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.
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.
\[\underset{\begin{smallmatrix}
\,\,\,\,\,\,| \\
\,\,\,\,\,\,\,Br
\end{smallmatrix}}{\mathop{{{H}_{2}}C}}\,-\underset{\begin{smallmatrix}
|\,\,\,\,\,\,\,\, \\
Br\,\,\,\,\,\,
\end{smallmatrix}}{\mathop{C{{H}_{2}}}}\,+KOH\xrightarrow{\text{ alcohol}}\] \[X\xrightarrow{NaN{{H}_{2}}}Y\]
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 $$AlC{l_3}.$$
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
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 ?
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.