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.
21.
Which one of the following on oxidation gives a ketone?
A
Primary alcohol
B
Secondary alcohol
C
Tertiary alcohol
D
All of these
Answer :
Secondary alcohol
Ketones can be prepared by the oxidation of secondary alcohols by using oxidising agent such as $$\frac{{{K_2}C{r_2}{O_7}}}{{{H_2}S{O_4}}}$$
22.
How many isomers of $${C_5}{H_{11}}OH$$ will be primary alcohols?
A
5
B
4
C
2
D
3
Answer :
4
The primary alcohols isomers of $${C_5}{H_{11}}OH$$ are
23.
Arrange the following alcohols in order of increasing reactivity towards sodium metal.
$$\eqalign{
& \left( {\text{i}} \right){\left( {C{H_3}} \right)_3}C - OH \cr
& \left( {{\text{ii}}} \right){\left( {C{H_3}} \right)_2}CH - OH \cr
& \left( {{\text{iii}}} \right)C{H_3}C{H_2}OH \cr} $$
A
(iii) < (ii) < (i)
B
(ii) < (i) < (iii)
C
(i) < (ii) > (iii)
D
(iii) < (i) < (ii)
Answer :
(i) < (ii) > (iii)
More the number of methyl groups ( electron releasing ) lesser will be the acidity. Hence the reactivity towards sodium metal will be $${3^ \circ } < {2^ \circ } < {1^ \circ }.$$
24.
The enzyme which can catalyse the conversion of glucose to ethanol is
A
invertase
B
zymase
C
maltase
D
diastase
Answer :
zymase
No explanation is given for this question. Let's discuss the answer together.
25.
Arrange the following in increasing order of their acidity ?
(a) $$o-$$ cresol, (b) salicylic acid, (c) phenol
A
c < a < b
B
b < c < a
C
a < b < a
D
a < c < b
Answer :
a < c < b
Electron releasing groups $$\left( { - C{H_3}, - OC{H_3}, - NC{H_3}\,{\text{etc}}} \right)$$ intensify the negative charge of phenoxide $$ion,$$ i.e., destablises it hence decreases ionization of parent phenol, therefore decreases acidity. While electron withdrawing groups $$\left( { - N{O_2}, - COOH, - CHO\,{\text{etc}}.} \right)$$ increases acidity.
26.
Consider the following reaction sequence,
\[C{{H}_{3}}CH\left( OH \right)C{{H}_{3}}\xrightarrow[\text{Heat}]{\text{Conc}.\,{{H}_{2}}S{{O}_{4}}}\] \[X\xrightarrow[{{H}_{2}}O]{\text{dil}\text{.}\,{{H}_{2}}S{{O}_{4}}}Y\]
$$X$$ and $$Y$$ in the reaction respectively are
A
$$C{H_3}CH = C{H_2},C{H_3}CH\left( {OH} \right)C{H_3}$$
B
$$C{H_3}CH = CHC{H_3},C{H_3}C{H_2}C{H_2}OH$$
C
$$C{H_3}C{H_2}CH = C{H_2},C{H_3}C{H_2}C{H_2}OH$$
D
$$C{H_3}C{H_2}C{H_2}C{H_3},$$ $$C{H_3}CH\left( {OH} \right)C{H_2}C{H_3}$$
27.
Diethyl ether when refluxed with excess of $$HI$$ gives two molecules of $$\underline {\,\,\left( {\text{i}} \right)\,\,} .$$ Ethers can be most commonly prepared by reaction of $$\underline {\,\,\left( {{\text{ii}}} \right)\,\,} $$ and $$\underline {\,\,\left( {{\text{iii}}} \right)\,\,} .$$ The method is called $$\underline {\,\,\left( {{\text{iv}}} \right)\,\,} .$$
(i), (ii), (iii) and (iv) respectively are
A
ethyl iodide, sodium alkoxide, alkyl halide, Williamson's synthesis
B
ethanol, alcohol, alkyl halide, substitution
C
methyl iodide, Grignards reagent, alkyl halide, Williamson's synthesis
D
ethyl iodide, phenol, ethyl iodide, esterification
No explanation is given for this question. Let's discuss the answer together.
28.
During dehydration of alcohols to alkenes by heating with conc. \[{{H}_{2}}S{{O}_{4}}\] the initiation step is
A
formation of carbocation
B
elimination of water
C
formation of an ester
D
protonation of alcohol molecule
Answer :
protonation of alcohol molecule
The dehydration of alcohol to form alkene occurs in following three step. Step (1) is initiation step.
Step (1) Formation of protonated alcohol.
Step (2) Formation of carbocation
Step (3) Elimination ofa proton to form ethene
29.
Diols ( I - IV ) which react with \[Cr{{O}_{3}}\] in aqueous \[{{H}_{2}}S{{O}_{4}}\] and yield products that readily under go dercarboxylation on heating, are :
A
I and II
B
II and III
C
II and IV
D
I and IV
Answer :
II and IV
30.
Benzoquinone is prepared by reaction of phenol with
A
$$N{a_2}C{r_2}{O_7},{H_2}S{O_4}$$
B
$$KMn{O_4},{H_2}S{O_4}$$
C
$$N{a_2}Cr{O_4},HCl$$
D
$${K_2}Mn{O_4},{H_2}S{O_4}$$
Answer :
$$N{a_2}C{r_2}{O_7},{H_2}S{O_4}$$
No explanation is given for this question. Let's discuss the answer together.