Chemistry 2019 Jamb Examination
Chemistry 2019 Jamb Examination
Question 1
A secondary alkanol can be oxidized to give an
A secondary alkanol has the hydroxyl group (−OH) attached to a carbon atom bonded to two other carbon atoms.
Upon oxidation, the secondary alkanol loses hydrogen atoms to form a carbonyl compound where the carbonyl carbon is attached to two alkyl groups.
This product is called an alkanone (ketone). Primary alkanols form alkanals, while tertiary alkanols do not undergo oxidation easily.
Therefore, oxidation of a secondary alkanol gives an alkanone.
Upon oxidation, the secondary alkanol loses hydrogen atoms to form a carbonyl compound where the carbonyl carbon is attached to two alkyl groups.
This product is called an alkanone (ketone). Primary alkanols form alkanals, while tertiary alkanols do not undergo oxidation easily.
Therefore, oxidation of a secondary alkanol gives an alkanone.
Not clear enough?
Question 2
Which of the following will give a precipitate with an aqueous solution of copper (I) chloride?
Copper (I) chloride reacts with terminal alkynes to form a white precipitate of copper (I) acetylide complex.
Among the options, only But-1-yne has a terminal triple bond (−C≡CH) that can react with CuCl.
Non-terminal alkynes and alkenes do not react with CuCl to form a precipitate.
Therefore, But-1-yne gives the precipitate.
Among the options, only But-1-yne has a terminal triple bond (−C≡CH) that can react with CuCl.
Non-terminal alkynes and alkenes do not react with CuCl to form a precipitate.
Therefore, But-1-yne gives the precipitate.
Not clear enough?
Question 3
The IUPAC name for CH3CH2COOCH2CH3 is
The compound has the ester functional group −COO−. The alkyl group attached to the oxygen atom forms the prefix, while the carboxylic acid part forms the suffix.
The parent acid is propanoic acid (CH3CH2COOH), and the alcohol part is ethanol (CH3CH2OH).
Thus, the ester name is ethylpropanoate.
The parent acid is propanoic acid (CH3CH2COOH), and the alcohol part is ethanol (CH3CH2OH).
Thus, the ester name is ethylpropanoate.
Not clear enough?
Question 4
A certain hydrocarbon on complete combustion at s.t.p produced 89.6dm3 of CO2 and 54g of water. The hydrocarbon should be
At s.t.p, 1 mole of gas occupies 22.4 dm3.
CO2 moles = 89.6 / 22.4 = 4 moles.
H2O moles = 54 / 18 = 3 moles.
The ratio of C:H is 4:6, simplifying to 2:3.
Hence, the empirical formula is C4H6, which corresponds to a hydrocarbon with two degrees of unsaturation such as an alkyne or diene.
CO2 moles = 89.6 / 22.4 = 4 moles.
H2O moles = 54 / 18 = 3 moles.
The ratio of C:H is 4:6, simplifying to 2:3.
Hence, the empirical formula is C4H6, which corresponds to a hydrocarbon with two degrees of unsaturation such as an alkyne or diene.
Not clear enough?
Question 5
2-methylprop-1-ene is an isomer of
2-methylprop-1-ene has the molecular formula C4H8.
But-2-ene also has the molecular formula C4H8 but a different structural arrangement of atoms.
Hence, both are structural isomers (same molecular formula, different structures).
But-2-ene also has the molecular formula C4H8 but a different structural arrangement of atoms.
Hence, both are structural isomers (same molecular formula, different structures).
Not clear enough?
Question 6
Hydrocarbons which will react with Tollen's reagent conform to the general formula
Tollen's reagent (ammoniacal silver nitrate) oxidises terminal alkynes, which conform to the general formula CnH2n−2.
Alkanes and alkenes do not react with Tollen's reagent because they lack the terminal triple bond necessary for the reaction.
Therefore, hydrocarbons that fit CnH2n−2 react with Tollen’s reagent.
Alkanes and alkenes do not react with Tollen's reagent because they lack the terminal triple bond necessary for the reaction.
Therefore, hydrocarbons that fit CnH2n−2 react with Tollen’s reagent.
Not clear enough?
Question 7
The IUPAC name of the compound CF3CHBrCl is
Numbering begins from the carbon attached to the fluorine atoms since fluorine has higher priority in numbering.
The substituents on the second carbon are bromo and chloro groups, listed alphabetically: bromo before chloro.
Hence, the correct IUPAC name is 2-bromo-2-chloro-1,1,1-trifluoroethane.
The substituents on the second carbon are bromo and chloro groups, listed alphabetically: bromo before chloro.
Hence, the correct IUPAC name is 2-bromo-2-chloro-1,1,1-trifluoroethane.
Not clear enough?
Question 8
A synthetic rubber is obtained from the polymerization of
Natural rubber is a polymer of isoprene (2-methyl-1,3-butadiene).
When isoprene undergoes polymerisation, it forms polyisoprene, a synthetic version of natural rubber.
Neoprene and butadiene also form rubbers but are different types (e.g., chloroprene rubber and buna rubber).
Therefore, isoprene is the monomer for synthetic rubber resembling natural rubber.
When isoprene undergoes polymerisation, it forms polyisoprene, a synthetic version of natural rubber.
Neoprene and butadiene also form rubbers but are different types (e.g., chloroprene rubber and buna rubber).
Therefore, isoprene is the monomer for synthetic rubber resembling natural rubber.
Not clear enough?
Question 9
Burning magnesium ribbon in air removes which of the following:
(i) oxygen (ii) nitrogen (iii) argon and (iv) carbon(IV) oxide?
(i) oxygen (ii) nitrogen (iii) argon and (iv) carbon(IV) oxide?
Magnesium burns in air producing magnesium oxide and magnesium nitride:
\(2Mg + O_2 \rightarrow 2MgO\)
\(3Mg + N_2 \rightarrow Mg_3N_2\)
Hence, it reacts with oxygen and nitrogen but not with argon or carbon dioxide.
Therefore, burning magnesium ribbon removes oxygen and nitrogen.
\(2Mg + O_2 \rightarrow 2MgO\)
\(3Mg + N_2 \rightarrow Mg_3N_2\)
Hence, it reacts with oxygen and nitrogen but not with argon or carbon dioxide.
Therefore, burning magnesium ribbon removes oxygen and nitrogen.
Not clear enough?
Question 10
Which of the following is the best starting material for the preparation of oxygen? Heating of trioxonitrate (V) with
Lead(II) trioxonitrate (V), Pb(NO3)2, decomposes on heating to give lead(II) oxide, nitrogen dioxide, and oxygen:\[2Pb(NO_3)_2 \rightarrow 2PbO + 4NO_2 + O_2\]Other trioxonitrates such as Zn(NO3)2 and NaNO3 do not yield oxygen as efficiently.
Thus, Pb(NO3)2 is the best starting material for preparing oxygen gas.
Thus, Pb(NO3)2 is the best starting material for preparing oxygen gas.
Not clear enough?
You might like
This question is based on Literary Appreciation. In William Shakespeare: Twelfth Night
'I have said too much unto a heart of stone,
And laid my honour too unchary on it',
There's something in me that
Reproves my fault. But such a headstrong potent fault it is That it but mocks reproof.'
A heart of stone in the lines above is an example of
A conductor of length 1 m moves with a velocity of 50 m/s at an angle of 30° to the direction of a uniform magnetic field of flux density 1.5 Wb/m². What is the e.m.f induced in the conductor?
Although clay soil is rich in nutrients, they are not good for most agricultural crop production because