Odpowiedź :
a) CH≡CH + HBr---->CH2=CH-Br
b) CH3-CH2-C(Br)2-CH(Br)2 + 2Zn---->CH3-CH2-C≡CH + 2ZnBr2
a) CH≡CH + HBr -> CH₂=CH-Br
b) CH₃-CH₂-C(Br)₂-CH(Br)₂ + 2Zn -> CH₃-CH₂-C≡CH + 2ZnBr₂
a) CH≡CH + HBr---->CH2=CH-Br
b) CH3-CH2-C(Br)2-CH(Br)2 + 2Zn---->CH3-CH2-C≡CH + 2ZnBr2
a) CH≡CH + HBr -> CH₂=CH-Br
b) CH₃-CH₂-C(Br)₂-CH(Br)₂ + 2Zn -> CH₃-CH₂-C≡CH + 2ZnBr₂