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Chemistry M.C.Q (1 Marks)

MODEL PAPER 10 · CBSE STD 12 Science (CBSE - English Medium). 12 questions.

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M.C.Q (1 Marks)

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12 questions · 11 auto-graded MCQ + 1 self-marked written.

MCQ 11 Mark
Aniline does not undergo Friedel $-$ Crafts reaction because$:$
  • A
    Anilium ion deactivates any further reaction
  • B
    Aluminium chloride reacts with Aniline
  • C
    All of these
  • ✓
    $AlCl_3$ act as a catalyst
Answer
Correct option: D.
$AlCl_3$ act as a catalyst
$AlCl _3$ being a lewis acid reacts with the lone pair of $- NH _2$ group of aniline forming an adduct $\left( C _6 H _5 NH _2{ }^{+} AlCl _3\right)$ which deactivates the benzene system hence no friedal craft reaction occurs.
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MCQ 21 Mark
Williamson's synthesis is used for the preparation of
  • A
    aldehydes
  • ✓
    ethers
  • C
    alkyl halides
  • D
    alcohols
Answer
Correct option: B.
ethers
The Williamson ether synthesis is an organic reaction, forming an ether from an organohalide and deprotonated alcohol $($alkoxide$)$. This reaction was developed by Alexander Williamson in $1850.$ Typically it involves the reaction of an alkoxide ion with a primary alkyl halide via an $S_N2$ reaction.
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MCQ 31 Mark
The reduction of ethanenitrile with sodium and alcohol gives:
  • ✓
    1-aminoethane
  • B
    Ethanamide
  • C
    1-aminopropane
  • D
    Ethanoic acid
Answer
Correct option: A.
1-aminoethane
(a) 1-aminoethane
Explanation: 1-aminoethane
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MCQ 41 Mark
The following experimental rate data were obtained for a reaction carried out at $25^{\circ} C :$
$A_{(g)}+B_{(g)} \rightarrow C_{(g)}+D_{(g)}$
Initial $[A_{(g)}]/mol \ dm^{-3}$ Initial $[B_{(g)}]/mol \ dm^{-3}$ Initial rate$/mol \ dm^{-3}S^{-1}$
$3.0 \times 10^{-2}$ $2.0 \times 10^{-2}$ $1.89 \times 10^{-4}$
$3.0 \times 10^{-2}$ $4.0 \times 10^{-2}$ $1.89 \times 10^{-4}$
$6.0 \times 10^{-2}$ $4.0 \times 10^{-2}$ $7.56 \times 10^{-4}$
  • A
    Order with respect to $A_{(g)} -$ Second Order with respect to $B_{(g)} -$ First
  • B
    Order with respect to $A_{(g)} -$ Zero Order with respect to $B_{(g)} -$ Second
  • C
    Order with respect to $A_{(g)} -$ First Order with respect to $B_{(g)} -$ Zero
  • ✓
    Order with respect to $A_{(g)} -$ Second Order with respect to $B_{(g)} -$ Zero
Answer
Correct option: D.
Order with respect to $A_{(g)} -$ Second Order with respect to $B_{(g)} -$ Zero
Order with respect to $A_{(g)}$
Second Order with respect to $B_{(g)} -$ Zero
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MCQ 51 Mark
Silver ornaments turn black by the presence of which gas in the atmosphere?
  • ✓
    $H_2S$
  • B
    $O_2$
  • C
    $Cl_2$
  • D
    $N_2$
Answer
Correct option: A.
$H_2S$
Silver ornaments turns black coming in contact with $H_2S$ due to formation of $A_{^g2}S.$ The chemical equation for this change can be represented as given below $:2Ag(s) + H_2S(g) A_{g2}S(s) + H_2(g)$
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MCQ 61 Mark
Image
  • A
    1-Bromo-2-ethyl-2-methylethane
  • ✓
    1-Bromo-2-methylbutane
  • C
    2-Methyl-1-bromobutane
  • D
    1-Bromo-2-ethylpropane
Answer
Correct option: B.
1-Bromo-2-methylbutane
(b) 1-Bromo-2-methylbutane
Explanation: First, we need to identify the longest carbon chain. Once we do that, the actual structure should read $CH _3- CH _2$ $- CH \left( CH _3\right)- CH _2- Br$. - Br , the functional halide group is attached to the first carbon atom (1- Bromo), so we start the numbering from that position. The methyl group branch is bond to the second carbon atom in the chain(2-methyl). The number of carbons in the unbranched parent chain is four, thus giving the name butane. The IUPAC is named 1-Bromo-2methylbutane.
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MCQ 71 Mark
Match the items given in column $I$ with that in column $II:$
Column $I$ Column $II$
$(a)$ Urea $(i) i < 1$
$(b) FeCl_3$ $(ii) i = 1$
$(c)$ Benzoic acid in Benzene $(iii) i = 2$
$(d) MgSO_4$ $(iv) i = 4$
  • A
    $(a) (i), (b) (ii), (c) - (iii), (d) - (iv)$
  • B
    $(a) (iv), (b) - (ii), (c) - (iii), (d) - (i)$
  • C
    $(a) (i), (b)(ii), (c) - (iii), (d) - (iv)$
  • ✓
    $(a) (ii), (b) - (iv), (c) - (i), (d) - (iii)$
Answer
Correct option: D.
$(a) (ii), (b) - (iv), (c) - (i), (d) - (iii)$
$(a)-(ii), (b)-(iv), (c) - (i), (d) - (iii)$
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MCQ 81 Mark
The reaction $A \rightarrow B$ is a second order process when the initial concentration of $A$ is $0.50 M$, the half life is $8.0$ minutes. What is the half life if the initial concentration of $A$ is $0.10 M?$
  • ✓
    $40.0$ minutes
  • B
    $1.6$ minutes
  • C
    $8.0$ minutes
  • D
    $16.0$ minutes
Answer
Correct option: A.
$40.0$ minutes
Explanation: For second-order reaction:
$t_{1 / 2}=\frac{2 k}{[R]}$
$\Rightarrow k=\frac{t_{1 / 2}[R]}{2}$
Applying this equation,
$\frac{t_{1 / 2}[R]}{2}=\frac{t_{1_{1 / 2}}\left[R^{\prime}\right]}{2}$
$\frac{8.0 \times 0.50}{2}=\frac{t_{1 / 2}^{\prime} \times 0.10}{2}$
$t_{1 / 2}^{\prime}=\frac{8 \times 0.50}{0.10}=40$
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MCQ 91 Mark
The compound which forms acetaldehyde when heated with dilute NaOH is:
  • A
    1, 2 dichloroethane
  • B
    1, 1, 1 trichloroethane
  • C
    1 chloroethane
  • ✓
    1, 1 dichloroethane
Answer
Correct option: D.
1, 1 dichloroethane
(d) 1, 1 dichloroethane
Explanation: $CH _3 CHCl _2+ OH ^{-} \rightarrow CH _3 CH ( OH )_2 \rightarrow CH _3 CHO + H _2 O$
Gem diols like $\left( CH _3 CH ( OH )_2\right.$ ) are generally not stable. The $2- OH$ group attached to the same C removes $H _2 O$ and forms carbonyl compounds.
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MCQ 101 Mark
$\ce{CH_3CONH_2}$ on reaction with $\ce{NaOH}$ and $\ce{Br_2}$ in alcoholic medium gives$:$
  • ✓
    $\ce{CH_3NH_2}$
  • B
    $\ce{CH_3CH_2NH_2}$
  • C
    $\ce{CH_3COONa}$
  • D
    $\ce{CH_3CH_2Br}$
Answer
Correct option: A.
$\ce{CH_3NH_2}$
Product formed is $\ce{CH_3NH_2}$
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MCQ 111 Mark
Which of the following reactions of glucose can be explained only by its cyclic structure?
  • A
    Glucose is oxidised by nitric acid to gluconic acid.
  • ✓
    Pentaacetate of glucose does not react with hydroxylamine
  • C
    Glucose reacts with hydroxylamine to form an oxime.
  • D
    Glucose forms pentaacetate.
Answer
Correct option: B.
Pentaacetate of glucose does not react with hydroxylamine
(b) Pentaacetate of glucose does not react with hydroxylamine.
Explanation: The pentaacetate of glucose does not react with the hydroxylamine indicating the absence of free -CHO group. This property of the glucose can be explained only by its own cyclic structure.
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MCQ 121 Mark
Chlorobenzene is formed by the reaction of chlorine with benzene in the presence of $AlCl_3.$ Which of the following species attacks the benzene ring in this reaction?
  • A
    $AlCl _3$
  • B
    $\left[ AlCl _4\right]^{-}$
  • C
    $Cl ^{+}$
  • D
    $Cl ^{-}$
Answer
Aluminum chloride $\left( AlCl _3\right)$ is a Lewis acid catalyst and works in the same way as $FeCl _3$ does. Benzene $\left( C _6 H _6\right)$ is converted into chlorobenzene by chlorination of benzene in the presence of $AlCl _3$. The reaction occurs by an electrophilic substitution reaction. $Cl _2$ forms a coordination complex with $AlCl _3$, forming $Cl ^{+} AlCl _4{ }^{-}$complex, which gives a slight positive charge to $Cl $, and $AlCl _4{ }^{-}$is negatively charged. This $Cl ^{+}$then reacts with the aromatic double bond of the benzene ring to form an additional product, followed by deprotonation to form chlorobenzene and $AlCl _3$ and $HCl$ as the side products.
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M.C.Q (1 Marks) - Chemistry STD 12 Science Questions - Vidyadip