Alcohol, Phenol and Ether MCQ Questions & Answers in Organic Chemistry | Chemistry
Learn Alcohol, Phenol and Ether MCQ questions & answers in Organic Chemistry are available for students perparing for IIT-JEE, NEET, Engineering and Medical Enternace exam.
101.
1-Phenylethanol can be prepared by the reaction of benzaldehyde with
A
methyl bromide
B
ethyl iodide and magnesium
C
methyl iodide and magnesium
D
methyl bromide and aluminium bromide
Answer :
methyl iodide and magnesium
102.
In the reaction :
Which of the following compounds will be formed ?
A
B
C
D
Answer :
In the cleavage of mixed ethers having two different alkyl groups, the alcohol and alkyl iodide formed depends on the nature of alkyl groups. When primary or secondary alkyl groups are present, it is the lower alkyl group that forms alkyl iodide therefore
103.
Which of the following compounds will give tribromo derivative on treatment with bromine water?
A
B
C
D
Answer :
Only $$m$$ - cresol has two $$ortho$$ and one $$para$$ - positions free with respect to $$-OH$$ group, hence it can form tribromo-derivative.
104.
Ethyl chloride is converted into diethyl ether by
A
Wurtz synthesis
B
Grignard reaction
C
Perkin’s reaction
D
Williamson's synthesis
Answer :
Williamson's synthesis
$${C_2}{H_5}Cl + Na\,O{C_2}{H_5} \to {C_2}{H_5} - O - {C_2}{H_5} + NaCl$$
It is an example of Williamson's synthesis. Limitations of Williamson’s Synthesis
$${3^ \circ }$$ alkyl halide does not gives Williamson's synthesis if $${3^ \circ }$$ alkyl halide used in reaction then alkene formed.
105.
Phenol reacts with bromine in carbon disulphide at low temperature to give
A
$$m$$ - bromophenol
B
$$o$$ - and $$p$$ - bromophenol
C
$$p$$ - bromophenol
D
2, 4, 6 - tribromophenol
Answer :
$$o$$ - and $$p$$ - bromophenol
NOTE : In absence of $$C{S_2},$$ polyhalogenation in $${\text{o - }}$$ and $${\text{p - }}$$ positions takes place.
106.
Which of the following compounds is aromatic alcohol?
A
I, II, III, IV
B
I, IV
C
II, III
D
I
Answer :
II, III
No explanation is given for this question. Let's discuss the answer together.
107.
Dehydration of the following in increasing order is
A
I < II < III < IV
B
II < III < IV < I
C
I < III < IV < II
D
None of these
Answer :
I < II < III < IV
Dehydration of IV is most facile since, it gives an aromatic compound. Dehydration of III gives a conjugated diene which is stabilised by resonance.
Dehydration of II gives only cyclohexene which is not stabilised by resonance.
In contrast, phenol (I) does not undergo dehydration. Thus, the ease of dehydration is: IV > III > II > I.
108.
Give IUPAC name of the compound given below :
\[C{{H}_{3}}\underset{\begin{smallmatrix}
|\,\,\,\,\, \\
Cl\,\,\,\,\,
\end{smallmatrix}}{\mathop{-CH-}}\,C{{H}_{2}}-C{{H}_{2}}\] \[\underset{\begin{smallmatrix}
|\,\,\,\,\, \\
OH\,\,
\end{smallmatrix}}{\mathop{-CH-}}\,C{{H}_{3}}\]
A
2-Chloro-5-hydroxyhexane
B
2-Hydroxy-5-chlorohexane
C
5-Chlorohexan-2-ol
D
2-Chlorohexan-5-ol
Answer :
5-Chlorohexan-2-ol
No explanation is given for this question. Let's discuss the answer together.
109.
The boiling point of $$p$$ - nitrophenol is higher than that of $$o$$ - nitrophenol because
A
$$N{O_2}$$ group at $$p$$ - position behaves in a different
way from that at $$o$$ - position
B
intramolecular hydrogen bonding exists in $$p$$ - nitrophenol
C
there is intermolecular hydrogen bonding in $$p$$ - nitrophenol
D
$$p$$ - nitrophenol has a higher molecular weight than $$o$$ - nitrophenol
Answer :
there is intermolecular hydrogen bonding in $$p$$ - nitrophenol
The boiling point of $$p$$ - nitrophenol is higher than that of $$o$$ - nitrophenol because $$p$$ - nitrophenol have intermolecular hydrogen bonding whereas $$o$$ - nitrophenol have intramolecular $$H$$ - bonding as given below.
110.
Methanol is industrially prepared by
A
oxidation of $$C{H_4}$$ by steam at $${900^ \circ }C$$
B
reduction of $$HCHO$$ using $$LiAl{H_4}$$
C
reaction of $$HCHO$$ with a solution of $$NaOH$$
D
reduction of $$CO$$ using $${H_2}$$ and $$ZnO - C{r_2}{O_3}$$
Answer :
reduction of $$CO$$ using $${H_2}$$ and $$ZnO - C{r_2}{O_3}$$
Commercially methanol is prepared from water gas which is a mixture of carbon monoxide and hydrogen. In this method, $$CO$$ gas is mixed with its half volume of hydrogen and is passed over heated $$C{r_2}{O_3} - ZnO$$ catalyst at $$673\,K$$ under high pressure.
\[CO+2{{H}_{2}}\xrightarrow[673\,K,\text{high pressure}]{C{{r}_{2}}{{O}_{3}}\,-ZnO}C{{H}_{3}}OH\]