
50 questions · timed · auto-graded

$\text{H}_2\text{C}=\text{CH}-\text{CH}-\text{CH}_2-\text{CH}_2-\text{CH}_3\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{OH}$

$\text{CH}_3-\text{CH}=\text{CH}_2\ \xrightarrow{\text{H}_2 \text{O}/\text{H}^+}$
Propane-2-ol
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\text{CH}_3-\text{CH}-\text{CH}-\text{C}-\text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3\ \ \ \ \text{OH}\ \ \ \ \text{CH}_3$
$\text{CH}_3\text{CH}_2\text{CH}_2\text{OCH}_3$
$\text{C}_6\text{H}_5-\text{O}-\text{C}_2\text{H}_5$

Methyl magnesium bromide → 2-Methylpropan-2-ol.
cyclohexylmethanol using an alkyl halide by an SN2 reaction.
$\text{HO}-\text{CH}_2-\text{CH}-\text{CH}_2-\text{OH}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{OH}$
Bromine in CS2 with phenol.
benzyl ethyl ether.
$\text{CH}_3-\text{CH}_2-\text{O}-\text{CH}-\text{CH}_2-\text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3$
$\text{CH}_3-\text{CH}-\text{CH}-\text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \ \text{OH}\ \ \ \ \ \text{OH}$

pentan-1-ol using a suitable alkyl halide?
$\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_2\text{Cl}+\text{NaOH}\xrightarrow{\ \ \ \ \ \ \ \ \ }\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_2\text{OH}+\text{NaCl}\\ \text{1-Chloropentane} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Pantan-1-ol}$
3-Chloromethylpentan-1-ol.
Benzyl chloride → Benzyl alcohol.
$\text{C}_6\text{H}_5-\text{O}-\text{C}_7\text{H}_{15}(\text{n}-)$
1-propoxypropane.
Explanation:
Amongst isomeric alcohols, the solubility increases with branching. This is due to the reason that as the branching increases, the surface area of the non-polar hydrocarbon and the solubility increases.
So, the sequence of solubility of alcohols are
Tertiary alcohol > secondary alcohol > primary alcohol

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_2\text{OH}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\text{CH}_3-\text{CH}-\text{CH}_2-\text{CH}-\text{CH}-\text{CH}_3\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ |\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\\\ \ \ \ \ \ \ \ \ \ \ \ \ \text{CH}_3 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{OH}$
$\text{CH}_3-\text{CH}_2-\text{CH}_2-\text{O}-\text{CH}_3+\text{HBr}\xrightarrow{\ \ \ }$
Explanation:
$\text{R}-\text{CH}_2-\text{OH}\xrightarrow{\text{Anhy Zncl}_2{\Delta}}\text{R}=\text{CH}_2$
Propene → Propan-2-ol.
Explanation:
The industrial method of preparation of methanol or methyl alcohol is by the action of CO+H2 as:
CO + H2 → CH3OH (230−400°C)
Carbon monoxide reacts with hydrogen to produce methanol/methyl alcohol.
This reaction takes place at a temperature of 230−400°C and at 50-100 atm. In this reaction, the catalyst is can be copper oxide on oxide chromium(III).