TIPS/Formulae :
The order of reactivity of alcohol with Lucas reagent is $$tert. > sec. > pri.$$
Lucas test is based on the difference in the three types of alcohols ( having 6 or less carbon ) towads Lucas reagent (a mixture of conc. hydrochloric acid and anhydrous zinc chloride) at room temperature.
\[ROH+HCl\xrightarrow{ZnC{{l}_{2}}}RCl+{{H}_{2}}O\]
The tertiary alcohols produce turbidity immediately,
the secondary alcohols give turbidity within 5 - 10
minutes, and the primary alcohols do not give turbidity
at all, at room temperature.
Hence \[\text{2 - methylpropan - 2 - ol}\] ( a $${3^ \circ }$$ alcohol ) reacts fastest.
34.
Acid catalysed dehydration of $$t$$ - butanol is faster than that of $$n$$ - butanol because
A
tertiary carbocation is more stable than primary carbocation
B
primary carbocation is more stable than tertiary carbocation
C
$$t$$ - butanol has a higher boiling point
D
rearrangement takes place during dehydration of $$t$$ - butanol.
Answer :
tertiary carbocation is more stable than primary carbocation
No explanation is given for this question. Let's discuss the answer together.
36.
A compound $$X\left( {{C_8}{H_{10}}O} \right)$$ upon treatment with alkaline solution of iodine gives a yellow precipitate. The filtrate on acidification gives a white solid $$Y\left( {{C_7}{H_6}{O_2}} \right).$$ Write the structures of $$X$$ and $$Y.$$
Since the compound $$X\left( {{C_8}{H_{10}}O} \right)$$ gives haloform reaction ( i.e., forms iodoform on reaction with $${I_2} + NaOH$$ ), so it must contain either \[-\underset{\begin{smallmatrix}
\parallel \\
O
\end{smallmatrix}}{\mathop{C}}\,-C{{H}_{3}}\] or \[\underset{\begin{smallmatrix}
|\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\
OH\,\,\,\,\,\,\,\,\,\,\,
\end{smallmatrix}}{\mathop{-CHC{{H}_{3}}}}\,\] group.
From the molecular formula of $$X$$ aromatic content of the compound is evident i.e., it contains a benzene ring
$${C_6}{H_5}CH\left( {OH} \right)C{H_3}$$ i.e.,
The given reactions are :
37.
The best reagent to convert pent-3-en-2-ol into pent-3-en-2-one is
\[\begin{align}
& C{{H}_{3}}COOH\xrightarrow[-{{H}_{2}}O]{NaOH}C{{H}_{3}}COONa \\
& C{{H}_{3}}CON{{H}_{2}}\xrightarrow{NaOH}C{{H}_{3}}COONa+N{{H}_{3}} \\
& {{C}_{6}}{{H}_{5}}OH\xrightarrow{NaOH}{{C}_{6}}{{H}_{5}}ONa+{{H}_{2}}O \\
\end{align}\]
But alcohols are weakly acidic even weaker than water and react with alkali metals to form alkoxides and not with alkalies.