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CBSE Questions for Class 12 Medical Chemistry Alcohols, Phenols And Ethers Quiz 7 - MCQExams.com
CBSE
Class 12 Medical Chemistry
Alcohols, Phenols And Ethers
Quiz 7
$$A$$ & $$B$$ respectively:
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Explanation
Refer to Image.
Firstly hydroxylation reaction occurs in presence of $$H_2O_2$$ to give syn-diol then, Pinacol-pinacolone reaction occurs in an acidic medium in which ring expansion occurs.
Phenol and benzoic acid is separated by:
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$${NaHCO}_{3}$$
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$$NaOH$$
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$$Na$$
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$${NaNH}_{2}$$
Explanation
$$(A)\ NaHCO_3$$
Both phenol and benzoic acid will be extracted as both can react with hydroxide.
Phenol reacts with acetone in the presence of conc. sulphuric acid to form a $$C_{15}H_{16}O_2$$ product. Which of the following compounds is this product?
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$$CH_3 - CH_2 - OH\mathop \to \limits^{PCl_3 } P\mathop \to \limits_\Delta ^{alc.KOH} Q$$
'Q' is:
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$$
H_2 C = CH_2
$$
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$$
CH_3 - CH_2 - OH
$$
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$$
CH_3 - CH_2 - Cl
$$
Explanation
$$CH_3-CH_2-OH\longrightarrow CH_3-CH_2-Cl\longrightarrow CH_2=CH_2+HCl$$
chlorination reaction followed by elimination.
The IUPAC name of the following compound is:
cyclopenta-1, 3-dien-1-ol
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True
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False
Explanation
It is fire membered ring.
Numbering will start from $$C$$ attach to $$-OH$$ group.
So,its name will be Cyclopenta-$$1,3-$$dien$$-1-$$ol
The product of acid catalyzed hydration of $$2-phenylpropene$$ is:
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$$3-phenyl-2-propanol$$
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$$1-phenyl-2-propanol$$
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$$2-phenyl-2-propanol$$
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$$2-phenyl-1-propanol$$
Predict the correct option for major product.
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Major product is an alcohol bearing two deuterium atoms per molecule
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Major product is alcohol bearing one deuterium atom per molecule
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Major product is an alkane bearing two deuterium atoms per molecule
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Major product is an alkane bearing one deuterium atom per molecule
Which among the following is corect IUPAC name ?
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1, 1 - Dimethyl cyclohexan - 3 - ol
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4 - Methyl bicyclo [3 . 2 . 0] heptane
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Butane - 1, 2 - dione
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4 - Ethyl -5 - methyl cyclohexene
The IUPAC name of the following compound is:
2 - bromo - 6 - chlorophenol
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True
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False
Explanation
Parent molecule is phenol bromine is given preference over chlorine while numbering due to fact that bromine comes first in alphabetical order.
So,numbering will be as $$2-$$bromo $$-6-$$chloro-phenol.
The IUPAC name of the following compound is
2, 4, 6 - tribromophenol.
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True
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False
Explanation
$$-OH$$ is given preference over bromine. $$OH$$ is a functional group so it will get the lowest number. i.e numbering will start from $$C$$ attach to $$-OH$$ group.Without bromine groups the structure is phenol and at 2,4, and 6 positions three bromo groups are attached. Hence the name will be 2,4,6-tribromophenol
So, the statement is true.
The conversation of m- nitrophenol to resorcinol involves _________ respectively.
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hydrolysis, diasotisation and reduction
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diasotisation, reduction and hydrolysis
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hydrolysis, reduction and diazotisation
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reduction,diazotisation and hydrolysis
Which of the following species show maximum volatility?
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$$CH_3CH_2OH$$
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$$CH_3-O-CH_3$$
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$$H_2O$$
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$$HF$$
Explanation
Due to the lack of any Hydrogen bonding in case of $$CH_3-O-CH_3$$, its boiling point is comparatively low. In other words, it is the most volatile among the given compounds.
Hence, Option B is correct.
What is its IUPAC name?
$$CH_3-\underset{\displaystyle CH_3}{\underset{|}{CH}}-CH_2-OH$$.
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2-methyl propanol
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2-ethyl ethanol
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3-methyl propanol
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3-ethyl ethanol
Explanation
$$\overset{3}{C}H_3-\overset{2}{\underset{CH_3}{\underset{|}{CH}}}-\overset{1\,\,\,\,\,}{CH_2}-OH$$
2-Methyl propanol.
On boiling with concentrated hydrobromic acid, phenyl ethyl ether will yield:
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phenol and ethyl bromide
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phenol and ethane
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bromobenzene and ethanol
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bromobenzene and ethane
Explanation
On boiling with concentrated hydrobromic acid, phenyl ethyl ether will yield phenol and ethyl bromide.
Hence, option $$A$$ is correct.
I is more acidic than II
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True
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False
Explanation
1 is more acidic than 2 due to fact that hyperconjugation of $$-CH_3$$ group show less electron releasing effect than resonance effect of $$-OCH_3$$ group.
What is the functional group in alcohol?
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$$-OH$$
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$$-CN$$
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$$-NC$$
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-$$NH_2$$
Explanation
$$-OH$$ (Hydroxyl group) is the functional group of alcohol. Because, if we add $$-OH$$ to a Methyl group $$(-CH_3)$$, it will be converted to Methyl alcohol $$(CH_3OH)$$
$$3$$-methyl-$$2$$-pentene on reaction with HOCl gives:
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$$3$$-chloro-$$3$$-methyl pentanol-$$2$$
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$$2, 3$$-dichloro-$$3$$-methyl pentane
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$$2$$-chloro-$$3$$-methyl pentanol-$$3$$
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$$2, 3$$ dimethyl butanol-$$2$$
Explanation
The given reaction follows Markownikoff's addition mechanism and add $$OH^-$$ and $$Cl^+$$ across carbon-carbon double bond.
Reaction of bromine water with phenol gives white ppt of:
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o - bromophenol
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p - bromophenol
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2,4,6 -tribromophenol
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m - bromophenol
Explanation
Phenol when treated with $$Br_2$$ water give polybromoderivative in which all H- atoms at ortho and para positions with respect to $$-OH$$ group are replace by Cr atoms. It is so because in cequeous medium, phenol iouizes to form pheroxide ion. Due to presence of negative charge, the oxygen of pheroxide ion donates $$e^-$$s to benzene erring to large extent. As a result, the ring gets highly activated and tribubstitution occurs. On the other hand, in non-polar solvents, the ionization of phenol is greatly supressed. As a result, oxygen of $$-OH$$ group donates $$e^-$$s to benzene ring to a small extent only which results in slight activation of ring and monosubstitution occurs. Furthur, due to steric hindrance at ortho position, para substitution products usually predominates.
In which of the following reactions, the product obtained is tert-butyl methyl ether?
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$$CH_3OH+HO-CH_2-CH_3\overset{Conc. H_2SO_4}{\rightarrow}$$
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$$H_3C-\overset{CH_3}{\overset{|}{\underset{CH_3}{\underset{|}{C}}}}-Br+CH_3OH\overset{OH^-Na^+}{\rightarrow}$$
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$$CH_3Br+Na^+O^--\overset{CH_3}{\overset{|}{\underset{CH_3}{\underset{|}{C}}}}-CH_3\rightarrow$$
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$$CH_3-O^-Na^++CH_3-\overset{CH_3}{\overset{|}{\underset{CH_3}{\underset{|}{C}}}}-Br\rightarrow$$
Explanation
The product can be obtained by Williamson's synthesis and it involves an $$S_N2$$ pathway.
Thus we can see that Option D is not possible as it will lead to $$E2$$ products.
Therefore, Option C is the correct option.
Which of the following statement is incorrrect:
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Phenol gives positive bromine water test
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Aniline gives foul smelling compound on reaction with $$CHCl_3$$+$$KOH$$
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Formic acid gives positive Tollen's test
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Nitrobenzene gives poitive Tollen's test
Explanation
Nitrobenzene gives negative Tollen's test.
Tollen's test is used to distinguish between aldehyde and ketone.
Phenol and bromine water test will decolourise colour and white precipitate will form.
formic acid or $$H-COOH$$ has an aldehyde linked to it and so like any other aldehyde it gives a reaction will Tollen's reagent, precipitating silver.
In the reaction for dinitration,
the major dinitrated product ($$X$$) is:
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$$Ph - C \equiv C - Ph \xrightarrow{Na/NH_3} A \xrightarrow{HOCl/H^+} B$$
$$B$$ is/are:
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Compound 'B' is?
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Explanation
Image $$1$$ refers that carboinic acid is stronger than phenols, so phenols cannot decompose carbonates to evolve $$CO_2$$ gas and $$H_2O$$ gas.
Therefore, pka for phenol is $$5$$ and pka for carbonic acid is $$7.$$
So now, refer to image $$2.$$
The compound A on treatment with Na gives B, and with $${ PCI }_{ 5 }$$ gives C. B and C react together to give diethyl ether. A, B and C are in the order.
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$${ C }_{ 2 }{ H }_{ 5 }{ OH },C_{ 2 }{ H }_{ 5 }ONa,{ C }_{ 2 }{ H }_{ 5 }Cl$$
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$${ C }_{ 2 }{ H }_{ 5 }Cl,{ C }_{ 2 }{ H }_{ 6 },{ C }_{ 2 }{ H }_{ 5 }OH$$
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$${ C }_{ 2 }{ H }_{ 5 }OH,{ C }_{ 2 }{ H }_{ 5 }Cl,{ C }_{ 2 }{ H }_{ 5 }ONa$$
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$${ C }_{ 2 }{ H }_{ 5 }OH,{ C }_{ 2 }{ H }_{ 6 },{ C }_{ 2 }{ H }_{ 5 }Cl$$
Which of the following leads to formation of $${ 1 }^{ 0 }$$ alcohol?
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$${ CH }_{ 3 }-CH={ CH }_{ 2 }\xrightarrow [ { H }^{ + } ]{ { H }_{ 2 }O } $$
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$${ CH }_{ 3 }-CH={ CH }_{ 2 }\xrightarrow [ Hg{ \left( OAc \right) }_{ 2 }/THF ]{ Na{ BH }_{ 4 }/{ OH }^{ \ominus } } $$
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$${ CH }_{ 3 }-CH={ CH }_{ 2 }\xrightarrow [ { B }_{ 2 }{ H }_{ 6 }/THF ]{ { H }_{ 2 }{ O }_{ 2 }/{ OH }^{ \ominus } } $$
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$${ CH }_{ 3 }-C\equiv C-{ CH }_{ 3 }\xrightarrow [ dil.Hg{ SO }_{ 4 } ]{ dil.{ H }_{ 2 }{ SO }_{ 4 } } $$
Explanation
Morbenzene rule $$\rightarrow H^+$$ goes to carbon having more hydrogen among the double bond ones.
Which of the following is the best reagent to convert cyclohexanol into cyclohexene?
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Conc. $$HCl$$
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Conc. $$HBr$$
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Conc. $${ H }_{ 3 }{ PO }_{ 4 }$$
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Conc. $$HCl$$ with $${ ZnCl }_{ 2 }$$
Explanation
Conc. $${ H }_{ 3 }{ PO }_{ 4 }$$ is a good dehydrating agent which converts an alcohal to an alkene while the other three reagents are nucleophiles which convert alcohals to alkyl halides.
An organic compound 'A' liberates carbon dioxide from aqueous sodium hydrogen carbonate. When 'A' is added to aqueous bromine, the colour of the bromine is rapidly discharged. What could be 'A' ?
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$$HOCH_2COOH$$
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$$CH_3COOH$$
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Explanation
Aliphatic carboxylic acids will not discharge the red colour of bromine water and Phenol is not acidic enough to liberate $$CO_2$$ from $$NaHCO_3$$ solution.
Hence, Option D is the correct answer.
The compound a on treatment with Na gives B, and with $${ PCI }_{ 5 }$$ gives C, B and C react together to give diethyl ether, A, B and C are in the order?
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$${ C }_{ 2 }{ H }_{ 5 }OH,{ C }_{ 2 }{ H }_{ 6 },{ C }_{ 2 }{ H }_{ 5 }Cl$$
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$${ C }_{ 2 }{ H }_{ 5 }OH,{ C }_{ 2 }{ H }_{ 5 }CI,{ C }_{ 2 }{ H }_{ 5 }ONa$$
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$${ C }_{ 2 }{ H }_{ 5 }OH,{ C }_{ 2 }{ H }_{ 5 }ONa,{ C }_{ 2 }{ H }_{ 5 }Cl$$
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$$C_{ 2 }{ H }_{ 5 }CI,{ C }_{ 2 }{ H }_{ 6 },{ C }_{ 2 }{ H }_{ 5 }OH$$
Explanation
$$A \xrightarrow{Na} B \xrightarrow{PCls} C$$
One of $$B$$ & $$C$$ must be $$Et- O Na$$ & other $$Et - Cl$$
$$Et\ ONa$$ on reaction $$PCl_3$$ gives $$Et\ Cl$$
$$\therefore B = Et \, ONa$$
$$C = Et - Cl$$
Hence A must be $$Et - OH$$
$$Phenol+X\xrightarrow [ { H }_{ 2 }O ]{ NaOH } Phenyl benzoate + NaCl$$. Identify $$X$$ in the given:
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Benzoic acid
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Benzoyl chloride
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Anisole
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Benzyl alcohol
Which of the following ether is not cleaved by HI even at 525K ?
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$${ C }_{ 6 }{ H }_{ 5 }-O-{ CH }_{ 3 }$$
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$${ C }_{ 6 }{ H }_{ 5 }-O-{ C }_{ 6 }{ H }_{ 5 }$$
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$${ C }_{ 6 }{ H }_{ 5 }-O-{ C }_{ 3 }{ H }_{ 7 }$$
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Explanation
Will not cleaned by HI even at $$525k$$ temperature.Cyclic ether don't reaction by HI.
Which of the following is the $$A$$ in the given reaction?
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Benzene diazonium chloride
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Benzene
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Nitro benzene
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Amino benzene
Explanation
Benzene diazonium chloride when undergoes hydration it forms phenol as a product,
Thus the compound A is benzene diazonium chloride.
$$1-$$Bromopentane on boiling with alcoholic potassium hydroxide gives:
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Pentan$$-1-ol$$
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Pentan$$-2-ol$$
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pent$$-1-ene$$
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pent$$-2-ene$$
Explanation
When halo alkaline is treated with conc.alcoholic solution of KOH, a molecule of hydrogen halide is eliminated to form alkene. The eliminated hydrogen atom comes from B-carbon atom, so it is called B-elimination reaction. Most highly substituted alkene is major product .
The IUPAC name of the compound is:
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3, 4 - dimethyl - 2 - butene - 4 - ol
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1, 2 - dimethyl - 2- butenol
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3 - methylpent - 3 - en - 2- ol
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2, 3 - dimethyl - 3- Pentenol
Explanation
For naming a compound, the longest carbon chain is considered and the carbon atom near the functional group is given the first number or priority.
Hence, $$-OH$$ is at second carbon and a double bond is between third carbon. Also, a methyl group is present at the third carbon.
So, the IUPAC name of the compound will be $$3-methyl pent-3-en-2-ol$$.
Hence, option $$C$$ is correct.
Which of the following is the best reagent to convert $$1-$$ Methylcyclohexene into $$2-$$ methylcyclohexanol?
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dil.$$H_2SO_4$$
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$$Hg(OAc)_2/NaBH_4, H_2O$$
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$$B_2H_6/H_2O_2, \overset{\ominus}{O}H$$
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conc. $$H_2SO_4$$
Identify $$Z$$ in the sequence of reaction, $$CH_3CH_2CH = CH_2 \xrightarrow{HBr/H_2O_2} Y \xrightarrow{C_2H_5ON_a} Z$$
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$$(CH_3)_2CH_2 - O - CH_2CH_3$$
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$$CH_3(CH_2)_4 - O - CH_3$$
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$$CH_3CH_2 - CH(CH_3) - O - CH_2CH_3$$
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$$CH_3 - (CH_2)_3 - O - CH_2CH_3$$
$$\underrightarrow { { H }^{ 2 } } $$
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Explanation
solution is given
Phenol $$\overset { (i)\quad NaOH }{ \underset { (ii)\quad { CO }_{ 2 }/{ 140 }^{ 0 }C }{ \longrightarrow } } (A)\overset { { H }^{ + }/{ H }_{ 2 }O }{ \longrightarrow } (B)\overset { { AI }_{ 2 }{ O }_{ 3 } }{ \underset { { CH }_{ 3 }COOH }{ \longrightarrow } } (C)$$ In this reaction, the end product (C) is?
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salicyladehyde
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salicylic acid
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phenyl acetate
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aspirin
Propene is allowed to react with HBr in the presence of benzoyl peroxide, and the product is subsequently heated with aqueous KOH. The final product obtained is:
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propan-$$2$$-ol
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propane-$$1, 2$$- diol
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propan-$$1$$-ol
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propane-$$1, 3$$- diol
Explanation
Propene is allowed to react with $$HBr$$ in the presence of benzoyl peroxide, and the product is subsequently heated with aqueous $$KOH$$. The final product obtained is $$propan-1-ol$$
Phenol is a weak acid than ethanol.
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True
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False
Explanation
Both alcohols and phenol are weak acid, the alcohols are less acidic than phenol because it is very tough to remove $$'H'$$ ion from alcohol. Phenol can lose ion easily because phenoxide ion formed is stabilized to some extent. This is as the negative charge on the oxygen atom is delocalized around the ring.
The final product 'X' is:
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What is $$A$$?
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Explanation
Option A is the correct answer.
As we can see from the given mechanism, the compound at Option A will be formed.
State the product formed during the reaction between sodium phenoxide and ethyl iodide on heating.
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Phenol
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Phenetole
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Benzyl alcohol
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None of the above
Explanation
solution is given
The number of carbon atoms present in a molecule of simple ether is:
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always even
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always odd
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either even or odd
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unpredictable
Explanation
Simple ethers have the same alkyl groups on both sides of the Oxygen atom.
Like, $$CH_3-O-CH_3$$
Thus, the number of carbon atoms is always even.
Hence, Option A is correct.
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Ethers can be prepared by?
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dehydration of acids
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hydrolysis of alkyl halides
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dehydrogenation of alcohols
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dehydration of alcohols
Explanation
Ethers can be prepared by dehydration of alcohols.
Eg: Ethanol is dehydrated in presence of $$H_SO_4$$ at $$413K$$ to give ethoxyethane at $$413K$$ . It is a neucliophilic bimolecular reaction $$(S_n2)$$.
The formation of oxanium salt of ether is due to:
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protonation of O atom in ether
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donation of an e pair by O atom in ether
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the attachment of a proton to O atom in ether by co-ordination
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all the above
Explanation
Protonation of oxygen atom in ether as oxygen in ether act as nucleophile.
The compounds A and B are mixed in equimolar proportion to from the product, $$A+B \rightleftharpoons C +D$$. AT equilibrium, one third of A and B are consumed.The equilibrium constant for the reaction is _______.
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0.5
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4.0
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2.5
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0.25
Explanation
Products A and B respectively are:
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Reaction is not possible
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None of these
Explanation
Hence, the correct option is $$A$$
Among the following sets of reactants which one produces anisole?
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$$C_{6}H_{5}OH$$; neutral $$FeCl_{3}$$
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$$C_{6}H_{5}-CH_{3}$$;$$CH_{3}COCl$$:$$AlCl_{3}$$
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$$CH_{3}CHO;RMgX$$
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$$C_{6}H_{5}OH; NaOH; CH_{3}I$$
Explanation
Back attack of phenoxide ion to $$CH_3-I$$
Phenolphthalein is obtained by the coupling of phenol with ?
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aniline
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benzene diazonium chloride
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phthalic anhydride
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any of these
Explanation
Hence, Option C is the right answer.
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