REDOX REACTIONS Quiz-15

Redox Reaction Quiz-15

Dear Readers,

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Q1. 

     

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Solution
(b). (A) must contain (MeCO-) group to give iodoform test. The reaction (C) to CH3COOH shows that (C) must be β-keto acid because on heating it easily undergoes decarboxylation, (or) (C) must be a dibasic acid. On heating it also loses CO_2. Therefore, (C) must be a dibasic acid because after iodoformreaction, (A) is converted to sodium salt of acid. So the answer is (b)

Q2.

     

•  No reaction
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Solution
(a). The isolated double bonds are not reduced by Birch reduction. So the answer is (a).

Q3. ,

    

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Solution (d) Red P + HI reduces both alcohols and acids to alkanes. So the answer is (d)

Q4. 

    

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Solution
(b). The reduction of ArNO_2 group under neutral conditions with (Zn dust +NH_4 Cl) or

Q5.In the reaction K+O2⟶KO2

•  O2 acts as an oxidizing agent
•  Both K and O2 are oxidized
•  O2 is oxidized while K is reduced
•  K acts as an oxidizing agent

Solution
 (a). K⟶K^⊕+e^- (oxidation,acts as reducing agent) e^-+O2⟶O2^(1-) (reduction,acts as oxidizing agent)

Q6. Which of the following is not a redox reaction?

•  CaCO3⟶CaO+CO2
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  O2+2H2⟶2H2O
• Na+H2O⟶NaOH+1/2H2
•  MnCl3⟶MnCl2+(1/2)Cl2

Solution (a). a. Oxidation states of Ca and C are +2 and +4, respectively, in both reactant and product: hence, not redox, b.4e^-+O2⟶2O^(2-) (Reduction) H2⟶2H^⊕+2e^- (Oxidation) Hence, redox c. Na ⟶Na^⊕+e^- (Oxidation) 2H^⊕+2e^-⟶H_2 (Reduction) Hence, redox d. e^-+Mn^(3+)⟶Mn^(2+) (Reduction) 2Cl^⊖⟶Cl_2+2e^-(Oxidation) Hence, redox

Q7.

    

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Solution
(b). (i) (H_2+Ni) converts (-CHO) group to (-CH2 OH) group. (ii) (H_2 + Ni) at high pressure also reduces double bond of benzene ring along with reduction of (-CHO) group to (-CH_2 OH) group. (iii) Birch reduction (Na + liq. NH3 + C2H5OH) reduces one double bond of benzene ring to give product with isolated double bond in which EWG (in problem (-CHO) group) ends up allylic to both double bonds. So the answer is (b).

Q8.The oxidation state of chromium in Cr(CO)6 is

•  0
•  +2
•  -2
•  +6

Solution
(a). Cr^0(CO^0)6 Oxidation State Of Cr=0

Q9.The oxidation number of S in H2SO5 is

•  +8
•  +6
•  +4
•  +2

Solution

(b). H2SO5 Oxidation state of S = +6

Q10. 

    

•  CH3NH2CH3NH2
•  CH3NH2 No reaction
•  No reaction CH3NH2
• No reaction No reaction

Solution
(a) LAH and NaBH4 both reduce azides (CH3N3) to 1°amine, so the answer is (a).