Hydrocarbons (Alkane, Alkene and Alkyne) MCQ Questions & Answers in Organic Chemistry | Chemistry
Learn Hydrocarbons (Alkane, Alkene and Alkyne) MCQ questions & answers in Organic Chemistry are available for students perparing for IIT-JEE, NEET, Engineering and Medical Enternace exam.
231.
Acidic hydrogen is present in :
A
ethyne
B
ethene
C
benzene
D
ethane
Answer :
ethyne
Acidic hydrogen is present in alkynes, attached to the triply bonded $$C-$$ atoms. They can be easily removed by means of a strong base.
232.
Identify $$(A), (B)$$ and $$(C).$$
A
$$A \to C{H_3}COCl,\,B \to {C_6}{H_5}Cl,$$ $$C \to NaCN$$
B
$$A \to C{H_3}Cl,\,B \to {C_6}{H_5}C{H_2}Cl,$$ $$C \to KCN$$
C
$$A \to C{H_4},\,B \to {C_6}{H_5}C{H_2}Cl,$$ $$C \to AgCN$$
D
$$A \to C{H_3}Cl,\,B \to {C_6}{H_5}COCl,$$ $$C \to KCN$$
235.
In Friedel-Craft’s alkylation, besides $$AlC{l_3}$$ the other reactants are
A
$${C_6}{H_6} + N{H_2}$$
B
$${C_6}{H_6} + C{H_4}$$
C
$${C_6}{H_6} + C{H_3}Cl$$
D
$${C_6}{H_6} + C{H_3}COCl$$
Answer :
$${C_6}{H_6} + C{H_3}Cl$$
Friedel-Craft’s alkylation When benzene reacts with alkyl halide in presence of anhy. $$AlC{l_3},$$ toluene is obtained. It is called Friedel-Craft's alkylation reaction. Friedel-Craft’s reaction is type of an electrophilic substitution reaction.
\[{{C}_{6}}{{H}_{6}}+C{{H}_{3}}Cl\xrightarrow{Anhy.\,AlC{{l}_{3}}}\underset{\text{Toluene}}{\mathop{{{C}_{6}}{{H}_{5}}C{{H}_{3}}}}\,+HCl\]
236.
Arrange the following carbanions in order of their decreasing stability.
$$\eqalign{
& \left( {\text{i}} \right){H_3}C - C \equiv {C^ - } \cr
& \left( {{\text{ii}}} \right)H - C \equiv {C^ - } \cr
& \left( {{\text{iii}}} \right){H_3}C - CH_2^ - \cr} $$
A
(i) > (ii) > (iii)
B
(ii) > (i) > (iii)
C
(iii) > (ii) > (i)
D
(iii) > (i) > (ii)
Answer :
(ii) > (i) > (iii)
The order of decreasing stability of carbanions is :
$$\mathop {H - C \equiv {C^ - }}\limits_{\left( {{\text{ii}}} \right)} > \mathop {C{H_3} - C \equiv {C^ - }}\limits_{\left( {\text{i}} \right)} $$ $$ > \mathop {C{H_3} - CH_2^ - }\limits_{\left( {{\text{iii}}} \right)} $$
$$sp$$ - hybridised carbon atom is more electronegative than $$s{p^3}$$ - hybridised carbon atom and hence, can accommodate the negative charge more effectively. $$ - C{H_3}$$ group has $$+I$$ effect, therefore, it intensifies the negative charge and hence, destabilises the carbanion
237.
The major product formed in the reaction is :
A
B
C
D
Answer :
Hyperconjugation → more $$\alpha \,H,$$ more reactive $$o, p$$ site.
238.
In the reaction with $$HCl,$$ an alkene reacts in accordance with the Markownikoff’s rule, to give a product $$1-chloro-1-methylcyclohexane.$$ The possible alkane is
A
B
C
D
Answer :
For structure A,
For structure B,
The rearrangement of carbocation occur because $${3^ \circ }$$ - carbocation is more stable than $${2^ \circ }$$ - carbocation.
239.
The hottest region of Bunsen flame shown in the figure
below is :
A
region 3
B
region 4
C
region 1
D
region 2
Answer :
region 2
Region 2 ( blue flame ) will be the hottest region of Bunsen flame shown in given figure
240.
The correct statement regarding the comparison of staggered and eclipsed conformations of ethane, is
A
The eclipsed conformation of ethane is more stable than staggered conformation, because
eclipsed conformation has no torsional strain
B
The eclipsed conformation of ethane is more stable than staggered conformation even
though the eclipsed conformation has torsional strain
C
The staggered conformation of ethane is more stable than eclipsed conformation, because
staggered conformation has no torsional strain
D
The staggered conformation of ethane is less stable than eclipsed conformation, because
staggered conformation has torsional strain
Answer :
The staggered conformation of ethane is more stable than eclipsed conformation, because
staggered conformation has no torsional strain
Due to the absence of torsional strain staggered conformation of ethane is more stable than eclipsed conformation of it.