CH2Br-CH2Br→b)NaNH2a)alc.KOHA→Li/liq.NH3H2B→XCH2Br-CH2Br ; 'X' in this conversion is

CH2Br-CH2Br→alc.KOHCH2=CH-Br→NaNH2CH≡CH; CH≡CH→Li/liq.NH3H2CH2=CH2→CCl4Br2CH2Br-CH2Br ;

CH2Br-CH2Br→b)NaNH2a)alc.KOHA→Li/liq.NH3H2B→XCH2Br-CH2Br ; 'X' in this conversion is

CH2Br-CH2Br→alc.KOHCH2=CH-Br→NaNH2CH≡CH; CH≡CH→Li/liq.NH3H2CH2=CH2→CCl4Br2CH2Br-CH2Br ;

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${CH}_{2} Br - {CH}_{2} Br \to_{b) {NaNH}_{2}}^{a) alc . KOH} A \to_{Li / liq . {NH}_{3}}^{H_{2}} B \overset{X}{\to} {CH}_{2} Br - {CH}_{2} Br;$ 'X' in this conversion is

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PBr3

Br2/CCl4

HBr

NBS/hν

answer is B.

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${CH}_{2} Br - {CH}_{2} Br \overset{alc . KOH}{\to} {CH}_{2} = CH - Br \overset{{NaNH}_{2}}{\to} CH \equiv CH; CH \equiv CH \to_{Li / liq . {NH}_{3}}^{H_{2}} {CH}_{2} = {CH}_{2} \to_{{CCl}_{4}}^{{Br}_{2}} {CH}_{2} Br - {CH}_{2} Br;$

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CH2Br-CH2Br→b)NaNH2a)alc.KOHA→Li/liq.NH3H2B→XCH2Br-CH2Br ; 'X' in this conversion is

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${CH}_{2} Br - {CH}_{2} Br \to_{b) {NaNH}_{2}}^{a) alc . KOH} A \to_{Li / liq . {NH}_{3}}^{H_{2}} B \overset{X}{\to} {CH}_{2} Br - {CH}_{2} Br;$ 'X' in this conversion is