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Question of 147

Q.Context: A chloro compound (A) on reduction with Zn–Cu and ethanol give hydrocarbon (B) with two carbon-atoms. When compound (A) is dissolved in dry ether and sodium-metal it gives hydrocarbon with four carbon atoms. When compound (A) is treated with alcoholic KCN it gives compound (C).
Answer following questions:

(a) Identify A, B, C.
(2)
(b) Write short note on Wurtz-reaction. Also write chemical equation. (2)
Himachal HpboseHPBOSE Plus Two Board 2025Subjective· 4mImportance★★★★★
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The clues (2-carbon reduction product, 4-carbon Wurtz product, and a nitrile from KCN) identify A as ethyl chloride; Wurtz reaction couples two alkyl halide molecules using sodium metal in dry ether to form a longer-chain alkane.

(a) Identifying A, B, C:

  • A is reduced by Zn-Cu couple/ethanol to give a hydrocarbon B with 2 carbon atoms — this is a simple reduction (replacing −Cl-Cl with −H-H), so if B is C2H6C_2H_6 (ethane), A must itself have 2 carbons: A = ethyl chloride, CH3CH2ClCH_3CH_2Cl.
  • A dissolved in dry ether + sodium metal gives a hydrocarbon with 4 carbon atoms — this is the Wurtz reaction, where two molecules of the 2-carbon alkyl halide couple to form a 4-carbon alkane: 2CH3CH2Cl+2Na→CH3CH2CH2CH32CH_3CH_2Cl + 2Na \rightarrow CH_3CH_2CH_2CH_3 (n-butane). This confirms A = ethyl chloride.
  • A treated with alcoholic KCN gives C: alkyl halides react with KCNKCN (via the carbon of CN−CN^-, which is a better nucleophile through carbon) to form alkyl cyanides (nitriles): CH3CH2Cl+KCN→CH3CH2CN (C)+KClCH_3CH_2Cl + KCN \rightarrow CH_3CH_2CN\ (C) + KCl. C = propanenitrile (ethyl cyanide).

A=CH3CH2Cl,B=CH3CH3,C=CH3CH2CNA = CH_3CH_2Cl,\quad B = CH_3CH_3,\quad C = CH_3CH_2CN

(b) Wurtz Reaction: …

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