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amulye

CH₃COOH+Br₂/P $\rightarrow$ Y+KCN $\rightarrow$ Z HERE Z IS

D ACID PRODUCED IN D SEQUENCE

C₂H₅I+alc.KOH $\rightarrow$ X+Br₂/CCl₄Y+KCN $\rightarrow$ Z+H₃O+ $\rightarrow$ A

Terminator.HCV

Y= CH2BrCOOH
Z=CNCH2COOH
X=C2H4
Y=BrCH2CH2Br

Plz state ur Q. also state how many moles of KCN(alcoholic) is supplied in the latter sequence.............

ayshwarya

sorry i think u didnot understand d questions i think d first sentance is d 1st question & frm d 2nd sentance onwards d second question

amulye

how cud b Z b CNCH2COOH according to 2nd reaction as der is no oxygens so plz reply it soon with ur soln

tarinbansal

For the first reaction-

Y --> CH₂BrCOOH( It is HVZ reaction or alpha

halogenation of acids)

Z--> CH₂CNCOOH (CN replaces Br)

For the second reaction-

X--> C₂H₄ (elimination takes place)

Y--> BrCH₂-CH₂Br (bromination of alkene)

Z--> BrCH₂-CH₂CN (If 1 equivalent KCN is provided)

CNCH₂-CH₂CN (If 2 equivalents KCN is provided)

A--> BrC₂H₄COOH or HOOC-C₂H₄-COOH (depending

upon the product Z)

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