NEET Practice Questions (MCQs) with Answers & Solutions

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The reduction of an alkyne to alkene using Lindlar's catalyst results into:

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Explanation

Lindlar's catalyst (Pd-CaCO3 poisoned with Pb(OAc)2 and quinoline) is used for the semi-hydrogenation of alkynes to cis-alkenes. It adds hydrogen atoms in a cis fashion across the triple bond, selectively producing the cis-alkene product.

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Acetylene can be converted to higher alkyne using the following sequence of reactions:

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Explanation

(c)CH≡CH+Na→CH≡CNa→RXCH≡CR      CH≡CH+RMgX→CH≡CMgX→RXCH≡CR

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R→COOH→RCH2OH. This mode of reduction can be effected only by:

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Explanation

(d) Na/C2H5OH, LiAlH4 or NaBH4 are used for this purpose.

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Propene, CH3-CH=CH2 can be converted into 1-propanol by oxidation. Which set of reagents among the following is ideal to effect the conversion?

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Explanation

(b) CH3CH=CH2 →B2H6(CH3CH2CH2)3B 

                        →H2O2CH3CH2CH2OH + H3BO3;

The process is called hydroboration.

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CH3CH=CHCHO is oxidized to CH3CH=CHCOOH using:

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Explanation

(b) A characteristic reaction of aldehyde with Tollen's reagent. Rest all also attacks C=C.

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Propene on reaction with diazomethane in presence of UV radiations gives:

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Explanation

Propene (CH3CH=CH2) reacts with diazomethane (CH2N2) in the presence of UV radiation via a carbene insertion reaction. The methylene carbene (:CH2) generated from diazomethane inserts into the C=C bond of propene, forming methyl cyclopropane.

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The product formed during the reaction,

CH≡CH+NaOCl→0°Cis:

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Explanation

(d)CH≡CH→0°CNaOClCCl≡CCl

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Identify Z in the following series,

CH2=CH2 →HBr X →Hydrolysis Y →I2 excessNa2CO3 Z :

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Explanation

(c) CH2=CH2 + HBr →           C2H5Br →HOH C2H5OH →+ I2 (excess)Na2CO3 Iodoform

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Zinc-copper couple that can be used as a reducing agent is obtained by:

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Explanation

(B) It is how Zn-Cu couple is used.

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Methane reacts with conc. HNO3 at high temperature to yield:

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Explanation

(d) CH4+HNO3→∆CH3NO2; nitration reaction

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