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7. A+Barrow NC HCequiv CH(NaNH_(2))/(.)?xrightarrow [C](CH_(3)CH_(2)Cl)]_(2)?Bxrightarrow [Delta ](NaH_(2)C)C_(2)CH_(3)I_(2)I_(2)

Вопрос

7.
A+Barrow NC
HCequiv CH(NaNH_(2))/(.)?xrightarrow [C](CH_(3)CH_(2)Cl)]_(2)?Bxrightarrow [Delta ](NaH_(2)C)C_(2)CH_(3)I_(2)I_(2)

7. A+Barrow NC HCequiv CH(NaNH_(2))/(.)?xrightarrow [C](CH_(3)CH_(2)Cl)]_(2)?Bxrightarrow [Delta ](NaH_(2)C)C_(2)CH_(3)I_(2)I_(2)

Решения

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Шота
Экспертная проверкаЭкспертная проверка
элита · Репетитор 8 лет

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1. A+B+HC≡CH→NC 2. NC+NaNH2→CH3CH2C≡CCH3 3. CH3CH2C≡CCH3+CH3CH2Cl→B 4. B+NaC2→CH3CH2C≡CCH3

Изложение

This question is about organic chemistry, specifically about the reactions of alkynes. The first reaction is the addition of hydrogen cyanide (HC≡CH) to a compound A+B to form a new compound NC. The second reaction is the reaction of the compound formed in the first reaction with sodium amide (NaNH2) in the presence of a catalyst (C). The product of this reaction is a compound with a triple bond between two carbon atoms (CH3CH2C≡CCH3). The third reaction is the reaction of this compound with chloroethane (CH3CH2Cl) to form a new compound B. The fourth reaction is the reaction of compound B with sodium acetylide (NaC2) to form a compound with a triple bond between two carbon atoms (CH3CH2C≡CCH3). The final product is a compound with a triple bond between two carbon atoms (CH3CH2C≡CCH3).
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