3.
In the presence of platinum catalyst , hydrocarbon $$A$$ adds hydrogen to form $$n$$ - hexane. When hydrogen bromide is added to $$A$$ instead of hydrogen only a single bromo compound is formed. Which of the following is $$A?$$
4.
The ozonolysis product$$(s)$$ of the following reaction is(are)
\[C{{H}_{3}}C{{H}_{2}}-C\equiv CH\xrightarrow[\left( \text{ii} \right)\,{{H}_{2}}O]{\left( \text{i} \right)\,{{O}_{3}}}\] \[\text{Product}\left( s \right)\]
In 1 - butyne terminal hydrogen is acidic where as in 2 -butyne there is no terminal hydrogen. Thus 2 - butyne will not react with ammonical $$C{u_2}C{l_2}.$$ While 1 - butyne, being terminal alkyne, will give red ppt. with ammonical cuprous chloride
7.
\[\xrightarrow[Zn]{{{O}_{3}}}A\xrightarrow[\text{Or}\,\,LiAl{{H}_{4}}]{{{H}_{3}}Ni}B\xrightarrow[\Delta ]{H+}\left( C \right);\] Product $$(C)$$ of the reaction is
An aromatic compound have cyclic clouds of delocalised $$\left( {4n + 2} \right)\pi $$ electrons above and below the plane of the molecule. Among the given three compounds, only compound (C) satisfies these conditions.
Carbon bearing $$ - N{O_2}$$ ( or $$ - C{H_3}$$ ) group is $$s{p^3}$$ hybridised and does not have a $$p$$ orbital with the result delocalisation of $$\pi $$ electrons over the complete ring is interrupted.
$$4\pi $$ electrons and 2 non-bonding electrons ( present in $$p$$ orbital of $$N$$ ) form a cyclic cloud of $$6\pi $$ electrons
Greater the $$s$$ - character of $$C$$ - atom in hydrocarbons, greater the electronegativity of that carbon and thus greater the acidic nature of the $$H$$ attached to electronegative carbon.