
2
H
6
[A]
dil. H
2
SO
4
[B]
3
CH (CH
3
)
2
2 3
(ii) H
2
O
2
,OH
(1)
CH
3
C CH CH
2
CH
3
CH (CH )
(2)
CH
3
C
–
CH
2
CH
2
CH
3
C
3 2
(3) CH
2
C CH
2
CH
2
CH
3
CH(CH
3
)
2
H
3
C
(4)
CH
3
CH
3
CH
CH CH
CH
3
CH(CH
3
)
2
Ans. (2)
2
2
2
3 2

X
Y
Sol.
H
3
C
H
3
C
CH
3

1
.B
2
H
6
2.H
2
O
2,¯ OH
OH

Dil.H
2
S
O
4
88. An unsaturated hydrocarbon X absorbs two hydrogen molecules on catalytic hydrogenation, and also
gives following reaction: [JEE Mains online 2020]
O
3
X
Zn/H O
A
X will be :
[Ag(NH ) ]
+
B (3-oxo-hexanediacarboxylic acid)
O
(1) (2) (3) (4)
Ans. (1)
Sol.
O
i
O
3
ii
Zn
CHO
CHO
O

Tol
len'
s
re
ag
ent
COOH
COOH
89. Kjeldahl’s method cannot be used to estimate nitrogen for which of the following compounds ?
[JEE Mains online 2020]
O
(1)
(2) C
H NH
(3) C H NO
(4) CH CH
— C
N
Ans. (3)
NH
2
C NH
2
6 5 2 6 5 2 3 2
Sol. Khjeldhal method is not applicable with nitro, diazo and ring nitrogen.
90. The major product (Y) in the following reactions is : [JEE Mains online 2020]
CH
3
CH
3
– CH – C
CH
HgSO
4
, H
2
SO
4
H
2
O
(i) C
2
H
5
MgBr, H
2
O
(ii) Conc.H
2
SO
4
/
CH
3
(1)
CH
3
– C = C – CH
3
CH
2
CH
3
CH
3
(2)
CH
3
– CH – C = CH – CH
3
CH
3
CH
3
(3)
CH
3
– CH – C = CH
2
CH
2
CH
3
CH
2
(4)
H
3
C – C – CH – CH
3
C
2
H
5
Ans. (1)
CH
3
OH
CH
3
CH
CH
3
Sol.
H
3
C
H
3
C
HgSO
4
, H
2
SO
4
CH
H
2
O
H
3
C
H
3
C
O
CH
3

i
C
2
H
5
M
gBr
ii
H
/H
2
O
H
3
C
H
3
C
3
(iii) H
2
SO
4
,
H
3
C
H
3
C
CH
3