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.
81.
Compound $${C_2}{H_6}O$$ has two isomers $$X$$ and $$Y.$$ On reaction with $$HI, X$$ gives alkyl iodide and water while $$Y$$ gives alkyl iodide and alcohol. Compounds $$X$$ and $$Y$$ are respectively
A
$${C_2}{H_5}O{C_2}{H_5}{\text{ and }}C{H_3}O{C_2}{H_5}$$
B
$$C{H_3}OC{H_3}{\text{ and }}{C_2}{H_5}OC{H_3}$$
C
$${C_2}{H_5}OH{\text{ and }}C{H_3}OC{H_3}$$
D
$$C{H_3}OH{\text{ and }}C{H_3}OC{H_3}$$
Answer :
$${C_2}{H_5}OH{\text{ and }}C{H_3}OC{H_3}$$
$$\eqalign{
& \mathop {{C_2}{H_5}OH}\limits_{\left( X \right)} + HI \to {C_2}{H_5}I + {H_2}O \cr
& \mathop {C{H_3}OC{H_3}}\limits_{\left( Y \right)} + HI \to C{H_3}I + C{H_3}OH \cr} $$
82.
Which of the following reagents would distinguish $$cis$$ - cyclopenta -1, 2 - diol from the
$$trans$$ - isomer?
A
Ozone
B
$$Mn{O_2}$$
C
Aluminium isopropoxide
D
Acetone
Answer :
Acetone
$$cis$$ - cyclopenta - 1, 2 - diol when reacts with acetone, forms cyclic ketal whereas $$trans$$ - isomer of cyclopenta - 1, 2 - diol can not form cyclic ketal.
83.
Consider the following reaction,
\[\text{Phenol}\xrightarrow{Zn-\text{dust}}X\xrightarrow[\text{Anhy}\text{.}\,AlC{{l}_{3}}]{C{{H}_{3}}Cl}Y\xrightarrow{\begin{smallmatrix}
\text{Alk}\text{.} \\
KMn{{O}_{4}}
\end{smallmatrix}}Z\]
The product $$Z$$ is
A
toluene
B
benzaldehyde
C
benzoic acid
D
benzene
Answer :
benzoic acid
84.
Thiol group is present in :
A
Cysteine
B
Methionine
C
Cytosine
D
Cystine
Answer :
Cysteine
Among 20 naturally occuring amino acids "Cysteine" has $$' - SH'$$ or thiol functional group.
$$ \Rightarrow {\text{Value of}}\,R = - C{H_2} - SH\,\,{\text{in Cysteine}}{\text{.}}$$
85.
Phenol is heated with a solution of mixture of $$KBr$$ and$$KBr{O_3}.$$ The major product obtained in the above reaction is :
A
2 - Bromophenol
B
3 - Bromophenol
C
4 - Bromophenol
D
2, 4, 6 - Tribromophenol
Answer :
2, 4, 6 - Tribromophenol
$$5KBr + KBr{O_3} + 3{H_2}O \to 3B{r_2} + 6KOH$$
86.
Sodium phenoxide when heated with $$C{O_2}$$ under pressure at
125°C yields a product which on acetylation produces $$C.$$
The major product C would be
A
B
C
D
Answer :
87.
The $$ - OH$$ group of an alcohol or the $$-COOH$$ group of a carboxylic acid can be replaced by $$-Cl$$ using
A
phosphorus pentachloride
B
hypochlorous acid
C
chlorine
D
hydrochloric acid
Answer :
phosphorus pentachloride
The $$-OH$$ group of alcohol or the $$-COOH$$ group of a carboxylic acid is replaced by $$-Cl$$ using phosphorus penta chloride $$\left( {{\text{i}}{\text{.e}}{\text{.}}\,PC{l_5}} \right)$$
$$\eqalign{
& \mathop {ROH}\limits_{{\text{Alcohol}}} + PC{l_5} \to RCl + POC{l_3} + HCl \cr
& \mathop {RCOOH}\limits_{{\text{Acid}}} + PC{l_5} \to RCOCl + POC{l_3} + HCl \cr} $$
No explanation is given for this question. Let's discuss the answer together.
89.
The ionisation constant of phenol is higher than that of ethanol because
A
phenoxide ion is bulkier than ethoxide
B
phenoxide ion is stronger base than ethoxide
C
phenoxide ion is stabilised through delocalisation
D
phenoxide ion is less stable than ethoxide
Answer :
phenoxide ion is stabilised through delocalisation
Resonance stabilisation of phenoxide ion
\[{{C}_{2}}{{H}_{5}}-\underset{\text{Stable}}{\mathop{O}}\,-H\rightleftharpoons \underset{\begin{smallmatrix}
\text{ Unstable} \\
\text{ ethoxide ion}
\\
\text{(due to absence of resonance)}
\end{smallmatrix}}{\mathop{{{C}_{2}}{{H}_{5}}-{{O}^{-}}+{{H}^{+}}}}\,\]
Phenoxide ion is more stable than ethoxide ion due to resonance. Therefore, the ionisation constant of phenol is higher than ethanol.
90.
For the identification of $$\beta $$ - naphthol using dye test, it is necessary to use
A
Dichloromethane solution of $$\beta $$ - naphthol
B
Acidic solution of $$\beta $$ - naphthol
C
Neutral solution of $$\beta $$ - naphthol
D
Alkaline solution of $$\beta $$ - naphthol
Answer :
Alkaline solution of $$\beta $$ - naphthol
In dye test, phenolic $$ - OH$$ group is converted to $$ - {O^ - }$$ which activates the ring towards electrophilic aromatic substitution