ChemistryQ501–550 of 806 questions
Page 11 of 12 · English
501
ChemistryEasyMCQMHT CET · 2026
Identify the monomers used for the preparation of $Glyptal$.
A
$Styrene$ and $phosgene$
B
$Ethylene \ glycol$ and $phthalic \ acid$
C
$Butadiene$ and $ethylene \ glycol$
D
$Acrylamide$ and $urea$
Solution
(B) Step $1$: $Glyptal$ is a polyester resin formed by the condensation polymerization of $ethylene \ glycol$ and $phthalic \ acid$.
Step $2$: The reaction involves the elimination of water molecules to form a polymer chain.
Step $3$: The monomers are $HO-CH_2-CH_2-OH$ $(ethylene \ glycol)$ and $C_6H_4(COOH)_2$ $(phthalic \ acid)$.
Step $2$: The reaction involves the elimination of water molecules to form a polymer chain.
Step $3$: The monomers are $HO-CH_2-CH_2-OH$ $(ethylene \ glycol)$ and $C_6H_4(COOH)_2$ $(phthalic \ acid)$.
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502
ChemistryEasyMCQMHT CET · 2026
Which of the following polymers is used to manufacture shampoo bottles?
A
$HDPE$
B
$LDPE$
C
$PP$
D
$PS$
Solution
(A) $1$. $HDPE$ (High-Density Polyethylene) is a strong, durable, and chemical-resistant polymer.
$2$. Due to its high density and rigidity, it is widely used for manufacturing containers such as shampoo bottles, detergent bottles, and milk jugs.
$3$. $LDPE$ is more flexible and used for films, while $PP$ and $PS$ have different specific applications.
$2$. Due to its high density and rigidity, it is widely used for manufacturing containers such as shampoo bottles, detergent bottles, and milk jugs.
$3$. $LDPE$ is more flexible and used for films, while $PP$ and $PS$ have different specific applications.
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503
ChemistryEasyMCQMHT CET · 2026
Which of the following is a fully fluorinated polymer?
A
$PVC$
B
$Teflon$
C
$Neoprene$
D
$Buna-S$
Solution
(B) Step $1$: $Teflon$ is polytetrafluoroethylene $(PTFE)$.
Step $2$: The monomer of $Teflon$ is tetrafluoroethene $(CF_2=CF_2)$.
Step $3$: In this polymer, all hydrogen atoms of the ethene backbone are replaced by fluorine atoms, making it a fully fluorinated polymer.
Step $4$: $PVC$ is polyvinyl chloride, $Neoprene$ is polychloroprene, and $Buna-S$ is a copolymer of $1,3-butadiene$ and $styrene$, none of which are fully fluorinated.
Step $2$: The monomer of $Teflon$ is tetrafluoroethene $(CF_2=CF_2)$.
Step $3$: In this polymer, all hydrogen atoms of the ethene backbone are replaced by fluorine atoms, making it a fully fluorinated polymer.
Step $4$: $PVC$ is polyvinyl chloride, $Neoprene$ is polychloroprene, and $Buna-S$ is a copolymer of $1,3-butadiene$ and $styrene$, none of which are fully fluorinated.
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504
ChemistryEasyMCQMHT CET · 2026
$Buna-S$ is an example of:
A
elastomer
B
fibre
C
thermoplastic polymer
D
thermosetting polymer
Solution
(A) $Buna-S$ is a copolymer of $1,3-butadiene$ and $styrene$.
It possesses weak intermolecular forces of attraction between the polymer chains, which allows the polymer to be stretched.
Polymers that have weak van der Waals forces and can be stretched like rubber are classified as elastomers.
Therefore, $Buna-S$ is an example of an elastomer.
It possesses weak intermolecular forces of attraction between the polymer chains, which allows the polymer to be stretched.
Polymers that have weak van der Waals forces and can be stretched like rubber are classified as elastomers.
Therefore, $Buna-S$ is an example of an elastomer.
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505
ChemistryEasyMCQMHT CET · 2026
What type of polymer is natural rubber?
A
Fibre
B
Elastomer
C
Thermoplastic
D
Thermosetting plastic
Solution
(B) Natural rubber is a polymer of $isoprene$ $(2-methyl-1,3-butadiene)$.
It possesses weak intermolecular forces of attraction, such as van der Waals forces, which allow the polymer chains to be stretched.
Due to these weak forces, it can return to its original shape after being stretched, which is the characteristic property of an elastomer.
Therefore, natural rubber is classified as an elastomer.
It possesses weak intermolecular forces of attraction, such as van der Waals forces, which allow the polymer chains to be stretched.
Due to these weak forces, it can return to its original shape after being stretched, which is the characteristic property of an elastomer.
Therefore, natural rubber is classified as an elastomer.
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506
ChemistryMediumMCQMHT CET · 2026
Which of the following is a chain growth polymer?
A
$Nylon-6$
B
$Teflon$
C
$Nylon-66$
D
$Terylene$
Solution
(B) $1$. Chain growth polymers are formed by the addition of monomer units containing double or triple bonds without the loss of any small molecules.
$2$. $Teflon$ (polytetrafluoroethene) is formed by the addition polymerization of tetrafluoroethene $(CF_2=CF_2)$, which is a chain growth process.
$3$. $Nylon-6$, $Nylon-66$, and $Terylene$ are step growth (condensation) polymers formed by the loss of small molecules like $H_2O$ or $CH_3OH$.
$2$. $Teflon$ (polytetrafluoroethene) is formed by the addition polymerization of tetrafluoroethene $(CF_2=CF_2)$, which is a chain growth process.
$3$. $Nylon-6$, $Nylon-66$, and $Terylene$ are step growth (condensation) polymers formed by the loss of small molecules like $H_2O$ or $CH_3OH$.
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507
ChemistryEasyMCQMHT CET · 2026
Identify the polymer obtained by addition polymerization from the following.
A
$Nylon \ 6$
B
$Dacron$
C
$Polythene$
D
$Nylon \ 6,6$
Solution
(C) Step $1$: Addition polymerization involves the repeated addition of monomers containing double or triple bonds without the loss of any small molecules.
Step $2$: $Nylon \ 6$, $Dacron$, and $Nylon \ 6,6$ are condensation polymers formed by the elimination of small molecules like $H_2O$ or $CH_3OH$.
Step $3$: $Polythene$ is formed by the addition polymerization of $ethene$ $(CH_2=CH_2)$ monomers.
Step $4$: Therefore, $Polythene$ is the correct answer.
Step $2$: $Nylon \ 6$, $Dacron$, and $Nylon \ 6,6$ are condensation polymers formed by the elimination of small molecules like $H_2O$ or $CH_3OH$.
Step $3$: $Polythene$ is formed by the addition polymerization of $ethene$ $(CH_2=CH_2)$ monomers.
Step $4$: Therefore, $Polythene$ is the correct answer.
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508
ChemistryEasyMCQMHT CET · 2026
Which among the following is $NOT$ a polyester polymer?
A
Glyptal
B
Novolac
C
$PHBV$
D
Dacron
Solution
(B) Step $1$: Identify the chemical nature of each polymer.
Step $2$: $Glyptal$ is a polyester formed by the condensation of ethylene glycol and phthalic acid.
Step $3$: $PHBV$ (poly-$\beta$-hydroxybutyrate-co-$\beta$-hydroxyvalerate) is a biodegradable polyester.
Step $4$: $Dacron$ (Terylene) is a well-known polyester formed from ethylene glycol and terephthalic acid.
Step $5$: $Novolac$ is a linear polymer formed by the condensation of phenol and formaldehyde, which is a phenol-formaldehyde resin, not a polyester.
Step $6$: Therefore, $Novolac$ is $NOT$ a polyester polymer.
Step $2$: $Glyptal$ is a polyester formed by the condensation of ethylene glycol and phthalic acid.
Step $3$: $PHBV$ (poly-$\beta$-hydroxybutyrate-co-$\beta$-hydroxyvalerate) is a biodegradable polyester.
Step $4$: $Dacron$ (Terylene) is a well-known polyester formed from ethylene glycol and terephthalic acid.
Step $5$: $Novolac$ is a linear polymer formed by the condensation of phenol and formaldehyde, which is a phenol-formaldehyde resin, not a polyester.
Step $6$: Therefore, $Novolac$ is $NOT$ a polyester polymer.
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509
ChemistryEasyMCQMHT CET · 2026
Which of the following is not a basis of classification of polymers?
A
Source
B
Number of monomers
C
Method of preparation
D
Structure
Solution
(B) Polymers are classified based on several criteria such as:
$(1)$ Source (Natural, Synthetic, Semi-synthetic)
$(2)$ Structure (Linear, Branched, Cross-linked)
$(3)$ Method of polymerization (Addition, Condensation)
$(4)$ Molecular forces (Elastomers, Fibres, Thermoplastics, Thermosetting polymers)
$(5)$ Number of monomers is not a standard criterion for the classification of polymers. Therefore, option $B$ is the correct answer.
$(1)$ Source (Natural, Synthetic, Semi-synthetic)
$(2)$ Structure (Linear, Branched, Cross-linked)
$(3)$ Method of polymerization (Addition, Condensation)
$(4)$ Molecular forces (Elastomers, Fibres, Thermoplastics, Thermosetting polymers)
$(5)$ Number of monomers is not a standard criterion for the classification of polymers. Therefore, option $B$ is the correct answer.
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510
ChemistryEasyMCQMHT CET · 2026
Which of the following is a homopolymer?
A
Polyacrylonitrile
B
Buna-$N$
C
Urea formaldehyde resin
D
Polycarbonate
Solution
(A) $1$. $A$ homopolymer is a polymer formed from only one type of monomeric unit.
$2$. Polyacrylonitrile is formed by the polymerization of acrylonitrile $(CH_2=CH-CN)$ units only, hence it is a homopolymer.
$3$. Buna-$N$ is a copolymer of $1,3$-butadiene and acrylonitrile.
$4$. Urea formaldehyde resin is a copolymer of urea and formaldehyde.
$5$. Polycarbonate is a copolymer formed from bisphenol-$A$ and phosgene.
$6$. Therefore, Polyacrylonitrile is the correct answer.
$2$. Polyacrylonitrile is formed by the polymerization of acrylonitrile $(CH_2=CH-CN)$ units only, hence it is a homopolymer.
$3$. Buna-$N$ is a copolymer of $1,3$-butadiene and acrylonitrile.
$4$. Urea formaldehyde resin is a copolymer of urea and formaldehyde.
$5$. Polycarbonate is a copolymer formed from bisphenol-$A$ and phosgene.
$6$. Therefore, Polyacrylonitrile is the correct answer.
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511
ChemistryMediumMCQMHT CET · 2026
Which of the following polymers is obtained by addition polymerization? $[NH - (CH_2)_5 - CO]_n$
A
Teflon
B
Polyacrylonitrile
C
$LDP$
D
$PHBV$
Solution
(A) $1$. The given polymer $[NH - (CH_2)_5 - CO]_n$ is Nylon-$6$, which is a condensation polymer formed from caprolactam.
$2$. Teflon $(CF_2=CF_2)_n$, Polyacrylonitrile $(CH_2=CH-CN)_n$, and $LDP$ (Low Density Polyethylene) are formed by addition polymerization.
$3$. However, the question asks to identify which of the options is an addition polymer. Among the choices, Teflon, Polyacrylonitrile, and $LDP$ are all addition polymers. Given the structure provided in the question is Nylon-$6$ (a condensation polymer), the question likely asks for an addition polymer from the list. All options $A$, $B$, and $C$ are addition polymers. Assuming the question asks for a standard example of addition polymerization, Teflon is a primary example.
$2$. Teflon $(CF_2=CF_2)_n$, Polyacrylonitrile $(CH_2=CH-CN)_n$, and $LDP$ (Low Density Polyethylene) are formed by addition polymerization.
$3$. However, the question asks to identify which of the options is an addition polymer. Among the choices, Teflon, Polyacrylonitrile, and $LDP$ are all addition polymers. Given the structure provided in the question is Nylon-$6$ (a condensation polymer), the question likely asks for an addition polymer from the list. All options $A$, $B$, and $C$ are addition polymers. Assuming the question asks for a standard example of addition polymerization, Teflon is a primary example.
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512
ChemistryMediumMCQMHT CET · 2026
Which of the following polymers is $NOT$ obtained by the addition polymerisation method?
A
$Teflon$
B
$Polyacrylonitrile$
C
$LDP$
D
$PHBV$
Solution
(D) $1$. Addition polymers are formed by the repeated addition of monomer molecules possessing double or triple bonds without the loss of any small molecules.
$2$. $Teflon$, $Polyacrylonitrile$, and $LDP$ $(Low Density Polyethylene)$ are all addition polymers.
$3$. $PHBV$ $(Poly \beta-hydroxybutyrate-co-\beta-hydroxyvalerate)$ is a biodegradable aliphatic polyester, which is formed by condensation polymerisation of $3-hydroxybutanoic$ acid and $3-hydroxypentanoic$ acid.
$4$. Therefore, $PHBV$ is not obtained by addition polymerisation.
$2$. $Teflon$, $Polyacrylonitrile$, and $LDP$ $(Low Density Polyethylene)$ are all addition polymers.
$3$. $PHBV$ $(Poly \beta-hydroxybutyrate-co-\beta-hydroxyvalerate)$ is a biodegradable aliphatic polyester, which is formed by condensation polymerisation of $3-hydroxybutanoic$ acid and $3-hydroxypentanoic$ acid.
$4$. Therefore, $PHBV$ is not obtained by addition polymerisation.
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513
ChemistryEasyMCQMHT CET · 2026
What is the starting material used to obtain viscose rayon?
A
Styrene
B
Cellulose
C
Acrylamide
D
Bisphenol
Solution
(B) Viscose rayon is a regenerated fibre. It is obtained from naturally occurring polymers. The starting material used for the production of viscose rayon is purified cellulose, which is typically derived from wood pulp or cotton linters.
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514
ChemistryMediumMCQMHT CET · 2026
Which of the following alcohols reacts fastest with Lucas reagent and turns the solution turbid instantly?
A
$n$-Butyl alcohol
B
sec-Butyl alcohol
C
iso-Butyl alcohol
D
tert-Butyl alcohol
Solution
(D) $1$. Lucas reagent is a mixture of concentrated $HCl$ and anhydrous $ZnCl_2$.
$2$. The reaction follows an $S_N1$ mechanism, where the rate depends on the stability of the carbocation intermediate formed.
$3$. The order of reactivity of alcohols with Lucas reagent is: $3^\circ > 2^\circ > 1^\circ$.
$4$. $tert$-Butyl alcohol is a tertiary $(3^\circ)$ alcohol, which forms a stable tertiary carbocation, leading to an instantaneous reaction and turbidity.
$5$. Therefore, $tert$-Butyl alcohol reacts the fastest.
$2$. The reaction follows an $S_N1$ mechanism, where the rate depends on the stability of the carbocation intermediate formed.
$3$. The order of reactivity of alcohols with Lucas reagent is: $3^\circ > 2^\circ > 1^\circ$.
$4$. $tert$-Butyl alcohol is a tertiary $(3^\circ)$ alcohol, which forms a stable tertiary carbocation, leading to an instantaneous reaction and turbidity.
$5$. Therefore, $tert$-Butyl alcohol reacts the fastest.
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515
ChemistryMediumMCQMHT CET · 2026
Select the correct decreasing order of acid strength for the following compounds: $(a)$ $Ethanol$ $(b)$ $2-Methylpropan-2-ol$ $(c)$ $Phenol$ $(d)$ $p-Nitrophenol$.
A
$a > b > c > d$
B
$c > d > b > a$
C
$d > c > a > b$
D
$d > b > c > a$
Solution
(C) $1$. Acid strength depends on the stability of the conjugate base formed after the loss of a proton $(H^+)$.
$2$. $p-Nitrophenol$ $(d)$ is the strongest acid because the $-NO_2$ group exerts a strong $-I$ and $-M$ effect, which stabilizes the phenoxide ion significantly.
$3$. $Phenol$ $(c)$ is more acidic than alcohols because the phenoxide ion is stabilized by resonance.
$4$. Among alcohols, $Ethanol$ $(a)$ is more acidic than $2-Methylpropan-2-ol$ $(b)$ because the electron-donating inductive effect $(+I)$ of the three methyl groups in $2-Methylpropan-2-ol$ destabilizes the alkoxide ion more than the single methyl group in $Ethanol$.
$5$. Thus, the decreasing order of acid strength is $d > c > a > b$.
$2$. $p-Nitrophenol$ $(d)$ is the strongest acid because the $-NO_2$ group exerts a strong $-I$ and $-M$ effect, which stabilizes the phenoxide ion significantly.
$3$. $Phenol$ $(c)$ is more acidic than alcohols because the phenoxide ion is stabilized by resonance.
$4$. Among alcohols, $Ethanol$ $(a)$ is more acidic than $2-Methylpropan-2-ol$ $(b)$ because the electron-donating inductive effect $(+I)$ of the three methyl groups in $2-Methylpropan-2-ol$ destabilizes the alkoxide ion more than the single methyl group in $Ethanol$.
$5$. Thus, the decreasing order of acid strength is $d > c > a > b$.
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516
ChemistryDifficultMCQMHT CET · 2026
Identify product '$B$' in the following series of reactions: $2-methylpropan-2-ol \xrightarrow{20\% H_2SO_4, 363 K} A \xrightarrow{H_2O, \Delta, Conc. H_2SO_4} B$
A
$n-Butyl alcohol$
B
$sec-Butyl alcohol$
C
$Iso-butyl alcohol$
D
$tert-Butyl alcohol$
Solution
(D) Step $1$: Dehydration of $2-methylpropan-2-ol$ with $20\% H_2SO_4$ at $363 \ K$ gives $2-methylprop-1-ene$ $(A)$ as the major product via $E1$ mechanism.
$(CH_3)_3C-OH \xrightarrow{H^+} (CH_3)_3C^+ \xrightarrow{-H^+} CH_2=C(CH_3)_2$ $(A)$.
Step $2$: Acid-catalyzed hydration of $2-methylprop-1-ene$ $(A)$ follows Markovnikov's rule to give $2-methylpropan-2-ol$ $(B)$ as the major product.
$CH_2=C(CH_3)_2 + H_2O \xrightarrow{H^+} (CH_3)_3C-OH$ $(B)$.
Thus, the product '$B$' is $tert-Butyl alcohol$.
$(CH_3)_3C-OH \xrightarrow{H^+} (CH_3)_3C^+ \xrightarrow{-H^+} CH_2=C(CH_3)_2$ $(A)$.
Step $2$: Acid-catalyzed hydration of $2-methylprop-1-ene$ $(A)$ follows Markovnikov's rule to give $2-methylpropan-2-ol$ $(B)$ as the major product.
$CH_2=C(CH_3)_2 + H_2O \xrightarrow{H^+} (CH_3)_3C-OH$ $(B)$.
Thus, the product '$B$' is $tert-Butyl alcohol$.
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517
ChemistryMediumMCQMHT CET · 2026
Which of the following molecules has the maximum boiling point?
A
$CH_3 - CH(OH) - CH_2 - CH_2 - OH$
B
$CH_3 - CH_2 - Br$
C
$CH_3 - CH_2 - CH_2 - Br$
D
$CH_3 - C(CH_3)(Cl) - OH$
Solution
(A) $1$. Boiling point depends on intermolecular forces. Hydrogen bonding significantly increases the boiling point compared to dipole-dipole interactions or London dispersion forces.
$2$. Option $(A)$ is a diol $(CH_3 - CH(OH) - CH_2 - CH_2 - OH)$, which can form extensive intermolecular hydrogen bonding due to the presence of two $-OH$ groups.
$3$. Option $(B)$ and $(C)$ are alkyl halides, which exhibit dipole-dipole interactions.
$4$. Option $(D)$ is a chlorohydrin, which has only one $-OH$ group and is less effective at hydrogen bonding than the diol.
$5$. Therefore, the diol in option $(A)$ has the highest boiling point.
$2$. Option $(A)$ is a diol $(CH_3 - CH(OH) - CH_2 - CH_2 - OH)$, which can form extensive intermolecular hydrogen bonding due to the presence of two $-OH$ groups.
$3$. Option $(B)$ and $(C)$ are alkyl halides, which exhibit dipole-dipole interactions.
$4$. Option $(D)$ is a chlorohydrin, which has only one $-OH$ group and is less effective at hydrogen bonding than the diol.
$5$. Therefore, the diol in option $(A)$ has the highest boiling point.
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518
ChemistryMediumMCQMHT CET · 2026
The number of primary isomeric alcohols with the molecular formula $C_4H_{10}O$ is:
A
$1$
B
$2$
C
$3$
D
$4$
Solution
(B) Step $1$: Identify the possible structures for the molecular formula $C_4H_{10}O$. This formula corresponds to saturated alcohols or ethers. We are looking for primary alcohols $(R-CH_2OH)$.
Step $2$: Draw the isomers:
$(i)$ $CH_3-CH_2-CH_2-CH_2OH$ (Butan$-1-$ol) - This is a primary alcohol.
(ii) $(CH_3)_2CH-CH_2OH$ ($2$-Methylpropan$-1-$ol) - This is a primary alcohol.
(iii) $CH_3-CH_2-CH(OH)-CH_3$ (Butan$-2-$ol) - This is a secondary alcohol.
(iv) $(CH_3)_3C-OH$ ($2$-Methylpropan$-2-$ol) - This is a tertiary alcohol.
Step $3$: Count the primary alcohols. There are two: Butan$-1-$ol and $2-$Methylpropan$-1-$ol.
Therefore, the number of primary isomeric alcohols is $2$.
Step $2$: Draw the isomers:
$(i)$ $CH_3-CH_2-CH_2-CH_2OH$ (Butan$-1-$ol) - This is a primary alcohol.
(ii) $(CH_3)_2CH-CH_2OH$ ($2$-Methylpropan$-1-$ol) - This is a primary alcohol.
(iii) $CH_3-CH_2-CH(OH)-CH_3$ (Butan$-2-$ol) - This is a secondary alcohol.
(iv) $(CH_3)_3C-OH$ ($2$-Methylpropan$-2-$ol) - This is a tertiary alcohol.
Step $3$: Count the primary alcohols. There are two: Butan$-1-$ol and $2-$Methylpropan$-1-$ol.
Therefore, the number of primary isomeric alcohols is $2$.
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519
ChemistryMediumMCQMHT CET · 2026
Identify the alcohol obtained when an aldehyde other than formaldehyde $(HCHO)$ is treated with Grignard's reagent $(RMgX)$ followed by hydrolysis.
A
Primary alcohol
B
Secondary alcohol
C
Tertiary alcohol
D
Ketone
Solution
(B) $1$. The general reaction of an aldehyde with a Grignard reagent $(R'MgX)$ is: $RCHO + R'MgX \rightarrow RCH(OMgX)R'$.
$2$. Upon subsequent hydrolysis, the intermediate forms a secondary alcohol: $RCH(OMgX)R' + H_2O \rightarrow RCH(OH)R' + Mg(OH)X$.
$3$. Since formaldehyde $(HCHO)$ reacts with Grignard reagents to form primary alcohols, all other aldehydes ($RCHO$ where $R \neq H$) yield secondary alcohols.
$2$. Upon subsequent hydrolysis, the intermediate forms a secondary alcohol: $RCH(OMgX)R' + H_2O \rightarrow RCH(OH)R' + Mg(OH)X$.
$3$. Since formaldehyde $(HCHO)$ reacts with Grignard reagents to form primary alcohols, all other aldehydes ($RCHO$ where $R \neq H$) yield secondary alcohols.
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520
ChemistryMediumMCQMHT CET · 2026
Identify the structure of product $A$ in the following reaction:
$\text{Cyclohexanecarboxylic acid} \xrightarrow{LiAlH_4} A$
$\text{Cyclohexanecarboxylic acid} \xrightarrow{LiAlH_4} A$
A
B
C
D
Solution
(C) The reaction involves the reduction of a carboxylic acid using $LiAlH_4$ (Lithium aluminium hydride).
$LiAlH_4$ is a strong reducing agent that reduces carboxylic acids to primary alcohols.
The reaction is:
$C_6H_{11}COOH + 4[H] \xrightarrow{LiAlH_4} C_6H_{11}CH_2OH + H_2O$
Here, the carboxylic acid group $(-COOH)$ is converted into a primary alcohol group $(-CH_2OH)$.
Therefore, the product $A$ is cyclohexylmethanol, which corresponds to the structure shown in option $C$.
$LiAlH_4$ is a strong reducing agent that reduces carboxylic acids to primary alcohols.
The reaction is:
$C_6H_{11}COOH + 4[H] \xrightarrow{LiAlH_4} C_6H_{11}CH_2OH + H_2O$
Here, the carboxylic acid group $(-COOH)$ is converted into a primary alcohol group $(-CH_2OH)$.
Therefore, the product $A$ is cyclohexylmethanol, which corresponds to the structure shown in option $C$.
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521
ChemistryMediumMCQMHT CET · 2026
Which of the following is the correct decreasing order of boiling points for the given organic compounds?
A
$Aldehyde > Ketone > Alcohol > Carboxylic acid$
B
$Carboxylic acid > Alcohol > Ketone > Aldehyde$
C
$Ketone > Aldehyde > Carboxylic acid > Alcohol$
D
$Alcohol > Aldehyde > Ketone > Carboxylic acid$
Solution
(B) $1$. Boiling point depends on the strength of intermolecular forces.
$2$. $Carboxylic \ acids$ have the highest boiling point due to the formation of stable intermolecular hydrogen-bonded dimers.
$3$. $Alcohols$ have high boiling points due to intermolecular hydrogen bonding, but it is weaker than the dimer formation in $carboxylic \ acids$.
$4$. $Aldehydes$ and $Ketones$ have dipole-dipole interactions, which are weaker than hydrogen bonding.
$5$. Between $Aldehydes$ and $Ketones$, $Ketones$ generally have higher boiling points due to higher dipole moments.
$6$. Thus, the decreasing order is: $Carboxylic \ acid > Alcohol > Ketone > Aldehyde$.
$2$. $Carboxylic \ acids$ have the highest boiling point due to the formation of stable intermolecular hydrogen-bonded dimers.
$3$. $Alcohols$ have high boiling points due to intermolecular hydrogen bonding, but it is weaker than the dimer formation in $carboxylic \ acids$.
$4$. $Aldehydes$ and $Ketones$ have dipole-dipole interactions, which are weaker than hydrogen bonding.
$5$. Between $Aldehydes$ and $Ketones$, $Ketones$ generally have higher boiling points due to higher dipole moments.
$6$. Thus, the decreasing order is: $Carboxylic \ acid > Alcohol > Ketone > Aldehyde$.
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522
ChemistryMediumMCQMHT CET · 2026
Which among the following has the lowest boiling point?
A
$n$-Butyl alcohol
B
Isobutyl alcohol
C
$sec$-Butyl alcohol
D
$tert$-Butyl alcohol
Solution
(D) $1$. Boiling point of alcohols depends on the extent of intermolecular hydrogen bonding and the surface area of the molecule.
$2$. As the branching in the alkyl chain increases, the surface area of the molecule decreases, which reduces the magnitude of van der Waals forces.
$3$. Among the given isomers, $n$-Butyl alcohol is a straight chain (highest surface area), while $tert$-Butyl alcohol is highly branched (lowest surface area).
$4$. Therefore, $tert$-Butyl alcohol has the weakest intermolecular forces and the lowest boiling point.
$2$. As the branching in the alkyl chain increases, the surface area of the molecule decreases, which reduces the magnitude of van der Waals forces.
$3$. Among the given isomers, $n$-Butyl alcohol is a straight chain (highest surface area), while $tert$-Butyl alcohol is highly branched (lowest surface area).
$4$. Therefore, $tert$-Butyl alcohol has the weakest intermolecular forces and the lowest boiling point.
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523
ChemistryMediumMCQMHT CET · 2026
Which of the following isomers of $C_4H_9OH$ has the lowest boiling point?
A
$n$-Butyl alcohol
B
Isobutyl alcohol
C
$sec$-Butyl alcohol
D
$tert$-Butyl alcohol
Solution
(D) $1$. Boiling point of alcohols depends on the surface area of the molecule and the extent of intermolecular hydrogen bonding.
$2$. As the branching in the alkyl chain increases, the surface area of the molecule decreases, which reduces the magnitude of van der Waals forces.
$3$. Among the given isomers, $n$-Butyl alcohol is a straight chain, while $tert$-Butyl alcohol is highly branched.
$4$. Due to maximum branching, $tert$-Butyl alcohol has the smallest surface area and the weakest intermolecular forces, resulting in the lowest boiling point.
$2$. As the branching in the alkyl chain increases, the surface area of the molecule decreases, which reduces the magnitude of van der Waals forces.
$3$. Among the given isomers, $n$-Butyl alcohol is a straight chain, while $tert$-Butyl alcohol is highly branched.
$4$. Due to maximum branching, $tert$-Butyl alcohol has the smallest surface area and the weakest intermolecular forces, resulting in the lowest boiling point.
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524
ChemistryDifficultMCQMHT CET · 2026
Identify the compound obtained when ethyl methyl ketone is treated with $CH_3MgBr$ followed by hydrolysis.
A
$(CH_3)_3C - OH$
B
$C_2H_5 - C(CH_3)_2 - OH$
C
$(C_2H_5)_3C - OH$
D
$CH_3 - C(C_2H_5)_2 - OH$
Solution
(B) Step $1$: Ethyl methyl ketone is $CH_3-CO-C_2H_5$.
Step $2$: Grignard reagent $CH_3MgBr$ acts as a nucleophile, where $CH_3^-$ attacks the carbonyl carbon.
Step $3$: The reaction is: $CH_3-CO-C_2H_5 + CH_3MgBr \rightarrow CH_3-C(OMgBr)(CH_3)-C_2H_5$.
Step $4$: Upon hydrolysis, the intermediate forms $CH_3-C(OH)(CH_3)-C_2H_5$, which is $2-methylbutan-2-ol$ or $C_2H_5-C(CH_3)_2-OH$.
Step $2$: Grignard reagent $CH_3MgBr$ acts as a nucleophile, where $CH_3^-$ attacks the carbonyl carbon.
Step $3$: The reaction is: $CH_3-CO-C_2H_5 + CH_3MgBr \rightarrow CH_3-C(OMgBr)(CH_3)-C_2H_5$.
Step $4$: Upon hydrolysis, the intermediate forms $CH_3-C(OH)(CH_3)-C_2H_5$, which is $2-methylbutan-2-ol$ or $C_2H_5-C(CH_3)_2-OH$.
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525
ChemistryMediumMCQMHT CET · 2026
Which of the following hydroxy compounds contains only one $-OH$ group in a molecule?
A
Crotonyl alcohol
B
Ethylene glycol
C
Pyrogallol
D
Hydroquinone
Solution
(A) $1$. Crotonyl alcohol is $CH_3-CH=CH-CH_2OH$, which contains only one $-OH$ group.
$2$. Ethylene glycol is $HO-CH_2-CH_2-OH$, which contains two $-OH$ groups.
$3$. Pyrogallol is $1,2,3-trihydroxybenzene$, which contains three $-OH$ groups.
$4$. Hydroquinone is $1,4-dihydroxybenzene$, which contains two $-OH$ groups.
$5$. Therefore, only Crotonyl alcohol contains one $-OH$ group.
$2$. Ethylene glycol is $HO-CH_2-CH_2-OH$, which contains two $-OH$ groups.
$3$. Pyrogallol is $1,2,3-trihydroxybenzene$, which contains three $-OH$ groups.
$4$. Hydroquinone is $1,4-dihydroxybenzene$, which contains two $-OH$ groups.
$5$. Therefore, only Crotonyl alcohol contains one $-OH$ group.
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526
ChemistryMediumMCQMHT CET · 2026
Which of the following pairs of reagents is used for the conversion of a carboxylic acid to a primary alcohol?
A
$LiAlH_4 / H_3O^+$
B
$H_2 / Ni - \text{heat}$
C
$B_2H_6 / H_2O_2, OH^-$
D
$H_2 / Pd$
Solution
(A) Step $1$: Carboxylic acids are resistant to reduction by catalytic hydrogenation ($H_2/Ni$ or $H_2/Pd$) because the $C=O$ bond in the carboxyl group is less reactive than in aldehydes or ketones.
Step $2$: $LiAlH_4$ (Lithium aluminium hydride) is a strong reducing agent capable of reducing carboxylic acids to primary alcohols.
Step $3$: The reaction proceeds as: $RCOOH \xrightarrow{LiAlH_4 / H_3O^+} RCH_2OH$.
Step $4$: Therefore, $LiAlH_4$ is the correct reagent for this transformation.
Step $2$: $LiAlH_4$ (Lithium aluminium hydride) is a strong reducing agent capable of reducing carboxylic acids to primary alcohols.
Step $3$: The reaction proceeds as: $RCOOH \xrightarrow{LiAlH_4 / H_3O^+} RCH_2OH$.
Step $4$: Therefore, $LiAlH_4$ is the correct reagent for this transformation.
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527
ChemistryMediumMCQMHT CET · 2026
Which of the following alcohols contains the hydroxy group, at the side chain of the aromatic ring?
A
Allylic alcohol
B
Vinylic alcohol
C
Aromatic alcohols
D
Aliphatic alcohol
Solution
(C) $1$. In aromatic alcohols, the hydroxyl group $(-OH)$ is attached to a carbon atom of the side chain of an aromatic ring.
$2$. An example is benzyl alcohol $(C_6H_5CH_2OH)$, where the $-OH$ group is attached to the side chain carbon, not directly to the benzene ring.
$3$. Phenols have the $-OH$ group directly attached to the aromatic ring, whereas aromatic alcohols have it on the side chain.
$4$. Therefore, the correct classification is aromatic alcohols.
$2$. An example is benzyl alcohol $(C_6H_5CH_2OH)$, where the $-OH$ group is attached to the side chain carbon, not directly to the benzene ring.
$3$. Phenols have the $-OH$ group directly attached to the aromatic ring, whereas aromatic alcohols have it on the side chain.
$4$. Therefore, the correct classification is aromatic alcohols.
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528
ChemistryEasyMCQMHT CET · 2026
Which of the following is found to raise the $HDL$ level in the blood?
A
Omega-6 fatty acids
B
Omega-3 fatty acids
C
Trans fats
D
Cis fats
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529
ChemistryMediumMCQMHT CET · 2026
Which of the following statements are correct?
$A$) $\alpha$-amino acids present in proteins are $L$-amino acids.
$B$) Amino acids contain both $-NH_2$ and $-COOH$ groups.
$C$) The number of amino groups and carboxylic groups is always the same in amino acids.
$D$) Tyrosine was first obtained from cheese.
Choose the correct alternatives from below:
$A$) $\alpha$-amino acids present in proteins are $L$-amino acids.
$B$) Amino acids contain both $-NH_2$ and $-COOH$ groups.
$C$) The number of amino groups and carboxylic groups is always the same in amino acids.
$D$) Tyrosine was first obtained from cheese.
Choose the correct alternatives from below:
A
$A$ and $B$
B
$A$, $B$ and $D$
C
$B$ and $C$
D
$A$, $B$ and $C$
Solution
(B) Step $1$: Statement $A$ is correct. Naturally occurring $\alpha$-amino acids in proteins are of the $L$-configuration.
Step $2$: Statement $B$ is correct. By definition, amino acids contain at least one amino group $(-NH_2)$ and one carboxylic acid group $(-COOH)$.
Step $3$: Statement $C$ is incorrect. Amino acids can be acidic (more $-COOH$ than $-NH_2$, e.g., aspartic acid) or basic (more $-NH_2$ than $-COOH$, e.g., lysine).
Step $4$: Statement $D$ is correct. The name 'Tyrosine' is derived from the Greek word 'tyros', meaning cheese, as it was first isolated from casein in cheese.
Step $5$: Therefore, statements $A$, $B$, and $D$ are correct.
Step $2$: Statement $B$ is correct. By definition, amino acids contain at least one amino group $(-NH_2)$ and one carboxylic acid group $(-COOH)$.
Step $3$: Statement $C$ is incorrect. Amino acids can be acidic (more $-COOH$ than $-NH_2$, e.g., aspartic acid) or basic (more $-NH_2$ than $-COOH$, e.g., lysine).
Step $4$: Statement $D$ is correct. The name 'Tyrosine' is derived from the Greek word 'tyros', meaning cheese, as it was first isolated from casein in cheese.
Step $5$: Therefore, statements $A$, $B$, and $D$ are correct.
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530
ChemistryMediumMCQMHT CET · 2026
Which of the following molecules is capable of forming a Zwitter ion?
A
$NH_2CH_2COOH$
B
$CH_3CH_2NH_2$
C
$C_2Cl_3NO_2$
D
None of these
Solution
(A) $1$. $A$ Zwitter ion is a molecule that contains both a positive and a negative charge, making it electrically neutral overall.
$2$. Amino acids, which contain both an acidic carboxyl group $(-COOH)$ and a basic amino group $(-NH_2)$, are capable of internal proton transfer.
$3$. In $NH_2CH_2COOH$ (Glycine), the proton from the $-COOH$ group is transferred to the $-NH_2$ group, forming $^+NH_3CH_2COO^-$, which is a Zwitter ion.
$4$. $CH_3CH_2NH_2$ is a simple amine and $C_2Cl_3NO_2$ does not possess the required acidic and basic functional groups to form a Zwitter ion.
$2$. Amino acids, which contain both an acidic carboxyl group $(-COOH)$ and a basic amino group $(-NH_2)$, are capable of internal proton transfer.
$3$. In $NH_2CH_2COOH$ (Glycine), the proton from the $-COOH$ group is transferred to the $-NH_2$ group, forming $^+NH_3CH_2COO^-$, which is a Zwitter ion.
$4$. $CH_3CH_2NH_2$ is a simple amine and $C_2Cl_3NO_2$ does not possess the required acidic and basic functional groups to form a Zwitter ion.
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531
ChemistryEasyMCQMHT CET · 2026
Which of the following amino acids does $NOT$ contain a chiral carbon?
A
Valine
B
Alanine
C
Glycine
D
Isoleucine
Solution
(C) $1$. $A$ chiral carbon is a carbon atom bonded to four different groups.
$2$. The general structure of an amino acid is $R-CH(NH_2)-COOH$.
$3$. For $Glycine$, the side chain $R$ is a hydrogen atom $(-H)$.
$4$. Thus, the structure of $Glycine$ is $H-CH(NH_2)-COOH$, which means the central carbon is bonded to two identical hydrogen atoms.
$5$. Since it does not have four different groups attached to the central carbon, $Glycine$ is achiral.
$2$. The general structure of an amino acid is $R-CH(NH_2)-COOH$.
$3$. For $Glycine$, the side chain $R$ is a hydrogen atom $(-H)$.
$4$. Thus, the structure of $Glycine$ is $H-CH(NH_2)-COOH$, which means the central carbon is bonded to two identical hydrogen atoms.
$5$. Since it does not have four different groups attached to the central carbon, $Glycine$ is achiral.
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532
ChemistryMediumMCQMHT CET · 2026
Which of the following amino acids has a unique structure such that the side chain connects to the backbone of the amino acid at two points?
A
Histidine
B
Proline
C
Tryptophan
D
Tyrosine
Solution
(B) Step $1$: Analyze the structure of amino acids. Most amino acids have a primary amine group $(-NH_2)$ attached to the $\alpha$-carbon.
Step $2$: Identify the structure of $Proline$. $Proline$ is a cyclic amino acid where the side chain (the alkyl group) is bonded to both the $\alpha$-carbon and the nitrogen atom of the amino group.
Step $3$: This forms a five-membered pyrrolidine ring, effectively connecting the side chain to the backbone at two points (the $\alpha$-carbon and the nitrogen atom).
Step $4$: Therefore, $Proline$ is the unique amino acid described.
Step $2$: Identify the structure of $Proline$. $Proline$ is a cyclic amino acid where the side chain (the alkyl group) is bonded to both the $\alpha$-carbon and the nitrogen atom of the amino group.
Step $3$: This forms a five-membered pyrrolidine ring, effectively connecting the side chain to the backbone at two points (the $\alpha$-carbon and the nitrogen atom).
Step $4$: Therefore, $Proline$ is the unique amino acid described.
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533
ChemistryEasyMCQMHT CET · 2026
Which of the following amino acids is an essential amino acid?
A
Phenylalanine
B
Tyrosine
C
Glutamine
D
Alanine
Solution
(A) Essential amino acids are those that cannot be synthesized by the human body and must be obtained through the diet. Among the given options, $Phenylalanine$ is an essential amino acid. $Tyrosine$, $Glutamine$, and $Alanine$ are non-essential amino acids as they can be synthesized by the body.
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534
ChemistryEasyMCQMHT CET · 2026
Which of the following enzymes cleaves the glycosidic bond in $sucrose$?
A
$Amylose$
B
$Trypsin$
C
$Invertase$
D
$Maltase$
Solution
(C) $1$. $Sucrose$ is a disaccharide composed of $glucose$ and $fructose$ units linked by a glycosidic bond.
$2$. The enzyme $Invertase$ (also known as $sucrase$) specifically catalyzes the hydrolysis of the glycosidic bond in $sucrose$ to yield $glucose$ and $fructose$.
$3$. $Amylose$ is a polysaccharide, $Trypsin$ is a protease, and $Maltase$ acts on $maltose$.
$4$. Therefore, the correct enzyme is $Invertase$.
$2$. The enzyme $Invertase$ (also known as $sucrase$) specifically catalyzes the hydrolysis of the glycosidic bond in $sucrose$ to yield $glucose$ and $fructose$.
$3$. $Amylose$ is a polysaccharide, $Trypsin$ is a protease, and $Maltase$ acts on $maltose$.
$4$. Therefore, the correct enzyme is $Invertase$.
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535
ChemistryEasyMCQMHT CET · 2026
Identify the glycosidic linkage present in lactose.
A
$\alpha - 1, 4$
B
$\beta - 1, 4$
C
$\alpha - 1, 6$
D
$\beta - 2, 6$
Solution
(B) Lactose is a disaccharide composed of one molecule of $D-(+)$-galactose and one molecule of $D-(+)$-glucose.
The two monosaccharide units are linked together through a $\beta - 1, 4$-glycosidic linkage.
This means the $C-1$ of the galactose unit is connected to the $C-4$ of the glucose unit via an oxygen atom in the $\beta$-configuration.
The two monosaccharide units are linked together through a $\beta - 1, 4$-glycosidic linkage.
This means the $C-1$ of the galactose unit is connected to the $C-4$ of the glucose unit via an oxygen atom in the $\beta$-configuration.
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536
ChemistryEasyMCQMHT CET · 2026
What type of glycosidic bond is present in maltose?
A
$\beta - 1, 4$
B
$\alpha - 1, 4$
C
$\alpha - 1, 6$
D
$\beta - 2, 4$
Solution
(B) Maltose is a disaccharide formed by the condensation of two molecules of $\alpha - D - \text{glucose}$.
The glycosidic linkage is formed between the $C1$ of one glucose unit and the $C4$ of the other glucose unit.
Since both glucose units are in the $\alpha$ configuration, the bond is an $\alpha - 1, 4 - \text{glycosidic linkage}$.
The glycosidic linkage is formed between the $C1$ of one glucose unit and the $C4$ of the other glucose unit.
Since both glucose units are in the $\alpha$ configuration, the bond is an $\alpha - 1, 4 - \text{glycosidic linkage}$.
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537
ChemistryMediumMCQMHT CET · 2026
The reaction of glucose with acetic anhydride confirms the:
A
straight chain structure of glucose
B
presence of carbonyl group
C
presence of keto group
D
presence of five hydroxyl groups
Solution
(D) $1$. Glucose reacts with acetic anhydride to form glucose pentaacetate.
$2$. This reaction indicates that there are $5$ hydroxyl $(-OH)$ groups present in the glucose molecule.
$3$. The reaction is: $C_6H_{12}O_6 + 5(CH_3CO)_2O \rightarrow C_6H_7O(OCOCH_3)_5 + 5CH_3COOH$.
$2$. This reaction indicates that there are $5$ hydroxyl $(-OH)$ groups present in the glucose molecule.
$3$. The reaction is: $C_6H_{12}O_6 + 5(CH_3CO)_2O \rightarrow C_6H_7O(OCOCH_3)_5 + 5CH_3COOH$.
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538
ChemistryMediumMCQMHT CET · 2026
Which of the following is confirmed by the reaction of glucose with hydroxylamine $(NH_2OH)$?
A
Straight chain of six carbon atoms
B
Presence of a carbonyl group
C
Presence of a primary alcoholic group
D
Presence of a secondary alcoholic group
Solution
(B) Glucose reacts with hydroxylamine $(NH_2OH)$ to form an oxime. The formation of an oxime indicates the presence of a carbonyl group (aldehyde or ketone) in the glucose molecule.
Reaction: $CHO(CHOH)_4CH_2OH + NH_2OH \rightarrow CH=NOH(CHOH)_4CH_2OH + H_2O$.
Reaction: $CHO(CHOH)_4CH_2OH + NH_2OH \rightarrow CH=NOH(CHOH)_4CH_2OH + H_2O$.
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539
ChemistryEasyMCQMHT CET · 2026
Identify the product formed when $D-(+)-glucose$ reacts with bromine water ($Br_2$ water).
A
Saccharic acid
B
Cyanohydrin
C
Gluconic acid
D
Nucleic acid
Solution
(C) Step $1$: Bromine water ($Br_2$ water) is a mild oxidizing agent.
Step $2$: It specifically oxidizes the aldehyde group $(-CHO)$ of $glucose$ to a carboxylic acid group $(-COOH)$.
Step $3$: The reaction is: $CHO(CHOH)_4CH_2OH + [O] \xrightarrow{Br_2/H_2O} COOH(CHOH)_4CH_2OH$.
Step $4$: The resulting product is $Gluconic \ acid$.
Step $2$: It specifically oxidizes the aldehyde group $(-CHO)$ of $glucose$ to a carboxylic acid group $(-COOH)$.
Step $3$: The reaction is: $CHO(CHOH)_4CH_2OH + [O] \xrightarrow{Br_2/H_2O} COOH(CHOH)_4CH_2OH$.
Step $4$: The resulting product is $Gluconic \ acid$.
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540
ChemistryEasyMCQMHT CET · 2026
What type of saccharide is $maltose$?
A
Polysaccharide
B
Disaccharide
C
Trisaccharide
D
Monosaccharide
Solution
(B) $Maltose$ is a carbohydrate formed by the condensation of two molecules of $D-glucose$. Since it yields two molecules of monosaccharides upon hydrolysis, it is classified as a disaccharide.
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541
ChemistryMediumMCQMHT CET · 2026
Which carbon atom of glucose, numbered from $1$ to $6$, forms a hemiacetal structure by reacting with the $-CHO$ group to close the ring?
A
$C-2$
B
$C-3$
C
$C-4$
D
$C-5$
Solution
(D) $1$. In the open-chain structure of glucose, the aldehyde group is at $C-1$.
$2$. The hydroxyl $(-OH)$ group at $C-5$ attacks the carbonyl carbon $(C-1)$ to form a stable six-membered pyranose ring.
$3$. This intramolecular reaction between the aldehyde group at $C-1$ and the hydroxyl group at $C-5$ results in the formation of a hemiacetal structure.
$4$. Therefore, the $C-5$ carbon atom is involved in the ring closure.
$2$. The hydroxyl $(-OH)$ group at $C-5$ attacks the carbonyl carbon $(C-1)$ to form a stable six-membered pyranose ring.
$3$. This intramolecular reaction between the aldehyde group at $C-1$ and the hydroxyl group at $C-5$ results in the formation of a hemiacetal structure.
$4$. Therefore, the $C-5$ carbon atom is involved in the ring closure.
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542
ChemistryEasyMCQMHT CET · 2026
Which of the following compounds forms starch on polymerization?
A
$\alpha-D-glucose$
B
$\beta-D-glucose$
C
$\alpha-L-glucose$
D
$\beta-L-glucose$
Solution
(A) Step $1$: Starch is a polysaccharide composed of glucose units.
Step $2$: Starch consists of two components, amylose and amylopectin, both of which are polymers of $\alpha-D-glucose$.
Step $3$: These units are linked together by $\alpha-glycosidic$ linkages.
Step $4$: Therefore, the monomer unit of starch is $\alpha-D-glucose$.
Step $2$: Starch consists of two components, amylose and amylopectin, both of which are polymers of $\alpha-D-glucose$.
Step $3$: These units are linked together by $\alpha-glycosidic$ linkages.
Step $4$: Therefore, the monomer unit of starch is $\alpha-D-glucose$.
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543
ChemistryEasyMCQMHT CET · 2026
The letter '$D$' in carbohydrates signifies
A
dextrorotatory
B
configuration
C
diamagnetic nature
D
mode of synthesis
Solution
(B) In the nomenclature of carbohydrates, the prefix '$D$' or '$L$' is used to specify the configuration of the chiral carbon atom furthest from the carbonyl group. It is based on the configuration of glyceraldehyde. Therefore, '$D$' signifies the configuration of the molecule.
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544
ChemistryEasyMCQMHT CET · 2026
Which carbon atom (numbered from $1'$ to $5'$) of ribose lacks the oxygen to form deoxyribose ?
A
$5'$
B
$3'$
C
$2'$
D
$1'$
Solution
(C) $1$. Ribose is a pentose sugar with the formula $C_5H_{10}O_5$, where each carbon atom is attached to a hydroxyl $(-OH)$ group.
$2$. Deoxyribose is a derivative of ribose with the formula $C_5H_{10}O_4$.
$3$. The prefix 'deoxy' indicates the removal of an oxygen atom.
$4$. In deoxyribose, the hydroxyl group at the $2'$ position of the ribose sugar is replaced by a hydrogen atom, meaning the oxygen atom is missing at the $2'$ carbon.
$5$. Therefore, the $2'$ carbon atom lacks the oxygen atom.
$2$. Deoxyribose is a derivative of ribose with the formula $C_5H_{10}O_4$.
$3$. The prefix 'deoxy' indicates the removal of an oxygen atom.
$4$. In deoxyribose, the hydroxyl group at the $2'$ position of the ribose sugar is replaced by a hydrogen atom, meaning the oxygen atom is missing at the $2'$ carbon.
$5$. Therefore, the $2'$ carbon atom lacks the oxygen atom.
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545
ChemistryMediumMCQMHT CET · 2026
Which of the following compounds is used to prepare adipic acid enzymatically in the Green technology developed by $Darth$ and $Frost$?
A
Sucrose
B
Glucose
C
Lactose
D
Maltose
Solution
(B) Step $1$: $Darth$ and $Frost$ developed a green chemical process to synthesize adipic acid from renewable resources.
Step $2$: The process utilizes the biocatalytic conversion of $D-glucose$ into muconic acid using genetically engineered $Escherichia \ coli$.
Step $3$: The resulting muconic acid is then hydrogenated to produce adipic acid.
Step $4$: Therefore, the starting material used is glucose.
Step $2$: The process utilizes the biocatalytic conversion of $D-glucose$ into muconic acid using genetically engineered $Escherichia \ coli$.
Step $3$: The resulting muconic acid is then hydrogenated to produce adipic acid.
Step $4$: Therefore, the starting material used is glucose.
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546
ChemistryDifficultMCQMHT CET · 2026
What is the total mass of products obtained when one gram mole of sucrose is hydrolysed (in $\text{ g}$)?
A
$180$
B
$342$
C
$170$
D
$360$
Solution
(D) The hydrolysis reaction of sucrose is: $C_{12}H_{22}O_{11} + H_2O \rightarrow C_6H_{12}O_6 (\text{glucose}) + C_6H_{12}O_6 (\text{fructose})$.
One mole of sucrose $(342 \text{ g})$ reacts with one mole of water $(18 \text{ g})$ to produce one mole of glucose $(180 \text{ g})$ and one mole of fructose $(180 \text{ g})$.
Total mass of products = Mass of glucose + Mass of fructose = $180 \text{ g} + 180 \text{ g} = 360 \text{ g}$.
One mole of sucrose $(342 \text{ g})$ reacts with one mole of water $(18 \text{ g})$ to produce one mole of glucose $(180 \text{ g})$ and one mole of fructose $(180 \text{ g})$.
Total mass of products = Mass of glucose + Mass of fructose = $180 \text{ g} + 180 \text{ g} = 360 \text{ g}$.
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547
ChemistryEasyMCQMHT CET · 2026
Which of the following sugars is an aldohexose?
A
Ribose
B
Fructose
C
Lactose
D
Glucose
Solution
(D) $1$. An aldohexose is a monosaccharide containing six carbon atoms and an aldehyde group $(-CHO)$.
$2$. $Ribose$ is an aldopentose ($5$ carbons).
$3$. $Fructose$ is a ketohexose ($6$ carbons with a ketone group).
$4$. $Lactose$ is a disaccharide.
$5$. $Glucose$ $(C_6H_{12}O_6)$ contains six carbon atoms and an aldehyde group, making it an aldohexose.
$2$. $Ribose$ is an aldopentose ($5$ carbons).
$3$. $Fructose$ is a ketohexose ($6$ carbons with a ketone group).
$4$. $Lactose$ is a disaccharide.
$5$. $Glucose$ $(C_6H_{12}O_6)$ contains six carbon atoms and an aldehyde group, making it an aldohexose.
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548
ChemistryEasyMCQMHT CET · 2026
Which of the following compounds is classified as an oligosaccharide?
A
Ribose
B
Sucrose
C
Fructose
D
Glucose
Solution
(B) $1$. Carbohydrates are classified based on the number of sugar units produced upon hydrolysis.
$2$. Monosaccharides like $Ribose$, $Fructose$, and $Glucose$ cannot be further hydrolyzed into simpler carbohydrates.
$3$. Oligosaccharides are carbohydrates that yield $2$ to $10$ monosaccharide units upon hydrolysis.
$4$. $Sucrose$ is a disaccharide (a type of oligosaccharide) which on hydrolysis yields one molecule of $Glucose$ and one molecule of $Fructose$ $(C_{12}H_{22}O_{11} + H_2O \rightarrow C_6H_{12}O_6 + C_6H_{12}O_6)$.
$2$. Monosaccharides like $Ribose$, $Fructose$, and $Glucose$ cannot be further hydrolyzed into simpler carbohydrates.
$3$. Oligosaccharides are carbohydrates that yield $2$ to $10$ monosaccharide units upon hydrolysis.
$4$. $Sucrose$ is a disaccharide (a type of oligosaccharide) which on hydrolysis yields one molecule of $Glucose$ and one molecule of $Fructose$ $(C_{12}H_{22}O_{11} + H_2O \rightarrow C_6H_{12}O_6 + C_6H_{12}O_6)$.
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549
ChemistryMediumMCQMHT CET · 2026
Which of the following pairs of carbohydrates contains $galactose$ in both of them as one of the constituents?
A
$Sucrose$ and $maltose$
B
$Maltose$ and $lactose$
C
$Lactose$ and $Raffinose$
D
$Sucrose$ and $lactose$
Solution
(C) $1$. $Sucrose$ is a disaccharide composed of $glucose$ and $fructose$.
$2$. $Maltose$ is a disaccharide composed of two units of $glucose$.
$3$. $Lactose$ is a disaccharide composed of $glucose$ and $galactose$.
$4$. $Raffinose$ is a trisaccharide composed of $galactose$, $glucose$, and $fructose$.
$5$. Since both $lactose$ and $raffinose$ contain $galactose$ as a constituent, the correct pair is $Lactose$ and $Raffinose$.
$2$. $Maltose$ is a disaccharide composed of two units of $glucose$.
$3$. $Lactose$ is a disaccharide composed of $glucose$ and $galactose$.
$4$. $Raffinose$ is a trisaccharide composed of $galactose$, $glucose$, and $fructose$.
$5$. Since both $lactose$ and $raffinose$ contain $galactose$ as a constituent, the correct pair is $Lactose$ and $Raffinose$.
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550
ChemistryEasyMCQMHT CET · 2026
Which of the following polymers is also known as $Dacron$?
A
$Teflon$
B
$PVC$
C
$Bakelite$
D
$Terylene$
Solution
(D) $Terylene$ is a polyester fibre obtained by the condensation polymerization of ethylene glycol and terephthalic acid. It is also known as $Dacron$ or $Mylar$.
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How many Chemistry questions are in MHT CET 2026?
There are 806 Chemistry questions from the MHT CET 2026 paper on Vedclass, each with a detailed step-by-step solution in English.
Are MHT CET 2026 Chemistry solutions available in English?
Yes. All solutions on this page are in English. You can also switch to English or Hindi using the language buttons above the questions.
Can I practice MHT CET 2026 Chemistry as a timed test?
Yes. Use the Vedclass Test Series to attempt a full MHT CET mock test covering Chemistry with time limits and instant score analysis.
Can teachers create Chemistry papers from MHT CET previous year questions?
Yes. The Vedclass Exam Paper Generator lets teachers mix MHT CET Chemistry questions and generate Set A/B/C/D papers in minutes.
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Pick MHT CET 2026 Chemistry questions, set difficulty, and generate Set A/B/C/D in 2 minutes.