Viết ptpu: CH2=C(CH3)-CH3+HBr CH2=CH-CH2-CH3+H2O(đk H+) CH3-CH=CH-CH3+ HBr 23/10/2021 Bởi Alice Viết ptpu: CH2=C(CH3)-CH3+HBr CH2=CH-CH2-CH3+H2O(đk H+) CH3-CH=CH-CH3+ HBr
$CH_2=C(CH_3)-CH_3 + HBr \to CH_3-CBr(CH_3)-CH_3$ $CH_2=CH-CH_2-CH_3 + HOH \buildrel{{H^+, t^o}}\over\to CH_3-CHOH-CH_2-CH_3$ $CH_3-CH=CH-CH_3+ HBr \to CH_3-CH_2-CHBr-CH_3$ Bình luận
Đáp án: Giải thích các bước giải: ( Áp dụng quy tắc Mac-cop-nhi-cop để viết PT ) CH2=C(CH3)-CH3 + HBr → CH3-C(CH3)Br -CH3 CH2=CH-CH2-CH3 + H-OH → CH3-CH(OH)-CH2-CH3 CH3-CH=CH-CH3 + HBr→ CH3-CH2-CHBr-CH3 Bình luận
$CH_2=C(CH_3)-CH_3 + HBr \to CH_3-CBr(CH_3)-CH_3$
$CH_2=CH-CH_2-CH_3 + HOH \buildrel{{H^+, t^o}}\over\to CH_3-CHOH-CH_2-CH_3$
$CH_3-CH=CH-CH_3+ HBr \to CH_3-CH_2-CHBr-CH_3$
Đáp án:
Giải thích các bước giải:
( Áp dụng quy tắc Mac-cop-nhi-cop để viết PT )
CH2=C(CH3)-CH3 + HBr → CH3-C(CH3)Br -CH3
CH2=CH-CH2-CH3 + H-OH → CH3-CH(OH)-CH2-CH3
CH3-CH=CH-CH3 + HBr→ CH3-CH2-CHBr-CH3