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.
131.
The cylindrical shape of an alkyne is due to
A
three sigma $$C-C$$ bonds
B
three $$\pi \,C - C$$ bonds
C
two sigma $$C-C$$ and one $$\pi \,C - C$$ bonds
D
one sigma $$C-C$$ and two $$\pi \,C - C$$ bonds
Answer :
one sigma $$C-C$$ and two $$\pi \,C - C$$ bonds
In alkynes $$C \equiv C$$ is present, out of these three bonds one sigma and two $$\pi $$ - bonds are present. Sigma bond is formed by $$sp$$ - hybrid orbitals whereas $$\pi $$ - bonds are formed by unhybridised orbitals.
Hence, it shows cylindrical shape.
132.
The nodal plane in the $$\pi $$ - bond of ethene is located in
A
the molecular plane
B
a plane parallel to the molecular plane
C
a plane perpendicular to the molecular plane which bisects the carbon - carbon $$\sigma $$ - bond at right angle
D
plane perpendicular to the molecular plane which contains the carbon - carbon $$\sigma $$ - bond.
Answer :
the molecular plane
TIPS/Formulae :
The $$\pi $$ bond is formed by the sideways overlapping of two $$p $$ - orbitals of the two carbon atoms.
The molecular plane does not have any $$\pi $$ electron density as the $$p $$ - orbitals are perpendicular to the plane containing the ethene molecule. The nodal plane in the $$\pi $$ - bond of ethene is located in the molecular plane.
133.
Identify $$X$$ and $$Y$$ in the following reaction.
\[\underset{\begin{smallmatrix}
\,\,\,\,\,\,| \\
\,\,\,\,\,\,\,Br
\end{smallmatrix}}{\mathop{{{H}_{2}}C}}\,-\underset{\begin{smallmatrix}
|\,\,\,\,\,\,\,\, \\
Br\,\,\,\,\,\,
\end{smallmatrix}}{\mathop{C{{H}_{2}}}}\,+KOH\xrightarrow{\text{ alcohol}}\] \[X\xrightarrow{NaN{{H}_{2}}}Y\]
134.
During halogenation of alkanes the halogens and alkane show a specific trend. Which of the following statements is not correct?
A
The reactivity of halogens is in the order $${F_2} > C{l_2} > B{r_2} > {I_2}.$$
B
For a given halogen the reactivity of hydrocarbon is in the order of $${3^ \circ } > {2^ \circ } > {1^ \circ }.$$
C
Bromine is less reactive than chlorine towards a particular alkane.
D
On chlorination monosubstituted product is formed while on bromination disubstituted products are formed.
Answer :
On chlorination monosubstituted product is formed while on bromination disubstituted products are formed.
Number of monohalogen products of alkane depends upon the number of different types of replaceable hydrogen atoms in the alkane.
135.
In Friedel-Craft’s synthesis of toluene, the reactants in addition to anhydrous $$AlC{l_3}$$ are
A
$${C_6}{H_5}Cl + C{H_4}$$
B
$${C_6}{H_5}Cl + C{H_3}Cl$$
C
$${C_6}{H_6} + C{H_4}$$
D
$${C_6}{H_6} + C{H_3}Cl$$
Answer :
$${C_6}{H_6} + C{H_3}Cl$$
Friedel-Craft’s alkylation of benzene $$\left( {Ar - H} \right)$$ Mechanism of this reaction is represented as follows :
Thus, $${C_6}{H_6}$$ and $$C{H_3}Cl$$ are required in addition to $$AlC{l_3}.$$
136.
Which of the following change is correct
A
B
C
D
Answer :
Both (A) and (B) are correct. Wet $$C{H_3}COOH$$ gives $$cis$$ addition and dry $$C{H_3}COOH$$ gives trans addition products.
137.
In electrophilic aromatic substitution reaction, the nitro group is $$meta$$ directing because it
A
decreases electron density at $$meta$$ position
B
increases electron density at $$meta$$ position
C
increases electron density at $$ortho$$ and $$para$$ positions
D
decreases electron density at $$ortho$$ and $$para$$ positions.
Answer :
decreases electron density at $$ortho$$ and $$para$$ positions.
No explanation is given for this question. Let's discuss the answer together.
138.
How many geometrical isomers are possible for the given compound?
$$C{H_3} - CH = CH - CH = CH - {C_2}{H_5}$$
A
Four
B
Three
C
Two
D
Five
Answer :
Four
No explanation is given for this question. Let's discuss the answer together.
139.
In the reaction,
\[\begin{align}
& H-C\equiv CH\xrightarrow[\text{(ii)}\,C{{H}_{3}}C{{H}_{2}}Br]{\text{(i)}\,\frac{NaN{{H}_{2}}}{liq.N{{H}_{3}}}} \\
& X\xrightarrow[\text{(ii)}\,C{{H}_{3}}C{{H}_{2}}Br]{\text{(i)}\,\frac{NaN{{H}_{2}}}{liq.N{{H}_{3}}}}Y \\
\end{align}\]
$$X$$ and $$Y$$ are
A
$$X = 2 - butyne;Y = 3 - hexyne$$
B
$$X = 2 - butyne;Y = 2 - hexyne$$
C
$$X = 1 - butyne;Y = 2 - hexyne$$
D
$$X = 1 - butyne;Y = 3 - hexyne$$
Answer :
$$X = 1 - butyne;Y = 3 - hexyne$$
Since, $${\frac{{NaN{H_2}}}{{liq.N{H_3}}}}$$ behaves as a base, so it abstracts proton from acetylene to form acetylide anion followed by alkylation to give compound $$(X)$$
i.e. $$1-butyne.$$ $$(X)$$ further reacts with $${\frac{{NaN{H_2}}}{{liq.N{H_3}}}}$$ followed by alkylation with ethyl bromide yields
$$3-hexyne$$ $$(Y).$$
140.
Which one of the following will react fastest with $${H_2}$$ under
catalytic hydrogenation condition ?
A
B
C
D
Answer :
TIPS/Formulae :
The relative rates of hydrogenation decreases with increase of steric hinderance.
$${R_2}C = C{H_2} > RCH = CHR > {R_2}C = CHR > {R_2}C = C{R_2}$$
Among the four olefins, (a) and (b) are less stable ( Saytzeff rule ). Further in (a), the bulky alkyl groups are on same side ( $$cis$$ - isomer ), hence it is less stable.